Transformada de Fourier & Laplace Antwoorden

Answers to selected exercises for chapter 1

1.1

Apply cos( + ) = cos cos sin sin , then f1 (t) + f2 (t) = A1 cos t cos 1 A1 sin t sin 1 + A2 cos t cos 2 A2 sin t sin 2 = (A1 cos 1 + A2 cos 2 ) cos t (A1 sin 1 + A2 sin 2 ) sin t = C1 cos t C2 sin t, where C1 = A1 cos 1 + A2 cos 2 and C2 = A1 sin 1 + A2 sin 2 . Put A = p 2 2 C1 + C2 and take such that cos = C1 /A and sin = C2 /A (this is possible since (C1 /A)2 +(C2 /A)2 = 1). Now f1 (t)+f2 (t) = A(cos t cos sin t sin ) = A cos(t + ).

1.2

Put c1 = A1 ei1 and c2 = A2 ei2 , then f1 (t) + f2 (t) = (c1 + c2 )eit . Let c = c1 + c2 , then f1 (t) + f2 (t) = ceit . The signal f1 (t) + f2 (t) is again a time-harmonic signal with amplitude | c | and initial phase arg c. The power P is given by Z / Z A2 / P = A2 cos2 (t + 0 ) dt = (1 + cos(2t + 20 )) dt 2 / 4 / A2 . = 2 The energy-content is E = The power P is given by P =3 1X | cos(n/2) |2 = 1 . 2 4 n=0

1.5

1.6 1.7

R0

e2t dt = 1 . 2

1.8 1.9

The energy-content is E = sum 1/(1 e2 ).

P

n=0

e2n , which is a geometric series with

a If u(t) is real, then the integral, and so y(t), is also real. b Since Z Z u( ) d | u( ) | d, it follows from the boundedness of u(t), so | u( ) | K for some constant K, that y(t) is also bounded. c The linearity follows immediately from the linearity of integration. The time-invariance follows from the substitution = t0 in the integral Rt u( t0 ) d representing the response to u(t t0 ). t1 Rt d Calculating t1 cos( ) d gives the following response: (sin(t) sin(t ))/ = 2 sin(/2) cos(t /2)/. Rt e Calculating t1 sin( ) d gives the following response: ( cos(t) + cos(t ))/ = 2 sin(/2) sin(t /2)/. f From the response to cos(t) in d it follows that the amplitude response is | 2 sin(/2)/ |. g From the response to cos(t) in d it follows that the phase response is /2 if 2 sin(/2)/ 0 and /2 + if 2 sin(/2)/ < 0. From

1

2


phase and amplitude response the frequency response follows: H() = 2 sin(/2)ei/2 /. 1.11 a The frequency response of the cascade system is H1 ()H2 (), since the reponse to eit is rst H1 ()eit and then H1 ()H2 ()eit . b The amplitude response is | H1 ()H2 () | = A1 ()A2 (). c The phase response is arg(H1 ()H2 ()) = 1 () + 2 (). a The amplitude response is | 1 + i | e2i = 2. b The input u[n] = 1 has frequency = 0, initial phase 0 and amplitude 1. Since ein H(ei )ein , the response is H(e0 )1 = 1 + i for all n. c Since u[n] = (ein + ein )/2 we can use ein H(ei )ein to obtain that y[n] = (H(ei )ein + H(ei )ein )/2, so y[n] = (1 + i) cos((n 2)). d Since u[n] = (1 + cos 4n)/2, we can use the same method as in b and c to obtain y[n] = (1 + i)(1 + cos(4(n 2)))/2. a The power is the integral of f 2 (t) over [/ | | , / | |], times | | /2. Now cos2 (t + 0 ) integrated over [/ | | , / | |] equals / | | and cos(t) cos(t + 0 ) integrated over [/ | | , / | |] is (/ | |) cos 0 . Hence, the power equals (A2 + 2AB cos(0 ) + B 2 )/2. R1 b The energy-content is 0 sin2 (t) dt = 1/2. The power is the integral of | f (t) |2 over [/ | | , / | |], times | | /2, which in this case equals | c |2 . a The amplitude response is | H() | = 1/(1 + 2 ). The phase response is arg H() = . b The input has frequency = 1, so it follows from eit H()eit that the response is H(1)ieit = iei(t+1) /2. a The signal is not periodic since sin(2N ) = 0 for all integer N . i b The frequency response H(ei ) equals A(ei )eie , hence, we obtain i i 2 that H(e ) = e /(1 + ). The response to u[n] = (e2in e2in )/2i is then y[n] = (e2i(n+1) e2i(n+1) )/(10i), so y[n] = (sin(2n + 2))/5. The amplitude is thus 1/5 and the initial phase 2 /2. a If u(t) = 0 for t < 0, then the integral occurring in y(t) is equal to 0 for t < 0. For t0 0 the expression u(t t0 ) is also causal. Hence, the system is causal for t0 0. b It follows from the boundedness of u(t), so | u( ) | K for some constant K, that y(t) is also bounded (use the triangle inequality and the inequality 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, the system is real. d The response is Z t y(t) = sin((t t0 )) + sin( ) d = sin((t t0 )) 2(cos t)/.t1

1.12

1.13

1.14 1.16

1.17

1.18

1.19

a If u[n] = 0 for n < 0, then y[n] is also equal to 0 for n < 0 whenever n0 0. Hence, the system is causal for n0 0. b It follows from the boundedness of u[n], so | u[n] | K for some constant K and all n, that y[n] is also bounded (use the triangle inequality): n n n X X X u[l] K + | u[l] | K + K, | y[n] | | u[n n0 ] | + l=n2 l=n2 l=n2


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 expression for y[n] is real, hence, y[n] is real. This means that the system is real. d The response to u[n] = cos n = (1)n is y[n] = (1)nn0 +n X l=n2

(1)l = (1)nn0 + (1)n (1 1 + 1)

= (1)n (1 + (1)n0 ).


2.1

p a The absolute values follow from x2 + y 2 and are given by 2, 2, 3, 2 respectively. The arguments follow from standard angles and are given by 3/4, /2, , 4/3 respectively. b Calculating modulus and argument gives 2 + 2i = 2 2ei/4 , 3 + i = 5i/6 3i/2 2e and 3i = 3e . In the proof of theorem 2.1 it was shown that | Re z | | z |, which implies that | 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 = 7/4, arg z2 = 2/3. Hence, | z1 /z2 | = |1 | / | z2 | = 2 and arg(z1 /z2 ) = arg(z1 ) arg(z2 ) = 13/12, z 2 3 2 3 so z1 /z2 = 2e13i/12 . Similarly we obtain z1 z2 = 2048e3i/2 and z1 /z2 = 1 3i/2 e . 2 The solutions are given in a separate gure on the website. a The four solutions 1 i are obtained by using the standard technique to solve this binomial equation (as in example 2.3). b As part a; we now obtain the six solutions 6 2(cos(/9 + k/3) + i sin(/9 + k/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 z 5 z 4 + z 1 as (z 1)(z 4 + 1) and then solve z 4 = 1 to nd the roots 2(1 i)/2. Combining linear factors with complex conjugate roots we obtain z 5 z 4 + z 1 = (z 1)(z 2 + 2z + 1)(z 2 2z + 1). Since 2i = 2ei/2 the solutions are z = ln 2 + i(/2 + 2k), where k Z.1 Split F (z) as A/(z 2 ) + B/(z 2) and multiply by the denominator of F (z) to obtain the values A = 1/3 and B = 4/3 (as in example 2.6).

