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).

Answers to selected exercises for chapter 2

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

2 Answers to selected exercises for chapter 2

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

Answers to selected exercises for chapter 3

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

4 Answers to selected exercises for chapter 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

Answers to selected exercises for chapter 3 5

+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.

6 Answers to selected exercises for chapter 3

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

Answers to selected exercises for chapter 4

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

Answers to selected exercises for chapter 4 13

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

14 Answers to selected exercises for chapter 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:

Answers to selected exercises for chapter 4 15

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:

16 Answers to selected exercises for chapter 4

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

Answers to selected exercises for chapter 4 17

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.

Answers to selected exercises for chapter 5

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

Answers to selected exercises for chapter 5 19

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).

20 Answers to selected exercises for chapter 5

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.

Answers to selected exercises for chapter 5 21

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).

22 Answers to selected exercises for chapter 5

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.

Answers to selected exercises for chapter 6

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

24 Answers to selected exercises for chapter 6

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.

Answers to selected exercises for chapter 6 25

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.

26 Answers to selected exercises for chapter 6

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.

Answers to selected exercises for chapter 7

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

Answers to selected exercises for chapter 16 69

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)

70 Answers to selected exercises for chapter 16

= 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.

Answers to selected exercises for chapter 17

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

72 Answers to selected exercises for chapter 17

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.

Answers to selected exercises for chapter 17 73

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 .

74 Answers to selected exercises for chapter 17

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].

Answers to selected exercises for chapter 18

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

76 Answers to selected exercises for chapter 18

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+ · · ·

«.

Answers to selected exercises for chapter 18 77

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):

78 Answers to selected exercises for chapter 18

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

Answers to selected exercises for chapter 18 79

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.

Answers to selected exercises for chapter 19

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

Answers to selected exercises for chapter 19 81

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

82 Answers to selected exercises for chapter 19

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

Answers to selected exercises for chapter 19 83

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 −

84 Answers to selected exercises for chapter 19

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

Answers to selected exercises for chapter 19 85

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.