[Solutions Manual] Fourier and Laplace Transform - Antwoorden

Answers to selected exercises for chapter 1

Apply cos(α + β) = cos α cos β − sin α sin β, then1.1

f1(t) + f2(t)= A1 cos ωt cos φ1 −A1 sin ωt sin φ1 + A2 cos ωt cos φ2 −A2 sin ωt sin φ2

= (A1 cos φ1 + A2 cos φ2) cos ωt− (A1 sin φ1 + A2 sin φ2) sin ωt= C1 cos ωt− C2 sin ωt,

where C1 = A1 cos φ1 + A2 cos φ2 and C2 = A1 sin φ1 + A2 sin φ2. Put A =pC2

1 + C22 and take φ such that cos φ = C1/A and sin φ = C2/A (this is

possible since (C1/A)2+(C2/A)2 = 1). Now f1(t)+f2(t) = A(cos ωt cos φ−sin ωt sin φ) = A cos(ωt + φ).

Put c1 = A1eiφ1 and c2 = A2e

iφ2 , then f1(t) + f2(t) = (c1 + c2)eiωt. Let1.2

c = c1 + c2, then f1(t) + f2(t) = ceiωt. The signal f1(t) + f2(t) is again atime-harmonic signal with amplitude | c | and initial phase arg c.

The power P is given by1.5

P =ω

2π

Z π/ω

−π/ω

A2 cos2(ωt + φ0) dt =A2ω

4π

Z π/ω

−π/ω

(1 + cos(2ωt + 2φ0)) dt

=A2

2.

The energy-content is E =R∞0

e−2t dt = 12.1.6

The power P is given by1.7

P =1

4

3Xn=0

| cos(nπ/2) |2 = 12.

The energy-content is E =P∞

n=0 e−2n, which is a geometric series with1.8sum 1/(1− e−2).

a If u(t) is real, then the integral, and so y(t), is also real.1.9b Since˛ Z

u(τ) dτ

˛≤

Z|u(τ) | dτ,

it follows from the boundedness of u(t), so |u(τ) | ≤ K for some constantK, that y(t) is also bounded.c The linearity follows immediately from the linearity of integration. Thetime-invariance follows from the substitution ξ = τ − t0 in the integralR t

t−1u(τ − t0) dτ representing the response to u(t− t0).

d CalculatingR t

t−1cos(ωτ) dτ gives the following response: (sin(ωt) −

sin(ωt− ω))/ω = 2 sin(ω/2) cos(ωt− ω/2)/ω.e Calculating

R t

t−1sin(ωτ) dτ gives the following response: (− cos(ωt) +

cos(ωt− ω))/ω = 2 sin(ω/2) sin(ωt− ω/2)/ω.f From the response to cos(ωt) in d it follows that the amplitude responseis | 2 sin(ω/2)/ω |.g From the response to cos(ωt) in d it follows that the phase responseis −ω/2 if 2 sin(ω/2)/ω ≥ 0 and −ω/2 + π if 2 sin(ω/2)/ω < 0. From

1

2 Answers to selected exercises for chapter 1

phase and amplitude response the frequency response follows: H(ω) =2 sin(ω/2)e−iω/2/ω.

a The frequency response of the cascade system is H1(ω)H2(ω), since the1.11reponse to eiωt is first H1(ω)eiωt and then H1(ω)H2(ω)eiωt.b The amplitude response is |H1(ω)H2(ω) | = A1(ω)A2(ω).c The phase response is arg(H1(ω)H2(ω)) = Φ1(ω) + Φ2(ω).

a The amplitude response is | 1 + i |˛e−2iω

˛=√

2.1.12b The input u[n] = 1 has frequency ω = 0, initial phase 0 and amplitude1. Since eiωn 7→ H(eiω)eiωn, the response is H(e0)1 = 1 + i for all n.c Since u[n] = (eiωn + e−iωn)/2 we can use eiωn 7→ H(eiω)eiωn to obtainthat y[n] = (H(eiω)eiωn +H(e−iω)e−iωn)/2, so y[n] = (1+ i) cos(ω(n−2)).d Since u[n] = (1 + cos 4ωn)/2, we can use the same method as in b andc to obtain y[n] = (1 + i)(1 + cos(4ω(n− 2)))/2.

a The power is the integral of f2(t) over [−π/ |ω | , π/ |ω |], times |ω | /2π.1.13Now cos2(ωt + φ0) integrated over [−π/ |ω | , π/ |ω |] equals π/ |ω | andcos(ωt) cos(ωt + φ0) integrated over [−π/ |ω | , π/ |ω |] is (π/ |ω |) cos φ0.Hence, the power equals (A2 + 2AB cos(φ0) + B2)/2.b The energy-content is

R 1

0sin2(πt) dt = 1/2.

The power is the integral of | f(t) |2 over [−π/ |ω | , π/ |ω |], times |ω | /2π,1.14which in this case equals | c |2.

a The amplitude response is |H(ω) | = 1/(1 + ω2). The phase response1.16is arg H(ω) = ω.b The input has frequency ω = 1, so it follows from eiωt 7→ H(ω)eiωt thatthe response is H(1)ieit = iei(t+1)/2.

a The signal is not periodic since sin(2N) 6= 0 for all integer N .1.17

b The frequency response H(eiω) equals A(eiω)eiΦeiω

, hence, we obtainthat H(eiω) = eiω/(1 + ω2). The response to u[n] = (e2in − e−2in)/2i isthen y[n] = (e2i(n+1) − e−2i(n+1))/(10i), so y[n] = (sin(2n + 2))/5. Theamplitude is thus 1/5 and the initial phase 2− π/2.

a If u(t) = 0 for t < 0, then the integral occurring in y(t) is equal to 0 for1.18t < 0. For t0 ≥ 0 the expression u(t− t0) is also causal. Hence, the systemis causal for t0 ≥ 0.b It follows from the boundedness of u(t), so |u(τ) | ≤ K for some con-stant K, that y(t) is also bounded (use the triangle inequality and theinequality from exercise 1.9b). Hence, the system is stable.c If u(t) is real, then the integral is real and so y(t) is real. Hence, thesystem is real.d The response is

y(t) = sin(π(t− t0)) +

Z t

t−1

sin(πτ) dτ = sin(π(t− t0))− 2(cos πt)/π.

a If u[n] = 0 for n < 0, then y[n] is also equal to 0 for n < 0 whenever1.19n0 ≥ 0. Hence, the system is causal for n0 ≥ 0.b It follows from the boundedness of u[n], so |u[n] | ≤ K for some constantK and all n, that y[n] is also bounded (use the triangle inequality):

| y[n] | ≤ |u[n− n0] |+

˛˛

nXl=n−2

u[l]

˛˛ ≤ K +

nXl=n−2

|u[l] | ≤ K +

nXl=n−2

K,

Answers to selected exercises for chapter 1 3

which equals 4K. Hence, the system is stable.c If u[n] is real, then u[n− n0] is real and also the sum in the expressionfor y[n] is real, hence, y[n] is real. This means that the system is real.d The response to u[n] = cos πn = (−1)n is

y[n] = (−1)n−n0 +

nXl=n−2

(−1)l = (−1)n−n0 + (−1)n(1− 1 + 1)

= (−1)n(1 + (−1)n0).


a The absolute values follow fromp

x2 + y2 and are given by√

2, 2, 3, 22.1respectively. The arguments follow from standard angles and are given by3π/4, π/2, π, 4π/3 respectively.b Calculating modulus and argument gives 2+2i = 2

√2eπi/4, −

√3+ i =

2e5πi/6 and −3i = 3e3πi/2.

In the proof of theorem 2.1 it was shown that |Re z | ≤ | z |, which implies2.2that − | z | ≤ ± |Re z | ≤ | z |. Hence,

| z ± w |2 = (z ± w)(z ± w) = zz ± zw ± wz + ww= | z |2 ± 2Re(zw) + |w |2≥ | z |2 − 2 | z | |w |+ |w |2 = (| z | − |w |)2.

This shows that | z ± w |2 ≥ (| z | − |w |)2.

We have | z1 | = 4√

2, | z2 | = 4 and arg z1 = 7π/4, arg z2 = 2π/3. Hence,2.4| z1/z2 | = | z1 | / | z2 | =

√2 and arg(z1/z2) = arg(z1) − arg(z2) = 13π/12,

so z1/z2 =√

2e13πi/12. Similarly we obtain z21z3

2 = 2048e3πi/2 and z21/z3

2 =12e3πi/2.

The solutions are given in a separate figure on the website.2.5

a The four solutions ±1± i are obtained by using the standard technique2.6to solve this binomial equation (as in example 2.3).b As part a; we now obtain the six solutions 6

√2(cos(π/9 + kπ/3) +

i sin(π/9 + kπ/3)) where k = 0, 1 . . . , 5.c By completing the square as in example 2.4 we obtain the two solutions−1/5± 7i/5.

Write z5 − z4 + z − 1 as (z − 1)(z4 + 1) and then solve z4 = −1 to find2.7the roots

√2(±1± i)/2. Combining linear factors with complex conjugate

roots we obtain z5 − z4 + z − 1 = (z − 1)(z2 +√

2z + 1)(z2 −√

2z + 1).

Since 2i = 2eπi/2 the solutions are z = ln 2 + i(π/2 + 2kπ), where k ∈ Z.2.8

Split F (z) as A/(z − 12) + B/(z − 2) and multiply by the denominator of2.9

F (z) to obtain the values A = −1/3 and B = 4/3 (as in example 2.6).

a Split F (z) as A/(z + 1) + B/(z + 1)2 + C/(z + 3) and multiply by2.11the denominator of F (z) to obtain the values C = 9/4, B = 1/2 and, bycomparing the coefficient of z2, A = −5/4 (as in example 2.8).

Trying the first few integers we find the zero z = 1 of the denominator. A2.12long division gives as denominator (z− 1)(z2− 2z +5). We then split F (z)as A/(z − 1) + (Bz + C)/(z2 − 2z + 5). Multiplying by the denominatorof F (z) and comparing the coefficients of z0 = 1, z and z2 we obtain thatA = 2, B = 0 and C = −1.

a Using the chain rule we obtain f ′(t) = −i(1 + it)−2.2.13

Use integration by parts twice and the fact that a primitive of eiω0t is2.14eiω0t/iω0. The given integral then equals 4π(1− πi)/ω3

0 , since e2πi = 1.

Since˛1/(2− eit)

˛= 1/

˛2− eit

˛and

˛2− eit

˛≥ 2−

˛eit

˛= 1, the result2.15

follows from˛ R 1

0u(t) dt

˛≤

R 1

0|u(t) | dt.

1


a Use that | an | = 1/√

n6 + 1 ≤ 1/n3 and the fact thatP∞

n=1 1/n3 con-2.16verges (example 2.17).b Use that | an | ≤ 1/n2 and the fact that

P∞n=1 1/n2 converges.

c Use that | an | = 1/˛neneni

˛= 1/(nen) ≤ 1/en and the fact thatP∞

n=1 1/en converges since it is a geometric series with ratio 1/e.

a Use the ratio test to conclude that the series is convergent:2.17

limn→∞

˛n!

(n + 1)!

˛= lim

n→∞

1

n + 1= 0.

b The series is convergent; proceed as in part a:

limn→∞

˛2n+1 + 1

3n+1 + n + 1

3n + n

2n + 1

˛= lim

n→∞

2 + 1/2n

3 + (n + 1)/3n

1 + n/3n

1 + 1/2n=

2

3.

Determine the radius of convergence as follows:2.19

limn→∞

˛2n+1z2n+2

(n + 1)2 + 1

n2 + 1

2nz2n

˛= lim

n→∞2

˛z2

˛ 1 + 1/n2

1 + 2/n + 2/n2= 2

˛z2

˛.

This is less than 1 if˛z2

˛< 1/2, that is, if | z | <

√2/2. Hence, the radius

of convergence is√

2/2.

This is a geometric series with ratio z−i and so it converges for | z − i | < 1;2.20the sum is (1/(1− i))(1/(1− (z − i))), so 1/(2− z(1− i)).

b First solving w2 = −1 leads to z2 = 0 or z2 = −2i. The equation2.23z2 = −2i has solutions −1 + i and 1 − i and z2 = 0 has solution 0 (withmultiplicity 2).c One has P (z) = z(z4 + 8z2 + 16) = z(z2 + 4)2 = z(z − 2i)2(z + 2i)2, so0 is a simple zero and ±2i are two zeroes of multiplicity 2.

Split F (z) as (Az+B)/(z2−4z+5)+(Cz+D)/(z2−4z+5)2 and multiply2.25by the denominator of F (z). Comparing the coefficient of z0, z1, z2 and z3

leads to the values A = 0, B = 1, C = −2 and D = 2.

Replace cos t by (eit +e−it)/2, then we have to calculateR 2π

0(e2it +1)/2 dt,2.26

which is π.

a Using the ratio test we obtain as limit√

5/3. This is less than 1 and so2.27the series converges.b Since (n + in)/n2 = (1/n) + (in/n2) and the series

P∞n=1 1/n diverges,

this series is divergent.

The seriesP∞

n=0 cn(z2)n converges for all z with˛z2

˛< R, so it has radius2.29

of convergence√

R.

a Determine the radius of convergence as follows:2.30

limn→∞

˛(1 + i)2n+2zn+1

n + 2

n + 1

(1 + i)2nzn

˛= lim

n→∞| z | n + 1

n + 2

˛(1 + i)2

˛= 2 | z | .

This is less than 1 if | z | < 1/2, so the radius of convergence is 1/2.b Calculate f ′(z) by termwise differentiation of the series and multiplythis by z. It then follows that

zf ′(z) + f(z) =

∞Xn=0

(1 + i)2nzn =

∞Xn=0

(2iz)n.

This is a geometric series with ratio 2iz and so it has sum 1/(1− 2iz).

3

a

2

b

4 5

c

1

d

3

e

32

1 + 2i

2

f

–2

1

g

2 3

2 12

0


A trigonometric polynomial can be written as3.2

f(t) =a0

2+

kXm=1

(am cos(mω0t) + bm sin(mω0t)).

Now substitute this for f(t) in the right-hand side of (3.4) and use thefact that all the integrals in the resulting expression are zero, except for

the integralR T/2

−T/2sin(mω0t) sin(nω0t) dt with m = n, which equals T/2.

Hence, one obtains bn.

The function g(t) = f(t) cos(nω0t) has period T , so3.4 Z T

0

g(t)dt =

Z T

T/2

g(t)dt +

Z T/2

0

g(t)dt

=

Z T

T/2

g(t− T )dt +

Z T/2

0

g(t)dt =

Z 0

−T/2

g(τ)dτ +

Z T/2

0

g(t)dt

=

Z T/2

−T/2

g(t)dt.

Multiplying by 2/T gives an.

From a sketch of the periodic function with period 2π given by f(t) = | t |3.6for t ∈ (−π, π) we obtain

cn =1

2π

Z 0

−π

(−t)e−int dt +1

2π

Z π

0

te−int dt.

As in example 3.2 these integrals can be calculated using integration byparts for n 6= 0. Calculating c0 separately (again as in example 3.2) weobtain

c0 =π

2, cn =

(−1)n − 1

n2π

Substituting these values of cn in (3.10) we obtain the Fourier series. Onecan also write this as a Fourier cosine series:

π

2− 4

π

∞Xk=0

cos((2k + 1)t)

(2k + 1)2.

From the description of the function we obtain that3.7

cn =1

2

Z 1

0

e−(1+inπ)t dt.

This integral can be evaluated immediately and leads to

cn =inπ − 1

2(n2π2 + 1)

`(−1)ne−1 − 1

´.

The Fourier series follows from (3.10) by substituting cn.

The Fourier coefficients are calculated by splitting the integrals into a real3.9and an imaginary part. For c0 this becomes:

3


c0 =1

2

Z 1

−1

t2 dt +i

2

Z 1

−1

t dt =1

3.

For n 6= 0 we have that

cn =1

2

Z 1

−1

t2e−inπt dt +i

2

Z 1

−1

te−inπt dt.

The second integral can be calculated using integration by parts. To cal-culate the first integral we apply integration by parts twice. Adding theresults and simplifying somewhat we obtain the Fourier coefficients (andthus the Fourier series):

cn =(−1)n(2− nπ)

n2π2.

From the values of the coefficients cn calculated earlier in exercises 3.6, 3.73.10and 3.9, one can immediately obtain the amplitude spectrum | cn | and thephase spectrum arg cn (note e.g. that arg cn = π if cn > 0, arg cn = −π ifcn < 0, arg cn = π/2 if cn = iy with y > 0 and arg cn = −π/2 if cn = iywith y < 0). This results in three figures that are given separately on thewebsite.

a By substituting a = T/4 in (3.14) it follows that3.11

cn =sin(nπ/4)

nπfor n 6= 0, c0 =

1

4.

b As in a, but now a = T and we obtain

c0 = 1, cn = 0 for n 6= 0.

Hence, the Fourier series is 1 (!). This is no surprise, since the function is1 for all t.

By substituting a = T/2 in (3.15) it follows that3.12

c0 =1

2, cn = 0 for n 6= 0 even, cn =

2

n2π2for n odd.

We have that f(t) = 2p2,4(t)− q1,4(t) and so the Fourier coefficients follow3.14by linearity from table 1:c0 = 3/4, cn = (2nπ sin(nπ/2)− 4 sin2(nπ/4))/(n2π2) for n 6= 0.

Note that f(t) can be obtained from the sawtooth z(t) by multiplying the3.15shifted version z(t − T/2) by the factor T/2 and then adding T/2, thatis, f(t) = T

2z(t− T

2) + T

2. Now use the Fourier coefficients of z(t) (table 1

e.g.) and the properties from table 2 to obtain that

c0 =T

2, cn =

iT

2πnfor all n 6= 0.

Shifts over a period T (use the shift property and the fact that e−2πin = 13.17for all n).

In order to determine the Fourier sine series we extend the function to an3.19odd function of period 8. We calculate the coefficients bn as follows (thean are 0):

bn =1

4

Z −2

−4

(−2 sin(nπt/4)) dt +1

4

Z 2

−2

t sin(nπt/4) dt


+1

4

Z 4

2

2 sin(nπt/4) dt.

The second integral can be calculated by an integration by parts and onethen obtains that

bn =8

n2π2sin(nπ/2)− 4

nπcos(nπ),

which gives the Fourier sine series. For the Fourier cosine series we extendthe function to an even function of period 8. As above one can calculatethe coefficients an and a0 (the bn are 0). The result is

a0 = 3, an =8

n2π2(cos(nπ/2)− 1) for all n 6= 0.

In order to determine the Fourier cosine series we extend the function to3.21an even function of period 8. We calculate the coefficients an and a0 asfollows (the bn are 0):

a0 =1

2

Z 4

0

(x2 − 4x) dx = −16

3,

while for n ≥ 1 we have

an =1

4

Z 0

−4

(x2 + 4x) cos(nπx/4) dx +1

4

Z 4

0

(x2 − 4x) cos(nπx/4) dx

=1

2

Z 4

0

x2 cos(nπx/4) dx− 2

Z 4

0

x cos(nπx/4) dx.

The first integral can be calculated by applying integration by parts twice;the second integral can be calculated by integration by parts. Combiningthe results one then obtains that

an =64(−1)n

n2π2− 32((−1)n − 1)

n2π2=

32((−1)n + 1)

n2π2,

which also gives the Fourier cosine series. One can write this series as

−8

3+

16

π2

∞Xn=1

1

n2cos(nπx/2).

For the Fourier sine series we extend the function to an odd function ofperiod 8. As above one can calculate the coefficients bn (the an are 0). Theresult is

bn =64((−1)n − 1)

n3π3for all n ≥ 1.

If f is real and the cn are real, then it follows from (3.13) that bn =3.240. A function whose Fourier coefficients bn are all 0 has a Fourier seriescontaining cosine functions only. Hence, the Fourier series will be even. If,on the other hand, f is real and the cn are purely imaginary, then (3.13)shows that an = 0. The Fourier series then contains sine functions onlyand is thus odd.

Since sin(ω0t) = (eiω0t − e−iω0t)/2i we have3.25

cn =1

2iT

Z T/2

0

ei(1−n)ω0t dt− 1

2iT

Z T/2

0

e−i(1+n)ω0t dt.


The first integral equals T/2 for n = 1 while for n 6= 1 it equals i((−1)n +1)/((1 − n)ω0). The second integral equals T/2 for n = −1 while forn 6= −1 it equals i((−1)n+1 − 1)/((1 + n)ω0). The Fourier coefficients arethus c1 = 1/(4i), c−1 = −1/(4i) and ((−1)n+1)/(2(1−n2)π) for n 6= 1,−1;the Fourier series follows immediately from this.

b The even extension has period 2a, but it has period a as well. We can3.27thus calculate the coefficients an and a0 as follows (the bn are 0):

a0 =2

a

Z a/2

0

2bt/a dt− 2

a

Z 0

−a/2

2bt/a dt = b.

while for n ≥ 0 we obtain from an integration by parts that

an =2

a

Z a/2

0

(2bt/a) cos(2nπt/a) dt− 2

a

Z 0

−a/2

(2bt/a) cos(2nπt/a) dt

=2b((−1)n − 1)

n2π2,

which gives the Fourier cosine series. It can also be determined using theresult of exercise 3.6 by applying a multiplication and a scaling.

The odd extension has period 2a and the coefficients bn are given by (thean are 0):

bn =1

a

Z −a/2

−a

(−2bt

a− 2b) sin(nπt/a) dt +

1

a

Z a/2

−a/2

2bt

asin(nπt/a) dt

+1

a

Z a

a/2

(−2bt

a+ 2b) sin(nπt/a) dt

=8b

n2π2sin(nπ/2),

where we used integration by parts.

0 n2 4

a

–2–4 0 n

π

2 4

b

–2–4

π/2

0 n2 4

a

–2–4 0 n2 4

b

–2–4

π2

π2

12

–

0 n2 4

a

–2–4 0 n2 4

b

–2–4

12

π


a The periodic block function from section 3.4.1 is a continuous function4.1on [−T/2, T/2], except at t = ±a/2. At these points f(t+) and f(t−) exist.Also f ′(t) = 0 for t 6= ±a/2, while f ′(t+) = 0 for t = ±a/2 and t = −T/2and f ′(t−) = 0 for t = ±a/2 and t = T/2. Hence f ′ is piecewise continuousand so the periodic block function is piecewise smooth. Existence of theFourier coefficients has already been shown in section 3.4.1. The periodictriangle function is treated analogously.b For the periodic block function we have

∞Xn=−∞

| cn |2 ≤a2

T 2+

8

T 2ω20

∞Xn=1

1

n2

since sin2(nω0a/2) ≤ 1. The seriesP∞

n=11

n2 converges, soP∞

n=−∞ | cn |2converges. The periodic triangle function is treated analogously.

This follows immediately from (3.11) (for part a) and (3.8) (for part b).4.2

Take t = T/2 in the Fourier series of the sawtooth from example 4.2 and4.4use that sin(nω0T/2) = sin(nπ) = 0 for all n. Since (f(t+)+f(t−))/2 = 0,this agrees with the fundamental theorem.

a If we sketch the function, then we see that it is a shifted block function.4.6Using the shift property we obtain

c0 =1

2, cn = 0 even n 6= 0, cn =

−i

nπodd n.

The Fourier series follows by substituting the cn. One can write the serieswith sines only (split the sum in two pieces: one from n = 1 to ∞ andanother from n = −1 to −∞; change from n to −n in the latter):

1

2+

2

π

∞Xk=0

sin(2k + 1)t

2k + 1.

b The function is piecewise smooth and it thus satisfies the conditions ofthe fundamental theorem. At t = π/2 the function f is continuous, so theseries converges to f(π/2) = 1. Since sin((2k + 1)π/2) = (−1)k, formula(4.11) follows:

∞Xk=0

(−1)k

2k + 1=

π

4.

a We have that c0 = (2π)−1R π

0t dt = π/4, while the Fourier coefficients4.7

for n 6= 0 follow from an integration by parts:

cn =1

2π

Z π

0

te−int dt =(−1)ni

2n+

(−1)n − 1

2n2π.

The Fourier series follows by substituting these cn:

π

4+

1

2

∞Xn=−∞,n6=0

„(−1)ni

n+

(−1)n − 1

n2π

«eint.

12


b From the fundamental theorem it follows that the series will convergeto 1

2(f(π+) + f(π−)) = π/2 at t = π (note that at π there is a jump). If

we substitute t = π into the Fourier series, take π4

to the other side of the=-sign, then multiply by 2, and finally split the sum into a sum from n = 1to ∞ and a sum from n = −1 to −∞, then it follows that

π

2= 2

∞Xn=1

(−1)n − 1

n2π(−1)n

(the terms with (−1)ni/n cancel each other). For even n we have (−1)n −1 = 0 while for odd n this will equal −2, so (4.10) results:

π2

8=

∞Xk=1

1

(2k − 1)2.

a From f(0+) = 0 = f(0−) and f(1−) = 0 = f((−1)+) it follows that f4.9is continuous. We have that f ′(t) = 2t+1 for−1 < t < 0 and f ′(t) = −2t+1for 0 < t < 1. Calculating the defining limits for f ′ from below and fromabove at t = 0 we see that f ′(0) = 1 and since f ′(0+) = 1 = f ′(0−)it follows that f ′ is continuous at t = 0. Similarly it follows that f ′ iscontinuous at t = 1. Since f ′′(t) = 2 for −1 < t < 0 and f ′′(t) = −2 for0 < t < 1 we see that f ′′ is discontinuous.b The function f is the sum of g and h with period 2 defined for −1 < t ≤1 by g(t) = t and h(t) = t2 for −1 < t ≤ 0 and h(t) = −t2 for 0 < t ≤ 1.Since g is a sawtooth, the Fourier coefficients are cn = (−1)ni/πn (seesection 3.4.3). The function h is the odd extension of −t2 on (0, 1] andits Fourier coefficients have been determined in the first example of section3.6. By linearity one obtains the Fourier coefficients of f . In terms of thean and bn they become an = 0 and bn = 4(1− (−1)n)/π3n3. Hence, theydecrease as 1/n3.c Use e.g. the fundamental theorem for odd functions to obtain

f(t) =8

π3

∞Xk=0

sin(2k + 1)πt

(2k + 1)3.

Now substitute t = 1/2 and use that f(1/2) = 1/4 and sin((2k + 1)π/2) =(−1)n to obtain the required result.

Use (3.8) to write the right-hand side of (4.14) as a20/4+ 1

2

P∞n=1(a

2n + b2

n).4.10

a The Fourier coefficients of f and g are (see table 1 or section 3.4.1),4.12respectively, fn = (sin na)/nπ for n 6= 0 and f0 = a/π and gn = (sin nb)/nπfor n 6= 0 and g0 = b/π. Substitute into Parseval (4.13) and calculate the

integral (1/π)R a/2

−a/21 dt (note that a ≤ b). Take all constants together and

then again (as in exercise 4.7) split the sum into a sum from n = 1 to ∞and a sum from n = −1 to −∞. The required result then follows.b Use that sin2(nπ/2) = 1 for n odd and 0 for n even, then (4.10) follows.

a The Fourier coefficients are (see table 1 or section 3.4.2 and use that4.13sin2(nπ/2) = 1 for n odd and 0 for n even): cn = 2/n2π2 for n odd, 0 forn 6= 0 even and c0 = 1/2. From Parseval for f = g, so from (4.14), it thenfollows that (calculate the integral occurring in this formula):

1

3=

1

4+

8

π2

∞Xk=1

1

(2k − 1)4


(again we split the sum in a part from n = 1 to ∞ and from n = −1 to−∞). Take all constants together and multiply by π2/8, then the requiredresult follows.b Since

S =

∞Xn=1

1

n4=

∞Xk=1

1

(2k)4+

∞Xk=0

1

(2k + 1)4,

it follows from part a that

S =1

16

∞Xk=1

1

k4+

π4

96=

1

16S +

π4

96.