2.2

2.4

2.5 2.6

2.7

2.8 2.9 2.11

a Split F (z) as A/(z + 1) + B/(z + 1)2 + C/(z + 3) and multiply by the denominator of F (z) to obtain the values C = 9/4, B = 1/2 and, by comparing the coecient of z 2 , A = 5/4 (as in example 2.8). Trying the rst few integers we nd the zero z = 1 of the denominator. A long division gives as denominator (z 1)(z 2 2z + 5). We then split F (z) as A/(z 1) + (Bz + C)/(z 2 2z + 5). Multiplying by the denominator of F (z) and comparing the coecients of z 0 = 1, z and z 2 we obtain that A = 2, B = 0 and C = 1. a Using the chain rule we obtain f (t) = i(1 + it)2 .

2.12

2.13 2.14 2.15

Use integration by parts twice and the fact that a primitive of ei0 t is 3 ei0 t /i0 . The given integral then equals 4(1 i)/0 , since e2i = 1. it Since 1/(2 eit ) = 1/ 2 eit and 2 eit 2 e = 1, the result R R 1 1 follows from 0 u(t) dt 0 | u(t) | dt. 1

2


2.16

P a Use that | an | = 1/ n6 + 1 1/n3 and the fact that 1/n3 conn=1 verges (example 2.17). P b Use that | an | 1/n2 and the fact that n=1 1/n2 converges. c Use that | an | = 1/ nen eni = 1/(nen ) 1/en and the fact that P n n=1 1/e converges since it is a geometric series with ratio 1/e. a Use the ratio test to conclude that the series is convergent: n! 1 = lim lim = 0. n (n + 1)! n n + 1 b The series is convergent; proceed as in part a: 2n+1 + 1 3n + n 2 + 1/2n 1 + n/3n 2 = lim = . lim n+1 3 n +1 n 1 + 1/2n n 3 + (n + 1)/3 n +n+1 2 3

2.17

2.19

Determine the radius of convergence as follows: n+1 2n+2 2 2 1 + 1/n2 z n +1 = lim 2 z 2 lim = 2 z2 . 2 + 1 2n z 2n n (n + 1) n 1 + 2/n + 2/n2 This is less than 1 if z 2 < 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; the sum is (1/(1 i))(1/(1 (z i))), so 1/(2 z(1 i)). b First solving w2 = 1 leads to z 2 = 0 or z 2 = 2i. The equation z 2 = 2i has solutions 1 + i and 1 i and z 2 = 0 has solution 0 (with multiplicity 2). c One has P (z) = z(z 4 + 8z 2 + 16) = z(z 2 + 4)2 = z(z 2i)2 (z + 2i)2 , so 0 is a simple zero and 2i are two zeroes of multiplicity 2. Split F (z) as (Az + B)/(z 2 4z + 5) + (Cz + D)/(z 2 4z + 5)2 and multiply by the denominator of F (z). Comparing the coecient of z 0 , z 1 , z 2 and z 3 leads to the values A = 0, B = 1, C = 2 and D = 2. R 2 Replace cos t by (eit + eit )/2, then we have to calculate 0 (e2it + 1)/2 dt, which is . a Using the ratio test we obtain as limit 5/3. This is less than 1 and so the series converges. P b Since (n + in )/n2 = (1/n) + (in /n2 ) and the series 1/n diverges, n=1 this series is divergent. P The series cn (z 2 )n converges for all z with z 2 < R, so it has radius n=0 of convergence R. a Determine the radius of convergence as follows: (1 + i)2n+2 z n+1 n+1 = lim | z | n + 1 (1 + i)2 = 2 | z | . lim n n+2 (1 + i)2n z n n n+2 This is less than 1 if | z | < 1/2, so the radius of convergence is 1/2. b Calculate f (z) by termwise dierentiation of the series and multiply this by z. It then follows that zf (z) + f (z) = X n=0

2.20 2.23

2.25

2.26 2.27

2.29 2.30

(1 + i)2n z n =

X n=0

(2iz)n .

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

a

b

c

3

2

4 5

d

e

f

3 2

1 + 2i

1

0

3

2

2

g

1

2 21 2

3


3.2

A trigonometric polynomial can be written as f (t) =k X a0 + (am cos(m0 t) + bm sin(m0 t)). 2 m=1

Now substitute this for f (t) in the right-hand side of (3.4) and use the fact that all the integrals in the resulting expression are zero, except for R T /2 the integral T /2 sin(m0 t) sin(n0 t) dt with m = n, which equals T /2. Hence, one obtains bn . 3.4 The function g(t) = f (t) cos(n0 t) has period T , so Z T /2 Z T Z T g(t)dt g(t)dt + g(t)dt = 0 T /2 0 Z T Z T /2 Z 0 Z T /2 g(t T )dt + g(t)dt = g( )d + g(t)dt = T /2 0 T /2 0 Z T /2 = g(t)dt.T /2

Multiplying by 2/T gives an . 3.6 From a sketch of the periodic function with period 2 given by f (t) = | t | for t (, ) we obtain Z 0 Z 1 1 cn = (t)eint dt + teint dt. 2 2 0 As in example 3.2 these integrals can be calculated using integration by parts for n = 0. Calculating c0 separately (again as in example 3.2) we obtain c0 = , 2 cn = (1)n 1 n2

Substituting these values of cn in (3.10) we obtain the Fourier series. One can also write this as a Fourier cosine series: 4 X cos((2k + 1)t) . 2 (2k + 1)2 k=0

3.7

From the description of the function we obtain that Z 1 1 (1+in)t cn = e dt. 2 0 This integral can be evaluated immediately and leads to cn = in 1 ` (1)n e1 1 . 2(n2 2 + 1)

The Fourier series follows from (3.10) by substituting cn . 3.9 The Fourier coecients are calculated by splitting the integrals into a real and an imaginary part. For c0 this becomes:

3

4


c0 =

1 2

Z

1

t2 dt +1

i 2

Z

1

t dt =1

1 . 3

For n = 0 we have that Z Z 1 1 2 int i 1 int cn = t e dt + te dt. 2 1 2 1 The second integral can be calculated using integration by parts. To calculate the rst integral we apply integration by parts twice. Adding the results and simplifying somewhat we obtain the Fourier coecients (and thus the Fourier series): cn = 3.10 (1)n (2 n) . n2 2

From the values of the coecients cn calculated earlier in exercises 3.6, 3.7 and 3.9, one can immediately obtain the amplitude spectrum | cn | and the phase spectrum arg cn (note e.g. that arg cn = if cn > 0, arg cn = if cn < 0, arg cn = /2 if cn = iy with y > 0 and arg cn = /2 if cn = iy with y < 0). This results in three gures that are given separately on the website. a By substituting a = T /4 in (3.14) it follows that sin(n/4) n for n = 0, c0 = 1 . 4

3.11

cn =

b As in a, but now a = T and we obtain c0 = 1, cn = 0 for n = 0.