Solving for S we obtainP∞

n=11

n4 = π4

90.

SinceR b

af(t) dt =

R b

−T/2f(t) dt −

R a

−T/2f(t) dt, we can apply theorem 4.94.15

twice. Two of the infinite sums cancel out (the ones representing h0 intheorem 4.9), the other two can be taken together and lead to the desiredresult.

This follows from exercise 4.15 by using (3.8), so cn = (an − ibn)/2 and4.16c−n = (an + ibn)/2 (n ∈ N).

a The Fourier series is given by4.17

4

π

∞Xn=0

sin(2n + 1)t

2n + 1.

b SinceZ t

−π

sin(2n + 1)τ dτ = −cos(2n + 1)t

2n + 1− 1

2n + 1,

the integrated series becomes

− 4

π

∞Xn=0

1

(2n + 1)2− 4

π

∞Xn=0

cos(2n + 1)t

(2n + 1)2.

From (4.10) we see that the constant in this series equals −π/2.c The series in part b represents the function

R t

−πf(τ) dτ (theorem 4.9 or

better still, exercise 4.16). Calculating this integral we obtain the functiong(t) with period 2π given for −π < t ≤ π by g(t) = | t | − π.d Subtracting π from the Fourier series of | t | in exercise 3.6 we obtain aFourier series for g(t) which is in accordance with the result from part b.

This again follows as in exercise 4.16 from (3.8).4.19

Since f ′ is piecewise smooth, f ′′ is piecewise continuous and so the Fourier4.20coefficients c′′n of f ′′ exist. Since f ′ is continuous, we can apply integrationby parts, as in the proof of theorem 4.10. It then follows that c′′n = inω0c

′n,

where c′n are the Fourier coefficients of f ′. But c′n = inω0cn by theorem4.10, so c′′n = −n2ω2

0cn. Now apply the Riemann-Lebesgue lemma to c′′n,then it follows that limn→±∞ n2cn = 0.

a The Fourier coefficients have been determined in exercise 3.25: c1 =4.221/(4i), c−1 = −1/(4i) and ((−1)n + 1)/(2(1 − n2)π) for n 6= 1,−1. Tak-ing positive and negative n in the series together, we obtain the followingFourier series:


1

2sin t +

1

π+

2

π

∞Xk=1

1

1− 4k2cos 2kt.

b The derivative f ′ exists for all t 6= nπ (n ∈ Z) and is piecewise smooth.According to theorem 4.10 we may thus differentiate f by differentiatingits Fourier series for t 6= nπ:

f ′(t) =1

2cos t− 4

π

∞Xk=1

k

1− 4k2sin 2kt.

At t = nπ the differentiated series converges to (f ′(t+) + f ′(t−))/2, whichequals 1/2 for t = 0, while it equals −1/2 for t = π. Hence, the differen-tiated series is a periodic function with period 2π which is given by 0 for−π < t < 0, 1

2for t = 0, cos t for 0 < t < π, − 1

2for t = π.

Write down the expression for Si(−x) and change from the variable t to4.25−t, then it follows that Si(−x) = −Si(−x).

a From the definition of Si(x) it follows that Si′(x) = sin x/x. So Si′(x) =4.260 if sin x/x = 0. For x > 0 we thus have Si′(x) = 0 for x = kπ with k ∈ N.A candidate for the first maximum is thus x = π. Since sin x/x > 0 for0 < x < π and sin x/x < 0 for π < x < 2π, it follows that Si(x) indeed hasits first maximum at x = π.b The value at the first maximum is Si(π). Since Si(π) = 1.852 . . . andπ/2 = 1.570 . . ., the overshoot is 0.281 . . .. The jump of f at x = 0 isπ = 3.141 . . ., so the overshoot is 8.95 . . .%, so about 9%.

a The function f is continuous for t 6= (2k + 1)π (k ∈ Z) and it then4.28converges to f(t), which is 2t/π for 0 ≤ | t | < π/2, 1 for π/2 ≤ t < π and −1for −π < t ≤ −π/2. For t = (2k+1)π it converges to (f(t+)+f(t−))/2 = 0.b Since f is odd we have an = 0 for all n. The bn can be found using anintegration by parts:

bn =4

π2

Z π/2

0

t sin nt dt +2

π

Z π

π/2

sin nt dt =4

n2π2sin(nπ/2)− 2(−1)n

nπ.

Since sin(nπ/2) = 0 if n even and (−1)k if n = 2k + 1, the Fourier series is

− 2

π

∞Xn=1

(−1)n

nsin nt +

4

π2

∞Xn=0

(−1)n

(2n + 1)2sin(2n + 1)t.

Substituting t = 0 and t = π it is easy to verify the fundamental theoremfor these values.c We cannot differentiate the series; the resulting series is divergent be-cause limn→∞(−1)n cos nt 6= 0. Note that theorem 4.10 doesn’t apply sincef is not continuous.d We can integrate the series since theorem 4.9 can be applied (note thatc0 = 0). If we put g(t) =

R t

−πf(τ) dτ , then g is even, periodic with period

2π and given by (t2/π)−(3π/4) for 0 ≤ t < π/2 and by t−π for π/2 ≤ t ≤ π.

Use table 1 to obtain the Fourier coefficients and then apply Parseval, that4.29is, (4.13). Calculating the integral in Parseval’s identity will then give thefirst result; choosing a = π/2 gives the second result.

a The Fourier series has been determined in the last example of section4.303.6. Since f is continuous (and piecewise smooth), the Fourier series con-verges to f(t) for all t ∈ R:


f(t) =2

π− 4

π

∞Xn=1

1

4n2 − 1cos 2nt.

b First substitute t = 0 in the Fourier series; since f(0) = 0 and cos 2nt =1 for all n, the first result follows. Next substitute t = π/2 in the Fourierseries; since f(π/2) = 1 and cos 2nt = (−1)n for all n, the second resultfollows.c One should recognize the squares of the Fourier coefficients here. Hencewe have to apply Parseval’s identity (4.14), or the alternative form givenin exercise 4.10. This leads to

1

2π

Z π

−π

sin2 t dt =4

π2+

1

2π2

∞Xn=1

16

(4n2 − 1)2.

SinceR π

−πsin2 t dt = π, the result follows.

a Since f1 is odd it follows that4.31

(f1 ∗ f2)(−t) = − 1

T

Z T/2

−T/2

f1(t + τ)f2(τ) dτ.

Now change the variable from τ to −τ and use that f2 is odd, then itfollows that (f1 ∗ f2)(−t) = (f1 ∗ f2)(t).b The convolution product equals

(f ∗ f)(t) =1

2

Z 1

−1

τf(t− τ) dτ.

Since f is odd, part a implies that f ∗ f is even. It is also periodic withperiod 2, so it is sufficient to calculate (f ∗ f)(t) for 0 ≤ t ≤ 1. First notethat f is given by f(t) = t − 2 for 1 < t ≤ 2. Since −1 ≤ τ ≤ 1 and0 ≤ t ≤ 1 we see that t − 1 ≤ t − τ ≤ t + 1. From 0 ≤ t ≤ 1 it followsthat −1 ≤ t − 1 ≤ 0, and so close to τ = 1 the function f(t − τ) is givenby t − τ . Since 1 ≤ t + 1 ≤ 2, the function f(t − τ) is given by t − τ − 2close to τ = −1. Hence, we have to split the integral precisely at the pointwhere t− τ gets larger than 1, because precisely then the function changesfrom t− τ to t− τ − 2. But t− τ ≥ 1 precisely when τ ≤ t− 1, and so wehave to split the integral at t− 1:

(f ∗ f)(t) =1

2

Z t−1

−1

τ(t− τ − 2) dτ +1

2

Z 1

t−1

τ(t− τ) dτ.

It is now straightforward to calculate the convolution product. The resultis (f ∗ f)(t) = −t2/2 + t− 1/3.c From section 3.4.3 or table 1 we obtain the Fourier coefficients cn ofthe sawtooth f and applying the convolution theorem gives the Fouriercoefficients of (f ∗ f)(t), namely c2

0 = 0 and c2n = −1/π2n2 (n 6= 0).

d Take t = 0 in part c; since f is odd and real-valued we can write(f ∗ f)(0) = 1

2

R 1

−1| f(τ) |2 dτ , and so we indeed obtain (4.13).

e For −1 < t < 0 we have (f ∗ f)′(t) = −t − 1, while for 0 < t < 1we have (f ∗ f)′(t) = −t + 1. Since f ∗ f is given by −t2/2 + t − 1/3 for0 < t < 2, (f ∗ f)′(t) is continuous at t = 1. Only at t = 0 we have thatf ∗ f is not differentiable. So theorem 4.10 implies that the differentiatedseries represents the function (f ∗ f)′(t) on [−1, 1], except at t = 0. Att = 0 the differentiated series converges to ((f ∗f)′(0+)+(f ∗f)′(0−))/2 =(1− 1)/2 = 0.f The zeroth Fourier coefficient of f ∗ f is given by


1

2

Z 1

−1

(f ∗ f)(t) dt =

Z 1

0

(−t2/2 + t− 1/3) dt = 0.

This is in agreement with the result in part c since c20 = 0. Since this

coefficient is 0, we can apply theorem 4.9. The function represented by theintegrated series is given by the (periodic) function

R t

−1(f ∗ f)(τ) dτ . It is

also odd, since f is even and for 0 ≤ t ≤ 1 it equalsZ 0

−1

(−τ2/2− τ − 1/3) dτ +

Z t

0

(−τ2/2 + τ − 1/3) dτ = −t(t− 1)(t− 2)/6.


For a stable LTC-system the real parts of the zeroes of the characteristic5.1polynomial are negative. Fundamental solutions of the homogeneous equa-tions are of the form x(t) = tlest, where s is such a zero and l ≥ 0some integer. Since

˛tlest

˛= | t |l e(Re s)t and Re s < 0 we have that

limt→∞ x(t) = 0. Any homogeneous solution is a linear combination ofthe fundamental solutions.

The Fourier coefficients of u are5.2

u0 =1

2, u2k = 0, u2k+1 =

(−1)k

(2k + 1)π

(u = pπ,2π, so use table 1 and the fact that sin(nπ/2) = (−1)k for n = 2k+1odd and 0 for n even). Since H(ω) = 1/(iω + 1) and yn = H(nω0)un =H(n)un it then follows that

y0 =1

2, y2k = 0, y2k+1 =

(−1)k

(1 + (2k + 1)i)(2k + 1)π.

a The frequency response is not a rational function, so the system cannot5.3be described by a differential equation (5.3).b Since H(nω0) = H(n) = 0 for |n | ≥ 4 (because 4 > π), we only need toconsider the Fourier coefficients of y with |n | ≤ 3. From Parseval it thenfollows that P =

P3n=−3 | yn |2 with yn as calculated in exercise 5.2. This

sum is equal to P = 14

+ 209π2 .

Note that u has period π and that the integral to be calculated is thus the5.4zeroth Fourier coefficient of y. Since y0 = H(0ω0)u0 = H(0)u0 and H(0) =−1 (see example 5.6 for H(ω)), it follows that y0 = −u0 = − 1

π

R π

0u(t) dt =

− 2π.

a According to (5.4) the frequency response is given by5.5

H(ω) =−ω2 + 1

−ω2 + 4 + 2iω.

Since H(ω) = 0 for ω = ±1, the frequencies blocked by the system areω = ±1.b Write u(t) = e−4it/4 − e−it/2i + 1/2 + eit/2i + e4it/4. It thus followsthat the Fourier coefficients unequal to 0 are given by u−4 = u4 = 1/4,u−1 = −1/2i, u1 = 1/2i and u0 = 1/2. Since yn = H(nω0)un = H(n)un

and H(1) = H(−1) = 0 we thus obtain that

y(t) = y−4e−4it + y−1e

−it + y0 + y1eit + y4e

4it

=15

12 + 8i· 1

4e−4it +

1

4· 1

2+

15

12− 8i· 1

4e4it.

It is a good exercise to write this with real terms only:

y(t) =45

104cos 4t +

30

104sin 4t +

1

8.

We have that5.6

H(ω) =1

−ω2 + ω20

.

18


Since |ω0 | is not an integer, there are no homogeneous solutions havingperiod 2π, while u does have period 2π. There is thus a uniquely determinedperiodic solution y corresponding to u. Since u(t) = πqπ,2π(t) the Fouriercoefficients of u follow immediately from table 1:

u0 =π

2, u2k = 0(k 6= 0), u2k+1 =

2

(2k + 1)2π2.

Since yn = H(nω0)un = H(n)un = 1−n2+ω2

0un, the line spectrum of y

follows.

For the thin rod the heat equation (5.8) holds on (0, L), with initial condi-5.7tion (5.9). This leads to the fundamental solutions (5.15), from which thesuperposition (5.16) is build. The initial condition leads to a Fourier serieswith coefficients

An =2

L

Z L/2

0

x sin(nπx/L) dx +2

L

Z L

L/2

(L− x) sin(nπx/L) dx,

which can be calculated using an integration by parts. The result is: An =(4L/n2π2) sin(nπ/2) (which is 0 for n even). We thus obtain the (formal)solution

u(x, t) =4L

π2

∞Xn=0

(−1)n

(2n + 1)2e−(2n+1)2π2kt/L2

sin((2n + 1)πx/L).

a The heat equation and initial conditions are as follows:5.9

ut = kuxx for 0 < x < L, t > 0,ux(0, t) = 0, u(L, t) = 0 for t ≥ 0,u(x, 0) = 7 cos(5πx/2L) for 0 ≤ x ≤ L.

b Separation of variables leads to (5.12) and (5.13). The function X(x)should satisfy X ′′(x)− cX(x) = 0 for 0 < x < L, X ′(0) = 0 and X(L) = 0.For c = 0 we obtain the trivial solution. For c 6= 0 the characteristicequation s2 − c = 0 has two distinct roots ±s1. The general solution isthen X(x) = αes1x + βe−s1x, so X ′(x) = s1αes1x − s1βe−s1x. The firstboundary condition X ′(0) = 0 gives s1(α − β) = 0, so β = α. Nextwe obtain from the second boundary condition X(L) = 0 the equationα(es1L + e−s1L) = 0. For α = 0 we get the trivial solution. So we musthave es1L + e−s1L = 0, implying that e2s1L = −1. From this it followsthat s1 = i(2n + 1)π/2L. This gives us eigenfunctions Xn(x) = cos((2n +1)πx/2L) (n = 0, 1, 2, 3, . . .). Since Tn(t) remains as in the textbook (forother parameters), we have thus found the fundamental solutions

un(x, t) = e−(2n+1)2π2kt/4L2cos((2n + 1)πx/2L).

Superposition gives

u(x, t) =

∞Xn=0

Ane−(2n+1)2π2kt/4L2cos((2n + 1)πx/2L).

Substituting t = 0 (and using the remaining initial condition) leads to

u(x, 0) =

∞Xn=0

An cos((2n + 1)πx/2L) = 7 cos(5πx/2L).


Since the right-hand side consists of one harmonic only, it follows thatA2 = 7 and An = 0 for all n 6= 2. The solution is thus u(x, t) =

7e−25π2kt/4L2cos(5πx/2L).

a The equations are5.11

ut = kuxx for 0 < x < L, t > 0,u(0, t) = 0, ux(L, t) = 0 for t ≥ 0,u(x, 0) = f(x) for 0 ≤ x ≤ L.

b Going through the steps one obtains the same fundamental solutions asin exercise 5.9. The coefficients An cannot be determined explicitly here,since f(x) is not given explicitly.

The equations are given by (5.17) - (5.20), where we only need to substitute5.12the given initial condition in (5.19), so u(x, 0) = 0.05 sin(4πx/L) for 0 ≤x ≤ L. All steps to be taken are the same as in section 5.2.2 of the textbookand lead to the solution

u(x, t) =

∞Xn=1

An cos(nπat/L) sin(nπx/L).

Substituting t = 0 (and using the remaining initial condition) gives

u(x, 0) =

∞Xn=1

An sin(nπx/L) = 0.05 sin(4πx/L).

Since the right-hand side consists of one harmonic only, it follows thatA4 = 0.05 and An = 0 for all n 6= 4. The solution is thus u(x, t) =0.05 cos(4πat/L) sin(4πx/L).

Separation of variables leads to X ′′(x)− cX(x) = 0 for 0 < x < π, X ′(0) =5.15X ′(π) = 0. For c = 0 we obtain the constant solution, so c = 0 is aneigenvalue with eigenfunction X(x) = 1. For c 6= 0 the characteristicequation s2 − c = 0 has two distinct roots ±s1. The general solutionis then X(x) = αes1x + βe−s1x, so X ′(x) = s1αes1x − s1βe−s1x. Theboundary condition X ′(0) = 0 gives s1(α − β) = 0, so β = α. Fromthe boundary condition X ′(π) = 0 we obtain s1α(es1π − e−s1π) = 0. Forα = 0 we get the trivial solution. So we must have es1π − e−s1π = 0,implying that e2s1π = 1. From this it follows that s1 = ni. This gives useigenfunctions Xn(x) = cos(nx) (n = 0, 1, 2, 3, . . .). For T (t) we get theequation T ′′(t) + n2a2T (t) = 0. From the initial condition ut(x, 0) = 0we obtain T ′(0) = 0. The non-trivial solution are Tn(t) = cos(nat) (n =0, 1, 2, 3, . . .) and we have thus found the fundamental solutions

un(x, t) = cos(nat) cos(nx).

Superposition gives

u(x, t) =

∞Xn=0

An cos(nat) cos(nx).

Substituting t = 0 (and using the remaining initial condition) leads to

u(x, 0) =

∞Xn=0

An cos(nx) = kx for 0 < x < π.


We have A0 = (2/π)R π

0kx dx = kπ and An = (2/π)

R π

0kx cos(nx) dx for

n 6= 0, which can be calculated by an integration by parts: An = 0 for neven (n 6= 0) and An = −4k/n2π for n odd. The solution is thus

u(x, t) =kπ

2− 4k

π

∞Xn=0

1

(2n + 1)2cos((2n + 1)at) cos((2n + 1)x).

a From H(−ω) = H(ω) and yn = H(nω0)un follows that the response5.16y(t) to a real signal u(t) is real: since u−n = un we also have y−n = yn.b Since we can write sin ω0t = (eiω0t − e−iω0t)/2i, the response is equalto (H(ω0)e

iω0t − H(−ω0)e−iω0t)/2i, which is ((1 − e−2iω0)2eiω0t − (1 −

e2iω0)2e−iω0t)/2i. This can be rewritten as sin ω0t − 2 sin(ω0(t − 2)) +sin(ω0(t− 4)).c A signal with period 1 has Fourier series of the form

P∞n=−∞ une2πint.

The response isP∞

n=−∞H(2πn)une2πint, which is 0 since H(2πn) = 0 forall n.

a The characteristic equation is s3 + s2 + 4s + 4 = (s2 + 4)(s + 1) = 05.18and has zeroes s = −1 and s = ±2i. The zeroes on the imaginary axiscorrespond to periodic eigenfrequencies with period π and so the responseto a periodic signal is not always uniquely determined. But see part b!b Since here the input has period 2π/3, we do have a unique response.From Parseval and the relation yn = H(nω0)un we obtain that the poweris given by

P =3

2π

Z 2π/3

0

| y(t) |2 dt =

∞Xn=−∞

| yn |2 =

∞Xn=−∞

|H(nω0)un |2 .

We have that

H(ω) =1 + iω

4− ω2 + iω(4− ω2).

Now use that only u3 = u−3 = 12

and that all other un are 0, then it followsthat P = 1/50.

For the rod we have equations (5.8) - (5.10), where we have to take f(x) =5.19u0 in (5.10). The solution is thus given by (5.16), where now the An are theFourier coefficients of the function u0 on [0, L]. These are easy te determine(either by hand or using tables 1 and 2): An = 0 for n even, An = 4u0/nπfor n odd. This gives

u(x, t) =4u0

π

∞Xn=0

1

(2n + 1)2e−(2n+1)2π2kt/L2

sin((2n + 1)πx/L).

Substituting x = L/2 in the x-derivative and using the fact that cos((2n +1)π/2) = 0 for all n leads to ux(L/2, t) = 0.

a As in the previous exercise the solution is given by (5.16). The An are5.20

given by (2/L)R L/2

0a sin(nπx/L) dx = 2a(1 − cos(nπ/2))/nπ, which gives

the (formal) solution

u(x, t) =2a

π

∞Xn=1

1

n(1− cos(nπ/2))e−n2π2kt/L2

sin(nπx/L).


b The two rods together form one rod and so part a can be applied withL = 40, k = 0.15 and a = 100. Substituting t = 600 in u(x, t) from parta then gives the temperature distribution. On the boundary between therods we have x = 20, so we have to calculate u(20, 600); using only thecontibution from the terms n = 1, 2, 3, 4 we obtain u(20, 600) ≈ 36.4.c Take k = 0.005, a = 100, L = 40, substitute x = 20 in u(x, t) frompart a, and now use only the first two terms of the series to obtain theequation u(20, t) ≈ 63.662e−0.0000308t = 36 (terms of the series tend to 0very rapidly, so two terms suffice). We then obtain 18509 seconds, whichis approximately 5 hours.


We have to calculate (the improper integral)R∞−∞ e−iωt dt. Proceed as in6.1

eaxample 6.1, but we now have to determine limB→∞ e−iωB . This limitdoes not exist.

a We have to calculate G(ω) =R∞0

e−(a+iω)t dt, which can be done pre-6.2

cisely as in section 6.3.3 if we write a = α + iβ and use that e−(a+iω)R =e−αRe−i(β+ω)R. If we let R →∞ then this tends to 0 since α > 0.b The imaginary part of G(ω) is −ω/(a2 +ω2) and applying the substitu-tion rule gives

Rω/(a2 + ω2) dω = 1

2ln(a2 + ω2), so this improper integral,

which is the Fourier integral for t = 0, does not exits (limA→∞ ln(a2 + A2)does not exist e.g.).c We have lima→0 g(t) = lima→0 ε(t)e−at = ε(t), while for ω 6= 0 we havethat lima→0 G(ω) = −i/ω.

To calculate the spectrum we split the integral at t = 0:6.4

G(ω) =

Z 1

0

te−iωt dt−Z 0

−1

te−iωt dt.

Changing from the variable t to −t in the second integral we obtain thatG(ω) = 2

R 1

0t cos ωt dt, which can be calculated for ω 6= 0 using an integ-

ration by parts. The result is:

G(ω) =2 sin ω

ω+

2(cos ω − 1)

ω2.

For ω = 0 we have that G(0) = 2R 1

0t dt = 1. Since limω→0 sin ω/ω = 1 and

limω→0(cos ω − 1)/ω2 = − 12

(use e.g. De l’Hopital’s rule), we obtain thatlimω→0 G(ω) = G(0), so G is continuous.

a Calculating the integral we have that6.5

F (ω) = 2icos(aω/2)− 1

ωfor ω 6= 0, F (0) = 0.

b Using Taylor or De l’Hopital it follows that limω→0 F (ω) = 0 = F (0),so F is continuous.

From the linearity and table 3 it follows that6.7

F (ω) =12

4 + ω2+ 8i

sin2(aω/2)

aω2.

Use (6.17) and table 3 for the spectrum of e−7| t |, then6.8

F (ω) =7

49 + (ω − π)2+

7

49 + (ω + π)2.

a From the shift property in the frequency domain (and linearity) it fol-6.9lows that the spectrum of f(t) sin at is F (ω − a)/2i− F (ω + a)/2i.b Write f(t) = p2π(t) sin t, obtain the spectrum of p2π(t) from table 3 andapply part a (and use the fact that sin(πω ± π) = − sin(ωπ)), then

F (ω) =2i sin(πω)

ω2 − 1.

23


Use section 6.3.3 (or exercise 6.2) and the modulation theorem 6.17, and6.10write the result as one fraction, then

(Fε(t)e−at cos bt)(ω) =a + iω

(a + iω)2 + b2.

Similarly it follows from section 6.3.3 (or exercise 6.2) and exercise 6.9athat

(Fε(t)e−at sin bt)(ω) =b

(a + iω)2 + b2.

Write6.12

F (ω) =

Z ∞

0

f(t)e−iωt dt +

Z 0

−∞f(t)e−iωt dt

and change from t to −t in the second integral, then it follows that F (ω) =−2i

R∞0

f(t) sin ωt dt.

a We have F (−ω) = F (ω) and F (ω) is even, so F (ω) = F (ω), and thus6.13F (ω) is real.b We have F (−ω) = F (ω) (by part a) and since |F (ω) | = (F (ω)F (ω))1/2,it follows that |F (ω) | = |F (−ω) |.

Calculate the spectrum in a direct way using exactly the same techniques6.14as in example 6.3.3 (or use (6.20) and twice an integration by parts):

F (ω) =−2iω

1 + ω2.

The spectrum is given byR a/2

−a/2te−iωt dt, which can be calculated using an6.16

integration by parts. The result is indeed equal to the formula given inexample 6.3.

a From the differentiation rule (and differentiating the Fourier transform6.17

of the Gauss function, of course) it follows that −iω√

πe−ω2/4a/(2a√

a) isthe spectrum of tf(t).b If we divide the Fourier transform of −f ′(t) by 2a, then we indeedobtain the same result as in part a.

Two examples are the constant function f(t) = 0 (k arbitrary), and the6.18

Gauss function e−t2/2 with k =√

2π. Using exercise 6.17a we obtain the

function te−t2/2 with k = −i√

2π.

Use table 3 for ε(t)e−at and then apply the differentiation rule in the fre-6.19quency domain, then the result follows: (a+iω)−2. (Differentiate (a+iω)−1

just as one would differentiate a real function.)

The function e−a| t | is not differentiable at t = 0. The function t3(1+ t2)−16.20e.g. is not bounded.

Use the fact that limx→∞ xae−x = 0 for all a ∈ R and change to the variable6.21

x = at2 in tk/eat2 (separate the cases t ≥ 0 and t < 0). Then part a followsand, hence, part b also follows since we have a finite sum of these terms.

Apply the product rule repeatedly to get an expression in terms of the de-6.22rivatives of f and g (this involves the binomial coefficients and is sometimescalled Leibniz rule). Since f and g belong to S, tn(f(t)g(t))(m) will be asum of terms belonging to S, and so the result follows.


We have that (ε ∗ ε)(t) =R∞0

ε(t − τ) dτ . Now treat the cases t > 0 and6.23t ≤ 0 separately, then it follows that (ε ∗ ε)(t) = ε(t)t. (If t ≤ 0, thent − τ < 0 for τ > 0 and so ε(t − τ) = 0; if t > 0 then ε(t − τ) = 0 forτ > t and the integral

R t

01 dτ = t remains.) Since ε(t)t is not absolutely

integrable, the function (ε ∗ ε)(t) is not absolutely integrable.

From the causality of f it follows that (f ∗ g)(t) =R∞0

f(τ)g(t− τ) dτ . For6.25

t < 0 this is 0. For t ≥ 0 it equalsR t

0f(τ)g(t− τ) dτ .

a We use the definition of convolution and then split the integral at τ = 0:6.26

(e−| v | ∗ e−| v |)(t) =

Z ∞

0

e−τe−| t−τ | dτ +

Z 0

−∞eτe−| t−τ | dτ.

First we take t ≥ 0. Then − | t− τ | = τ−t for τ < 0. Furthermore we havefor τ > t that − | t− τ | = t − τ and for 0 ≤ τ < t that − | t− τ | = τ − t.Hence,

(e−| v | ∗ e−| v |)(t) =

Z t

0

e−t dτ +

Z ∞

t

et−2τ dτ +

Z 0

−∞e2τ−t dτ.

A straightforward calculation of these integrals gives (1 + t)e−t.Next we take t < 0. Then − | t− τ | = t−τ for τ > 0. Furthermore we havefor τ < t that − | t− τ | = τ − t and for t ≤ τ < 0 that − | t− τ | = t − τ .Hence,

(e−| v | ∗ e−| v |)(t) =

Z ∞

0

et−2τ dτ +

Z 0

t

et dτ +

Z t

−∞e2τ−t dτ.

A straightforward calculation of these integrals gives (1− t)et.b Use the result from section 6.3.3 and the convolution theorem to obtainthe spectrum (2(1 + ω2)−1)2 = 4/(1 + ω2)2.c Since (1 + | t |)e−| t | = e−| t | + | t | e−| t | and the spectrum of e−| t | is2(1 + ω2)−1, we only need to determine the spectrum of f(t) = | t | e−| t |.But f(t) = tg(t) with g(t) the function from exercise 6.14, whose spec-trum we’ve already determined: G(ω) = −2iω(1 + ω2)−1. Apply theorem6.8 (differentiation rule in the frequency domain): the spectrum of f(t)is −G′(ω)/i. Calculating this and taking the results together we obtain4/(1 + ω2)2, in agreement with part b.

a From the differentiation rule in the frequency domain we obtain that6.28

the spectrum of tg(t) is iG′(ω) = −iω√

2πe−ω2/2. Since (Ftg(t))(0) = 0,

we may apply the integration rule to obtain that F1(ω) = −√

2πe−ω2/2.b Apply the differentiation rule in the frequency domain with n = 2, then

F2(ω) =√

2π(1− ω2)e−ω2/2.c Since f3(t) = f2(t− 1), it follows from the shift property that F3(ω) =e−iωF2(ω).

d From part a and exercise 6.9 it follows that F4(ω) = (−√

2πe−(ω−4)2/2+√2πe−(ω+4)2/2)/2i.

e Use the scaling property from table 4 with c = 4, then F5(ω) =G(ω/4)/4.

b Since p1(τ) = 0 for | τ | > 12

and 1 for | τ | < 12, we have6.29

(p1 ∗ p3)(t) =

Z 1/2

−1/2

p3(t− τ) dτ.


Here p3(t − τ) 6= 0 only if t − 3/2 ≤ τ ≤ t + 3/2. Moreover, we havethat −1/2 ≤ τ ≤ 1/2, and so we have to separate the cases as indicatedin the textbook: if t > 2, then (p1 ∗ p3)(t) = 0; if t < −2, then also

(p1 ∗ p3)(t) = 0; if −1 ≤ t ≤ 1, then (p1 ∗ p3)(t) =R 1/2

−1/21 dτ = 1; if

1 < t ≤ 2, then (p1 ∗ p3)(t) =R 1/2

t−3/21 dτ = 2 − t; finally, if −2 ≤ t < −1,

then (p1 ∗ p3)(t) =R t+3/2

−1/21 dτ = 2 + t.

c Apply the convolution theorem to T (t) = (p1∗p3)(t), then the spectrumof T (t) follows: 4 sin(ω/2) sin(3ω/2)/ω2.


From the spectra calculated in exerices 6.2 to 6.5 it follows immediately7.1that the limits for ω → ±∞ are indeed 0: they are all fractions with abounded numerator and a denominator that tends to ±∞. As an examplewe have from exercise 6.2 that limω→±∞ 1/(a + iω) = 0.

Use table 3 with a = 2A and substitute ω = s− t.7.2

Take C > 0, then it follows by first changing from the variable Au to v and7.3then applying (7.3) that

limA→∞

Z C

0

sin Au

udu = lim

A→∞

Z AC

0

sin v

vdv =

π

2.

Split 1/(a+iω) into the real part 1/(1+ω2) and the imaginary part −ω/(1+7.4ω2). The limit of A →∞ of the integrals over [−A, A] of these parts giveslimA→∞ 2 arctan A = π for the real part and limA→∞(ln(1 + A2)− ln(1 +(−A)2)) = 0 for the imaginary part.

a In exercise 6.9b it was shown that F (ω) = 2i sin(πω)/(ω2 − 1). The7.6function f(t) is absolutely integrable since

R∞−∞ | f(t) | dt =

R π

−π| sin t | dt <

∞. Moreover, f(t) is piecewise smooth, so all conditions of the fundamentaltheorem are satisfied. We now show that the improper integral of F (ω)exists. First, F (ω) is continuous on R according to theorem 6.10, so itis integrable over e.g. [−2, 2]. Secondly, the integrals

R∞2

F (ω) dω andR −2

−∞ F (ω) dω both exist. For the former integral this can be shown asfollows (the other integral can be treated similarly):˛ Z ∞

2

F (ω) dω

˛≤Z ∞

2

2

ω2 − 1dω

since | 2i sin(πω) | ≤ 2 (and ω2 − 1 > 0 for ω > 2). The integral in theright-hand side is convergent.b Apply the fundamental theorem, then

f(t) =1

2π

Z ∞

−∞

2i sin(πω)

ω2 − 1eiωt dω

for all t ∈ R (f is continuous). Now use that F (ω) is an odd function andthat 2 sin πω sin ωt = cos(π − t)ω − cos(π + t)ω, then

f(t) =1

π

Z ∞

0

cos(π − t)ω − cos(π + t)ω

1− ω2dω.

a In exercise 6.15b it was shown that Fs(ω) = (1 − cos aω)/ω. This7.8exercise used the odd extension to R. So f(t) is odd and using (7.12) wethus obtain

2

π

Z ∞

0

1− cos aω

ωsin ωt dω =

1

2(f(t+) + f(t−)).

Since f(t) is continuous for t > 0 and t 6= a we have for these values that

f(t) =2

π

Z ∞

0

1− cos aω

ωsin ωt dω.

27


b At t = a the function is discontinuous, so we have convergence to12(f(a+) + f(a−)) = 1

2.

If we take g(t) = 0 in theorem 7.4, then G(ω) = 0 and so we get the7.10statement: if F (ω) = 0 on R, then f(t) = 0 at all points where f(t) iscontinuous. We now prove the converse. Take f(t) and g(t) as in theorem7.4 with spectra F (ω) and G(ω) and assume that F (ω) = G(ω) on R.Because of the linearity of the Fourier transform, (F−G)(ω) is the spectrumof (f − g)(t); but (F −G)(ω) = F (ω)−G(ω) = 0. From our assumption itnow follows that (f − g)(t) = 0 at all points where (f − g)(t) is continuous.Hence f(t) = g(t) at all points where f(t) and g(t) are continuous, whichis indeed theorem 7.4.

The spectrum of pa(t) is 2 sin(aω/2)/ω (table 3). From duality it then7.11follows that the spectrum of sin(at/2)/t is πpa(ω) at the points wherepa(t) is continuous; at ω = ±a/2 we should take the value π/2. (We canapply duality since the Fourier integral exists as improper integral; this isexercise 7.5b).

The spectrum of qa(t) is F (ω) = 4 sin2(aω/2)/(aω2) (see table 3). From du-7.12ality it then follows that the spectrum of sin2(at/2)/t2 is (aπ/2)qa(ω). (Wecan apply duality since qa is continuous, piecewise smooth, and absolutelyintegrable and since its Fourier integral exists as improper integral; thislatter fact follows immediately if we use that F (ω) is even and continuousand that e.g. F (ω) ≤ 1/ω2 for ω ≥ 1).

The function 1/(a+ iω) is not integrable on R (see exercise 6.2), so duality7.14cannot be applied.

These results follow immediately from duality (and calculating the right7.15

constants). For example:p

π/ae−t2/4a ↔ 2πe−aω2, now divide by 2π.

This is an important exercise: it teaches to recognize useful properties.7.16a Complete the square, then one can apply a shift in time: f(t) = 1/(1 +(t − 1)2). Since the spectrum of 1/(1 + t2) is πe−|ω |, the result follows:πe−iωe−|ω |.b Here we have a shift from t to t− 3; from the spectrum of sin 2πt/t theresult follows: πe−3iωp4π(ω) with value 1

2at ω = ±2π.

c We now have 1/(t2 − 4t + 7), multiplied by a sine function. The sinefunction is easy to deal with using exercise 6.9 (a variant of the modulationtheorem). As in a we complete the square and note that 1/(3 + (t − 2)2)

has spectrum F (ω) = πe−2iωe−√

3|ω |/√

3. Hence, the result is now (F (ω−4)− F (ω + 4))/2i.d We use that 3πq6(ω) is the spectrum of sin2(3t)/t2 and apply a shift intime from t to t− 1, then the result is 3πe−iωq6(ω).

Again, this is an important exercise: it teaches to recognize useful proper-7.17ties for the inverse transform.a We immediately use table 3 to obtain that 1/(4 + ω2) is the spectrumof f(t) = e−2| t |/4.b Apply a shift in the frequency domain to the spectrum πp2a(ω) ofsin(at)/t, then it follows that f(t) = (eiω0t + e−iω0t) sin(at)/(πt), so f(t) =2 cos(ω0t) sin(at)/(πt).c As in part b it follows that f(t) = 3e9it/(π(t2 + 9)).

From the convolution theorem it follows that F(Pa ∗ Pb)(ω) = (FPa)(ω) ·7.19


(FPb)(ω) = e−(a+b)|ω |, where we also used table 3. But also (FPa+b)(ω) =e−(a+b)|ω |, and since F is one-to-one (theorem 7.4) it then follows thatPa+b = Pa ∗ Pb.

a Use the result of exercise 6.14 (G(ω) = −2iω/(1+ω2)), the fundamental7.21theorem and the fact that the spectrum is odd to change from

R∞−∞ to

R∞0

.It then follows that (use x instead of ω)Z ∞

0

x sin xt

1 + x2dt =

π

2e−t.

Since g is not continuous at t = 0, this result is not correct at t = 0. Hereone should take the average of the jump, which is 0.b We apply Parseval (formula (7.19)) and calculate

R∞−∞ | g(t) |2 dt =R 0

−∞ e2t dt +R∞0

e−2t dt, which is 1. InR∞−∞ |G(ω) |2 dω we can use the

fact that the integrand is even. Writing x instead of ω, the result follows.

Use Parseval (7.18) with f(t) = e−a| t | and g(t) = e−b| t | and calculate7.22 R∞−∞ f(t)g(t) dt = 2

R∞0

e−(a+b)t dt = 2/(a + b). The spectra of f and g are

2a/(a2 + ω2) and 2b/(b2 + ω2) (table 3).

a Since sin4 t/t4 is the square of sin2 t/t2 and (F sin2 t/t2)(ω) = πq2(ω)7.23(table 3), it follows from the convolution theorem in the frequency domainthat (F sin4 t/t4)(ω) = (π/2)(q2 ∗ q2)(ω).b The integral

R∞−∞ sin4 t/t4 dt is the Fourier transform of sin4 x/x4 cal-

culated at ω = 0, henceR∞−∞ sin4 x/x4 dx = (π/2)(q2 ∗ q2)(0). Using that

q2 is an even function we obtain that

(q2 ∗ q2)(0) =

Z ∞

−∞q2(t)q2(−t) dt = 2

Z 2

0

(1− t/2)2 dt.

This integral equals 4/3 and soR∞−∞ sin4 x/x4 dx = 2π/3.

From table 3 we know that e−| t |/2 ↔ 1/(1 + ω2). By the convolution7.24theorem we then know that the spectrum of f(t) = (e−| v |/2∗e−| v |/2)(t) is1/(1+ω2)2. Calculating this convolution product at t = 0 gives f(0) = 1/4(or use exercise 6.26a, where it was shown that f(t) = (1 + | t |)e−| t |/4).Now apply the fundamental theorem (formula (7.9)) at t = 0 and use thatthe integrand is even. We then obtain

1

π

Z ∞

0

1

(1 + ω2)2dω = f(0) =

1

4,

which is indeed the case a = b = 1 from exercise 7.22.

The Gauss function f(t) = e−at2 belongs to S and so we can apply Poisson’s7.26∗

summation formula. Since F (ω) =p

π/ae−ω2/4a (see table 3), it followsfrom (7.23) with T = 1 that

∞Xn=−∞

e−an2=p

π/a

∞Xn=−∞

e−π2n2/a.

Replacing a by πx the result follows.

Take f(t) = a/(a2+t2), then F (ω) = πe−a|ω | (see table 3); we can then use7.27∗

(7.22) with T = 1 (in example 7.8 the conditions were verified) to obtain


∞Xn=−∞

a

a2 + (t + n)2= π

1 +

∞Xn=1

e−2πn(a+it) +

∞Xn=1

e−2πn(a−it)

!.

Here we have also split a sum in terms with n = 0, n > 1 and n < −1, andthen changed from n to −n in the sum with n < −1. The sums in the right-hand side are geometric series with ratio r = e−2π(a+it) and r = e−2π(a−it)

respectively. Note that | r | < 1 since a > 0. Using the formula for thesum of an infinite geometric series (example 2.16), then writing the resultwith a common denominator, and finally multiplying everything out andsimplifying, it follows that

a

π

∞Xn=−∞

1

a2 + (t + n)2=

1− e−4πa

1 + e−4πa − e−2πa(e2πit + e−2πit).

Multiplying numerator and denominator by e2πa the result follows.

a To determine the spectrum we write sin t = (eit− e−it)/2 and calculate7.28the integral defining F (ω) in a direct way:

F (ω) =1

2i

„Z π

0

ei(1−ω)t dt−Z π

0

e−i(1+ω)t dt

«.

Writing the result with a common denominator and using the fact thateπi = e−πi = −1 gives F (ω) = (1 + e−iωπ)/(1 − ω2). From theorem 6.10we know that F (ω) is continuous, so we do not have to calculate F (ω) atthe exceptional points ω = ±1.b Apply the fundamental theorem, so (7.9), noting that f(t) is continuouson R. We then obtain

f(t) =1

2π

Z ∞

−∞

1 + e−iωπ

1− ω2eiωt dω.

Split the integral at t = 0 and change from ω to −ω in the integral over(−∞, 0]. Then

f(t) =1

2π

Z ∞

0

eiωt + e−iωt + eiω(t−π) + e−iω(t−π)

1− ω2dω,

which leads to the required result.c Take t = π/2 in part b and use that f(π/2) = 1, then the result follows.d Apply Parseval’s identity (7.19) to f and use that

R π

0sin2 t dt = π/2,

then it follows that

1

2π

Z ∞

−∞|F (ω) |2 dω =

π

2.

Since F (ω) can be rewritten as 2e−iωπ/2 cos(ωπ/2)/(1 − ω2) and we have

that˛e−iωπ/2

˛= 1, it follows that |F (ω) |2 = 4 cos2(ωπ/2)/(1−ω2)2. This

integrand being even, the result follows.

a We know from table 3 that p2a(t) ↔ 2 sin aω/ω and e−| t | ↔ 2/(ω2 +1).7.29From the convolution theorem it then follows that p2a(v)∗e−| v | ↔ 4f(ω) =G(ω).b We now determine g explicitly by calculating the convolution product(use the definition of p2a):

(p2a(v) ∗ e−| v |)(t) =

Z a

−a

e−| t−τ | dτ =

Z t+a

t−a

e−|u | du


where we changed to the variable u = t − τ . Now if −a ≤ t ≤ a, thent− a ≤ 0 ≤ t + a and so

(p2a(v) ∗ e−| v |)(t) =

Z 0

t−a

eu du +

Z t+a

0

e−u du = 2− 2e−a cosh t.

If t > a, then t− a > 0 and so

(p2a(v) ∗ e−| v |)(t) =

Z t+a

t−a

e−u du = 2e−t sinh a.

Finally, if t < −a, then t + a < 0 and so

(p2a(v) ∗ e−| v |)(t) =

Z t+a

t−a

eu du = 2et sinh a.

c The function g from part b is continuous at t = a since limt↓a g(t) =2e−a sinh a = e−a(ea − e−a) = 1 − e−2a and g(a) = limt↑a g(t) = 2 −2e−a cosh a = 2−e−a(ea +e−a) = 1−e−2a. In the same way it follows thatg(t) is continuous at t = −a. So g(t) is a piecewise smooth function whichmoreover is continuous. Also, g(t) is certainly absolutely integrable sincee−| t | is absolutely integrable over | t | > a. Finally, the Fourier integral ex-ists as improper Riemann integral since G(ω) is even absolutely integrable:|G(ω) | ≤ 4/

˛ω(1 + ω2)

˛. We can now apply the duality rule (theorem

7.5) and it then follows that G(−t) ↔ 2πg(ω), so f(−t) ↔ πg(ω)/2. Sincef(−t) = f(t) we thus see that F (ω) = π − πe−a cosh ω for |ω | ≤ a andF (ω) = πe−|ω | sinh a for |ω | > a.


b For t 6= 0 we have that lima↓0 Pa(t) = 0, while for t = 0 we have8.1that lima↓0 Pa(t) = ∞. Since

R ∞−∞ Pa(t) dt = 1, we see that Pa(t) fits the

description of the delta function.c From table 3 it follows that Pa(t) ↔ e−a|ω | and lima↓0 e

−a|ω | = 1.Combining this with part b shows that it is reasonable to expect that thespectrum of δ(t) is 1.

a Since φ(a) ∈ C for all φ ∈ S, it follows from (8.10) that δ(t − a) is a8.2mapping from S to C. For c ∈ C and φ ∈ S we have that

〈δ(t− a), cφ〉 = (cφ)(a) = c 〈δ(t− a), φ〉 ,

and for φ1, φ2 ∈ S we have

〈δ(t− a), φ1 + φ2〉 = (φ1 + φ2)(a) = 〈δ(t− a), φ1〉+ 〈δ(t− a), φ2〉 .

So δ(t− a) is a linear mapping from S to C, hence a distribution.b Taking the limit inside the integral in (8.1) givesZ ∞

−∞

„1

2πlim

a→∞

2 sin aω

ω

«f(t− ω) dω = f(t)

for any absolutely integrable and piecewise smooth function f(t) on R thatis continuous at t. Using (8.3) this can symbolically be written as (taket = a)Z ∞

−∞δ(ω)f(a− ω) dω = f(a)

and by changing from ω to a− t we then obtainZ ∞

−∞δ(a− t)f(t) dt = f(a).

Using δ(a− t) = δ(t−a), which by (8.3) is reasonable to expect (see section8.4 for a proof), this indeed leads to (8.11).

Since 〈1, φ〉 =R ∞−∞ φ(t) dt ∈ C for all φ ∈ S, it follows that 1 is a mapping8.4

from S to C. The linearity of this mapping follows from the linearity ofintegration: for c ∈ C and φ ∈ S we have that

〈1, cφ〉 =

Z ∞

−∞(cφ)(t) dt = c

Z ∞

−∞φ(t) dt = c 〈1, φ〉 ,

and for φ1, φ2 ∈ S we have

〈1, φ1 + φ2〉 =

Z ∞

−∞(φ1 + φ2)(t) dt =

Z ∞

−∞φ1(t) dt+

Z ∞

−∞φ2(t) dt,

so 〈1, φ1 + φ2〉 = 〈1, φ1〉 + 〈1, φ2〉. This proves that 1 is a linear mappingfrom S to C, hence a distribution.

For φ ∈ S there exists a constantM > 0 such that (e.g.) (1+t2) |φ(t) | ≤M8.5for all t ∈ R. Hence,˛ Z ∞

0

φ(t) dt

˛≤

Z ∞

0

|φ(t) | dt ≤M

Z ∞

0

1

1 + t2dt <∞

32


(the latter integral equals [arctan]∞0 = π/2). The integralR ∞0φ(t) dt thus

exists and one can now show that ε is indeed a distribution precisely as inexercise 8.4 (linearity of integration).

In example 8.4 it was already motivated why the integralR ∞−∞ | t |φ(t) dt8.7

exists: there exists a constant M > 0 such that (e.g.) (1 + t2) | tφ(t) | ≤Mfor all t ∈ R. Hence,˛ Z ∞

−∞| t |φ(t) dt

˛≤

Z ∞

−∞| tφ(t) | dt ≤M

Z ∞

−∞

1

1 + t2dt <∞.

So 〈| t | , φ〉 exists and one can now show that | t | is indeed a distributionprecisely as in exercise 8.4 (linearity of integration).

a For the integral over [−1, 1] we have8.9 Z 1

−1

| t |−1/2 dt = 2

Z 1

0

t−1/2 dt = [4t1/2]10 = 4,

hence, | t |−1/2 is integrable over [−1, 1]. SinceR ∞0t−1/2 dt = 2 limR→∞

√R

does not exist, | t |−1/2 is not integrable over R.

b We first show thatR ∞−∞ | t |−1/2 φ(t) dt exists for φ ∈ S. To do so, we

split the integral in an integral over [−1, 1] and over | t | ≥ 1. For the firstintegral we note that |φ(t) | ≤M1 for some constant M1 > 0. From part awe then get˛ Z 1

−1

| t |−1/2 φ(t) dt

˛≤

Z 1

−1

| t |−1/2 |φ(t) | dt ≤M1

Z 1

−1

| t |−1/2 dt <∞.

For the second integral we use that | t |−1/2 ≤ 1 for | t | ≥ 1. Hence,˛˛

Z| t |≥1

| t |−1/2 φ(t) dt

˛˛ ≤

Z| t |≥1

|φ(t) | dt ≤Z ∞

−∞|φ(t) | dt.

In example 8.1 it has been shown that the latter integral exists. This shows

thatD| t |−1/2 , φ

Eexists and one can now show that | t |−1/2 is indeed a

distribution (linearity of integration; see e.g. exercise 8.4).

a For φ ∈ S there exists a constant M > 0 such that (e.g.) (1 +8.10t2) | tφ(t) | ≤M for all t ∈ R. Hence,˛ Z ∞

−∞tφ(t) dt

˛≤

Z ∞

−∞| tφ(t) | dt ≤M

Z ∞

−∞

1

1 + t2dt <∞.

As in exercise 8.3 this shows that t defines a distribution.

a From the linearity for distributions it follows immediately that the com-8.12plex number 2φ(0) + i

√3φ′(0) + (1 + i)

R ∞0

((φ(t)− φ(−t))dt is assigned.b This defines a distribution if 1, t and t2 are distributions. The firstone is known from example 8.1, the other two from exercise 8.10. Fromdefinition 8.15 it thus follows that f(t) defines a distribution as well.

a From (8.17) it follows that the complex number −φ(3)(0) is assigned.8.14b This number −φ(3)(0) is meaningfull for all functions that are 3 timescontinuously differentiable.

a First apply example 8.3, then definition 8.4, and finally integration by8.15parts, then


˙(sgn t)′, φ

¸=

Z ∞

0

(φ′(−t)− φ′(t)) dt = [φ(t)]0−∞ − [φ(t)]∞0 = 2φ(0),

hence, (sgn t)′ = 2δ(t).b Since sgn t = 2ε(t) − 1 (verify this), it follows from the linearity ofdifferentiation that (sgn t)′ = 2ε′(t) = 2δ(t). Here we used that 1′ = 0 andthat ε′(t) = δ(t) (see (8.18)).c Since | t |′ = sgn t it follows from part a that | t |′′ = (sgn t)′ = 2δ(t).

Since the function | t | from example 8.9 is continuously differentiable out-8.16side t = 0, it follows from the jump formula that | t |′ = sgn t (at t = 0there is no jump and outside t = 0 this equality holds for the ordinaryderivatives). The function from example 8.10 has a jump of magnitude 1at t = 0, while for t < 0 the derivative is 0 and for t > 0 the derivative is− sin t. Hence, the jump formula implies that (ε(t) cos t)′ = δ(t)− ε(t) sin t.

a The function pa has a jump of magnitude 1 at t = −a/2 and of mag-8.17nitude −1 at t = a/2. Outside t = 0 the ordinary derivative is 0, so itfollows from the jump formula that pa(t)′ = δ(t+ a/2)− δ(t− a/2).b The function ε(t) sin t has no jump at t = 0, for t < 0 the ordinaryderivative is 0 and for t > 0 the ordinary derivative is cos t, so it followsfrom the jump formula that (ε(t) sin t)′ = ε(t) cos t.

a This is entirely analagous to exercises 8.10 and 8.12b.8.18b The function is differentiable outside t = 1 and the ordinary derivativeis 1 for t < 1 and 2t − 2 for t > 1. We denote this derivative as thedistribution Tf ′ . At t = 1 the jump is 2, so according to the jump formulathe derivative is Tf ′ + 2δ(t− 1).

From the jump formula it follows that the derivative as distribution is given8.19by aε(t)eat + δ(t), so f ′(t)− af(t) = δ(t) as distributions.

Subsequently apply definition 8.6 and the definition of δ(t − a) in (8.10),8.22then

〈p(t)δ(t− a), φ〉 = (pφ)(a) = p(a)φ(a) = p(a) 〈δ(t− a), φ〉 .

Now use definition 8.5, then 〈p(t)δ(t− a), φ〉 = 〈p(a)δ(t− a), φ〉, whichshows that p(t)δ(t− a) = p(a)δ(t− a).

a The definition becomes: 〈f(t)δ′(t), φ〉 = 〈δ′(t), fφ〉. The product fφ of8.23two continuously differentiable functions is continuously differentiable, sothis definition is correct and it gives a mapping from S to C. The linearityfollows immediately from the linearity of δ′(t), so f(t)δ′(t) is a distribution.b According to part a we have 〈f(t)δ′(t), φ〉 = −(fφ)′(0), where we alsoapplied δ′(t) to fφ. Now apply the product rule for differentiation and writethe result as −f ′(0) 〈δ(t), φ〉+ f(0) 〈δ′(t), φ〉 = 〈f(0)δ′(t)− f ′(0)δ(t), φ〉.c If f(t) = t, then f(0) = 0 and f ′(0) = 1, so tδ′(t) = −δ(t); if f(t) = t2

then f(0) = 0 and f ′(0) = 0, so t2δ′(t) = 0.

First apply definition 8.6 and then the definition of pv(1/t) from example8.258.5 to obtain

〈t · pv(1/t), φ〉 = limα↓0

Z| t |≥α

tφ(t)

tdt = lim

α↓0

Z| t |≥α

φ(t) dt.

Since φ ∈ S is certainly integrable over R, the limit exists and it will beequal to

R ∞−∞ φ(t) dt. Hence, t · pv(1/t) = 1.

Let T be an even distribution, then T (−t) = T (t) (definiton 8.8), so8.26


〈T (t), φ(t)〉 = 〈T (t), φ(−t)〉 for all φ ∈ S, where we used definition 8.7.Similarly for odd T .

a From the definition of sgn t in example 8.3 it follows that 〈sgn t, φ(t)〉 =8.27−〈sgn t, φ(−t)〉 for all φ ∈ S. This shows that sgn t is odd according toexercise 8.26. Similarly for pv(1/t) (change from t to −t in the integralsdefining pv(1/t)).b From the definition of | t | in example 8.4 it follows that 〈| t | , φ(t)〉 =〈| t | , φ(−t)〉 for all φ ∈ S (change from t to −t in the integral defining | t |).This shows that | t | is even according to exercise 8.26.

a Applying (8.12) to f(t) gives8.29

〈Tf , φ〉 =

Z 0

−∞2tφ(t) dt+

Z ∞

0

t2φ(t) dt

and in e.g. exercises 8.10, 8.12b and 8.18a we have seen that such integralsare well-defined for φ ∈ S. This gives a mapping from S to C and thelinearity of this mapping follows precisely as in e.g. exercise 8.3 or 8.4.Hence, f indeed defines a distribution Tf .b Apply the jump formula (8.21): outside t = 0 the function f is continu-ously differentiable with derivative f ′(t) = 2t for t > 0 and f ′(t) = 2 fort < 0. Note that f ′ again defines a distribution Tf ′ . At t = 0 the functionhas no jump, hence (8.21) implies that T ′f = Tf ′ .c Again we have that the function f ′ is continuously differentiable out-side t = 0 and f ′′(t) = 2 for t > 0 and f ′′(t) = 0 for t < 0. Let Tf ′′

be the distribution defined by f ′′. At t = 0 the function f ′ has a jumpf ′(0+)− f ′(0−) = 0− 2 = −2, and according to (8.21) (applied to Tf ′ andusing that T ′f = Tf ′ and so T ′′f = T ′f ′) we have that

T ′′f = T ′f ′ = Tf ′′ + (f ′(0+)− f ′(0−))δ(t) = Tf ′′ − 2δ(t).