Hence, the Fourier series is 1 (!). This is no surprise, since the function is 1 for all t. 3.12 By substituting a = T /2 in (3.15) it follows that c0 = 3.14 1 , 2 cn = 0 for n = 0 even, cn = 2 n2 2 for n odd.

We have that f (t) = 2p2,4 (t) q1,4 (t) and so the Fourier coecients follow by linearity from table 1: c0 = 3/4, cn = (2n sin(n/2) 4 sin2 (n/4))/(n2 2 ) for n = 0. Note that f (t) can be obtained from the sawtooth z(t) by multiplying the shifted version z(t T /2) by the factor T /2 and then adding T /2, that is, f (t) = T z(t T ) + T . Now use the Fourier coecients of z(t) (table 1 2 2 2 e.g.) and the properties from table 2 to obtain that c0 = T , 2 cn = iT 2n for all n = 0.

3.15

3.17 3.19

Shifts over a period T (use the shift property and the fact that e2in = 1 for all n). In order to determine the Fourier sine series we extend the function to an odd function of period 8. We calculate the coecients bn as follows (the an are 0): Z Z 1 2 1 2 (2 sin(nt/4)) dt + t sin(nt/4) dt bn = 4 4 4 2


5

+

1 4

Z2

4

2 sin(nt/4) dt.

The second integral can be calculated by an integration by parts and one then obtains that 8 4 bn = 2 2 sin(n/2) cos(n), n n which gives the Fourier sine series. For the Fourier cosine series we extend the function to an even function of period 8. As above one can calculate the coecients an and a0 (the bn are 0). The result is a0 = 3, 3.21 an = 8 (cos(n/2) 1) n2 2 for all n = 0.

In order to determine the Fourier cosine series we extend the function to an even function of period 8. We calculate the coecients an and a0 as follows (the bn are 0): Z 1 4 2 16 a0 = (x 4x) dx = , 2 0 3 while for n 1 we have Z Z 1 0 2 1 4 2 an = (x + 4x) cos(nx/4) dx + (x 4x) cos(nx/4) dx 4 4 4 0 Z 4 Z 4 1 = x2 cos(nx/4) dx 2 x cos(nx/4) dx. 2 0 0 The rst integral can be calculated by applying integration by parts twice; the second integral can be calculated by integration by parts. Combining the results one then obtains that an = 32((1)n 1) 32((1)n + 1) 64(1)n = , 2 2 2 2 n n n2 2

which also gives the Fourier cosine series. One can write this series as 8 16 X 1 + 2 cos(nx/2). 3 n=1 n2

For the Fourier sine series we extend the function to an odd function of period 8. As above one can calculate the coecients bn (the an are 0). The result is bn = 3.24 64((1)n 1) n3 3 for all n 1.

If f is real and the cn are real, then it follows from (3.13) that bn = 0. A function whose Fourier coecients bn are all 0 has a Fourier series containing 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 only and is thus odd. Since sin(0 t) = (ei0 t ei0 t )/2i we have Z T /2 Z T /2 1 1 cn = ei(1n)0 t dt ei(1+n)0 t dt. 2iT 0 2iT 0

3.25

6


The rst integral equals T /2 for n = 1 while for n = 1 it equals i((1)n + 1)/((1 n)0 ). The second integral equals T /2 for n = 1 while for n = 1 it equals i((1)n+1 1)/((1 + n)0 ). The Fourier coecients are thus c1 = 1/(4i), c1 = 1/(4i) and ((1)n +1)/(2(1n2 )) for n = 1, 1; the Fourier series follows immediately from this. 3.27 b The even extension has period 2a, but it has period a as well. We can thus calculate the coecients an and a0 as follows (the bn are 0): Z Z 2 a/2 2 0 a0 = 2bt/a dt 2bt/a dt = b. a 0 a a/2 while for n 0 we obtain from an integration by parts that Z Z 2 a/2 2 0 an = (2bt/a) cos(2nt/a) dt (2bt/a) cos(2nt/a) dt a 0 a a/2 n 2b((1) 1) = , n2 2 which gives the Fourier cosine series. It can also be determined using the result of exercise 3.6 by applying a multiplication and a scaling. The odd extension has period 2a and the coecients bn are given by (the an are 0): Z Z 1 a/2 2bt 1 a/2 2bt 2b) sin(nt/a) dt + sin(nt/a) dt bn = ( a a a a a/2 a Z a 1 2bt + ( + 2b) sin(nt/a) dt a a/2 a 8b = 2 2 sin(n/2), n where we used integration by parts.

a /2

b

4

2

0

2

4

n

4

2

0

2

4

n

a 1 2

b

2

4

2

0

2

4

n

4

2

0

2

4

n

2

a 1 2

b

4

2

0

2

4

n

4

2

0

2

4

n


4.1

a The periodic block function from section 3.4.1 is a continuous function on [T /2, T /2], except at t = a/2. At these points f (t+) and f (t) exist. Also f (t) = 0 for t = a/2, while f (t+) = 0 for t = a/2 and t = T /2 and f (t) = 0 for t = a/2 and t = T /2. Hence f is piecewise continuous and so the periodic block function is piecewise smooth. Existence of the Fourier coecients has already been shown in section 3.4.1. The periodic triangle function is treated analogously. b For the periodic block function we have X n=

| cn |2

a2 8 X 1 + 2 2 2 T T 0 n=1 n2

P P 2 1 since sin2 (n0 a/2) 1. The series n2 converges, so n= | cn | n=1 converges. The periodic triangle function is treated analogously. 4.2 4.4 This follows immediately from (3.11) (for part a) and (3.8) (for part b). Take t = T /2 in the Fourier series of the sawtooth from example 4.2 and use that sin(n0 T /2) = sin(n) = 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. Using the shift property we obtain c0 = 1 , 2 cn = 0 even n = 0, cn = i n odd n.

4.6

The Fourier series follows by substituting the cn . One can write the series with sines only (split the sum in two pieces: one from n = 1 to and another from n = 1 to ; change from n to n in the latter): 1 2 X sin(2k + 1)t + . 2 2k + 1 k=0

b The function is piecewise smooth and it thus satises the conditions of the fundamental theorem. At t = /2 the function f is continuous, so the series converges to f (/2) = 1. Since sin((2k + 1)/2) = (1)k , formula (4.11) follows: X (1)k = . 2k + 1 4

k=0

4.7

R a We have that c0 = (2)1 0 t dt = /4, while the Fourier coecients for n = 0 follow from an integration by parts: Z (1)n i (1)n 1 1 cn = teint dt = + . 2 0 2n 2n2 The Fourier series follows by substituting these cn : X (1)n i (1)n 1 1 + + eint . 4 2 n n2 n=,n=0

12


13

b From the fundamental theorem it follows that the series will converge to 1 (f (+) + f ()) = /2 at t = (note that at there is a jump). If 2 we substitute t = into the Fourier series, take to the other side of the 4 =-sign, then multiply by 2, and nally split the sum into a sum from n = 1 to and a sum from n = 1 to , then it follows that X (1)n 1 =2 (1)n 2 n2 n=1