The second derivative of f considered as distribution is the same as thesecond derivative of f outside t = 0, minus the distribution 2δ(t) at t = 0.

a Since δ′′(t) can be defined for all twice continuously differentiable func-8.30tions, the product f(t)δ′′(t) can also be defined for all twice continuouslydifferentiable functions f(t) by 〈f(t)δ′′(t), φ(t)〉 = 〈δ′′(t), f(t)φ(t)〉. This isbecause it follows from the product rule that the product f(t)φ(t) is againtwice continuously differentiable.b From part b and the definition of the second derivative of a distribution(formula (8.17) for k = 2) we obtain 〈f(t)δ′′(t), φ(t)〉 = 〈δ(t), (f(t)φ(t))′′〉.Since (f(t)φ(t))′′ = f ′′(t)φ(t) + 2f ′(t)φ′(t) + f(t)φ′′(t) we thus obtainthat 〈f(t)δ′′(t), φ(t)〉 = f ′′(0)φ(0) + 2f ′(0)φ′(0) + f(0)φ′′(0), which equals〈f ′′(0)δ(t)− 2f ′(0)δ′(t) + f(0)δ′′(t), φ(t)〉 (φ ∈ S). This proves the iden-tity.c Apply part b to the function f(t) = t2 and use that f(0) = f ′(0) = 0and f ′′(0) = 2, then t2δ′′(t) = 2δ(t). Next apply b to f(t) = t3 and usethat f(0) = f ′(0) = f ′′(0) = 0, then it follows that t3δ′′(t) = 0.d According to definition 8.7 we have that˙δ′′(at), φ(t)

¸= | a |−1 ˙

δ′′(t), φ(a−1t)¸.

Now put ψ(t) = φ(a−1t), then the right-hand side equals | a |−1 ψ′′(0).Next we use the chain rule twice to obtain that ψ′′(0) = a−2φ′′(0). Hence〈δ′′(at), φ(t)〉 = | a |−1 a−2 〈δ′′, φ〉.


Let φ ∈ S. From theorem 6.12 it follows that the spectrum Φ belongs9.1to S. Since T is a distribution, we then have that 〈T, Φ〉 ∈ C, and so〈FT, φ〉 = 〈T, Φ〉 ∈ C as well. So FT is a mapping from S to C. Thelinearity of FT follows from the linearity of T and F ; we will only givethe necessary steps for 〈FT, cφ〉, since the rule for 〈FT, φ1 + φ2〉 followssimilarly.

〈FT, cφ〉 = 〈T,F(cφ)〉 = 〈T, cΦ〉 = c 〈T, Φ〉 = c 〈FT, φ〉 .

a Use table 5 to obtain that δ(t− 4) ↔ e−4iω.9.3b Again use table 5 to obtain that e3it ↔ 2πδ(ω − 3).c First write the sine function as combination of exponentials, so sin at =(eiat − e−iat)/2i. From linearity and table 5 it then follows that sin at ↔−πi(δ(ω − a)− δ(ω + a)).d First determine the spectrum of pv(1/t) and 4 cos 2t = 2e2it + 2e−2it

using table 5 and then (again) apply linearity to obtain the spectrum4π(δ(ω − 2) + δ(ω + 2)) + 2πsgn ω.

a From example 9.1 (or table 5) we obtain the result e−5it/2π.9.4b See example 9.2: 2 cos 2t.c The spectrum of pv(1/t) is −πisgn ω (table 5). Note that 2 cos ω =eiω + e−iω and that the spectrum of δ(t− a) is e−iaω (table 5). Hence theanswer is iπ−1pv(1/t) + δ(t− 1) + δ(t + 1).

Let T be an even distribution with spectrum U . We have to show that9.5U(−ω) = U(ω), so 〈U, φ(t)〉 = 〈U, φ(−t)〉 for all φ ∈ S (see exercise8.26). But 〈U, φ(−t)〉 = 〈T,Fφ(−t)〉 and from table 4 we know that(Fφ(−t))(ω) = Φ(−ω) if Φ is the spectrum of φ. Since T is even, wehave that 〈T, Φ(−ω)〉 = 〈T, Φ(ω)〉. From these observations it follows that〈U, φ(−t)〉 = 〈T, Φ(ω)〉 = 〈U, φ(t)〉, which shows that U is even. Similarlyfor odd T .

It is obvious that ε(t) = (1 + sgn t)/2 by looking at the cases t > 0 and9.7t < 0. Since 2πδ(ω) is the spectrum of 1 and −2ipv(1/ω) is the spectrumof ε(t), it follows that ε(t) has spectrum πδ(ω)− ipv(1/ω).

a Let φ ∈ S have spectrum Φ. From definition 9.1 and the action of δ′′ it9.8follows that 〈Fδ′′, φ〉 = 〈δ′′, Φ〉 = 〈δ, Φ′′〉 = Φ′′(0). From the differentiationrule in the frequency domain (table 4) with k = 2 we see that Φ′′(ω) =F((−it)2φ(t))(ω), and hence δ′′ ↔ −ω2 is proven as follows:˙Fδ′′, φ

¸= F(−t2φ(t))(0) = −

Z ∞

−∞t2φ(t) dt = −

Z ∞

−∞ω2φ(ω) dω,

so 〈Fδ′′, φ〉 =˙−ω2, φ

¸for all φ ∈ S, proving the required result.

Parts b and c can be proven using similar steps.

a Subsequently apply definitions 9.1 and 8.7: 〈FT (at), φ〉 = 〈T (at), Φ〉 =9.9| a |−1 ˙

T, Φ(a−1ω)¸

(φ ∈ S having spectrum Φ). From table 4 we see thatΦ(a−1ω) = | a | (Fφ(at))(ω), so it follows that 〈FT (at), φ〉 = 〈T,Fφ(at)〉 =〈U, φ(at)〉, where we again used definition 9.1 in the final step. Now againapply definition 8.7, then 〈FT (at), φ〉 = | a |−1 ˙

U(a−1ω), φ¸, which proves

T (at) ↔ | a |−1 U(a−1ω).

36


b We have that δ(4t+3) is the distribution δ(t+3) scaled by 4. Accordingto the shift rule in the time domain (see table 6) it follows from δ(t) ↔ 1that δ(t+3) ↔ e3iω. From part a it then follows that δ(4t+3) ↔ 4−1U(ω/4)with U(ω) = e3iω. Hence, δ(4t + 3) ↔ 4−1e3iω/4. (This can also be solvedby considering δ(4t + 3) as the distribution δ(4t) shifted over −3/4.)

From table 5 it follows that δ′(t) ↔ iω. Using (9.12) we then obtain that9.11−itδ′(t) ↔ (iω)′ = i, so tδ′(t) ↔ −1. Exercise 8.23c gives: tδ′(t) = −δ(t)and since δ(t) ↔ 1 we indeed get tδ′(t) ↔ −1 again. Similarly we gettδ′′(t) ↔ −2iω using (9.12) or using exercise 8.30b: tδ′′(t) = −2δ′(t).

From iωT = 1 we may not conclude that T = 1/(iω) since there exist9.12distributions S 6= 0 such that ωS = 0 (e.g. δ(ω)).

The linearity follows as in definition 8.6. The main point is that one has9.13to show that eiatφ(t) ∈ S whenever φ ∈ S. So we have to show that for

any m, n ∈ Z+ there exists an M > 0 such that˛tn(eiatφ(t))(m)

˛< M .

From the product rule for differentiation it follows that (eiatφ(t))(m) is asum of terms of the form ceiatφ(k)(t) (k ∈ Z+). It is now sufficient to show

that˛tneiatφ(k)(t)

˛< M for some M > 0 and all k, n ∈ Z+. But since˛

eiat˛= 1 this means that we have to show that

˛tnφ(k)(t)

˛< M for some

M > 0 and all k, n ∈ Z+, which indeed holds precisely because φ ∈ S.

From definition 9.1 and the definition of eiatT (see exercise 9.13) it follows9.15that

˙FeiatT, φ

¸=

˙eiatT, Φ

¸=

˙T, eiatΦ

¸(φ ∈ S having spectrum Φ). Ac-

cording to the shift property in the frequency domain (table 4) we have thateiatΦ(t) = F(φ(ω+a))(t) (note that for convenience we’ve interchanged therole of the variables ω and t). Hence,

˙FeiatT, φ

¸= 〈T,F(φ(ω + a))(t)〉 =

〈U, φ(ω + a)〉 = 〈U(ω − a), φ〉, where we used definition 9.2 in the last step.So we indeed have eiatT ↔ U(ω − a).

a Use table 5 for ε(t) and apply a shift in the time domain, then it follows9.16that ε(t− 1) ↔ e−iω(πδ(ω)− ipv(1/ω)).b Use table 5 for ε(t) and apply a shift in the frequency domain, then itfollows that eiatε(t) ↔ πδ(ω − a)− ipv(1/(ω − a)).c We have ε(t) ↔ πδ(ω) − ipv(1/ω) and if we now write the cosine asa combination of exponentials, then we can use a shift in the frequencydomain (as in part b) to obtain that ε(t) cos at ↔ 1

2(πδ(ω−a)− ipv(1/(ω−

a)) + πδ(ω + a)− ipv(1/(ω + a))).d Use that 1 ↔ 2πδ(ω) and δ′(t) ↔ iω (table 5), so 3i ↔ 6iπδ(ω) and(apply a shift) δ′(t− 4) ↔ e−4iωiω; the sum of these gives the answer.e First note that ε(t)sgn t = ε(t) and the spectrum of this is known;furthermore we have that t3 ↔ 2πi3δ(3)(ω) (table 5), so the result is2π2δ(3)(ω) + πδ(ω)− ipv(1/ω).

a Use table 5 for the sign function and apply a shift: 12ieitsgn t.9.17

b Write sin t as a combination of exponentials and apply a shift to 12isgn t,

then we obtain the result 14(sgn(t + 3)− sgn(t− 3)).

c Apply reciprocity to ε(t), then we obtain (πδ(−t) − ipv(−1/t))/2π ↔ε(ω). Now δ(−t) = δ(t) and pv(−1/t) = −pv(1/t), hence, the result is:12iπ−1pv(1/t) + 1

2δ(t).

d Apply the scaling property (table 6) to 1 ↔ 2πδ(ω) to obtain 1 ↔6πδ(3ω). Next we apply a shift in the frequency domain (table 6), which


results in e2it/3 ↔ 6πδ(3ω − 2). Using the differentiation rule in the fre-quency domain (table 6) we obtain from 1/2π ↔ δ(ω) that (−it)2/2π ↔δ′′(ω). From linearity it then follows that (3−1e2it/3 − t2)/2π ↔ δ(3ω −2) + δ′′(ω).

We know that δ′ ∗ T = T ′, so δ′ ∗ | t | = | t |′ = sgn t (by example 8.9).9.19

According to definition 9.3 we have that9.20∗

〈T (t) ∗ δ(t− a), φ〉 = 〈T (τ), 〈δ(t− a), φ(t + τ)〉〉 .

Since 〈δ(t− a), φ(t + τ)〉 = φ(a + τ), the function τ → 〈δ(t− a), φ(t + τ)〉belongs to S, so T (t) ∗ δ(t− a) exists and

〈T (t) ∗ δ(t− a), φ〉 = 〈T (τ), φ(a + τ)〉 = 〈T (τ − a), φ(τ)〉

(the last step uses definition 9.2). This proves that T (t)∗δ(t−a) = T (t−a).

Use exercise 9.20 with T (t) = δ(t − b). The convolution theorem leads to9.21∗

the obvious e−ibωe−iaω = e−i(a+b)ω.

a Use table 5: δ(t− 3) ↔ e−3iω.9.24b Since δ(t + 4) ↔ e4iω (as in part a) and cos t = (eit + e−it)/2 we applya shift in the frequency domain: cos tδ(t + 4) ↔ (e4i(ω−1) + e4i(ω+1))/2.c From table 5 we have ε(t) ↔ πδ(ω)−ipv(1/ω). Apply the differentiationrule in the time domain (with n = 2), then we obtain t2ε(t) ↔ −πδ′′(ω) +ipv(1/ω)′′.d Apply the differentiation rule in the time domain to the result obtainedin exercise 9.16c, then it follows that (2ε(t) cos t)′ ↔ iω(πδ(ω−1)+πδ(ω +1)− ipv(1/(ω − 1))− ipv(1/(ω + 1))).e Since δ(t) ↔ 1 it follows from first the scaling property and then ashift in the time domain that δ(7(t − 1/7)) ↔ e−iω/77. Finally apply thedifferentiation rule in the time domain to obtain the result: (δ(7t− 1))′ ↔iωe−iω/77.f This is a convergent Fourier series and so we can determine the spectrumterm-by-term. Since 1 ↔ 2πδ(ω) and e(2k+1)it ↔ 2πδ(ω − (2k + 1)) weobtain the following result:

π2δ(ω)− 4

∞Xk=−∞

(2k + 1)−2δ(ω − (2k + 1)).

a From table 5 we know that eit ↔ 2πδ(ω − 1) and similarly for e−it.9.25Hence, iπ−1 sin t ↔ δ(ω − 1)− δ(ω + 1).b Apply the differentiation rule in the time domain (with n = 2) to δ(t) ↔1, then −δ′′(t) ↔ ω2.c From table 5 we obtain that δ(t + 1

2)/4 ↔ eiω/2/4.

d From table 5 (and linearity) we obtain that (δ(t + 1)− δ(t− 1))/2i ↔(eiω−e−iω)/2i, which is sin ω. Now apply differentiation in the time domain(with n = 3), then we obtain that (δ(3)(t + 1)− δ(3)(t− 1))/2 ↔ ω3 sin ω.e From exercise 9.4c and a shift in the frequency domain it follows thate4it(δ(t + 1) + δ(t− 1))/2 ↔ cos(ω − 4).

a In exercise 9.25b it was shown that −δ(t)′′ ↔ ω2. Applying a shift in9.26the frequency domain leads to −eitδ′′(t) ↔ (ω − 1)2.b From the differentiation rule in the time domain and table 5 it followsas in exercise 9.25b that δ′(t) ↔ iω and δ′′(t) ↔ −ω2, so −δ′′(t)+2iδ′(t)+δ(t) ↔ ω2 − 2ω + 1.


c Since (ω − 1)2 = ω2 − 2ω + 1, the results in part a and b should bethe same. Using exercise 8.30b with f(t) = eit we indeed obtain thateitδ′′(t) = δ′′(t)− 2iδ′(t)− δ(t).

a From exercise 9.25b it follows that δ′′(t) ↔ −ω2. The convolution9.27theorem then implies that T ∗ δ′′(t) ↔ −ω2U where U is the spectrum ofT . This also follows by applying the differentiation rule in the time domainto T ′′, which equals T ∗ δ′′(t) by (9.21).b As noted in part a we have that δ′′ ∗ | t | = | t |′′. In exercise 8.15c itwas shown that | t |′′ = 2δ, so we indeed get δ′′ ∗ | t | = 2δ. Now let V bethe spectrum of | t |. Since δ′′ ↔ −ω2 and δ ↔ 1 it then follows as in parta from the convolution theorem that ω2V = −2.


a When the system is causal, then the response to the causal signal δ(t)10.1is again causal, so h(t) is causal. On the other hand, if h(t) is causal, thenit follows that y(t) = (u ∗ h)(t) =

R t

−∞ h(t− τ)u(τ) dτ and if we now havea causal input u, then the integral will be 0 for t < 0 and so y(t) is causalas well, proving that the system is causal.b When the system is real, then the response to the real signal δ(t) isagain real, so h(t) is real. On the other hand, if h(t) is real, then it followsfrom the integral for y(t) = (u ∗ h)(t) that if u(t) is real, then y(t) is alsoreal, proving that the system is real.

a If we substitute u(t) = δ(t) then it follows that h(t) = δ(t−1)+ε(t)e−2t,10.2so h(t) is causal and real and according to exercise 10.1 the system is thencausal and real.b We substitute u(t) = ε(t), then it follows for t ≥ 0 that a(t) = ε(t −1) +

R t

0e−2(t−τ) dτ = ε(t − 1) + 1

2(1 − e−2t), while for t < 0 the integral

is 0 and so a(t) = ε(t − 1). This result can be written for all t as a(t) =ε(t− 1) + 1

2ε(t)(1− e−2t).

We can express p2(t) as p2(t) = ε(t + 1) − ε(t − 1). Since a(t) is (by10.3definition) the response to ε(t) and we have a linear system, the responseto p2(t) = ε(t + 1)− ε(t− 1) is a(t + 1)− a(t− 1).

a We differentiate a(t) in distribution sense, which results in h(t) =10.5δ(t) − ε(t)e−3t(2 sin 2t + 3 cos 2t), since a(t) has a jump at t = 0 of mag-nitude 1 and we can differentiate in ordinary sense outside t = 0.b We use theorem 10.1, which implies that we may ignore the delta com-ponent and only have to show that ε(t)e−3t(2 sin 2t + 3 cos 2t) is absolutelyintegrable. Since both

R ∞0

˛e−3t sin 2t

˛dt and

R ∞0

˛e−3t cos 2t

˛dt exist

(e.g., both are smaller thanR ∞0

e−3t dt), this is indeed the case and hencethe system is stable.

a The impulse response is h1 ∗ h2.10.6b If the input for the first system is bounded, then the output is boundedsince the system is stable. This output is then used as input for the secondsystem, which is again stable. So the output of the second system is againbounded. This means that the cascade system itself is stable: the responseto a bounded input is bounded.

a The spectrum of h is H(ω) = 1+1/(1+iω)2 since δ ↔ 1 and te−tε(t) ↔10.71/(1 + iω)2.b Since eiωt 7→ H(ω)eiωt it follows that the response is given by eiωt(1 +1/(1 + iω)2).

a We need to determine the inverse Fourier transform of the function10.8H(ω) = cos ω/(ω2 + 1). First note that 1

2e−| t | ↔ 1/(ω2 + 1). Writing the

cosine as a combination of exponentials we obtain from the shift rule thath(t) = 1

2(e−| t+1 | + e−| t−1 |).

b It now suffices to use the time-invariance of the system: the reponse toδ(t) is h(t), so the response to δ(t− 1) is h(t− 1).

a We need to determine the inverse Fourier transform of the function from10.9figure 10.3, which is H(ω) = qωc(ω). This is h(t) = 2 sin2(ωct/2)/(πωct

2).

40


b The function u has a Fourier series with terms cneint (note that ω0 = 1).But the response to eint is H(n)eint and H(n) = 0 for n > 1 and n < −1.So we only have to determine c0, c1 and c−1. These can easily be calculatedfrom the defining integrals: c0 = 1

2and c−1 = c1 = −1/π. Hence, the

response follows: y(t) = H(0)c0+H(1)c1eit +H(−1)c−1e

−it = 12− 2

3πcos t.

a Put s = iω and apply partial fraction expansion to the system function10.11(s+1)(s− 2)/(s− 1)(s+2). A long division results in 1− 2s/(s− 1)(s+2)and a partial fraction expansion then gives

(s + 1)(s− 2)

(s− 1)(s + 2)= 1− 2

3

1

s− 1− 4

3

1

s + 2= 1− 2

3

1

iω − 1− 4

3

1

iω + 2.

Now δ(t) ↔ 1 and ε(t)e−2t ↔ 1/(iω + 2) (table 3, no. 7) and from timereversal (scaling with a = −1 from table 4, no. 5) it follows for 1

iω−1=

−1i(−ω)+1

that −ε(−t)et ↔ 1iω−1

. Hence, h(t) = δ(t) + 23ε(−t)et − 4

3ε(t)e−2t.

b The impulse reponse h(t) is not causal, so the system is not causal.c The modulus of H(ω) is 1, so it is an all-pass system and from Parsevalit then follows that the energy-content of the input is equal to the energy-content of the output (if necessary, see the textbook, just above example10.7).

a From the differential equation we immediately obtain the frequency10.13response:

H(ω) =ω2 − ω2

0

ω2 − i√

2ω0ω − ω20

.

b Write the cosine as a combination of exponentials, then it follows fromeiωt 7→ H(ω)eiωt that y(t) = (H(ω0)e

iω0t + H(−ω0)e−iω0t)/2. However

H(±ω0) = 0, so y(t) = 0 for all t.c Note that we cannot use the method from part b. Instead we use(10.6) to determine the spectrum of the response y(t). From table 5 weobtain that ε(t) ↔ pv(1/iω) + πδ(ω). Write the cosine as a combinationof exponentials, then it follows from the shift rule that the spectrum ofu(t) = cos(ω0t)ε(t) is given by U(ω) = pv(1/(2i(ω − ω0)) + pv(1/(2i(ω +ω0)) + (π/2)δ(ω − ω0) + (π/2)δ(ω + ω0). To determine Y (ω) = H(ω)U(ω)we use that H(ω)δ(ω ± ω0) = H(±ω0)δ(ω ± ω0) = 0. Hence, writingeverything with a common denominator, Y (ω) = ω/i(ω2 − i

√2ω0ω − ω2

0).Put s = iω and apply partial fraction expansion to obtain that 2Y (ω) =(1+ i)/(s+ω0(1− i)/

√2)+(1− i)/(s+ω0(1+ i)/

√2). The inverse Fourier

transform of this equals e−ω0t/√

2(cos(ω0t/√

2)− sin(ω0t/√

2))ε(t).

a From the differential equation we immediately obtain the frequency10.14response:

H(ω) =iω + 1 + α

2iω + α=

1

2+

1 + α/2

2iω + α,

where we also used a long division. Using the tables it then follows thath(t) = 1

2(δ(t) + (1 + α/2)ε(t)e−tα/2).

b We have to interpret the differential equation ‘the other way around’, sowith input and output interchanged. This means that the system functionis now 1/H(ω), so

2iω + α

iω + 1 + α= 2− 2 + α

iω + 1 + α


where we also used a long division. Using the tables it then follows thath1(t) = 2δ(t)− (2 + α)ε(t)e−t(1+α).c Note that the spectrum of (h ∗ h1)(t) is the function H(ω) · (1/H(ω)),which is 1. Since δ(t) ↔ 1, it follows that (h ∗ h1)(t) = δ(t).

We use separation of variables, so we substitute u(x, y) = X(x)Y (y) into10.16uyy + uxx = 0. This gives for some arbitrary constant c (the separationconstant) that X ′′ + cX = 0, Y ′′ − cY = 0. In order to satisfy the linearhomogeneous condition as well, X(x)Y (y) has to be bounded, and thisimplies that both X(x) and Y (y) have to be bounded functions. Solvingthe differential equations we obtain from the boundedness condition thatX(x) = 1 if c = 0 and ei

√cx, e−i

√cx if c > 0. Similarly Y (y) = 1 if c = 0 and

e√

cy, e−√

cy if c > 0. But e√

cy is not bounded for y > 0 so Y (y) = e−√

cy forc ≥ 0. We put c = s2, then it follows that the class of functions satisfyingthe differential equation and being bounded, can be described by

X(x)Y (y) = eisxe−| s |y, where s ∈ R.

By superposition we now try a solution u(x, y) of the form

u(x, y) =

Z ∞

−∞e−| s |yF (s)eisx ds.

If we substitute y = 0 in this integral representation, then we obtain that

u(x, 0) =1

1 + x2=

Z ∞

−∞F (s)eisx ds.

Since 12e−| t | ↔ 1/(1 + ω2) this means that

1

1 + ω2=

1

2

Z ∞

−∞e−| t |e−iωt dt.

The (formal) solution is thus given by

u(x, y) = 12

Z ∞

−∞e−| s |(1+y)eisx ds =

Z ∞

0

e−| s |(1+y) cos(sx) ds.

a We substitute u(t) = δ(t),then10.18

h(t) =

Z t

t−1

e−(t−τ)δ(τ) dτ =

Z ∞

−∞(ε(τ + 1− t)− ε(τ − t))e−(t−τ)δ(τ) dτ,

so h(t) = (ε(t) − ε(t − 1))e−t. Applying table 3, no. 7 and a shift in thetime domain gives H(ω) = (1− e−(1+iω))/(1 + iω).b The impulse response is causal, so the system is causal.c It is straightforward to verify that the impulse response is absolutelyintegrable, so the system is stable.d The response y(t) to the block function p2(t) is equal to the convolutionof h(t) with p2(t), which equals

y(t) =

Z 1

−1

h(t− τ) dτ =

Z t+1

t−1

h(τ) dτ.

This is 0 for t < −1 or for t > 2. For −1 < t < 0 it equals 1− e−(t+1), for0 ≤ t < 1 it equals 1− e−1, and for 1 ≤ t < 2 it equals e−(t−1) − e−1.

a The impulse response is the derivative in the sense of distributions of10.19a(t) = e−tε(t), which is δ(t)− e−tε(t).


b The frequency response is H(ω) = iω/(1 + iω). (Apply e.g. the differ-entiation rule to h(t) = a′(t).)c We have Y (ω) = H(ω)U(ω) (using obvious notations), so Y (ω) =iω/(1 + iω)2 = 1/(1 + iω) − 1/(1 + iω)2. The response is the inverseFourier transform of Y (ω): y(t) = (1− t)e−tε(t).

a The frequency response H(ω) is the triangle function qωc(ω). The in-10.21verse Fourier transform follows from table 3: h(t) = 2 sin2(ωct/2)/(πωct2).b Since a′(t) = h(t) and h(t) ≥ 0, the function a(t) is a monotone increas-ing function.

The frequency response follows immediately from the differential equation:10.22

H(ω) =1 + iω

−ω2 + 2iω + 2.

Applying partial fraction expansion (use s = iω) we obtain

H(ω) =1

2(iω + 1− i)+

1

2(iω + 1 + i).

The inverse Fourier transform is then h(t) = (e−t cos t)ε(t). Integratingthis over (−∞, t] gives the step response a(t) = 1

2(1+e−t(sin t− cos t))ε(t).

b We have Y (ω) = H(ω)U(ω) (using obvious notations), so Y (ω) =1/(−ω2 + 2iω + 2). Applying partial fraction expansion we obtain

Y (ω) =1

2i(iω + 1− i)− 1

2i(iω + 1 + i).

The response is the inverse Fourier transform of Y (ω), which gives asrepsonse the function y(t) = (e−t sin t)ε(t).

a Since | iω − 1− i | = | iω + 1− i | and˛e−iωt0

˛= 1 we have |H(ω) | = 110.23

and so L is an all-pass system.b First write (iω − 1− i)/(iω + 1− i) as 1− 2/(iω + 1− i) and then usethe inverse Fourier transform (the tables) to obtain δ(t)− 2e−(1−i)tε(t) ↔1− 2/(iω + 1− i). From the shift property in the time domain we obtainh(t) = δ(t− t0)− 2e−(1−i)(t−t0)ε(t− t0).c Ignoring the delta function it is easy to verify that h(t) is absolutely

integrable (˛ei(t−t0)

˛= 1), hence the system is stable.

d Write u(t) = 1 + eit + e−it and use that eiωt 7→ H(ω)eiωt, then y(t) =H(0)+H(1)eit +H(−1)e−it, which equals −i−e−it0eit +(4i−3)eit0e−it/5.

We use separation of variables, so we substitute u(x, y) = X(x)Y (y) into10.24uxx − 2uy = 0. This gives for some arbitrary constant c (the separationconstant) that X ′′+cX = 0, 2Y ′+cY = 0. Here X(x) and Y (y) have to bebounded functions. Solving the differential equations, we obtain from theboundedness condition that X(x) = 1 if c = 0 and ei

√cx, e−i

√cx if c > 0.