(the terms with (1)n i/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: X 2 1 = . 8 (2k 1)2k=1

4.9

a From f (0+) = 0 = f (0) and f (1) = 0 = f ((1)+) it follows that f is continuous. We have that f (t) = 2t+1 for 1 < t < 0 and f (t) = 2t+1 for 0 < t < 1. Calculating the dening limits for f from below and from above 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 is continuous at t = 1. Since f (t) = 2 for 1 < t < 0 and f (t) = 2 for 0 < t < 1 we see that f is discontinuous. b The function f is the sum of g and h with period 2 dened 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 coecients are cn = (1)n i/n (see section 3.4.3). The function h is the odd extension of t2 on (0, 1] and its Fourier coecients have been determined in the rst example of section 3.6. By linearity one obtains the Fourier coecients of f . In terms of the an and bn they become an = 0 and bn = 4(1 (1)n )/ 3 n3 . Hence, they decrease as 1/n3 . c Use e.g. the fundamental theorem for odd functions to obtain f (t) = 8 X sin(2k + 1)t . 3 (2k + 1)3 k=0

4.10 4.12

Now substitute t = 1/2 and use that f (1/2) = 1/4 and sin((2k + 1)/2) = (1)n to obtain the required result. P 1 Use (3.8) to write the right-hand side of (4.14) as a2 /4 + 2 (a2 + b2 ). 0 n n=1 n a The Fourier coecients of f and g are (see table 1 or section 3.4.1), respectively, fn = (sin na)/n for n = 0 and f0 = a/ and gn = (sin nb)/n for n = 0 and g0 = b/. Substitute into Parseval (4.13) and calculate the R a/2 integral (1/) a/2 1 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 (n/2) = 1 for n odd and 0 for n even, then (4.10) follows. a The Fourier coecients are (see table 1 or section 3.4.2 and use that sin2 (n/2) = 1 for n odd and 0 for n even): cn = 2/n2 2 for n odd, 0 for n = 0 even and c0 = 1/2. From Parseval for f = g, so from (4.14), it then follows that (calculate the integral occurring in this formula): 1 1 8 X 1 = + 2 3 4 (2k 1)4 k=1

4.13

14


(again we split the sum in a part from n = 1 to and from n = 1 to ). Take all constants together and multiply by 2 /8, then the required result follows. b Since S= X 1 X 1 X 1 = + , n4 (2k)4 (2k + 1)4 n=1 k=1 k=0

it follows from part a that S= 1 4 4 1 X 1 = S+ . + 4 16 k 96 16 96 k=1

4.15

P 4 1 Solving for S we obtain n4 = . n=1 90 Rb Rb Ra Since a f (t) dt = T /2 f (t) dt T /2 f (t) dt, we can apply theorem 4.9 twice. Two of the innite sums cancel out (the ones representing h0 in theorem 4.9), the other two can be taken together and lead to the desired result. This follows from exercise 4.15 by using (3.8), so cn = (an ibn )/2 and cn = (an + ibn )/2 (n N). a 4 The Fourier series is given by X n=0

4.16 4.17

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

b Since Z t cos(2n + 1)t 1 sin(2n + 1) d = , 2n + 1 2n + 1 the integrated series becomes 4X 1 4 X cos(2n + 1)t . n=0 (2n + 1)2 n=0 (2n + 1)2

From (4.10) we see that the constant in this series equals /2. Rt c The series in part b represents the function f ( ) d (theorem 4.9 or better still, exercise 4.16). Calculating this integral we obtain the function g(t) with period 2 given for < t by g(t) = | t | . d Subtracting from the Fourier series of | t | in exercise 3.6 we obtain a Fourier series for g(t) which is in accordance with the result from part b. 4.19 4.20 This again follows as in exercise 4.16 from (3.8). Since f is piecewise smooth, f is piecewise continuous and so the Fourier coecients cn of f exist. Since f is continuous, we can apply integration by parts, as in the proof of theorem 4.10. It then follows that cn = in0 cn , where cn are the Fourier coecients of f . But cn = in0 cn by theorem 2 4.10, so cn = n2 0 cn . Now apply the Riemann-Lebesgue lemma to cn , then it follows that limn n2 cn = 0. a The Fourier coecients have been determined in exercise 3.25: c1 = 1/(4i), c1 = 1/(4i) and ((1)n + 1)/(2(1 n2 )) for n = 1, 1. Taking positive and negative n in the series together, we obtain the following Fourier series:

4.22


15

1 1 2X 1 sin t + + cos 2kt. 2 1 4k2 k=1

b The derivative f exists for all t = n (n Z) and is piecewise smooth. According to theorem 4.10 we may thus dierentiate f by dierentiating its Fourier series for t = n: f (t) = 1 4X k cos t sin 2kt. 2 1 4k2 k=1

At t = n the dierentiated series converges to (f (t+) + f (t))/2, which equals 1/2 for t = 0, while it equals 1/2 for t = . Hence, the dierentiated series is a periodic function with period 2 which is given by 0 for 1 < t < 0, 2 for t = 0, cos t for 0 < t < , 1 for t = . 2 4.25 4.26 Write down the expression for Si(x) and change from the variable t to t, then it follows that Si(x) = Si(x). a From the denition of Si(x) it follows that Si (x) = sin x/x. So Si (x) = 0 if sin x/x = 0. For x > 0 we thus have Si (x) = 0 for x = k with k N. A candidate for the rst maximum is thus x = . Since sin x/x > 0 for 0 < x < and sin x/x < 0 for < x < 2, it follows that Si(x) indeed has its rst maximum at x = . b The value at the rst maximum is Si(). Since Si() = 1.852 . . . and /2 = 1.570 . . ., the overshoot is 0.281 . . .. The jump of f at x = 0 is = 3.141 . . ., so the overshoot is 8.95 . . .%, so about 9%. a The function f is continuous for t = (2k + 1) (k Z) and it then converges to f (t), which is 2t/ for 0 | t | < /2, 1 for /2 t < and 1 for < t /2. For t = (2k+1) 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 an integration by parts: Z /2 Z 2(1)n 4 2 4 . bn = 2 t sin nt dt + sin nt dt = 2 2 sin(n/2) 0 /2 n n Since sin(n/2) = 0 if n even and (1)k if n = 2k + 1, the Fourier series is 4 X (1)n 2 X (1)n sin nt + 2 sin(2n + 1)t. n=1 n n=0 (2n + 1)2

4.28

Substituting t = 0 and t = it is easy to verify the fundamental theorem for these values. c We cannot dierentiate the series; the resulting series is divergent because limn (1)n cos nt = 0. Note that theorem 4.10 doesnt apply since f is not continuous. d We can integrate the series since theorem 4.9 can be applied (note that Rt c0 = 0). If we put g(t) = f ( ) d , then g is even, periodic with period 2 and given by (t2 /)(3/4) for 0 t < /2 and by t for /2 t . 4.29 Use table 1 to obtain the Fourier coecients and then apply Parseval, that is, (4.13). Calculating the integral in Parsevals identity will then give the rst result; choosing a = /2 gives the second result. a The Fourier series has been determined in the last example of section 3.6. Since f is continuous (and piecewise smooth), the Fourier series converges to f (t) for all t R:

4.30

16


f (t) =

2 4X 1 cos 2nt. n=1 4n2 1

b First substitute t = 0 in the Fourier series; since f (0) = 0 and cos 2nt = 1 for all n, the rst result follows. Next substitute t = /2 in the Fourier series; since f (/2) = 1 and cos 2nt = (1)n for all n, the second result follows. c One should recognize the squares of the Fourier coecients here. Hence we have to apply Parsevals identity (4.14), or the alternative form given in exercise 4.10. This leads to Z 1 4 1 X 16 sin2 t dt = 2 + . 2 2 2 n=1 (4n2 1)2 R Since sin2 t dt = , the result follows. 4.31 a Since f1 is odd it follows that Z 1 T /2 (f1 f2 )(t) = f1 (t + )f2 ( ) d. T T /2

Now change the variable from to and use that f2 is odd, then it follows that (f1 f2 )(t) = (f1 f2 )(t). b The convolution product equals Z 1 1 (f f )(t) = f (t ) d. 2 1 Since f is odd, part a implies that f f is even. It is also periodic with period 2, so it is sucient to calculate (f f )(t) for 0 t 1. First note that f is given by f (t) = t 2 for 1 < t 2. Since 1 1 and 0 t 1 we see that t 1 t t + 1. From 0 t 1 it follows that 1 t 1 0, and so close to = 1 the function f (t ) is given by t . Since 1 t + 1 2, the function f (t ) is given by t 2 close to = 1. Hence, we have to split the integral precisely at the point where t gets larger than 1, because precisely then the function changes from t to t 2. But t 1 precisely when t 1, and so we have to split the integral at t 1: Z Z 1 1 1 t1 (t 2) d + (t ) d. (f f )(t) = 2 1 2 t1 It is now straightforward to calculate the convolution product. The result is (f f )(t) = t2 /2 + t 1/3. c From section 3.4.3 or table 1 we obtain the Fourier coecients cn of the sawtooth f and applying the convolution theorem gives the Fourier coecients of (f f )(t), namely c2 = 0 and c2 = 1/ 2 n2 (n = 0). 0 n d Take t = 0 in part c; since f is odd and real-valued we can write R1 (f f )(0) = 1 1 | f ( ) |2 d , and so we indeed obtain (4.13). 2 e For 1 < t < 0 we have (f f ) (t) = t 1, while for 0 < t < 1 we have (f f ) (t) = t + 1. Since f f is given by t2 /2 + t 1/3 for 0 < t < 2, (f f ) (t) is continuous at t = 1. Only at t = 0 we have that f f is not dierentiable. So theorem 4.10 implies that the dierentiated series represents the function (f f ) (t) on [1, 1], except at t = 0. At t = 0 the dierentiated series converges to ((f f ) (0+) + (f f ) (0))/2 = (1 1)/2 = 0. f The zeroth Fourier coecient of f f is given by


17

1 2

Z

1

Z (f f )(t) dt =0

1

(t2 /2 + t 1/3) dt = 0.

1

This is in agreement with the result in part c since c2 = 0. Since this 0 coecient is 0, we can apply theorem 4.9. The function represented by the Rt integrated series is given by the (periodic) function 1 (f f )( ) d . It is also odd, since f is even and for 0 t 1 it equals Z 0 Z t ( 2 /2 1/3) d + ( 2 /2 + 1/3) d = t(t 1)(t 2)/6.1 0


5.1

For a stable LTC-system the real parts of the zeroes of the characteristic polynomial are negative. Fundamental solutions of the homogeneous equations are of the form x(t) = tl est , where s is such a zero and l 0 some integer. Since tl est = | t |l e(Re s)t and Re s < 0 we have that limt x(t) = 0. Any homogeneous solution is a linear combination of the fundamental solutions. The Fourier coecients of u are u0 = 1 , 2 u2k = 0, u2k+1 = (1)k (2k + 1)

5.2

(u = p,2 , so use table 1 and the fact that sin(n/2) = (1)k for n = 2k+1 odd and 0 for n even). Since H() = 1/(i + 1) and yn = H(n0 )un = H(n)un it then follows that y0 = 5.3 1 , 2 y2k = 0, y2k+1 = (1)k . (1 + (2k + 1)i)(2k + 1)

a The frequency response is not a rational function, so the system cannot be described by a dierential equation (5.3). b Since H(n0 ) = H(n) = 0 for | n | 4 (because 4 > ), we only need to consider the Fourier coecients of y with | n | 3. From Parseval it then P 2 follows that P = 3 n=3 | yn | with yn as calculated in exercise 5.2. This 1 20 sum is equal to P = 4 + 92 . Note that u has period and that the integral to be calculated is thus the zeroth Fourier coecient of y. Since y0 = H(00 )u0 = H(0)u0 R and H(0) = 1 1 (see example 5.6 for H()), it follows that y0 = u0 = 0 u(t) dt = 2 . a According to (5.4) the frequency response is given by 2 + 1 . 2 + 4 + 2i

5.4

5.5

H() =

Since H() = 0 for = 1, the frequencies blocked by the system are = 1. b Write u(t) = e4it /4 eit /2i + 1/2 + eit /2i + e4it /4. It thus follows that the Fourier coecients unequal to 0 are given by u4 = u4 = 1/4, u1 = 1/2i, u1 = 1/2i and u0 = 1/2. Since yn = H(n0 )un = H(n)un and H(1) = H(1) = 0 we thus obtain that y(t) = y4 e4it + y1 eit + y0 + y1 eit + y4 e4it 15 1 1 1 15 1 = e4it + + e4it . 12 + 8i 4 4 2 12 8i 4 It is a good exercise to write this with real terms only: y(t) = 5.6 45 30 1 cos 4t + sin 4t + . 104 104 8

We have that 1 H() = . 2 2 + 0

18


19

Since | 0 | is not an integer, there are no homogeneous solutions having period 2, while u does have period 2. There is thus a uniquely determined periodic solution y corresponding to u. Since u(t) = q,2 (t) the Fourier coecients of u follow immediately from table 1: u0 = , 2 u2k = 0(k = 0), u2k+1 = 2 . (2k + 1)2 2 the line spectrum of y

Since yn = H(n0 )un = H(n)un = follows. 5.7

1 2u , n2 +0 n

For the thin rod the heat equation (5.8) holds on (0, L), with initial condition (5.9). This leads to the fundamental solutions (5.15), from which the superposition (5.16) is build. The initial condition leads to a Fourier series with coecients Z Z 2 L/2 2 L An = x sin(nx/L) dx + (L x) sin(nx/L) dx, L 0 L L/2 which can be calculated using an integration by parts. The result is: An = (4L/n2 2 ) sin(n/2) (which is 0 for n even). We thus obtain the (formal) solution u(x, t) = 4L X (1)n (2n+1)2 2 kt/L2 e sin((2n + 1)x/L). 2 n=0 (2n + 1)2