For c ≥ 0 we obtain for Y (y) the solution e−cy/2. We put c = s2 withs ∈ R. It then follows that the class of functions satisfying the differentialequation and being bounded, can be described by

X(x)Y (y) = eisxe−s2y/2, where s ∈ R.

By superposition we now try a solution u(x, y) of the form

u(x, y) =

Z ∞

−∞e−s2y/2F (s)eisx ds.


If we substitute y = 0 in this integral representation, then we obtain that

u(x, 0) = xe−xε(x) =

Z ∞

−∞F (s)eisx ds.

Since te−tε(t) ↔ 1/(1+iω)2 this means that F (s) = 12π(1+is)2

. The (formal)

solution is thus given by

u(x, y) =1

2π

Z ∞

−∞

1

(1 + is)2e−s2y/2eisx ds

=1

2π

Z ∞

−∞

(1− s2) cos sx + 2s sin sx

(1 + s2)2e−s2y/2 ds.


In parts a, b and c the domain is C and the range is C as well. In part d11.1the domain is C \ {−3} and the range is C \ {0}.

a If we write z = x + iy, then z = x − iy, so the real part is x and the11.2imaginary part is −y.b Expanding z3 = (x + iy)3 we see that the real part is x3 − 3xy2 andthat the imaginary part is 3x2y − y3.c The real part is x− 4 and the imaginary part is −y − 1.d The real part is (3y − 2x − 6)/((x + 3)2 + y2) and the imaginary partis (2y + 3x + 9)/((x + 3)2 + y2). To see this, write z = x + iy, thenz + 3 = x + 3 + iy and so

f(z) =(3i− 2)(x + 3− iy)

(x + 3 + iy)(x + 3− iy)=

3y − 2x− 6 + i(2y + 3x + 9)

(x + 3)2 + y2.

Apply definition 11.3 and expand the squares; several of the exponentials11.3cancel and only 1/2+1/2 remains, so sin2 z+cos2 z = 1. Similarly it followsby substitution that 2 sin z cos z = sin 2z.

a From definition 11.3 it follows that sin(iy) = (e−y − ey)/2i = i sinh y11.5and cos(iy) = (e−y + ey)/2 = cosh y.b Write z = x + iy; from exercise 11.4 with z = x and w = iy it followsthat sin(x + iy) = sin x cos(iy) + cos x sin(iy). Now apply part a, thensin(x + iy) = sin x cosh y + i cos x sinh y, so the real part is sin x cosh y andthe imaginary part is cos x sinh y.

The proofs can be copied from the real case; this is a straightforward mat-11.6ter. The same applies to exercise 11.7.

This rational function is continuous for all z ∈ C for which the denominator11.8is unequal to 0. But the denominator is 0 for z = 1, z = −i or z = 2i. Sog(z) is continuous on G = C \ {1,−i, 2i}.

The proof can be copied from the real case; this is a straightforward matter:11.9

limw→z

f(w)− f(z)

w − z= lim

w→z

w2 − z2

w − z= lim

w→z(w + z) = 2z.

a The derivative is 4(z − 1)3; the function is differentiable on C, so it is11.11analytic on C.b The derivative is 1 − 1/z2; since the function is not differentiable atz = 0, it is analytic on C \ {0}.c The derivative is ((z3 + 1)(2z − 3) − 3z2(z2 − 3z + 2))/(z3 + 1)2; thefunction is not differentiable when z3 = −1. Solving this equation oneobtains that it is analytic on C \ {−1, 1

2± 1

2i√

3}.d The derivative is 2zez2

; the function is analytic on C.

Using definition 11.3 and the chain rule it follows that (cos z)′ = (ieiz −11.12ie−iz)/2 = (−eiz + e−iz)/2i = − sin z.

The step√

wz =√

w√

z cannot be applied for non-real numbers (so e.g.11.13for w = −1, z = −1).

The real and imaginary part are u(x, y) = x2 − y2 and v(x, y) = 2xy11.14∗

45


(example 11.4). So ∂u/∂x = 2x, which equals ∂v/∂y, and ∂u/∂y = −2ywhich equals −∂v/∂x. Hence, the Cauchy-Riemann equations are satisfiedon the whole of R2.

a This is not true; take e.g. z = 2i, then cos 2i = (e−2 + e2)/2 > 3.11.16b Use definition 11.3; expand the exponentials in the resulting expressionfor cos z cos w − sin z sin w.c Write z = x + iy and use part b, then cos(x + iy) = cos x cos iy −sin x sin iy. But cos iy = cosh y and sin iy = i sinh y (see exercise 11.5a), socos(x+ iy) = cos x cosh y− i sin x sinh y, which gives the real and imaginaryparts.d Since ez is analytic on C, it follows from theorem 11.5 that cos z =(eiz + e−iz)/2 is also analytic on C.

a This function is analytic on C \ {1}. From the quotient rule it follows11.17that the derivative is given by (2z3 − 3z2 − 1)/(z − 1)2.b This function is analytic when z4 6= −16, so on C \ {±

√2± i

√2}. The

derivative is given by −40z3/(z4 + 16)11.c This function is analytic when z2 6= −3, so on C \ {±i

√3}. The deriv-

ative is given by ez(z2 − 2z + 3)/(z2 + 3)2 (use the quotient rule).d This function is analytic on C since both ez and sin w are analytic onC. The derivative is given by ez cos ez (use the chain rule).


a The complex number e−iωR lies on the unit circle for all R and so the12.1limit R →∞ does not exist.b Since limR→∞ e−σR = 0 only for σ > 0, and

˛e−iωR

˛= 1, it follows

that the limit exists precisely for σ > 0.

The integrand equals e(a−s)t and a primitive of this is given by e(a−s)t/(a−12.2s). The lower limit 0 leads to 1/(s−a), while the upper limit results in thelimit limR→∞ e(a−s)R (note that 1/(a−s) does not influence the outcome ofthis limit). Write s = σ + iω, a = α+ iβ and use that limR→∞ e(α−σ)R = 0only if α−σ < 0, so if σ > α. Hence, the Laplace transform equals 1/(s−a)for Re s > Re a.

b The function equals 1 for 0 ≤ t < b and is 0 elsewhere. It is easy to12.3calculate the Laplace transform using the definition and an integration byparts and it is given by (1− e−bs)/s for all (!) s 6= 0, while it equals b fors = 0.

One can use the method of example 12.9 and apply an integration by parts12.4for s 6= 0. It then follows that

F (s) = −1

slim

R→∞R2e−sR +

2

s

Z ∞

0

te−st dt.

The remaining integral is the Laplace transform of t for Re s > 0, whichis 1/s2 (example 12.9). Since limR→∞R2e−σR = 0 for σ > 0, it follows asbefore that limR→∞R2e−sR = 0 for Re s > 0. This shows that F (s) = 2

s3 .

a From examples 12.2 and 12.8 it follows that 1/(s + 2) is the Laplace12.5transform and that σa = σc = −2.b From example 12.7 it follows that e−4s/s is the Laplace transform andthat σa = σc = 0.c From exercise 12.2 it follows that 1/(s−2−3i) is the Laplace transformand that σa = σc = 2.

a From example 12.1 it follows that f(t) = 1 (that is, ε(t)).12.6b From example 12.7 it follows that f(t) = ε(t− 3).c From example 12.2 it follows that f(t) = e7t (that is, ε(t)e7t).d From exercise 12.4 it follows that f(t) = t2/2 (that is, ε(t)t2/2).

a Using the method of example 12.10 (write cos t as 12eit + 1

2e−it), we12.8

obtain the Laplace transform

1

2

„1

s− i+

1

s + i

«and this is equal to s/(s2 + 1) (for Re s > 0).b As in part a we now obtain the Laplace transform s/(s2 + a2).c Using the same method as in parts a and b it follows that the Laplacetransform of cosh at = (eat + e−at)/2 is given by

1

2

„1

s− a+

1

s + a

«=

s

s2 − a2.

Follow the hint, e.g. write cos(at+ b) = cos at cos b− sin at sin b, then apply12.9

47


linearity and the fact that we know the transforms of cos at and sin at. Intable 7, lines 8 and 9 the answers are given.

In all these exercises we have to use linearity and/or table 7.12.10a (20/s3)− (5/s2) + ((8i− 3)/s)b 4/(s2 + 16)c s/(s2 − 25)d (1/s2) + (1/s)− (s/(s2 + 1))e (1/(s− 2)) + (1/(s + 3))f (1/2s)− (s/2(s2 + 4))g (cos 2− s sin 2)/(s2 + 1)h 1/(s− ln 3) (use that 3t = et ln 3).

a The function is 1 for 0 ≤ t < 1, 3− 2t for 1 ≤ t < 2 and 1− t for t ≥ 2.12.11b The Laplace transform of 1 equals 1/s. Also, ε(t− 1)(2t− 2) = 2ε(t−1)(t − 1) and so we can apply a shift in the time-domain, which gives asLaplace transform 2e−s/s2. The same applies to ε(t−2)(t−2), which givesas Laplace transform e−2s/s2. Add these three results.

The function equals cos t for 0 ≤ t < 2π and is 0 elsewhere since cos(t −12.122π) = cos t. The Laplace transform of cos t is s/(s2 + 1). A shift inthe time-domain shows that se−2πs/(s2 + 1) is the Laplace transform ofε(t− 2π) cos(t− 2π). Adding these results leads to s(1− e−2πs)/(s2 + 1).

Apply the scaling property to f(t), then G(s) = (s2−2s+4)/(4(s+1)(s−2)).12.15

For all these exercises one first needs to recognize the basic form of the12.16function, then apply table 7, in combination with a shift in the time- ors-domain.a 1/(s− 2)2; use a shift in the s-domain.b 2e−s/s3; use a shift in the time-domain.c 5/((s + 3)2 + 25); use a shift in the s-domain.d (s− b)/((s− b)2 + a2); use a shift in the s-domain.e se−3s/(s2 − 1); use a shift in the time-domain.f 2e−3/(s− 1)3; use a shift in the s-domain.

a The function is 0 for 0 ≤ t < 1 and t− 1 for t ≥ 1. From table 7 and a12.17shift in the time-domain the Laplace transform e−s/s2 follows.b The function equals t − 1 for t ≥ 0. From table 7 (and linearity) theLaplace transform follows: (1/s2)− (1/s).c The function equals 0 for 0 ≤ t < 1 and t for t ≥ 1 (so there is a jumpat t = 1 of magnitude 1). In order to apply a shift in the time-domainwe write f(t) = ε(t − 1)(t − 1) + ε(t − 1). Using part a and table 7 (andlinearity of course), the Laplace transform follows: (e−s/s2) + (e−s/s).

Write f(t) = ε(t)t− ε(t− 1)t, then it follows from exercise 12.17c and table12.187 that the Laplace transform is given by (1− (1 + s)e−s)/s2.

For this exercise one again first has to recognize the basic form of the12.19Laplace transform, e.g. using table 7, and then, if necessary, combine itwith properties like a shift in the time- or s-domain.a 2e3t

b 3 sin tc 4 cos 2td (sinh 2t)/2e ε(t− 2)(t− 2); use a shift in the time-domain.f ε(t− 3) cos(t− 3); use a shift in the time-domain.


g (et sin 4t)/4; use a shift in the s-domain.h e−t(3 cos t− sin t); use a shift in the s-domain; in order to do so we firsthave to write F (s) as

3s + 1

(s + 1)2 + 1− 1

(s + 1)2 + 1,

which can be obtained by noting that 3s + 2 = 3(s + 1)− 1.i −3e3tt2; use a shift in the s-domain.j e2t cos 2t; use a shift in the s-domain (first complete the square in thedenominator, writing it as (s− 2)2 + 4).k 1

4ε(t− 1) cos(3(t− 1)/2); use a shift in the time-domain (first write the

function F (s) as 14e−ss/(s2 + (9/4)).

Apply De l’Hopital’s rule repeatedly (n times) to the limit t→∞ of tn/eαt12.21and use that limt→∞ e−αt = 0 for any α > 0. Theorem 12.3 implies thatthe Laplace transform of tn exists for Re s > 0 (it is easy to show thattn = ε(t)tn is of exponential order for α > 0 arbitrary).

We know that (L1)(s) = 1/s. According to (12.13) we then have for n ∈ N12.23(and Re s > 0) that

dn

dsn

1

s= (−1)n(Ltn)(s),

hence, (−1)nn!/sn+1 = (−1)n(Ltn)(s), that is, (Ltn)(s) = n!/sn+1.

a We know (e.g. from table 7) that (Ltn)(s) = n!/sn+1 and using the12.25shift rule in the s-domain we then obtain that F (s) = n!/(s− a)n+1.b We know (e.g. from table 7) that (Leat)(s) = 1/(s − a) and using thedifferentiation rule in the s-domain we then obtain that

F (s) = (−1)n dn

dsn

1

s− a=

n!

(s− a)n+1.

The causal function sinh at is continuous on R, so it follows from the integ-12.27ration rule (table 8) that„L

Z t

0

sinh aτ dτ

«(s) =

1

sF (s)

with F (s) = (L sinh at)(s) = a/(s2 − a2). Hence f(t) =R t

0sinh aτ dτ =

(cosh at− 1)/a.

a Use the differentiation rule in the s-domain and table 7 for the Laplace12.28transform of cos at. Then f has Laplace transform

d2

ds2

s

s2 + a2=

2s(s2 − 3a2)

(s2 + a2)3.

b Using the differentiation rule in the s-domain and table 7 for the Laplacetransform of sinh 3t one obtains that

(Lt sinh 3t)(s) = − d

ds

3

s2 − 9=

6s

(s2 − 9)2

and

(Lt2 sinh 3t)(s) =d2

ds2

3

s2 − 9=

18(s2 + 3)

(s2 − 9)3.


Finally apply linearity, then one obtains that

F (s) =18(s2 + 3)

(s2 − 9)3− 18s

(s2 − 9)2+

6

s2 − 9.

a The function equals 2ε(t)t for t < 1 and since t = 2t− t it follows that12.30f(t) = 2ε(t)t− ε(t− 1)t = 2ε(t)t− ε(t− 1)(t− 1)− ε(t− 1).b From part a, table 7 and the shift rule in the time domain it followsthat

F (s) =2

s2− e−s

s2− e−s

s.

c The function is not differentiable at t = 0 and t = 1. Apart from thesetwo points the derivative equals 0 for t < 0, 2 for 0 < t < 1 and 1 fort > 1. Hence, f ′(t) = 2ε(t) − ε(t − 1) for t 6= 0, 1. Since the Laplacetransform does not depend on the value at these points, we have that(Lf ′)(s) = (2/s)− (e−s/s).d According to the differentiation rule in the time domain one should have(Lf ′)(s) = s(Lf)(s), so

2− e−s

s= s

2− e−s(s + 1)

s2.

This is not correct since we cannot apply the differentiation rule in thepresent situation: f has a jump at t = 1 and so it isn’t differentiable on R.

a From the shift rule in the s-domain it follows that (Leibtf(t))(s) = F (s−12.31ib) and since sin at = (eiat − e−iat)/2i it follows that (Lf(t) sin at)(s) =(F (s− ia)− F (s + ia))/2i.b First apply the scaling property and then the shift rule in the s-domain,then (Le−2tf(3t))(s) = e−(s+2)/3/(s + 2).c We can apply the integration rule here (t3f(t) is continuous on R).Hence the Laplace transform of

R t

0τ3f(τ) dτ is given by G(s)/s with G(s) =

(Lt3f(t))(s). Now apply the differentiation rule in the s-domain, then itfollows that the Laplace transform of

R t

0τ3f(τ) dτ is given by

−1

s

d3

ds3F (s).

a Since 3et−2 = 3e−2et it follows that (L3et−2)(s) = 3e−2/(s − 1) (see12.32table 7). We also have that (Lε(t− 2))(s) = e−2s/s so (for Re s > 1)

F (s) =3e−2s + e−2s(s− 1)

s(s− 1).

b Apply linearity (and table 7) to (t− 1)2 = t2 − 2t + 1, then we obtainthat

F (s) =s2 − 2s + 2

s3.

c From (Lε(t − 4))(s) = e−4s/s and the shift property in the s-domain(table 8) it follows that F (s) = e−4(s−2)/(s− 2).d Note that f(t) = e2ite−t = e(−1+2i)t and applying table 7 we thus obtainthat F (s) = 1/(s + 1 − 2i). (One can also use the Laplace transforms ofsin 2t and cos 2t and a shift in the s-domain.)e The Laplace transform of sin t is 1/(s2+1). Applying a shift in the time-domain we obtain that (Lε(t− 2) sin(t− 2))(s) = e−2s/(s2 +1). Finally wenote that et+3 = e3et and so we apply a shift in the s-domain:


F (s) =e3e−2(s−1)

(s− 1)2 + 1.

f The Laplace transform of cos 2t is s/(s2 +4). Since 3t = et ln 3 we applya shift in the s-domain:

F (s) =s− ln 3

(s− ln 3)2 + 4.

g One can write f(t) as the following combination of shifted unit stepfunctions: f(t) = ε(t) − ε(t − 1) + ε(t − 2). From table 7 we then obtainthat

F (s) =1− e−s + e−2s − e−3s

s.

a Since (L1)(s) = 1/s and (Lε(t− 1))(s) = e−s/s (table 7) it follows that12.33f(t) = 1− ε(t− 1).b Since F (s) = 1/s + 3/s4 we obtain from table 7 that f(t) = 1 + t3/2.c Since (Lte−t)(s) = 1/(s+1)2 and (L sinh 2t)(s) = 2/(s2−4) (table 7) itfollows that te−t + 1

2sinh 2t has Laplace transform 1/(s + 1)2 + 1/(s2 − 4).

Furthermore we have that (L sin t)(s) = 1/(s2 + 1) (table 7) and from ashift in the time-domain it then follows that (Lε(t − π) sin(t − π))(s) =e−πs/(s2 + 1). Hence f(t) = te−t + 1

2sinh 2t + sin t + ε(t− π) sin(t− π).

d The denominator equals (s− 2)2 + 16 and if we now use that 3s− 2 =3(s− 2) + 4, then it follows that

F (s) = 3s− 2

(s− 2)2 + 16+

4

(s− 2)2 + 16.

From table 7 and a shift in the s-domain we then obtain that f(t) =3e2t cos 4t + e2t sin 4t.e The denominator equals (s+4)2 and if we now use that s+3 = (s+4)−1,then it follows that

F (s) =1

s + 4− 1

(s + 4)2.

From table 7 and a shift in the s-domain we then obtain that f(t) =e−4t(1− t).f Apply a shift in the time domain to (Lt2e2t)(s) = 2/(s− 2)3 (table 7),then we obtain that f(t) = 1

2ε(t− 4)e2t−8(t− 4)2.

g Applying the integration rule to (L sin 3t)(s) = 3/(s2 + 9) (the causalfunction sin t is continuous on R) we obtain that (L

R t

0sin 3τ dτ)(s) =

3/(s(s2 +9)). But the integral equals (1− cos 3t)/3, so (L(1− cos 3t))(s) =9/(s(s2 + 9)). From a shift in the time domain it then follows that f(t) =ε(t− 1)(1− cos 3(t− 1))/9.


The integral defining the convolution can be calculated by using the formula13.1for the product of two cosines. The convolution then equals 1

2t cos t +

12

sin t. Using the convolution theorem we obtain s2/(s2+1)2 as the Laplacetransform. On the other hand we obtain from the Laplace transforms ofcos t and sin t and the differentiation rule in the s-domain the Laplacetransform (s2 − 1)/(2(s2 + 1)2) + (1/(2(s2 + 1)), which agrees with theresult obtained from the convolution theorem.

a Table 7 gives f(t) = eat.13.2b The convolution theorem implies that g(t) = eav ∗ ebv; to determineg(t) explicitly, we need to calculate this convolution. From the definitionit follows that g(t) = ebt

R t

0eτ(a−b) dτ . If a = b, then g(t) = teat. If a 6= b

then g(t) = (eat−ebt)/(a−b). Next we can verify the convolution theorem.Write G(s) = (Lg)(s). If a = b, then G(s) = 1/(s − a)2 (table 7, no. 10).If a 6= b then G(s) = (1/(s− a)− 1/(s− b))/(a− b) (table 7, no. 2), whichequals 1/(s− a)(s− b).

a Consider this as the product of the Laplace transforms of t and e−t,13.4which gives t ∗ e−t as result.b Similarly we now obtain e−2t ∗ cos 2t.c sinh t ∗ cosh t.d 1

16sinh 4t ∗ sinh 4t.

This is not possible, since lims→∞ sn does not exist, contradicting theorem13.5theorem 13.2.

a From table 7 we obtain that F (s) = s/(s2 − 9). We indeed have13.6f(0+) = 1 = lims→∞ sF (s).b From table 7 we obtain that (L sin t)(s) = 1/(s2 + 1) and (L1)(s) =1/s. Applying the differentiation rule in the s-domain it follows that(Lt sin t)(s) = 2s/(s2 + 1)2 and so F (s) = (2/s) + (2s/(s2 + 1)2). Weindeed have f(0+) = 2 = lims→∞ sF (s).c From the integration rule (table 8) we obtain that F (s) = (Lg)(s)/sand so sF (s) = G(s) with G(s) the Laplace transform of g(t). Applyingtheorem 13.2 to g(t) we obtain that lims→∞ sF (s) = lims→∞G(s) = 0 andfor f we indeed have that f(0+) = f(0) = 0.

a From table 7 we obtain that F (s) = 1/(s + 3). Since f(∞) exists,13.8we may apply the final value theorem and we indeed have f(∞) = 0 =lims→0 sF (s).b From table 7 we obtain that (L sin 2t)(s) = 2/(s2 + 4). Applying theshift property in the s-domain we obtain that F (s) = 2/((s + 1)2 + 4).Since f(∞) exists, we may apply the final value theorem and we indeedhave f(∞) = 0 = lims→0 sF (s).c From table 7 we obtain that F (s) = (1/s)− (e−s/s). Since f(∞) exists,we may apply the final value theorem and we indeed have f(∞) = 0 =lims→0(1− e−s).

For the functions cos t and sinh t the value f(∞) does not exist and so the13.9final value theorem cannot be applied.

a For a periodic function f(∞) will in general not exist and so the final13.11value theorem cannot be applied.b Theorem 13.5 implies that

52


sF (s) =s

1− e−sT

Z T

0

f(t)e−st dt.

Taking the limit s→ 0 gives (note that the integral is over a bounded inter-

val) lims→0

R T

0f(t)e−st dt =

R T

0f(t) dt. From the definition of derivative

(definition 11.7) it follows that (e−zT )′(0) = lims→0(e−sT − 1)/s and since

(e−zT )′ = −Te−zT we thus obtain that

lims→0

s

1− e−sT=

1

T,

which shows that lims→0 sF (s) = 1T

R T

0f(t) dt.

c In example 13.4 we have sF (s) = 1/(1+ e−s) and so lims→0 sF (s) = 12.

The function has period 2 and so the integral is 12

R 2

0f(t) dt = 1

2, which

verifies the result of part b for f .

a For t < a we have φ(t) = ε(t); for t < 2a we then have φ(t) = ε(t) −13.122ε(t−a); finally, for all t we have φ(t) = ε(t)− 2ε(t−a)+ ε(t− 2a). If Φ(s)is the Laplace transform of φ(t), then it follows from table 7 that

Φ(s) =1

s− 2e−as

s+

e−2as

s.

Since f has period 2a we then obtain from theorem 13.5 (or table 8) that

F (s) =1− 2e−as + e−2as

s(1− e−2as).

b Multiply numerator and denominator by eas, write the denominator ass(eas/2 + e−as/2)(eas/2 − e−as/2), and use the definitions of the hyperbolicsine and cosine functions, then it follows that F (s) = tanh(as/2)/s.

b Let φ(t) denote the restriction of f(t) to one period, then φ(t) = tε(t)−13.13(t− 2)ε(t− 2)− 2ε(t− 1). Since (Lt)(s) = 1/s2, we obtain from a shift inthe time-domain that (L(t−2)ε(t−2))(s) = e−2s/s2. Also (Lε(t−1))(s) =e−s/s and so

Φ(s) =1

s2− e−2s

s2− 2e−s

s.

Theorem 13.5 (or table 8) then implies that

F (s) =1

s2− 2e−s

s(1− e−2s)=

1

s2− 1

s sinh s.

Apply (13.7) (or definition 13.2) and the definition of distribution derivat-13.16ive, then it follows that

(Lδ(n)(t− a))(s) =Dδ(n)(t− a), e−st

E= (−1)n

Dδ(t− a), (e−st)(n)

E.

Since (e−st)(n) = (−s)ne−st, we then obtain from the definition of theshifted delta function that (Lδ(n)(t− a))(s) = sne−as.

This follows immediately from table 9, no 2 and (9.21) together with (13.9).13.18

Consult tables 7 and 9 for this exercise and use linearity, and for part d13.19the convolution theorem.a 1 + (1/(s2 + 1)) = (s2 + 2)/(s2 + 1),b s + 3s2,


c 1/s + e−2s + 2is2e−4s,d s3e−as.

Consult tables 7 and 9 for this exercise, but now in the opposite direction,13.20and use mainly linearity.a δ′(t) + 3δ(t)− δ(t− 2),b δ′′(t)− 4δ′(t) + 4δ(t) + e2t (first write (s− 2)2 as s2 − 4s + 4),c ε(t− 2) sin(t− 2) + δ(3)(t− 2) (apply a shift in the time-domain to theLaplace transform of sin t),d δ(t)− sin t (first write s2/(s2 + 1) as 1− 1/(s2 + 1)).

This is an important exercise: it shows how to use partial fraction expan-13.22sions.a Factorise the denominator as (s + 2)(s + 3) and apply partial fractionexpansion to obtain 3/(s+3)−2/(s+2). From table 7 it then follows thatf(t) = 3e−3t − 2e−2t.b Completing the square in the denominator gives (s + 3)2 + 1 and fromtable 7 it then follows that f(t) = e−3t sin t.c Apply partial fraction expansion to obtain − 3

4(1/(s + 3)) + 2

3(1/(s +

2)) + 112

(1/(s − 1)). From table 7 it then follows that f(t) = −3e−3t/4 +2e−2t/3 + et/12.d First put y = s2 and then apply partial fraction expansion in the vari-able y, which gives 5

3(1/(y − 4)) − 2

3(1/(y − 1)). With y = s2 we obtain

from table 7 the inverse Laplace transform f(t) = (5 sinh 2t− 4 sinh t)/6.e A partial fraction expansion leads to −1/(s+1)+2/(s−1)−4/(s+1)2.From table 7 it then follows that f(t) = −e−t + 2et − 4te−t.f Since s2 − 1 = (s− 1)(s + 1) one could solve this using partial fractionexpansions. However, it is easier to note that

d

ds

s

s2 − 1= − 1

s2 − 1− 2

(s2 − 1)2

and so

1

(s2 − 1)2= −1

2

d

ds

s

s2 − 1− 1

2

1

s2 − 1.

We have (L sinh t)(s) = 1/(s2 − 1) and (L cosh t)(s) = s/(s2 − 1) and thedifferentiation rule in the s-domain then implies that

(Lt cosh t)(s) = − d

ds

s

s2 − 1.

Hence, f(t) = (t cosh t− sinh t)/2.g Since the degree of the numerator equals the degree of the denominatorwe first perform a long division:

s3 + 4

(s2 + 4)(s− 1)= 1 +

s2 − 4s + 8

(s2 + 4)(s− 1).

Applying a partial fraction expansion to the second term leads to 1/(s−1)−4/(s2+4). From tables 7 and 9 it then follows that f(t) = δ(t)+et−2 sin 2t.h The factor e−2s can be dealt with afterwards by applying a shift in thetime-domain. Since the degree of the numerator is greater than the degreeof the denominator we first perform a long division:

s6 + s2 − 1

s2(s2 − 1)= s2 + 1 +

2s2 − 1

s2(s2 − 1).