5.9

a

The heat equation and initial conditions are as follows: for 0 < x < L, t > 0, for t 0, for 0 x L.

ut = kuxx ux (0, t) = 0, u(L, t) = 0 u(x, 0) = 7 cos(5x/2L)

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 = 0 the characteristic equation s2 c = 0 has two distinct roots s1 . The general solution is then X(x) = es1 x + es1 x , so X (x) = s1 es1 x s1 es1 x . The rst boundary condition X (0) = 0 gives s1 ( ) = 0, so = . Next we obtain from the second boundary condition X(L) = 0 the equation (es1 L + es1 L ) = 0. For = 0 we get the trivial solution. So we must have es1 L + es1 L = 0, implying that e2s1 L = 1. From this it follows that s1 = i(2n + 1)/2L. This gives us eigenfunctions Xn (x) = cos((2n + 1)x/2L) (n = 0, 1, 2, 3, . . .). Since Tn (t) remains as in the textbook (for other parameters), we have thus found the fundamental solutions un (x, t) = e(2n+1)2

2 kt/4L2

cos((2n + 1)x/2L).

Superposition gives u(x, t) = X n=0

An e(2n+1)

2

2 kt/4L2

cos((2n + 1)x/2L).

Substituting t = 0 (and using the remaining initial condition) leads to u(x, 0) = X n=0

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

20


Since the right-hand side consists of one harmonic only, it follows that A2 = 7 and An = 0 for all n = 2. The solution is thus u(x, t) = 2 2 7e25 kt/4L cos(5x/2L). 5.11 a The equations are for 0 < x < L, t > 0, for t 0, for 0 x L.

ut = kuxx u(0, t) = 0, ux (L, t) = 0 u(x, 0) = f (x)

b Going through the steps one obtains the same fundamental solutions as in exercise 5.9. The coecients An cannot be determined explicitly here, since f (x) is not given explicitly. 5.12 The equations are given by (5.17) - (5.20), where we only need to substitute the given initial condition in (5.19), so u(x, 0) = 0.05 sin(4x/L) for 0 x L. All steps to be taken are the same as in section 5.2.2 of the textbook and lead to the solution u(x, t) = X n=1

An cos(nat/L) sin(nx/L).

Substituting t = 0 (and using the remaining initial condition) gives u(x, 0) = X n=1

An sin(nx/L) = 0.05 sin(4x/L).

Since the right-hand side consists of one harmonic only, it follows that A4 = 0.05 and An = 0 for all n = 4. The solution is thus u(x, t) = 0.05 cos(4at/L) sin(4x/L). 5.15 Separation of variables leads to X (x) cX(x) = 0 for 0 < x < , X (0) = X () = 0. For c = 0 we obtain the constant solution, so c = 0 is an eigenvalue with eigenfunction X(x) = 1. For c = 0 the characteristic equation s2 c = 0 has two distinct roots s1 . The general solution is then X(x) = es1 x + es1 x , so X (x) = s1 es1 x s1 es1 x . The boundary condition X (0) = 0 gives s1 ( ) = 0, so = . From the boundary condition X () = 0 we obtain s1 (es1 es1 ) = 0. For = 0 we get the trivial solution. So we must have es1 es1 = 0, implying that e2s1 = 1. From this it follows that s1 = ni. This gives us eigenfunctions Xn (x) = cos(nx) (n = 0, 1, 2, 3, . . .). For T (t) we get the equation T (t) + n2 a2 T (t) = 0. From the initial condition ut (x, 0) = 0 we 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) = X n=0

An cos(nat) cos(nx).

Substituting t = 0 (and using the remaining initial condition) leads to u(x, 0) = X n=0

An cos(nx) = kx

for 0 < x < .


21

R R We have A0 = (2/) 0 kx dx = k and An = (2/) 0 kx cos(nx) dx for n = 0, which can be calculated by an integration by parts: An = 0 for n even (n = 0) and An = 4k/n2 for n odd. The solution is thus u(x, t) = k 4k X 1 cos((2n + 1)at) cos((2n + 1)x). 2 n=0 (2n + 1)2

5.16

a From H() = H() and yn = H(n0 )un follows that the response y(t) to a real signal u(t) is real: since un = un we also have yn = yn . b Since we can write sin 0 t = (ei0 t ei0 t )/2i, the response is equal to (H(0 )ei0 t H(0 )ei0 t )/2i, which is ((1 e2i0 )2 ei0 t (1 e2i0 )2 ei0 t )/2i. This can be rewritten as sin 0 t 2 sin(0 (t 2)) + sin(0 (t 4)). P 2int . c A signal with period 1 has Fourier series of the form n= un e P 2int , which is 0 since H(2n) = 0 for The response is n= H(2n)un e all n. a The characteristic equation is s3 + s2 + 4s + 4 = (s2 + 4)(s + 1) = 0 and has zeroes s = 1 and s = 2i. The zeroes on the imaginary axis correspond to periodic eigenfrequencies with period and so the response to a periodic signal is not always uniquely determined. But see part b! b Since here the input has period 2/3, we do have a unique response. From Parseval and the relation yn = H(n0 )un we obtain that the power is given by Z 2/3 X X 3 P = | y(t) |2 dt = | yn |2 = | H(n0 )un |2 . 2 0 n= n= We have that H() = 1 + i . 4 2 + i(4 2 )1 2

5.18

Now use that only u3 = u3 = that P = 1/50. 5.19

and that all other un are 0, then it follows

For the rod we have equations (5.8) - (5.10), where we have to take f (x) = u0 in (5.10). The solution is thus given by (5.16), where now the An are the Fourier coecients 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 /n for n odd. This gives u(x, t) = 2 2 2 4u0 X 1 e(2n+1) kt/L sin((2n + 1)x/L). 2 n=0 (2n + 1)

Substituting x = L/2 in the x-derivative and using the fact that cos((2n + 1)/2) = 0 for all n leads to ux (L/2, t) = 0. 5.20 a As in the previous exercise the solution is given by (5.16). The An are R L/2 given by (2/L) 0 a sin(nx/L) dx = 2a(1 cos(n/2))/n, which gives the (formal) solution u(x, t) = 2 2 2 2a X 1 (1 cos(n/2))en kt/L sin(nx/L). n=1 n

22


b The two rods together form one rod and so part a can be applied with L = 40, k = 0.15 and a = 100. Substituting t = 600 in u(x, t) from part a then gives the temperature distribution. On the boundary between the rods we have x = 20, so we have to calculate u(20, 600); using only the contibution 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) from part a, and now use only the rst two terms of the series to obtain the equation u(20, t) 63.662e0.0000308t = 36 (terms of the series tend to 0 very rapidly, so two terms suce). We then obtain 18509 seconds, which is approximately 5 hours.


6.1

6.2

R We have to calculate (the improper integral) eit dt. Proceed as in eaxample 6.1, but we now have to determine limB eiB . This limit does not exist. R a We have to calculate G() = 0 e(a+i)t dt, which can be done precisely as in section 6.3.3 if we write a = + i and use that e(a+i)R = eR ei(+)R . If we let R then this tends to 0 since > 0. b The imaginary part of G() is /(a2 + 2 ) and applying the substituR 1 tion rule gives /(a2 + 2 ) d = 2 ln(a2 + 2 ), 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 lima0 g(t) = lima0 (t)eat = (t), while for = 0 we have that lima0 G() = i/. To calculate the spectrum we split the integral at t = 0: Z 1 Z 0 G() = teit dt teit dt.0 1

6.4

Changing R from the variable t to t in the second integral we obtain that 1 G() = 2 0 t cos t dt, which can be calculated for = 0 using an integration by parts. The result is: G() = 2(cos 1) 2 sin + . 2

R1 For = 0 we have that G(0) = 2 0 t dt = 1. Since lim0 sin / = 1 and 1 lim0 (cos 1)/ 2 = 2 (use e.g. De lHpitals rule), we obtain that o lim0 G() = G(0), so G is continuous. 6.5 a Calculating the integral we have that cos(a/2) 1 for = 0, F (0) = 0.

F () = 2i

b Using Taylor or De lHpital it follows that lim0 F () = 0 = F (0), o so F is continuous. 6.7 From the linearity and table 3 it follows that F () = 6.8 sin2 (a/2) 12 + 8i . 2 4+ a 2

Use (6.17) and table 3 for the spectrum of e7| t | , then F () = 7 7 + . 49 + ( )2 49 + ( + )2

6.9

a From the shift property in the frequency domain (and linearity) it follows that the spectrum of f (t) sin at is F ( a)/2i F ( + a)/2i. b Write f (t) = p2 (t) sin t, obtain the spectrum of p2 (t) from table 3 and apply part a (and use the fact that sin( ) = sin()), then F () = 2i sin() . 2 1

23

24


6.10

Use section 6.3.3 (or exercise 6.2) and the modulation theorem 6.17, and write the result as one fraction, then (F (t)eat cos bt)() = a + i . (a + i)2 + b2

Similarly it follows from section 6.3.3 (or exercise 6.2) and exercise 6.9a that (F (t)eat sin bt)() = 6.12 Write Z F () =0

b . (a + i)2 + b20

f (t)eit dt +

Z

f (t)eit dt

and R change from t to t in the second integral, then it follows that F () = 2i 0 f (t) sin t dt. 6.13 a We have F () = F () and F () is even, so F () = F (), and thus F () is real. b We have F () = F () (by part a) and since | F () | = (F ()F ())1/2 , it follows that | F () | = | F () |. Calculate the spectrum in a direct way using exactly the same techniques as in example 6.3.3 (or use (6.20) and twice an integration by parts): F () = 6.16 2i . 1 + 2

6.14

R a/2 The spectrum is given by a/2 teit dt, which can be calculated using an integration by parts. The result is indeed equal to the formula given in example 6.3. a From the dierentiation rule (and dierentiating the Fourier transform 2 of the Gauss function, of course) it follows that i e /4a /(2a a) is the spectrum of tf (t). b If we divide the Fourier transform of f (t) by 2a, then we indeed obtain the same result as in part a. Two examples are the constant function f (t) = 0 (k arbitrary), and the 2 Gauss function et /2 with k = 2. Using exercise 6.17a we obtain the 2 function tet /2 with k = i 2. Use table 3 for (t)eat and then apply the dierentiation rule in the frequency domain, then the result follows: (a+i)2 . (Dierentiate (a+i)1 just as one would dierentiate a real function.) The function ea| t | is not dierentiable at t = 0. The function t3 (1 + t2 )1 e.g. is not bounded. Use the fact that limx xa ex = 0 for all a R and change to the variable 2 x = at2 in tk /eat (separate the cases t 0 and t < 0). Then part a follows and, hence, part b also follows since we have a nite sum of these terms. Apply the product rule repeatedly to get an expression in terms of the derivatives of f and g (this involves the binomial coecients and is sometimes called Leibniz rule). Since f and g belong to S, tn (f (t)g(t))(m) will be a sum of terms belonging to S, and so the result follows.

6.17

6.18

6.19

6.20 6.21

6.22


25

6.23

6.25 6.26

R We have that ( )(t) = 0 (t ) d . Now treat the cases t > 0 and t 0 separately, then it follows that ( )(t) = (t)t. (If t 0, then t < 0 for > 0 and so (t ) = 0; if t > 0 then (t ) = 0 for Rt > t and the integral 0 1 d = t remains.) Since (t)t is not absolutely integrable, the function ( )(t) is not absolutely integrable. R From the causality of f it follows that (f g)(t) = 0 f ( )g(t ) d . For Rt t < 0 this is 0. For t 0 it equals 0 f ( )g(t ) d . a We use the denition of convolution and then split the integral at = 0: Z Z 0 (e| v | e| v | )(t) = e e| t | d + e e| t | d.0

First we take t 0. Then | t | = t for < 0. Furthermore we have for > t that | t | = t and for 0 < t that | t | = t. Hence, Z t Z Z 0 (e| v | e| v | )(t) = et d + et2 d + e2 t d.0 t

A straightforward calculation of these integrals gives (1 + t)et . Next we take t < 0. Then | t | = t for > 0. Furthermore we have for < t that | t | = t and for t < 0 that | t | = t . Hence, Z Z 0 Z t (e| v | e| v | )(t) = et2 d + et d + e2 t d.0 t

A straightforward calculation of these integrals gives (1 t)et . b Use the result from section 6.3.3 and the convolution theorem to obtain the spectrum (2(1 + 2 )1 )2 = 4/(1 + 2 )2 . c Since (1 + | t |)e| t | = e| t | + | t | e| t | and the spectrum of e| t | is 2(1 + 2 )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 spectrum weve already determined: G() = 2i(1 + 2 )1 . Apply theorem 6.8 (dierentiation rule in the frequency domain): the spectrum of f (t) is G ()/i. Calculating this and taking the results together we obtain 4/(1 + 2 )2 , in agreement with part b. 6.28 a From the dierentiation rule in the frequency domain we obtain that 2 the spectrum of tg(t) is iG () = i 2e /2 . Since (F tg(t))(0) = 0, 2 we may apply the integration rule to obtain that F1 () = 2e /2 . b Apply the dierentiation rule in the frequency domain with n = 2, then 2 F2 () = 2(1 2 )e /2 . c Since f3 (t) = f2 (t 1), it follows from the shift property that F3 () = ei F2 (). 2 d From part a and exercise 6.9 it follows that F4 () = ( 2e(4) /2 + 2 2e(+4) /2 )/2i. e Use the scaling property from table 4 with c = 4, then F5 () = G(/4)/4.1 1 b Since p1 ( ) = 0 for | | > 2 and 1 for | | < 2 , we have Z 1/2 (p1 p3 )(t) = p3 (t ) d. 1/2

6.29

26


Here p3 (t ) = 0 only if t 3/2 t + 3/2. Moreover, we have that 1/2 1/2, and so we have to separate the cases as indicated in the textbook: if t > 2, then (p1 p3 )(t) = 0; if t < 2, then also R 1/2 (p1 p3 )(t) = 0; if 1 t 1, then (p1 p3 )(t) = 1/2 1 d = 1; if R 1/2 1 < t 2, then (p1 p3 )(t) = t3/2 1 d = 2 t; nally, if 2 t < 1, R t+3/2 then (p1 p3 )(t) = 1/2 1 d = 2 + t. c Apply the convolution theorem to T (t) = (p1 p3 )(t), then the spectrum of T (t) follows: 4 sin(/2) sin(3/2)/ 2 .