To the rational part we apply partial fraction expansion (again we first puty = s2), which leads to (1/s2) + (1/(s2 − 1)). From table 7, a shift in thetime-domain and table 9 it then follows that f(t) = δ′′(t− 2) + δ(t− 2) +ε(t− 2)(t− 2 + sinh(t− 2)).

a First write F (s) as13.24

F (s) =1

2

s

s2 + 4

2

s2 + 4.

From table 7 and the convolution theorem (or table 8) it then follows that(g ∗ h)(t) = (cos 2v ∗ 1

2sin 2v)(t).

b The definition of convolution gives f(t) = 12

R t

0cos 2τ sin(2t−2τ) dτ and

using the trigonometric formula 2 cos a sin b = sin(a+b)−sin(a−b) this canbe written as f(t) = 1

4

R t

0(sin 2t− sin(4τ − 2t)) dτ . Caculating this integral

gives f(t) = 14t sin 2t.

c Applying the differentiation rule in the time-domain to (L sin 2t)(s) =2/(s2 + 4), we obtain that (Lt sin 2t)(s) = 4s/(s2 + 4)2, so f(t) = 1

4t sin 2t.

a The function f is piecesewise smooth and f(∞) = 2, so the final value13.25theorem implies that lims→0 sF (s) = 2. Since f(0+) = 0, the inital valuetheorem implies that lims→∞ sF (s) = 0.b Apply a shift in the time-domain to (Lt)(s) = 1/s2, then (Lε(t −2)(t − 2))(s) = e−2s/s2, so F (s) = (1 − e−2s)/s2. Hence, lims→0 sF (s) =lims→0(1 − e−2s)/s. But according to definition 11.7 this limit equals−(e−2z)′(0), which is −2, and so lims→0 sF (s) = 2. This agrees with parta. Finally, applying De l’Hopital’s rule we obtain that lims→∞ sF (s) =lims→∞(1− e−2s)/s = lims→∞ 2e−2s = 0, also in agreement with part a.

a Note that lims→0 sF (s) = 1. We now determine the inverse Laplace13.26transform of F (s) using partial fraction expansion. We have

F (s) =1

s− s + 1

(s + 1)2 + 4

and from table 7 and a shift in the s-domain it then follows that f(t) =1 − e−t cos 2t. We see that f(∞) = 1 = lims→0 sF (s), which verifies thefinal value theorem.b Note that lims→0 sF (s) = 0. We now determine the inverse Laplacetransform of F (s). Since F (s) is a function of y = s2, we use partialfraction expansion for y/(y − 1)(y + 4), which gives

F (s) =1

5

1

s2 − 1+

2

5

2

s2 + 4.

From table 7 we then obtain that f(t) = (sinh t + 2 sinh 2t)/5. Since f(∞)does not exits, the final value theorem cannot be applied.

a For 0 ≤ t < 1 we have f(t) = ε(t)t2; for 0 ≤ t < 2 we then have13.27f(t) = ε(t)t2 − ε(t− 1)t2.b Let φ(t) = (ε(t) − ε(t − 1))t2 for all t, and let Φ(s) be the Laplacetransform of φ(t), then it follows from theorem 13.5 (or table 8) that F (s) =Φ(s)/(1−e−2s). If we write φ(t) as φ(t) = ε(t)t2−ε(t−1)(t−1)2−ε(t−1)−2ε(t− 1)(t− 1), then it follows from table 7 and a shift in the time-domain(to get the Laplace transforms of ε(t− 1)(t− 1)2 and ε(t− 1)(t− 1)) that

Φ(s) = −e−s

s− 2e−s

s2+

2

s3− 2e−s

s3,


which then also gives F (s) = Φ(s)/(1− e−2s).

a A partial fraction expansion of G(s) = 1/s(s + 1) gives G(s) = (1/s)−13.28(1/(s+1)), which has inverse Laplace transform g(t) = 1−e−t. Combiningthis with a shift in the time-domain we obtain that f(t) = 1− e−t − ε(t−3)(1− e−t+3).b A partial fraction expansion gives

F (s) =2

s+

1

s2− 2

s− 1+

2

(s− 1)2.

From table 7 we then obtain that f(t) = 2 + t− 2et + 2tet.c Write F (s) = G(s) + H(s) where

G(s) =1

s2(s2 + 4)and H(s) = e−πs s5 − 4s4 − 8s + 64

s2(s2 + 4).

We first determine the inverse Laplace transform g(t) of G(s). Since G(s)is a function of y = s2, we apply a partial fraction expansion to 1/y(y +4),which gives G(s) = (1/4s2) − (1/4(s2 + 4)) and so (by table 7) g(t) =(2t − sin 2t)/8. Next we determine the inverse Laplace transform h(t) ofH(s). Since the degree of the numerator is larger than the degree of thedenominator, we first perform a long division:

s5 − 4s4 − 8s + 64

s2(s2 + 4)= s− 4− 4s3 − 16s2 + 8s− 64

s2(s2 + 4).

Partial fraction expansion of the rational function gives

4s3 − 16s2 + 8s− 64

s2(s2 + 4)=

2

s− 16

s2+

2s

s2 + 4.

Hence,

H(s) = se−πs − 4e−πs − e−πs

„2

s− 16

s2+

2s

s2 + 4

«.

From table 7 and the shift property in the time-domain it then follows thath(t) = δ′(t − π) − 4δ(t − π) − ε(t − π) (2− 16(t− π) + 2 cos 2(t− π)) andthen f(t) = g(t) + h(t).


a We have to determine the Laplace transform H(s) of h(t). From tables14.17 (for te−t) and 9 (for δ) it follows immediately that H(s) = 1+1/(s+1)2.b Using that H(s) exists for Re s > −1, we can substitute s = iω intoH(s) to obtain the frequency response 1 + 1/(iω + 1)2.c We first determine the response y(t) by calculating the convolutionproduct y(t) = (h∗u)(t). Since δ ∗u = u we only need to calculate te−t ∗u,that isZ t

0

u(τ)h(t − τ) dτ =

Z t

0

e−τ sin τe−(t−τ)(t − τ) dτ.

Here e−τ and eτ cancel each other and the integral that remains can becalculated by an integration by parts. This gives te−t − sin te−t and soy(t) = u(t) + te−t − sin te−t = te−t.Next we determine y(t) using the Laplace transform: Y (s) = H(s)U(s) andsince U(s) = 1/((s+1)2+1) (table 7 and a shift in the s-domain), it followsthat Y (s) = 1/(s + 1)2 and so (inverse Laplace transform) y(t) = te−t.

Again use the important formula Y (s) = H(s)U(s), so H(s) = Y (s)/U(s).14.2In this case U(s) = 1/s2 and Y (s) = 1/s2 − s/(s2 + 4), so H(s) = 1 −s3/(s2 +4) = 1− s+4s/(s2 +4) (divide s3 by s2 +4). The inverse Laplacetransform gives h(t) = δ(t)− δ′(t) + 4 cos 2t.

a It is clear that y(0) = 0. The derivative at t = 0 can be obtained from14.3the definition of the derivative (and e.g. using De l’Hopital’s rule) and weindeed obtain that y′(0) = 0. For t > 0 it is straightforward to check thaty′′ − y = 2t. So in ordinary sense y(t) does indeed satisfy the differentialequation.We now differentiate in distribution sense. Since y has no jump at t = 0and y′ also has no jump at t = 0, the jump formula (8.21) will give thesame result (outside t = 0 the derivative can be taken in ordinary sense).b Since h has no jump at t = 0 we have that h′(t) = 2e2t−et (in the senseof distributions). Now h′ has a jump 1 at t = 0, so h′′(t) = 4e2t − et + δ(t).Hence, we indeed have that h′′ − 3h′ + 2h = δ(t).c Similar as in part b: there is no jump at t = 1 and so y′(t) = 2ε(t −1)(2e2t−2−et−1). Now y′ has a jump 2 at t = 1, so y′′(t) = 2ε(t−1)(4e2t−2−et−1) + 2δ(t − 1). Thus, y′′ − 3y′ + 2y = 2δ(t − 1).

Taking Laplace transforms of the left- and right-hand side gives (note the14.4condition of initial rest) Y (s) = 48/((s2 + 4)(s2 + 16)). A partial fractionexpansion (in the variable y = s2) results in 4/(s2 + 4) − 4/(s2 + 16) andfrom table 7 it then follows that y(t) = 2 sin 2t − sin 4t.

a Using (14.9) we obtain that H(s) = 1/(s2 − 5s+4) = 1/((s− 1)(s− 4))14.5and its zeroes do not satisfy Re s < 0, so the system is not stable.b Partial fraction expansion of H(s) shows that H(s) = (1/3(s − 4)) −(1/3(s− 1)) and from table 7 we then obtain that h(t) = (e4t − et)/3; thisfunction is not absolutely integrable, which is in agreement with the factthat the system is not stable.c Integrating the impulse response over [0, t] gives the step response a(t) =(e4t − 4et + 3)/12.d Either calculate the convolution product (h∗u)(t) or use Laplace trans-forms; in general the latter is preferable and we will apply it here. Since

57


Y (s) = U(s)H(s) = 1/((s− 1)(s− 4)(s− 2)) we use partial fraction expan-sion:

Y (s) =1

3(s − 1)− 1

2(s − 2)+

1

6(s − 4).

From table 7 we then obtain that y(t) = (e4t + 2et − 3e2t)/6.e Use time-invariance (and linearity): 3δ(t− 1) has response 3h(t− 1) =ε(t − 1)(e4t−4 − et−1).

a From the differential equation we get H(s) = 1/(R(s + 1/RC)) (use14.7(14.9)) and so H(s) has only one zero s = −1/RC. Since this zero lies inRe s < 0 (because RC > 0), the system is stable.b Let Q and V be the Laplace transforms of q and v, then Q(s) =V (s)H(s) = E/(Rs(s+1/RC)) and partial fraction expansion gives Q(s) =EC(1/s − 1/(s + 1/RC)). The inverse transform of this is q(t) = EC(1 −e−t/RC) and since i(t) = q′(t) it follows that i(t) = Ee−t/RC/R.c As in part b it now follows that Q(s) = aE/(R(s + 1/RC)(s2 + a2)),and partial fraction expansion results in Q(s) = p(1/(s + 1/RC)− s/(s2 +a2) + 1/(RC(s2 + a2))), where p = aE/(R(a2 + 1/R2C2)). The inversetransform of this is q(t) = p(e−t/RC − cos at + (1/aRC) sin at).d Use time-invariance (and linearity): Eδ(t − 3) has as response Eh(t −3) = Eε(t − 3)e−(t−3)/RC/R (h(t) follows from part a).

a From the differential equation we get H(s) = 1/(L(s + R/L)) (use14.8(14.9)) and so H(s) has only one zero s = −R/L. Since this zero lies inRe s < 0 (because R/L > 0), the system is stable. The inverse transformgives h(t) = e−Rt/L/L.b Integrating the impulse response over [0, t] gives the step response a(t) =(1− e−Rt/L)/R.c Let V be the Laplace transform of v, then it follows that Y (s) =V (s)H(s) = e−as/(Ls(s + R/L)). Partial fraction expansion results inY (s) = (e−as/R)(1/s − 1/(s + R/L)). The inverse transform of this isy(t) = ε(t − a)(1 − e−(t−a)R/L)/R (use a shift in the time-domain). How-ever, it is much simpler here to use time-invariance: the response to ε(t−a)is a(t − a) = ε(t − a)(1− e−R(t−a)/L)/R.

Let Y (s) be the Laplace transform of y(t), then we obtain from the differen-14.10tial equation (and tables 7 and 8) that (s2Y (s)−sy(0)−y′(0))+Y (s) = 1/s2.Since y(0) = 0 and y′(0) = 1 it follows that s2Y (s) − 1 + Y (s) = 1/s2, so(s2 + 1)Y (s) = 1 + 1/s2 = (s2 + 1)/s2 and thus Y (s) = 1/s2. The inverseLaplace transform of Y (s) is y(t) = t.

Let Y (s) be the Laplace transform of y(t), then we obtain from the dif-14.12ferential equation and the initial conditions y(0) = 3 and y′(0) = 1 (andtables 7 and 8) that (s2Y − 3s− 1)− 4(sY − 3)− 5Y = 3/(s− 1). Solvingfor Y gives (s2 − 4s − 5)Y (s) + 11− 3s = 3/(s − 1), hence,

Y (s) =3

(s − 1)(s2 − 4s − 5)+

3s − 11

s2 − 4s − 5=

3s2 − 14s + 14

(s − 1)(s2 − 4s − 5).

Partial fraction expansion gives

Y (s) =31

12(s + 1)− 3

8(s − 1)+

19

24(s − 5).

From table 7 we obtain the inverse Laplace transform y(t) = (62e−t−9et +19e5t)/24.


Note that u(t) = t − ε(t − 2)(t − 2), so (shift in time-domain) U(s) =14.13(1− e−2s)/s2. From the differential equation and the initial conditions weobtain that s2Y − s + Y = (1 − e−2s)/s2, hence (apply partial fractionexpansion in s2 to the first term),

Y (s) =1− e−2s

s2(s2 + 1)+

s

s2 + 1= (1− e−2s)

„1

s2− 1

s2 + 1

«+

s

s2 + 1.

From tables 7 and 8 we obtain the inverse Laplace transform y(t) = t −sin t + cos t − ε(t − 2)(t − 2− sin(t − 2)).

From tables 7 and 9 we know that the Laplace transforms of 1 and δ(t− 2)14.14are given by 1/s and e−2s. From the differential equation and the initialconditions we obtain that (s2Y − 2s + 2) + 2(sY − 2) + 5Y = 2e−2s + 1/s,and thus (s2 + 2s + 5)Y (s) = 2s + 2 + 2e−2s + 1/s. Hence,

Y (s) =2(s + 1)

(s + 1)2 + 4+

2e−2s

(s + 1)2 + 4+

1

s(s2 + 2s + 5).

Using table 7 and a shift in the s-domain it is easy to get the inverse Laplacetransform of the first two terms since (Le−t cos 2t)(s) = (s+1)/((s+1)2+4)and (Le−t sin 2t)(s) = 2/((s + 1)2 + 4). For the third term we use partialfraction expansion:

1

s(s2 + 2s + 5)=

1

5s− (s + 2)

5(s2 + 2s + 5).

We write the second term as (s + 1)/5(s2 + 2s + 5) + 1/5(s2 + 2s + 5) andtaking everything together now we get

Y (s) =1

5s+

9(s + 1)

5((s + 1)2 + 4)− 1

5((s + 1)2 + 4)+ e−2s 2

(s + 1)2 + 4.

From our previous remarks and a shift in the time-domain it then followsthat y(t) = (2 + 18e−t cos 2t − e−t sin 2t)/10 + ε(t − 2)e−(t−2) sin 2(t − 2).

Let X(s) and Y (s) be the Laplace transforms of x(t) and y(t). From table14.167 we know that (L cos 2t)(s) = s/(s2 + 4) and (L sin 2t)(s) = 2/(s2 + 4).Applying the Laplace transform to the system and substituting the initialconditions x(0) = −1 and y(0) = 0 we obtain the algebraic system

sX + Y = −1 + 2s/(s2 + 4),X + sY = 2/(s2 + 4).

Next we solve this system of two linear equations in the unknowns X =X(s) and Y = Y (s). We can find Y (s) by multiplying the second equationby −s and adding it to the first equation; we then obtain

Y − s2Y =2s

s2 + 4− 1− 2s

s2 + 4= −1

and so Y (s) = 1/(s2 − 1). One similarly obtains

−s2X + X =−2s2

s2 + 4+ s +

2

s2 + 4

and so X(s) = 2/(s2 +4)− s/(s2−1). The inverse Laplace transform givesthe solution x(t) = sin 2t − cosh t, y(t) = sinh t.

Let X(s) and Y (s) be the Laplace transforms of x(t) and y(t). Applying14.17the Laplace transform to the system and substituting the initial conditions


x(0) = 0, x′(0) = 2, y(0) = −1 and y′(0) = 0 we obtain the algebraicsystem

s2X + sY = 1,X − sY = 1.

We can find X(s) by adding these two equations; this gives (s2 + 1)X = 2and thus X(s) = 2/(s2 + 1). Since sY = X − 1 this gives

Y =1

sX − 1

s=

2

s(s2 + 1)− 1

s=

1

s− 2

s

s2 + 1,

where we also applied partial fraction expansion to 2/(s(s2 + 1)). Theinverse Laplace transform gives the solution x(t) = 2 sin t, y(t) = 1−2 cos t.

As in exercise 14.16 we obtain the algebraic system14.18 (2s + 1)Y + (5s − 2)X = 3 + 2/(s + 1),−sY + (1− 2s)X = −1 + 1/(s2 + 1).

Next we solve this system of two linear equations in the unknowns X =X(s) and Y = Y (s). We can find X by multiplying the first equation by s,and adding it to the second equation multiplied by 2s + 1; we then obtain,after some simplification,

X(s) =2s

(s − 1)2(s + 1)+

2s + 1

(s2 + 1)(s − 1)2+

1

s − 1.

One similarly obtains

Y (s) =2− 4s

(s + 1)(s − 1)2− 5s − 2

(s2 + 1)(s − 1)2− 1

s − 1.

Four terms in X and Y need a partial fraction expansion, which leads to

X(s) = − 1

2(s + 1)+

1

s − 1+

5

2(s − 1)2+

s

2(s2 + 1)− 1

s2 + 1,

Y (s) =3

2(s + 1)− 7

2(s − 1)− 5

2(s − 1)2+

s

s2 + 1+

5

2(s2 + 1).

The inverse Laplace transform gives the solution x(t) = (−e−t+2et+5tet+cos t − 2 sin t)/2, y(t) = (3e−t − 7et − 5tet + 2 cos t + 5 sin t)/2.

Apply Laplace transform with respect to t to the partial differential equa-14.20tion and substitute the initial conditions u(x, 0) = 2 sin 2πx and ut(x, 0) =0, then it follows as in example 14.16 that

s2U(x, s)− (2 sin 2πx)s = 4Uxx,

where U(x, s) is the Laplace transform of u(x, t). Hence, we have obtainedthe ordinary differential equation

U ′′ − s2

4U = −1

2s sin 2πx.

The general solution of the homogeneous equation is

U(x, s) = Aesx/2 + Be−sx/2,

where A and B can still be functions of s. A particular solution can befound using the ’classical method’, so by trying U(x, s) = a sin 2πx. It thenfollows that a(4π2 + s2/4) = s/2 and so the general solution follows:


U(x, s) = Aesx/2 + Be−sx/2 +2s

s2 + (4π)2sin 2πx.

To determine A and B, we translate the remaining boundary conditions tothe s-domain by Laplace transforming them. From the conditions u(0, t) =u(1, t) = 0 it follows that U(0, s) = U(1, s) = 0. Using U(0, s) = 0 weobtain that A + B = 0 and using U(1, s) = 0 we obtain that A(es/2 −e−s/2) = 0. Hence, A = B = 0 and so

U(x, s) =2s

s2 + (4π)2sin 2πx.

The inverse Laplace transform gives the solution u(x, t) = 2 cos 4πt sin 2πx.

As in exercise 4.20, for example, one obtains the ordinary differential equa-14.22tion

U ′′ − (s + 6)U = − cos(x/2).

The general solution of the homogeneous equation is

U(x, s) = Aex√

s+6 + Be−x√

s+6,

where A and B can still be functions of s. A particular solution can befound by trying U(x, s) = a cos(x/2). The general solution then follows:

U(x, s) = Aex√

s+6 + Be−x√

s+6 +4

4s + 25cos(x/2).

To determine A and B, we translate the remaining boundary conditions tothe s-domain by Laplace transforming them, which results in U(π, s) = 0and U ′(0, s) = 0. This leads to A = B = 0 and so

U(x, s) =1

s + 25/4cos(x/2).

The inverse Laplace transform gives the solution u(x, t) = e−25t/4 cos(x/2).

a The impulse response is the derivative (in distribution sense if neces-14.23sary) of the step response. Since the step response has no jump at t = 0,it follows that h(t) = a′(t) = 2 sinh 2t + 2 sin t− e−t. The transfer functionis the Laplace transform of h(t) and from table 7 we obtain that

H(s) =4

s2 − 4+

2

s2 + 1− 1

s + 1.

b Since (Lδ(t − 1))(s) = e−s, it follows that Y (s) = 2e−sH(s), whereY (s) = (Ly)(s). From part a and the shift rule in the time domain itfollows that y(t) = 2ε(t− 1)(2 sinh(2t− 2) + 2 sin(t− 1)− e−t+1). One canalso use the time-invariance of the system.c Since (Lt)(s) = 1/s2 it follows that

Y (s) =4

s2(s2 − 4)+

2

s2(s2 + 1)− 1

s2(s + 1).

A partial fraction expansion of these terms leads to

Y (s) =1

s2 − 4− 2

s2 + 1+

1

s− 1

s + 1.

The inverse Laplace transform gives y(t) = 12

sinh 2t − 2 sin t + 1− e−t.

a From the differential equation it immediately follows that14.24


H(s) =1

L

1

s2 + ω20

,

where ω0 = (LC)−1/2. From table 7 we obtain: h(t) = (sin ω0t)/(Lω0).b The system is not stable since h(t) is not absolutely integrable (or: sincethe poles iω0 and −iω0 do not lie in the half-plane Re s < 0).c Let Q and V be the Laplace transforms of q and v, where v(t) = e−at

(a > 0). Then

Q(s) = V (s)H(s) =1

L

1

(s + a)(s2 + ω20)

.

A partial fraction expansion gives

L · Q(s) =1

a2 + ω20

1

s + a+

a

a2 + ω20

1

s2 + ω20

− 1

a2 + ω20

s

s2 + ω20

.

The inverse Laplace transform gives

q(t) =1

L(a2 + ω20)

„e−at +

a

ω0sin ω0t − cos ω0t

«.

d Note that ω0 6= a. Since V (s) = s/(s2 + a2) it follows as in part c that

Q(s) =1

L

s

(s2 + a2)(s2 + ω20)

=1

L(ω20 − a2)

„s

s2 + a2− s

s2 + ω20

«,

where we also applied a partial fraction expansion. The inverse Laplacetransform gives

q(t) =1

L(ω20 − a2)

(cos at − cos ω0t) .

e Since V (s) = 2(L cos ω0t)(s) = 2s/(s2 + ω20) it follows as in part c that

Q(s) =1

L

2s

(s2 + ω20)2

.

Since (L sin ω0t)(s) = ω0/(s2 + ω20), it follows from the differentiation rule

in the s-domain that

(Lt sin ω0t)(s) = − d

ds

ω0

s2 + ω20

=2sω0

(s2 + ω20)2

.

Hence, q(t) = (t sin ω0t)/(Lω0). This keeps increasing (in an oscillatingway) as t increases.

a One can write u(t) = cos t + ε(t − π) cos(t − π). Taking the Laplace14.25transform gives U(s) = (Lu)(s) = (1 + e−πs)s/(s2 + 1). Applying theLaplace transform to the differential equation and substituting the initialconditions gives (s2+s−2)Y −s−2 = U(s). Since s2+s−2 = (s−1)(s+2)we thus obtain, after a little simplifying, that

Y (s) =1

s − 1+

`1 + e−πs´ s

(s + 2)(s − 1)(s2 + 1).

For the second term we use a partial fraction expansion, which eventuallyleads to

Y (s) =1

s − 1+

`1 + e−πs´·

„2

15(s + 2)+

1

6(s − 1)− 3s

10(s2 + 1)+

1

10(s2 + 1)

«.


From table 7 and a shift in the time domain it follows by the inverseLaplace transform that y(t) = et + g(t)/30 + ε(t − π)g(t − π)/30 withg(t) = 4e−2t + 5et − 9 cos t + 3 sin t.b We have (Lδ(t−2))(s) = e−2s and (Lδ′(t−3))(s) = se−3s (table 9). Ap-plying the Laplace transform to the differential equation and substitutingthe initial conditions gives as in part a:

Y (s) =1

s − 1+

3e−2s + 6se−3s

(s + 2)(s − 1).

Partial fraction expansion leads to

Y (s) =1

s − 1+ e−2s

„1

s − 1− 1

s + 2

«+ 2e−3s

„1

s − 1+

2

s + 2

«.

From a shift in the time domain it follows by the inverse Laplace transformthat y(t) = et + ε(t − 2)(et−2 − e−2t+4) + 2ε(t − 3)(et−3 + 2e−2t+6).

a From table 7 we know that (LE)(s) = E/s. Applying the Laplace14.26transform to the system of differential equations and substituting the initialconditions gives

2RI2 + (3R + sL)I1 = E/s,2(R + sL)I2 − (R + sL)I1 = 0,

where I1 and I2 are the Laplace transforms of i1 and i2. From the secondequation we get I1 = 2I2. Substituting this into the first equation we obtain

I2 =E

2L

1

s(s + 4R/L).

After a partial fraction expansion we obtain from the inverse Laplace trans-form the solution

i1(t) = 2i2(t) =E

4R

“1− e−4Rt/L

”.

b Since (L sin 2t)(s) = 2/(s2 + 4) we obtain as in part a that

I2 =1

L

1

(s2 + 4)(s + 4R/L).

After a partial fraction expansion we obtain from the inverse Laplace trans-form the solution

i1(t) = i2(t) =L

8R2 + 2L2

„e−4Rt/L − cos 2t +

2R

Lsin 2t

«.

Apply the Laplace transform with respect to t to the partial differential14.27equation and substitute the initial conditions u(x, 0) = 0 and ut(x, 0) =2 sin πx, then it follows that

s2U(x, s)− 2 sin πx = 4Uxx,

where U(x, s) is the Laplace transform of u(x, t). Hence, U(x, s) satisfiesthe ordinary differential equation

U ′′ − s2

4U = −1

2sin πx.


The general solution of the homogeneous equation is U(x, s) = Aesx/2 +Be−sx/2, where A and B can still be functions of s. A particular solutioncan be found by trying U(x, s) = a sin πx. It then follows that a(π2 +s2/4) = 1/2 and so the general solution follows:

U(x, s) = Aesx/2 + Be−sx/2 +2

s2 + 4π2sin πx.

To determine A and B, we translate the remaining boundary conditions tothe s-domain by Laplace transforming them. From the conditions u(0, t) =u(2, t) = 0 it follows that U(0, s) = U(2, s) = 0. Using U(0, s) = 0 weobtain that A+B = 0 and using U(2, s) = 0 we obtain that A(es−e−s) = 0.Hence A = B = 0 and so

U(x, s) =2

s2 + 4π2sin πx.

The inverse Laplace transform gives u(x, t) = π−1 sin 2πt sin πx as solution.


For n ≥ 4 we have f [n] = 1− 1 = 0, for n < −1 we have f [n] = 0− 0 = 0,15.1and for −1 ≤ n < 4 we have f [n] = 0 − 1 = −1. Hence, f [n] = −(δ[n +1] + δ[n] + δ[n− 1] + δ[n− 2] + δ[n− 3]).

Since f [n] has period 5 we obtain that f [n] = δ5[n] + δ5[n− 2] + δ5[n− 3].15.2

Write z = eiπ/6, then z12 = 1, so f [n] = zn has period 12.15.3

Since ε[k] = 0 for k < 0 and ε[k] = 1 for k ≥ 0 we can write the sum in the15.4given right-hand side as

∞Xk=0

δ[n− k] =

∞Xk=−∞

ε[k]δ[n− k].

From theorem 15.1 it then follows that the sum equals ε[n].

Let Φ(ω) be the spectrum of φ(t), then Φ(ω) = 0 for |ω | > ω0 for some15.7ω0. According to the convolution theorem the spectrum of the convolutionis equal to the product of the spectra. This product is then also equal to 0for |ω | > ω0, hence the convolution is also band-limited.

Using Fourier series one can determine P (ω) explicitly as follows. We have15.9

P (ω) =

∞Xn=−∞

cneinω0ω = eiω0ω + e−iω0ω,

where ω0 = 2. Hence, F (ω) = (eiω0ω + e−iω0ω)pωs(ω). From table 3 weknow that p(t) = sin(ωst/2)/(πt) ↔ pωs(ω) and applying the shift rule wethen obtain that f(t) = p(t + 2) + p(t− 2).

We have to determine the Nyquist frequency of the convolution. By defin-15.10ition of convolution we have

F (ω) =

Z ∞

−∞pπ(ω − u)p2π(u) du =

Z π

−π

pπ(ω − u) du =

Z ω+π

ω−π

pπ(u) du.

Since pπ(ω) is 0 outside [−π/2, π/2] it follows that F (ω) = 0 outside[−3π/2, 3π/2]. The Nyquist frequency is thus equal to 3π and the samplingfrequency should satisfy 2π/T > 3π, hence T < 2/3.

We write f(t) = (ei(φ0+ω0t) + e−i(φ0+ω0t))/2. Then the spectrum F (ω) =15.11π(eiφ0δ(ω−ω0)+ e−iφ0δ(ω +ω0). In the proof of the sampling theorem wesee that the spectrum Fr(ω) of the reconstructed signal fr(t) equals

Fr(ω) =

∞Xk=−∞

F (ω − kωs)pωs(ω).

Hence,

Fr(ω) = πeiφ0

∞Xk=−∞

δ(ω − ω0 − kωs)pωs(ω)

+ πe−iφ0

∞Xk=−∞

δ(ω + ω0 − kωs)pωs(ω).

65


Since f(t)δ(t− t0) = f(t0)δ(t− t0) we thus have

Fr(ω) = πeiφ0

∞Xk=−∞

δ(ω − ω0 − kωs)pωs(ω0 + kωs)

+ πe−iφ0

∞Xk=−∞

δ(ω + ω0 − kωs)pωs(kωs − ω0).

Since ω0 = 2ωs/3 we have pωs(kωs ± ω0) = pωs((k ± 23)ωs). This means

that only the terms k = −1 and k = 1 contribute to the sum. Hence,

Fr(ω) = πeiφ0δ(ω + 12ω0) + πe−iφ0δ(ω − 1

2ω0).

Applying the inverse Fourier transform leads to the reconstructed signalfr(t) = cos(φ0 − 1

2ω0t).

a From T = 4/3 it follows that ωs = 3π/2. The sampling condition is15.12satisfied, so fr(t) = f(t).b From T = 2 it follows that ωs = π. Although the sampling condition isnot satisfied, it is easy to see that in this case Fs(ω) = 1 for all ω and soFs(ω) = pπ(ω) = F (ω). Again, fr(t) = f(t).c From T = 8/3 it follows that ωs = 3π/4. The sampling condition is notsatisfied. We have Fr(ω) = 2pωs(ω) − pα(ω) with α = ωs/3 = π/4. Usingtable 3 fr(t) follows:

fr(t) =2 sin(3πt/8)

πt− sin(πt/8)

πt.

a The impulse response follows from table 3:15.13

h(t) =2 sin2(πt/2)

π2t2.

b Let u(t) be an input with spectrum U(ω), then the spectrum Y (ω) ofthe output y(t) is given by Y (ω) = U(ω)H(ω). The Nyquist frequency ofu(t) is π, so U(ω) = 0 outside [−π/2, π/2]. Hence, also Y (ω) = 0 outside[−π/2, π/2].c For T = 1 the sampling frequency equals 2π. This is greater than theNyquist frequency of the input and so the sampling condition is satisfied.From the sampling theorem it then follows that

u(t) =1

π

1Xn=−1

sin(π(t− nT ))

t− nT.

To determine y(t) it suffices to know the response to the signal (sin πt)/tsince the system is linear and time-invariant. The spectrum of (sin πt)/(πt)is p2π(ω) and the spectrum of the corresponding output is then given byp2π(ω)qπ(ω) = qπ(ω). The response to (sin πt)/(πt) is thus the inverseFourier transform of qπ(ω), which is 2 sin2(πt/2)/(π2t2) (table 3). Hence,

y(t) =2 sin2(πt/2)

π2t2+

2 sin2(π(t− 1)/2)

π2(t− 1)2+

2 sin2(π(t + 1)/2)

π2(t + 1)2.

a We know that f [n] =P3

k=0 f [k]δ4[n− k] holds for all periodic discrete-15.15time signals with period 4. Hence, we only have to show that the samplinghas period 4, which is easy because

f [n] = f(nπ

2+ 2π) = f((n + 4)

π

2) = f [n + 4].


b The sampling frequency is 2π/T = 4 > 3, so the sampling conditionis satisfied. The signal can thus be reconstructed completely from thesampling.c Let f(t) =

P∞k=−∞ ckeikω0t be the Fourier series of f , where ω0 =

2π/T = 1. Since eiω0t ↔ 2πδ(ω − ω0) we can write the spectrum as

F (ω) = 2π

∞Xk=−∞

ckδ(ω − kω0).

This is a line spectrum with lines at kω0. The periodic signal is band-limitedwith Nyquist frequency 3. Hence, the spectrum is 0 outside [−3/2, 3/2].This implies that cn = 0 for |n | ≥ 2.d From part c follows that f(t) = c0+c1e

it +c−1e−it. Next substitute the

values t = 0,−π/2, π/2, π and use the given sampling from part b. Thenwe obtain the three equations c0 + c1 + c−1 = 0, c0 + ic1 − ic−1 = 1 andc0 − ic1 + ic−1 = 1. Hence, c0 = 1.

a The Nyquist frequency is 2π. The sampling frequency is 2π/T = 3π, so15.16the sampling condition is satisfied. The signal can thus be reconstructedcompletely from the sampling.b We know that (use table 3, no 2 and table 4, shift in the time domain,twice)

f(t) =sin π(t + 1)

2π(t + 1)+

sin π(t− 1)

2π(t− 1)=

t sin πt

π(1− t2).

c Using the sampling theorem we obtain thatZ ∞

−∞f(t) dt = T

∞Xn=−∞

f [n]1

π

Z ∞

−∞

sin π(t− nT )/T

t− nTdt.

The integral in the right-hand side is π, soR∞−∞ f(t) dt = T

P∞n=−∞ f [n].

d According to Parseval’s identity we have

E =

Z ∞

−∞| f(t) |2 dt =

1

2π

Z ∞

−∞|F (ω) |2 dω =

1

2π

Z π

−π

cos2 ω dω =1

2.


The formula for the 2-point DFT is16.1

F [k] =

1Xn=0

f [n]e−2πink/2.

Since e−πi = −1 we have

F [k] =

1Xn=0

f [n](−1)nk = f [0] + (−1)kf [1].

The signal f [n] = (−1)n has period 2 and, hence, period 4 as well. The16.22-point DFT F2[k] is given by 1− (−1)k (see 16.1). The 4-point DFT F4[k]is by definition given by

F4[k] =

3Xn=0

f [n]e−2πink/4 = 1− e−πik/2 + e−πik − e−3πik/2

= 1− (−i)k + (−1)k − ik = (1 + (−1)k)(1− ik).

The Fourier coefficients are given by16.3

ck =1

T

Z T/2

0

e−ikω0t dt =1

ikω0T(1− e−ikω0T/2).

Since ω0T = 2π we obtain ck = (1− (−1)k)/2πik for k 6= 0 and c0 = 1/2.The value F [0] follows immediately from the formula for F [k]:

F [0] =

N−1Xn=0

f [n] =

N−1Xn=0

f(nT/N).

When N is even then

F [0]

N=

1

N(f(0) + f(T/N) + · · ·+ f(T/2)) =

1

2= c0.

When N is odd then

F [0]

N=

N − 1

2N=

1

2− 1

2N= c0 −

1

2N.

Hence, we do not have c0 = F [0]/N for all N .

It is better to define the function value at the jumps as the average value16.5of the left-hand and right-hand limit. Using the DFT we then find a betterapproximation of the Fourier coefficients. (In exercise 16.4 we do havec0 = F [0]/N for all N .)

Apply the inverse 4-point DFT to F [k], then16.7

f [n] =“F [0] + F [1]eπin/2 + F [2]eπin + F [3]e3πin/2

”/4 = (1 + (−i)n) /4.

Since F [k] = |F [k] | ei arg(F [k]) we obtain from the given amplitude spec-16.8trum and phase spectrum that F [k] = 2eπik/2 and so F [0] = 2, F [1] = 2i,F [2] = −2, F [3] = −2i. Next apply the inverse DFT:

f [n] =“2 + 2ieπin/2 − 2eπin − 2ie3πin/2

”/4

68


and hence, f [0] = 0, f [1] = 0, f [2] = 0, f [3] = 2, that is, f [n] = 2δ4[n− 3].

a On [0, 2π] the periodic function f is given by16.10

f(t) = 1− | t− π |π

=

( tπ

for 0 ≤ t ≤ π,

2− tπ

for π ≤ t ≤ 2π.

Since f(t + π) = 1 − t/π for 0 < t ≤ π and f(t + π) = f(t − π) = t/π − 1for π ≤ t ≤ 2π we obtain that f(t) + f(t + π) = 1 for all t.b Since f [n] = f(2πn/N) we obtain from part a that f [n]+f [n+N/2] = 1.c Apply the N -point DFT to f [n] + f [n + N/2] = 1, using the shiftrule in the n-domain. Since 1 ↔ NδN [k] (table 11), we the obtain thatF [k] + e2πiNk/2NF [k] = NδN [k], hence (1 + eπik)F [k] = NδN [k]. When kis even and not a multiple of N then 2F [k] = NδN [k] = 0, so F [k] = 0.When k is a multiple of N then δN [k] = 1, hence, F [k] = N/2.

The signal is real, so F [−k] = F [k], which gives F [3] = F [−1] = F [1] = −i.16.12Applying the inverse DFT leads to f [n] = (1 + in+1 + (−i)n+1)/4.

Apply the definition of the cyclical convolution and use theorem 15.2, then16.13one obtains that (f ∗f)[n] = f [n]+f [n−1] = δN [n]+2δN [n−1]+δN [n−2].

Use the convolution theorem: first determine the functions f1 and f2 with16.14f1 ↔ cos(2πk/N) = (e2πik/N + e−2πik/N )/2 and f2 ↔ sin(4πk/N) =(e4πik/N − e−4πik/N )/2i and then calculate f [n] = (f1 ∗ f2)[n]. SinceδN [n − m] ↔ e−2πimk/N (table 11 and table 12, shift in the n-domain),we have f1[n] = (δN [n − 1] + δN [n + 1])/2 and f2[n] = (δN [n + 2] −δN [n − 2])/2i. From the convolution theorem and theorem 15.2 it thenfollows that f [n] = (f1 ∗ f2)[n] = (f2[n + 1] + f2[n − 1])/2, which equals(δN [n + 3] + δN [n + 1]− δN [n− 1]− δN [n− 3])/4i.

From table 11 we have that δN [n] ↔ 1 and so (table 12, shift in the n-16.16domain) δN [n− l] ↔ e−2πilk/N . Since cos2(πk/N) = (1+cos(2πk/N))/2 =(2+eπik/N +e−πik/N )/4 it follows that f [n] = (2δN [n]+δN [n−1]+δN [n+1])/4. The power equals

1

N

N−1Xn=0

| f [n] |2 .

Now | f [n] |2 = (4δN [n] + δN [n− 1] + δN [n + 1])/16 since δN [n]δN [n + 1] =δN [n]δN [n−1] = δN [n−1]δN [n+1] = 0. Also note that δN [n+1] = δN [n−(N − 1)] and hence, | f [0] |2 = 1/4, | f [1] |2 = 1/16, | f [N − 1] |2 = 1/16,while all other values are 0. This means that

P =1

N

„1

4+

1

16+

1

16

«=

3

8N.

a We calculate G[k] from the expression for the 5-point DFT. Note that16.18g[3] = g[−2] = c−2 = 1 and g[4] = g[−1] = c−1 = 2. Hence,

G[k] = g[0] + g[1]e−2πik/5 + g[2]e−4πik/5 + g[3]e−6πik/5 + g[4]e−8πik/5

= 1 + 2e−2πik/5 + e−4πik/5 + e−6πik/5 + 2e−8πik/5

= 1 + 4 cos(2πk/5) + 2 cos(4πk/5).

b Since the Fourier coefficients of f are known, we can express f as aFourier series: f(t) = c0 + c1e

iω0t + c−1e−iω0t + c2e

2iω0t + c−2e−2iω0t.

Hence,

f (2πm/5ω0)


= g[0] + g[1]e2πim/5 + g[2]e4πim/5 + g[3]e6πim/5 + g[4]e8πim/5 = G[−m].

c First use Parseval for Fourier series:

1

T

Z T

0

| f(t) |2 dt =

∞Xn=−∞

| cn |2 .

Since ck = 0 for | k | ≥ 3 we obtain thatP∞

n=−∞ | cn |2 =P2

n=−2 | g[n] |2 =P4n=0 | g[n] |2. Applying Parseval for the DFT one obtains the result.

a From table 11 we have that δN [n] ↔ 1 and so (table 12, shift in the16.19n-domain) f [n] ↔ e2πik/N − 1 + e−2πik/N = 2 cos(2πk/N)− 1.b Calculate the convolution using (16.14) and theorem 15.2, then

(f ∗ g)[n] =

N−1Xl=0

f [l]g[n− l] = g[n + 1]− g[n] + g[n− 1].

c First write g[n] as complex exponentials and then determine the N -point DFT using table 11 and the shift rule in the k-domain:

G[k] =N

2(δN [k − 2] + δN [k + 2]).

Finally apply (16.19) to calculate the power:

1

N

N−1Xn=0

| g[n] |2 =1

N2

N−1Xk=0

„N

2| δN [k − 2] + δN [k + 2] |

«2

=1

2.

a Since f(t) is real and even, the sampling f [n] is real and even since16.20f [−n] = f(−nT/5) = f(nT/5) = f [n].b Apply the inverse DFT, where the values of F [3] and F [4] are calcu-lated using the fact that F has period 5 and is even. Hence, f [n] = (1 +2e2πin/5 +2e8πin/5 + e4πin/5 + e6πin/5)/5, which equals (1+4 cos(2πn/5)+2 cos(4πn/5))/5.c The function f is band-limited with band-width 10π/T . The Fouriercoefficients ck of f contribute to the frequencies kω0 = 2πk/T . Hence,these are 0 for | k | ≥ 3. This means that f(t) is equal to the Fourier seriesc0 + c1e

iω0t + c−1e−iω0t + c2e

2iω0t + c−2e−2iω0t. Substituting t = nT/5

(and rearranging) we obtain that f [n] = c0 + c1e2πin/5 + c2e

4πin/5 +c−2e

6πin/5 + c−1e8πin/5. But this is precisely the expression for the in-

verse DFT, which implies that c0 = F [0]/5, c1 = F [1]/5, c2 = F [2]/5,c−2 = F [3]/5 = F [−2]/5, c−1 = F [4]/5 = F [−1]/5. Hence ck = F [k]/5 for| k | ≤ 2. Since F [k] has period 5 and ck = 0 for | k | > 2 we do not haveck = F [k]/5 for | k | > 2.


We can write (17.2) for N = 5 as follows: F [k] = f [0] + f [1]w−k5 + · · · +17.1

f [4]w−4k5 where w5 = e2πi/5 = w. Hence,

f [0] + f [1] + · · ·+ f [4] = F [0],

f [0] + f [1]w−1 + · · ·+ f [4]w−4 = F [1],

f [0] + f [1]w−2 + · · ·+ f [4]w−8 = F [2],

f [0] + f [1]w−3 + · · ·+ f [4]w−12 = F [3],

f [0] + f [1]w−4 + · · ·+ f [4]w−16 = F [4].

Using that w−5 = 1 we then obtain a system that is equal to the systemarising from the matrix representation.

As in exercise 17.1 we can write the formula for the inverse DFT in matrix17.2form:

1

5

0BBB@1 1 1 1 11 w w2 w3 w4

1 w2 w4 w w3

1 w3 w w4 w2

1 w4 w3 w2 w

1CCCA0BBB@

F [0]F [1]F [2]F [3]F [4]

1CCCA =

0BBB@f [0]f [1]f [2]f [3]f [4]

1CCCA .

The matrix in the left-hand side is thus the inverse of the matrix in exercise17.1.

Take N1 = 2 and N2 = 2. Note that f [3] = f [−1]. The matrix Mf now17.4looks as follows:

Mf =

„f [0] f [2]f [1] f [3]

«=

„2 20 1

«.

The 2-point DFT of the rows of this matrix gives

C =

„4 01 −1

«.

Multiplying this by the twiddle factors w−µν4 with w4 = eπi/2 = i gives

Ct =

„4 01 i

«.

Now calculate the 2-point DFT of the columns of this matrix to get the4-point DFT:

MF =

„5 i3 −i

«.

Hence, F [0] = 5, F [1] = i, F [2] = 3, F [3] = −i.

The matrix Mf looks as follows:17.6

Mf =

„f [0] f [2] . . . f [N − 2]1 1 . . . 1

«.

The N/2-point DFT of the first row of this matrix is A[k], while the N/2-point DFT of the second row follows from table 11. This gives the matrixC:

71


C =

„A[0] A[1] . . . A[N/2− 1]N/2 0 . . . 0

«.

Multiplying this by the twiddle factors will not change this matrix becauseof the zeroes in the second row of C and hence Ct = C. Now calculate the2-point DFT of the columns of Ct = C to get the matrix MF and, hence,the required N -point DFT of f [n]:

MF =

„A[0] + N/2 A[1] . . . A[N/2− 1]A[0]−N/2 A[1] . . . A[N/2− 1]

«=

„F [0] F [1] . . . F [N/2− 1]

F [N/2] F [N/2 + 1] . . . F [N − 1]

«.

Let A1[k] be the 2N -point DFT of f [2n] and B1[k] the 2N -point DFT of17.7f [2n + 1]. Then we have for the 4N -point DFT of f [n] (see (17.14)):

F [ν] = A1[ν] + w−ν4NB1[ν],

F [2N + ν] = A1[ν]− w−ν4NB1[ν],

where ν = 0, 1, . . . , 2N−1. The 2N -point DFT of f [2n] follows analogouslyfrom the N -point DFT of f [4n] and f [4n + 1]:

A1[ν] = A[ν] + w−ν2NC[ν],

A1[N + ν] = A[ν]− w−ν2NC[ν],

where ν = 0, 1, . . . , N − 1. Also, the 2N -point DFT of f [2n + 1] followsfrom the N -point DFT of f [4n + 1] and f [4n + 3]:

B1[ν] = B[ν] + w−ν2ND[ν],

B1[N + ν] = B[ν]− w−ν2ND[ν],

where ν = 0, 1, . . . , N − 1. Combining these results leads to the 4N -pointDFT of f [n]:

F [ν] = A[ν] + w−ν2NC[ν] + w−ν

4NB[ν] + w−3ν4N D[ν],

F [N + ν] = A[ν]− w−ν2NC[ν] + w−ν

4NB[ν]− w−3ν4N D[ν],

F [2N + ν] = A[ν] + w−ν2NC[ν]− w−ν

4NB[ν]− w−3ν4N D[ν],

F [3N + ν] = A[ν]− w−ν2NC[ν]− w−ν

4NB[ν] + w−3ν4N D[ν],

where ν = 0, 1, . . . , N − 1.

Since f(t) is causal we have FT (ω) =R T

0f(t)e−iωt dt. Apply the trapezium17.8

rule to the integral and substitute ω = (2k + 1)π/T , then it follows that

FT ((2k + 1)π/T ) ≈ T

N

N−1Xn=0

e−πin/Nf [n]e−2πin/N .

This shows that the spectrum at the frequencies ω = (2k + 1)π/T can beapproximated by the N -point DFT of e−πin/Nf [n].

Let F (ω) be the Fourier transform of f(t). Applying the trapezium rule to17.9 R T

0f(t)e−iωt dt and to

R T

0f(t)eiωt dt leads to (T/N)F [k] and (T/N)F [−k]

respectively, where F [k] denotes the N -point DFT of f [n]. Adding thisgives

F (2πk

T) ≈ T

N(F [k] + F [−k]) for | k | < N/2.


This means that we can efficiently approximate the spectrum of f at thefrequencies 2πk/T with | k | < N/2 by using an N -point DFT.

First we determine the DFT’s of the signals using table 11. Since f1[n] =17.11δN [n] + δN [n− 1] ↔ F1[k] = 1 + e−2πik/N and f2[n] = δN [n] + δN [n + 1] ↔F2[k] = 1 + e2πik/N we obtain the DFT of the cross-correlation as follows:

ρ12 ↔ F1[k]F2[k] = (1 + e2πik/N )2 = 1 + 2e2πik/N + e4πik/N .

Applying the inverse transform gives the cross-correlation ρ12 = δN [n] +2δN [n + 1] + δN [n + 2].

Let P (z) = f [0] + f [1]z + f [2]z2 and w3 = e2πi/3 = (−1 + i√

3)/2 = 1/w.17.13Then F [k] = P (w−k

3 ) = P (wk) = f [0]+f [1]wk +f [2]w2k = f [0]+wk(f [1]+wkf [2]).

Let Mf be the 3×N -matrix given by17.14

Mf =

0@ f [0] f [3] . . . f [3N − 3]f [1] f [4] . . . f [3N − 2]f [2] f [5] . . . f [3N − 1]

1A .

The N -point DFT of the rows of this matrix are given by A[k], B[k] andC[k] respectively. The matrix C is then given by

C =

0@ A[0] A[1] . . . A[N − 1]B[0] B[1] . . . B[N − 1]C[0] C[1] . . . C[N − 1]

1A .

Multiplying this by the twiddle factors w−νµ3N gives the matrix Ct. The

3-point DFT of the columns of Ct will give us the matrix MF :

MF =

0@ F [0] F [1] . . . F [N − 1]F [N ] F [N + 1] . . . F [2N − 1]F [2N ] F [2N + 1] . . . F [3N − 1]

1A .

Since the 3-point DFT of g[n] is given by G[k] = g[0] + g[1]w−k3 + g[2]w−2k

3

we conclude that

F [µN + ν] = A[ν] + w−(ν+Nµ)3N B[ν] + w

−2(ν+Nµ)3N C[ν].

Take N1 = 3 and N2 = 3m−1 and consider the N1 ×N2-matrix17.15

Mf =

0@ f [0] f [3] . . . f [N − 3]f [1] f [4] . . . f [N − 2]f [2] f [5] . . . f [N − 1]

1A .

Let the N2-point DFT of the rows f [3n], f [3n + 1] and f [3n + 2] be givenby A[k], B[k] and C[k], then the matrix C is given by

C =

0@ A[0] A[1] . . . A[N2 − 1]B[0] B[1] . . . B[N2 − 1]C[0] C[1] . . . C[N2 − 1]

1A .

Multiplying this by the twiddle factors and then applying the 3-point DFT

of the columns gives the matrix MF containing the N -point DFT of f [n]:

MF =

0@ F [0] F [1] . . . F [N2 − 1]F [N2] F [N2 + 1] . . . F [2N2 − 1]F [2N2] F [2N2 + 1] . . . F [N − 1]

1A .


To calculate the 3-point DFT we need 6 additions and 4 multiplications,hence 10 elementary operations. The N2-point DFT of f [3n], of f [3n +1], and of f [3n + 2] can be determined by repeatedly splitting this into(multiples of 3), (multiples of 3)+1, and (multiples of 3)+2. We thus onlyhave to determine 3-point DFT’s.

Let h[n] =PN

l=0 f [l]g[n − l]. Since g[n] is causal and g[n] = 0 for n > N17.16we have that h[n] = 0 for n < 0 or n > 2N . Hence, it suffices to calculateh[n] for n = 0, 1, . . . , 2N . Now let fp[n] and gp[n] be two periodic signalswith period 2N + 1 and such that fp[n] = f [n] and gp[n] = g[n] for n =0, 1, . . . , 2N . Then

h[n] =

2NXl=0

fp[l]gp[n− l] = (fp ∗ gp)[n] for n = 0, 1, . . . , 2N .

Now let fp[n] ↔ Fp[k] and gp[n] ↔ Gp[k]. Then we have

h[n] =1

2N + 1

2NXk=0

Fp[k]Gp[k]e2πink/(2N+1)

for n = 0, 1, . . . , 2N . Although these are really 2N + 1-point DFT’s, wecontinue for convenience our argument with 2N . If we assume that cal-culating a 2N -point DFT using the FFT requires 2N(2 log N) elementaryoperations, then the number of elementary operations to calculate h[n] will(approximately) equal 2N(2 log N)+2N(2 log N)+(2N+1)+2N(2 log N) =6N(2 log N) + 2N + 1. A direct calculation would require in the order N2

operations, which for large N is much less efficient.

Let P (z) = f [0] + f [1]z + f [2]z2 + f [3]z3 and w = e−2πi/4 = −i. Then17.17F [k] = P (wk), hence, F [0] = P (1), F [1] = P (−i), F [2] = P (−1), F [3] =P (i). Now f [n] is even and so f [3] = f [−1] = f [1]. Then F [k] is alsoeven, hence F [3] = F [1]. This gives F [0] = f [0] + f [1] + f [2] + f [3] =f [0] + 2f [1] + f [2], F [1] = f [0] − if [1] − f [2] + if [3] = f [0] − f [2], andF [2] = f [0]− f [1] + f [2]− f [3] = f [0]− 2f [1] + f [2].


When f [n] = 0 for |n | > N for some N > 0, then18.1

F (z) =

∞Xn=−∞

f [n]z−n =

−1Xn=−N

f [n]z−n +

NXn=0

f [n]z−n.

The anti-causal part converges for all z, while the causal part convergesfor all z 6= 0. The region of convergence is thus given by 0 < | z | < ∞. Iff [n] = 0 for n ≥ 1 then the z-transform converges for all z ∈ C.

The anti-causal part converges for all z. The causal part can be written as18.2

∞Xn=0

„„1

2z

«n

+

„1

3z

«n«.

Hence, the z-transform converges for | 2z | > 1 and | 3z | > 1, that is, for| z | > 1/2.

a A direct calculation of F (z) gives18.3

∞Xn=0

““z

2

”n

+“z

3

”n”.

This series converges for | z/2 | < 1 and | z/3 | < 1, hence, for | z | < 2.b The z-transform is given by

F (z) =

∞Xn=0

cos(πn/2)z−n.

Since | cos(πn/2) | ≤ 1 for all n and sinceP∞

n=0 | z |−n converges for | z | > 1,

the z-transform also converges for | z | > 1. Moreover,P∞

n=0 cos(πn/2)diverges since limn→∞ cos(πn/2) 6= 0. We conclude that the z-transformconverges for | z | > 1.c From parts a and b it follows immediately that the region of convergenceis the ring 1 < | z | < 2.

a We rewrite f [n] in order to apply a shift in the n-domain: f [n] =18.42(n− 2)ε[n− 2] + 4ε[n− 2]. Since ε[n] ↔ z/(z − 1) for | z | > 1, we obtainfrom the shift rule that 4ε[n − 2] ↔ 4z−1/(z − 1). It also follows fromε[n] ↔ z/(z − 1) and the differentiation rule that nε[n] ↔ z/(z − 1)2 andapplying the shift rule we then obtain that (n− 2)ε[n− 2] ↔ z−1/(z− 1)2.Combining these results gives the z-transform F (z) of f [n]:

F (z) =2z−1

(z − 1)2+

4z−1

z − 1=

4z − 2

z(z − 1)2for | z | > 1.

b We rewrite f [n] in order to apply a shift in the n-domain: f [n] =2(n + 2)ε[−(n + 2)]− 4ε[−(n + 2)]. From ε[n] ↔ z/(z − 1) we obtain fromtime reversal that ε[−n] ↔ (1/z)/((1/z)−1) = 1/(1−z) for | z | > 1. Fromthe shift rule it follows that 4ε[−(n+2)] ↔ 4z2/(1−z). It also follows fromε[−n] ↔ 1/(1− z) and the differentiation rule that nε[−n] ↔ −z/(1− z)2

and applying the shift rule we then obtain that 2(n + 2)ε[−(n + 2)] ↔−2z3/(1− z)2. Combining these results gives

75


F (z) =−2z3

(1− z)2− 4z2

1− z=

2z3 − 4z2

(1− z)2.

c We can, for example, calculate this z-transform in a direct way:

f [n] = (−1)nε[−n] ↔∞X

n=0

(−1)nzn =1

1 + zfor | z | < 1.

d Again we can, for example, calculate this z-transform in a direct way:

f [n] = ε[4− n] ↔∞X

n=−4

zn =z−4

1− zfor | z | < 1.

One can also use ε[−n] ↔ 1/(1− z) and apply a shift rule.e From example 18.6 it follows that (n2 − n)ε[n] ↔ 2z/(z − 1)3 andnε[n] ↔ z/(z − 1)2. Adding these results gives

n2ε[n] ↔ z(z + 1)

(z − 1)3for | z | > 1.

From example 18.2 we obtain that 4nε[n] ↔ z/(z − 4) for | z | > 4. Thedifferentiation rule implies that n4nε[n] ↔ 4z/(z−4)2 for | z | > 4. Togetherthese results give

F (z) =z(z + 1)

(z − 1)3+

4z

(z − 4)2for | z | > 4.

a A direct calculation of the z-transform gives (for | z | > 1):18.5

F (z) =

∞Xn=0

cos(nπ/2)z−n = 1− z−2 + z−4 − · · · = 1

1 + z−2=

z2

1 + z2.

b A direct calculation of the z-transform gives (for | z | > 1):

F (z) =

∞Xn=0

sin(nπ/2)z−n = z−1 − z−3 + · · · = z−1

1 + z−2=

z

1 + z2.

c A direct calculation of the z-transform of einφε[n] gives (for | z | > 1):

∞Xn=0

einφz−n = 1 + eiφz−1 + e2iφz−2 + · · · = 1

1− eiφ/z=

z

z − eiφ.

A direct calculation of the z-transform of 2neinφε[−n] gives (for | z | < 2):

0Xn=−∞

2neinφz−n =

∞Xn=0

2−ne−inφzn =1

1− e−iφz/2.

Hence we get for 1 < | z | < 2:

F (z) =z

z − eiφ+

1

1− e−iφz/2.

One could apply a partial fraction expansion here (see e.g. exercise 18.10).18.8However, in this case it is easy to obtain the z-transform in a direct wayby developing F (z) in a series expansion. Since the z-transform has toconverge for | z | > 2 (there are poles at z = ±2i and f [n] has a finiteswitch-on time) we develop F (z) as follows:

F (z) =1

z2 + 4=

1

z2

„1− 4

z2+

16

z4+ · · ·

«.


From the series we now obtain that f [n] = 0 for n ≤ 0 and that f [2n] =(−1)n−122n−2, f [2n + 1] = 0, for n > 0.

As in the previous exercise we obtain the z-transform in a direct way by18.9developing F (z) in a series expansion. Since the unit circle | z | = 1 has tobelong to the region of convergence (f [n] has to be absolute convergent)we develop F (z) for | z | < 2 as follows:

F (z) =1

4(1 + z2/4)=

1

4

„1− z2

4+

z4

16+ · · ·

«.

From the series we obtain that f [n] = 0 for n ≥ 1 and f [2n] = (−1)n22n−2,f [2n− 1] = 0, for n ≤ 0.

The poles are at z = −1/2 and z = −3; the signal f [n] must have a finite18.10switch-on time, hence, the z-transform has to converge for | z | > 3. Apartial fraction expansion of F (z)/z gives

F (z)

z= 1 +

1/10

z + 1/2− 18/5

z + 3.

Applying (18.14) to the expansion of F (z) gives

(−1/2)nε[n] ↔ z

z + 1/2for | z | > 1/2, (−3)nε[n] ↔ z

z + 3for | z | > 3.

Combining this gives f [n] = δ[n + 1] + ((−1/2)nε[n]− 36(−3)nε[n])/10.

In exercise 18.10 we obtained the partial fraction expansion. Now the18.11signal f [n] has to be absolutely convergent, which means that | z | = 1 hasto belong to the region of convergence. Applying (18.14) to z/(z + 1/2)gives

(−1/2)nε[n] ↔ z

z + 1/2for | z | > 1/2,

while applying (18.15) to z/(z + 3) gives

(−3)nε[−n− 1] ↔ −z

z + 3for | z | < 3.

Combining this gives f [n] = δ[n+1]+((−1/2)nε[n]+36(−3)nε[−n−1])/10.

A direct application of the definition of the convolution product gives18.12

(f ∗ g)[n] =

M1Xl=N1

f [l]g[n− l]

= f [N1]g[n−N1] + f [N1 + 1]g[n−N1 − 1] + · · ·+ f [M1 − 1]g[n−M1 + 1] + f [M1]g[n−M1].

Since g[n − N1] = 0 for n < N1 + N2 and, hence, also g[n − N1 − 1] = 0,g[n−N1 − 2] = 0, etc., we have that (f ∗ g)[n] = 0 for n < N1 + N2. Thismeans that the switch-on time is N1 + N2.Since g[n−M1] = 0 for n > M1 + M2 and, hence, also g[n−M1 + 1] = 0,g[n−M1 + 2] = 0, etc., we have that (f ∗ g)[n] = 0 for n > M1 + M2. Thismeans that the switch-off time is M1 + M2.

a The signal is causal and F (z) has poles at z = ±i. The z-transform18.13converges for | z | > 1. Write F (z) as the sum of a geometric series (or usea partial fraction expansion):


F (z) =z

z2 + 1=

1

z

„1− 1

z2+

1

z4+ · · ·

«.

From the series we obtain that f [n] = 0 for n < 0 and f [2n + 1] = (−1)n,f [2n] = 0, for n ≥ 0.b The convolution theorem gives (f ∗ f)[n] ↔ F 2(z). Hence,

h[n] = (f ∗ f)[n] =

∞Xl=−∞

f [l]f [n− l] =

nX

l=0

f [l]f [n− l]

!ε[n].

From this we obtain that h[n] = 0 for n < 0 while for m ≥ 0 we have:

h[2m + 1] = f [0]f [2m + 1] + f [1]f [2m] + · · ·+ f [2m + 1]f [0] = 0,h[2m] = f [0]f [2m] + f [1]f [2m− 1] + · · ·+ f [2m]f [0]

= 0 + (−1)0(−1)m−1 + 0 + · · ·+ (−1)m−1(−1)0 + 0 = m(−1)m−1.

Apply the definition of the convolution product and use that ε[n − l] = 018.14for l > n and ε[n− l] = 1 for l ≤ n.

Use that f [n] ↔ F (z) and δ[n] ↔ 1 and apply the convolution theorem18.15(assuming that the intersection of the regions of convergence is non-empty)then (f ∗ δ)[n] ↔ F (z) · 1 = F (z). Hence, f [n] = (f ∗ δ)[n].

Define a discrete-time signal h[n] by h[n] =Pn

l=−∞ 2l−nf [l], which is the18.17

convolution product of f [n] with 2−nε[n]. Now 2−nε[n] ↔ z/(z − 1/2) for| z | > 1/2 and f [n] ↔ F (z). Hence, h[n] ↔ zF (z)/(z − 1/2). Assumingthat | z | = 1 belongs to the region of convergence of the z-transform of f [n]we get

eiωF (eiω)

eiω − 1/2=

∞Xn=−∞

h[n]e−inω.

Using (18.22) to determine h[n] gives

h[n] =1

π

Z π

−π

eiωF (eiω)einω

2eiω − 1dω.

Applying (18.31) gives18.19

∞Xn=−∞

| f [n] |2 =1

2π

Z π

−π

˛F (eiω)

˛2dω =

1

2π

Z π

−π

cos2 ω dω =1

2.

(One can also determine f [n] first and then calculateP∞

n=−∞ | f [n] |2 dir-ectly.)

We have that ρ[n] = (g ∗ f)[n] with g[l] = f [−l]. The convolution theorem18.20and property (18.25) (or table 15, entry 2) imply that the spectrum of ρ[n]is given by G(eiω)F (eiω). But G(eiω) = F (eiω) (combine table 15, entries

2 and 5) and hence ρ[n] ↔ F (eiω)F (eiω) =˛F (eiω)

˛2.

a The signal f [n] is causal. The region of convergence is the exterior of a18.22circle. Applying the shift rule to 2−nε[n] ↔ z/(z−1/2) for | z | > 1/2 gives2−(n+2)ε[n+2] ↔ z3/(z−1/2) for | z | > 1/2. Hence, F (z) = 4z3/(z−1/2)for | z | > 1/2.b One could write g[n] = (f ∗ ε)[n], apply the convolution theorem andthen a partial fraction expansion. However, it is easier to do a directcalculation: g[n] =

Pnl=−∞ 2−lε[l+2] and hence, g[n] = 0 for n < −2 while

g[n] =Pn

l=−2 2−l = 8(1− 2−n−3) for n ≥ −2 (it is a geometric series). We


thus obtain that g[n] = (8− 2−n)ε[n + 2].c The z-transform of f [n] converges for | z | > 1/2, which contains theunit circle | z | = 1. Hence, the Fourier transform of f [n] equals F (eiω) =4e3iω/(eiω − 1/2).

a The signal f [n] is absolutely summable. The z-transform has one pole18.23at z = −2 and therefore the region of convergence is | z | < 2, since itmust contain | z | = 1. Since F (z)/z = 1 − 2/(z + 2) we have F (z) =z − 2z/(z + 2) for | z | < 2. The inverse transform of this gives f [n] =δ[n + 1]− (−2)n+1ε[−n− 1].b The Fourier transform of f [n] equals F (eiω) = e2iω/(eiω + 2).c We have f [n] ↔ F (z) for | z | < 2. The scaling property (table 14, entry5) gives 2nf [n] ↔ F (z/2) for | z | < 4. The spectrum of 2nf [n] is thus equalto F (eiω/2) = e2iω/(2eiω + 8).

a The signal f [n] is causal. The poles of F (z) are at ±i/2 and at 0. The18.24region of convergence is the exterior of the circle | z | = 1/2. This containsthe unit circle and so the signal is absolutely summable. The spectrum off [n] equals F (eiω) = 1/(eiω(4e2iω + 1)).b First apply a partial fraction expansion to F (z)/z (the denominatorequals z2(2z + i)(2z − i)):

F (z)

z=

1

z2+

i

z − i/2− i

z + i/2.

Applying (18.14) to F (z) gives f [n] = δ[n− 1] + (i(i/2)n − i(−i/2)n)ε[n].c If F (eiω) = F (e−iω), then the signal is real. Since

F (eiφ) = 1/(e−iφ(4e−2iφ + 1)) = F (e−iφ),

the signal is indeed real. (One can also write (i(i/2)n − i(−i/2)n) =in+12−n(1 − (−1)n), which equals 0 for n = 2k and 21−n(−1)k+1 forn = 2k + 1, showing clearly that it is real.)

For n < 0 we have thatPn

l=0 g[l]g[n− l] = 0 since g[n] is causal. Thus f [n]18.25is also causal. Since g[l] = 0 for l < 0 and g[n − l] = 0 for l > n we canwrite

f [n] =

∞Xl=−∞

g[l]g[n− l] = (g ∗ g)[n].

This implies that F (z) = G(z)2, so we need to determine G(z). To do so,we write G(eiω) = 1/(4 + cos 2ω) as a function of eiω:

1

4 + cos 2ω=

1

4 + 12(e2iω + e−2iω)

=2e2iω

e4iω + 8e2iω + 1.

Taking z = eiω we find that G(z) = 2z2/(z4 +8z2 +1), which gives F (z) =G(z)2.


a Substituting δ[n] for u[n] we find that19.2

h[n] =

n−1Xl=−∞

2l−nδ[l] = 2−nε[n− 1].

Use the definition of δ[n] and ε[n] to verify (19.3):

y[n] =

n−1Xl=−∞

2l−nu[l] =

∞Xl=−∞

2l−nε[n− 1− l]u[l] =

∞Xl=−∞

h[n− l]u[l].

b As in part a we obtain that h[n] = (δ[n + 1] + δ[n− 1])/2 and

∞Xl=−∞

h[n− l]u[l] =1

2

∞Xl=−∞

u[l] (δ[n− l + 1] + δ[n− l − 1])

=1

2(u[n + 1] + u[n− 1]) = y[n].

c We now have h[n] =P∞

l=n 2l−nδ[l] = 2−nε[−n] and

∞Xl=−∞

h[n− l]u[l] =

∞Xl=−∞

u[l]2l−nε[l − n] =

∞Xl=n

u[l]2l−n = y[n].

a An LTD-system is causal if and only if the impulse response is a causal19.3signal. So this system is causal. The system is stable if and only ifthe impulse response is absolutely summable. Since

P∞n=−∞ |h[n] | =P∞

n=1 2−n = 1 < ∞, this system is stable.b This system is not causal, but it is stable since

∞Xn=−∞

|h[n] | =∞X

n=−∞

(δ[n + 1] + δ[n− 1])/2 = 1 < ∞.

c This system is not causal, and it is not stable sinceP∞

n=−∞ |h[n] | =P∞n=0 2n = ∞.

a The step response is the response to ε[n] and can be calculated using19.4(19.3): a[n] = (h ∗ ε)[n] =

P∞l=−∞(δ[l] − 2δ[l − 1] + δ[l − 2])ε[n − l] =

ε[n]− 2ε[n− 1] + ε[n− 2].b The response to an arbitrary input also follows from (19.3): y[n] =(h ∗ u)[n] = u[n]− 2u[n− 1] + u[n− 2].

Since δ[n] = ε[n] − ε[n − 1], it follows from linearity and time-invariance19.5that h[n] = a[n] − a[n − 1]. Now use (19.3) to calculate the response y[n]to u[n] = 4−nε[n]:

y[n] = (h ∗ u)[n] =

∞Xl=−∞

a[l]u[n− l]−∞X

l=−∞

a[l − 1]u[n− l]

=

∞Xl=−∞

a[l]u[n− l]−∞X

l=−∞

a[l]u[n− 1− l].

80


We now calculate the first sum; the second one then follows by replacing nby n− 1.

∞Xl=−∞

a[l]u[n− l] =

∞Xl=−∞

“2−lε[l]− 3−lε[l − 1]

”u[n− l]

=

∞Xl=0

2−l4l−nε[n− l]−∞X

l=1

3−l4l−nε[n− l]

=

4−n

nXl=0

2l

!ε[n]− 4−n

nXl=1

(4/3)lε[n− 1]

= 4−n(2n+1 − 1)ε[n]− 3 · 4−n((4/3)n+1 − (4/3))ε[n− 1]= (21−n − 4−n)ε[n]− 4(3−n − 4−n)ε[n− 1].

Hence, replacing n by n − 1 and then taking terms together in the sum,y[n] = (21−n−4−n)ε[n]−4(3−n+2−n−2·4−n)ε[n−1]−4(31−n−41−n)ε[n−2].

The transfer function follows from anε[n] ↔ z/(z − a) for | z | > | a |; this19.8is because we can write cos nφ = (einφ + e−inφ)/2, and hence

h[n] ↔ 1

2

„z

z − eiφ/2+

z

z − e−iφ/2

«for | z | >

˛eiφ/2

˛= 1/2. We thus obtain:

H(z) =z(z − cos φ/2)

z2 − cos φz + 1/4

for | z | > 1/2. The poles are at eiφ/2 and e−iφ/2, which is inside the unitcircle. Therefore the system is stable.

a The impulse response h[n] can be determined by a partial fraction ex-19.9pansion of H(z)/z. Since

H(z)

z=

1

9

„1

z + 1/3− 1/3

(z + 1/3)2

«we have

H(z) =1

9

„z

z + 1/3− z/3

(z + 1/3)2

«.

The system is stable, so the region of convergence must contain | z | = 1.This region is thus given by | z | > 1/3. Using table 13 we find thath[n] = ((−1/3)n + n(−1/3)n)ε[n]/9.b Write u[n] = (einπ/2 − e−inπ/2)/2i and use (19.10): e±inπ/2 has re-sponse H(e±iπ/2)e±inπ/2, so the response to u[n] is (H(eiπ/2)einπ/2 −H(e−iπ/2)e−inπ/2)/2i, which is of the form (w − w)/2i = Im (w). Hencethe response is

Im

„−1

−9 + 6i + 1einπ/2

«= Im

„8 + 6i

100(cos(nπ/2) + i sin(nπ/2))

«,

which is (3 cos(nπ/2) + 4 sin(nπ/2))/50.

a The frequency response can be written as H(eiω) = 1 + e2iω + e−2iω.19.11Since H(eiω) =

P∞n=−∞ h[n]e−inω it follows that h[0] = 1, h[2] = h[−2] = 1

and h[n] = 0 for all other n. Hence, the impulse response is h[n] = δ[n] +δ[n− 2]+ δ[n+2]. The input is u[n] = δ[n− 2]. Since δ[n] 7→ h[n], we have


u[n] 7→ h[n− 2], so the response to u[n] is y[n] = δ[n− 2] + δ[n− 4] + δ[n].b The impulse response is not causal, so the system is not causal.

From (19.15) and the fact that H(eiω) is even we obtain that19.13

h[n] =1

2π

Z π

−π

H(eiω) cos(nω) dω =1

π

Z ωb

ωa

cos(nω) dω

=1

nπ(sin nωb − sin nωa),

which can be written as 2(sin 12n(ωb−ωa) cos 1

2n(ωb + ωa))/(nπ) for n 6= 0

and

h[0] =1

π

Z ωb

ωa

dω =ωb − ωa

π.

The spectrum Y (eiω) of y[n] is a periodic function with period 2π. Apply19.14Parseval for periodic functions and substitute Y (eiω) = H(eiω)U(eiω) toget the desired result.

a Apply the z-transform to the difference equation, using the shift rule19.15in the n-domain. We then obtain that (1 + 1

2z−1)Y (z) = U(z). Since

H(z) = Y (z)/U(z) it follows that

H(z) =1

1 + z−1/2=

z

z + 1/2.

From table 13 we get the impulse response h[n] = (−1/2)nε[n].b We could use z-transforms here: from Y (z) = H(z)U(z) we get Y (z) =z2/((z + 1/2)(z − 1/2)) (use table 13) and applying a partial fraction ex-pansion to Y (z)/z then leads to y[n] (again use table 13). However, in thiscase it is easier to follow the direct way:

(h ∗ u)[n] =

∞Xl=−∞

(−1/2)lε[l](1/2)n−lε[n− l].

Now if n < 0 then this is 0, while if n ≥ 0 then it equals (1/2)nPnl=0(−1)l =

(1/2)n+1(1 + (−1)n).c The transfer function H(z) has one pole at z = −1/2, which is insidethe unit circle, so the system is stable.

a From the difference equation we obtain that (1 − z−2/4)Y (z) = (1 +19.17z−1)U(z) and hence

H(z) =1 + z−1

1− z−2/4=

z(z + 1)

z2 − 1/4.

A partial fraction expansion of H(z)/z gives

H(z)

z=

−1/2

z + 1/2+

3/2

z − 1/2.

Hence, H(z) = (−z/2)/(z + 1/2) + (3z/2)/(z − 1/2). Using table 13 wefind that h[n] = ((−1/2)n+1 + 3(1/2)n+1)ε[n].b The (rational) transfer function H(z) has poles at z = ±1/2, which lieinside the unit circle, so the system is stable.c The z-transform of ε[n] is z/(z− 1) and therefore the z-transform A(z)of the step response a[n] is given by A(z) = H(z)U(z) = z2(z + 1)/((z −1)(z + 1/2)(z − 1/2)). A partial fraction expansion of A(z)/z gives


A(z)

z=

8/3

z − 1− 3/2

z − 1/2− 1/6

z + 1/2.

Multiply this by z and use table 13 to obtain a[n] = ((8/3)− 3(1/2)n+1 −(1/3)(−1/2)n+1)ε[n].d Since u[n] = ε[n]+ε[n−2] and ε[n] 7→ a[n] we get u[n] 7→ a[n]+a[n−2].

a To find the impulse response we first apply a partial fraction expansion19.18to H(z)/z, which gives

H(z)

z=

1

z− 1

z + 1/2− 1/4

(z + 1/2)2.

Multiply this by z and use table 13 to obtain that h[n] = δ[n]− ((−1/2)n−(1/2)n(−1/2)n)ε[n].b According to (19.9) we have that zn 7→ H(z)zn. Substituting z = −1we get the response to the input (−1)n. Since H(−1) = 0, the response isthe null-signal.c The (rational) transfer function H(z) has one pole at z = −1/2, whichis inside the unit circle, so the system is stable.d The impulse response is real, so the system is real.e Let u[n] 7→ y[n], then Y (eiω) = H(eiω)U(eiω). Furthermore we havethat U(eiω) =

P∞n=−∞ u[n]e−inω. Comparing this with U(eiω) = cos 2ω =

(e2iω + e−2iω)/2 we may conclude that u[2] = u[−2] = 1/2 and u[n] = 0for n 6= ±2, hence, u[n] = (δ[n− 2] + δ[n + 2])/2. By superposition it thenfollows from δ[n] 7→ h[n] that y[n] = (h[n− 2] + h[n + 2])/2.

a The response to u[n + N ] is y[n + N ] (time-invariance), but also u[n] =19.19u[n + N ] for n ∈ Z, so y[n] = y[n + N ] for n ∈ Z.b In (16.7) (with f replaced by u) the input u[n] is written as superpos-ition of the signals e2πink/N . Since zn 7→ H(z)zn, we have e2πink/N 7→H(e2πik/N )e2πink/N . Hence we obtain from (16.7) that

y[n] =1

N

N−1Xk=0

U [k]H(e2πik/N )e2πink/N .

On the other hand we have from the inverse DFT for y[n] that

y[n] =1

N

N−1Xk=0

Y [k]e2πink/N ,

where Y [k] is the N -point DFT of y[n]. Hence, Y [k] = U [k]H(e2πik/N ).

a Since H(eiω) = cos 2ω = (e2iω +e−2iω)/2 we see from definition (19.11)19.20of the frequency response that h[2] = h[−2] = 1/2 and h[n] = 0 for n 6= ±2.Hence, h[n] = (δ[n + 2] + δ[n− 2])/2.b Using the inverse DFT we can recover u[n] from the DFT F [k] of u[n]:

u[n] =1

4

3Xk=0

F [k]e2πink/4,

where F [k] is the 4-point DFT of u[n]. We have F [0] = 1, F [1] = −1,F [2] = 0, F [3] = 1 and since e2πink/4 7→ H(e2πik/4)e2πink/4, it follow bysuperposition that y[n] = (1 + 2i sin(nπ/2))/4.

a Since δ[n] = ε[n] − ε[n − 1] we have for the responses that h[n] =19.21a[n] − a[n − 1] and hence h[n] = n2(1/2)nε[n] − (n − 1)2(1/2)n−1ε[n −


1]. For the z-transforms we have (use the shift rule in the n-domain)H(z) = A(z) − A(z)/z = A(z)(z − 1)/z. From table 13 it follows that(1/2)nε[n] ↔ z/(z−1/2), n(1/2)nε[n] ↔ (z/2)/(z−1/2)2,

`n2

´(1/2)nε[n] ↔

(z/4)/(z − 1/2)3 and since n2 = 2`

n2

´+ n it then follows that

A(z) =z/2

(z − 1/2)3+

z/2

(z − 1/2)2=

1

2

z(z + 1/2)

(z − 1/2)3.

This means that the transfer function H(z) equals

H(z) =1

2

(z − 1)(z + 1/2)

(z − 1/2)3.

b The impulse response is causal, so the system is causal.c From einω 7→ H(eiω)einω it follows that

y[n] =1

2

(eiω − 1)(eiω + 1/2)

(eiω − 1/2)3einω.

a From the difference equation we obtain that (1− z−1/2)Y (z) = (z−1 +19.22z−2)U(z) and hence

H(z) =z−1 + z−2

1− z−1/2=

z + 1

z2 − z/2.

Since ε[n] ↔ z/(z − 1) = U(z) we have

a[n] ↔ A(z) = H(z)U(z) =z + 1

(z − 1/2)(z − 1).

A partial fraction expansion of A(z)/z gives

A(z)

z=

2

z+

4

z − 1− 6

z − 1/2.

Multiply this by z and use table 13 to obtain that a[n] = 2δ[n] + (4 −6(1/2)n)ε[n].b The (rational) transfer function H(z) has poles at z = 0 and z = 1/2,which lie inside the unit circle, so the system is stable.

a From the difference equation we obtain that (6 − 5z−1 + z−2)Y (z) =19.23(6− 6z−2)U(z) and hence

H(z) =6− 6z−2

6− 5z−1 + z−2=

6(z2 − 1)

6z2 − 5z + 1.

A partial fraction expansion of H(z)/z (note that the denominator equalsz(2z − 1)(3z − 1)) gives

H(z)

z= −6

z+

16

z − 1/3− 9

z − 1/2.

Multiply this by z and use table 13 to obtain that h[n] = −6δ[n] +(16(1/3)n − 9(1/2)n)ε[n].b The frequency response H(eiω) is obtained from the transfer functionH(z) by substituting z = eiω, so H(eiω) = 6(e2iω − 1)/(6e2iω − 5eiω + 1).c From (19.10) we know that Y (eiω) = H(eiω)U(eiω). Since y[n] isidentically 0, we know that Y (eiω) = 0 for all frequencies ω. Since H(eiω) 6=0 for e2iω 6= 1 we have U(eiω) = 0 for e2iω 6= 1. The frequencies ω = 0or ω = π may still occur in the input. These frequencies correspond to


the time-harmonic signals ei0n = 1 and eiπn = (−1)n respectively. Hence,the input consists of a linear combination of 1 and (−1)n, which meansthat the solution equals u[n] = A + B(−1)n, where A and B are complexconstants.

a The frequency response can be written as H(eiω) = 1+2 cos ω+cos 2ω =19.241 + eiω + e−iω + e2iω/2 + e−2iω/2. Since H(eiω) =

P∞n=−∞ h[n]e−inω it

follows that h[n] = δ[n] + δ[n + 1] + δ[n− 1] + δ[n + 2]/2 + δ[n− 2]/2.b If we write U(eiω) = 1+sin ω+sin 2ω as complex exponentials, as in parta, then it follows that u[0] = 1, u[1] = u[2] = i/2, u[−1] = u[−2] = −i/2and so u[n] = δ[n]+ i(δ[n−1]+ δ[n−2]− δ[n+1]− δ[n+2])/2. We have tocalculate E =

P∞n=−∞ | y[n] |2. In this case it is easiest to do this directly

(and so not using Parseval). From the expression for u[n] it follows (bylinearity and time-invariance) that y[n] = h[n] + i(h[n − 1] + h[n − 2] −h[n +1]−h[n +2])/2. Substituting h[n] from part a we get y[n] as a linearcombination of δ[n− 4], δ[n− 3], . . . , δ[n + 3], δ[n + 4]. The coefficients inthis combination are not hard to determine. In fact, they are −i/4, −3i/4,(1/2)− i, 1−3i/4, 1, 1+3i/4, (1/2)+ i, 3i/4, i/4. Their contribution to Eis 1/16, 9/16, 20/16, 25/16, 1, 25/16, 20/16, 9/16, 1/16. The sum of thesecontributions is 126/16, hence, E = 7 7

8.

[Solutions Manual] Fourier and Laplace Transform - Antwoorden

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