7.1

From the spectra calculated in exerices 6.2 to 6.5 it follows immediately that the limits for are indeed 0: they are all fractions with a bounded numerator and a denominator that tends to . As an example we have from exercise 6.2 that lim 1/(a + i) = 0. Use table 3 with a = 2A and substitute = s t. Take C > 0, then it follows by rst changing from the variable Au to v and then applying (7.3) that Z C Z AC sin Au sin v lim du = lim dv = . A 0 A 0 u v 2 Split 1/(a+i) into the real part 1/(1+ 2 ) and the imaginary part /(1+ 2 ). The limit of A of the integrals over [A, A] of these parts gives limA 2 arctan A = 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 () = 2i sin()/( 2 1). The R R function f (t) is absolutely integrable since | f (t) | dt = | sin t | dt < . Moreover, f (t) is piecewise smooth, so all conditions of the fundamental theorem are satised. We now show that the improper integral of F () exists. First, F () is continuous on R according to theorem 6.10, so it R is integrable over e.g. [2, 2]. Secondly, the integrals 2 F () d and R 2 F () d both exist. For the former integral this can be shown as follows (the other integral can be treated similarly): Z Z 2 F () d d 2 1 2 2 since | 2i sin() | 2 (and 2 1 > 0 for > 2). The integral in the right-hand side is convergent. b Apply the fundamental theorem, then Z 2i sin() it 1 f (t) = e d 2 2 1 for all t R (f is continuous). Now use that F () is an odd function and that 2 sin sin t = cos( t) cos( + t), then Z 1 cos( t) cos( + t) d. f (t) = 0 1 2

7.2 7.3

7.4

7.6

7.8

a In exercise 6.15b it was shown that Fs () = (1 cos a)/. This exercise used the odd extension to R. So f (t) is odd and using (7.12) we thus obtain Z 2 1 cos a 1 sin t d = (f (t+) + f (t)). 0 2 Since f (t) is continuous for t > 0 and t = a we have for these values that Z 2 1 cos a f (t) = sin t d. 0

27

28


b At t = a the function is discontinuous, so we have convergence to 1 (f (a+) + f (a)) = 1 . 2 2 7.10 If we take g(t) = 0 in theorem 7.4, then G() = 0 and so we get the statement: if F () = 0 on R, then f (t) = 0 at all points where f (t) is continuous. We now prove the converse. Take f (t) and g(t) as in theorem 7.4 with spectra F () and G() and assume that F () = G() on R. Because of the linearity of the Fourier transform, (F G)() is the spectrum of (f g)(t); but (F G)() = F () G() = 0. From our assumption it now 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, which is indeed theorem 7.4. The spectrum of pa (t) is 2 sin(a/2)/ (table 3). From duality it then follows that the spectrum of sin(at/2)/t is pa () at the points where pa (t) is continuous; at = a/2 we should take the value /2. (We can apply duality since the Fourier integral exists as improper integral; this is exercise 7.5b). The spectrum of qa (t) is F () = 4 sin2 (a/2)/(a 2 ) (see table 3). From duality it then follows that the spectrum of sin2 (at/2)/t2 is (a/2)qa (). (We can apply duality since qa is continuous, piecewise smooth, and absolutely integrable and since its Fourier integral exists as improper integral; this latter fact follows immediately if we use that F () is even and continuous and that e.g. F () 1/ 2 for 1). The function 1/(a + i) is not integrable on R (see exercise 6.2), so duality cannot be applied. These results follow immediately from duality (and calculating the right p 2 2 constants). For example: /aet /4a 2ea , now divide by 2. This is an important exercise: it teaches to recognize useful properties. a 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 e| | , the result follows: ei e| | . b Here we have a shift from t to t 3; from the spectrum of sin 2t/t the 1 result follows: e3i p4 () with value 2 at = 2. c We now have 1/(t2 4t + 7), multiplied by a sine function. The sine function is easy to deal with using exercise 6.9 (a variant of the modulation theorem). As in a we completethe square and note that 1/(3 + (t 2)2 ) has spectrum F () = e2i e 3| | / 3. Hence, the result is now (F ( 4) F ( + 4))/2i. d We use that 3q6 () is the spectrum of sin2 (3t)/t2 and apply a shift in time from t to t 1, then the result is 3ei q6 (). Again, this is an important exercise: it teaches to recognize useful properties for the inverse transform. a We immediately use table 3 to obtain that 1/(4 + 2 ) is the spectrum of f (t) = e2| t | /4. b Apply a shift in the frequency domain to the spectrum p2a () of sin(at)/t, then it follows that f (t) = (ei0 t + ei0 t ) sin(at)/(t), so f (t) = 2 cos(0 t) sin(at)/(t). c As in part b it follows that f (t) = 3e9it /((t2 + 9)). From the convolution theorem it follows that F (Pa Pb )() = (F Pa )()

7.11

7.12

7.14 7.15 7.16

7.17

7.19


29

(F Pb )() = e(a+b)| | , where we also used table 3. But also (F Pa+b )() = e(a+b)| | , and since F is one-to-one (theorem 7.4) it then follows that Pa+b = Pa Pb . 7.21 a Use the result of exercise 6.14 (G() = 2i/(1+ 2 )), the fundamental R R theorem and the fact that the spectrum is odd to change from to 0 . It then follows that (use x instead of ) Z x sin xt dt = et . 1 + x2 2 0 Since g is not continuous at t = 0, this result is not correct at t = 0. Here one should take the average of the jump, which is 0. R b We apply Parseval (formula (7.19)) and calculate | g(t) |2 dt = R0 R 2t R e2t dt + 0 e dt, which is 1. In | G() |2 d we can use the fact that the integrand is even. Writing x instead of , the result follows. 7.22a| t | Use and g(t) = eb| t | and calculate R Parseval (7.18)R with f (t) = e (a+b)t f (t)g(t) dt = 2 0 e dt = 2/(a + b). The spectra of f and g are 2a/(a2 + 2 ) and 2b/(b2 + 2 ) (table 3).

7.23

a Since sin4 t/t4 is the square of sin2 t/t2 and (F sin2 t/t2 )() = q2 () (table 3), it follows from the convolution theorem in the frequency domain that (F sin4 t/t4 )() = (/2)(q2 q2 )(). R b The integral sin4 t/t4 dt is the Fourier transform of sin4 x/x4 calR culated at = 0, hence sin4 x/x4 dx = (/2)(q2 q2 )(0). Using that q2 is an even function we obtain that Z Z 2 (q2 q2 )(0) = q2 (t)q2 (t) dt = 2 (1 t/2)2 dt. 0

This integral equals 4/3 and so 7.24

R

sin4 x/x4 dx = 2/3.

From table 3 we know that e| t | /2 1/(1 + 2 ). By the convolution theorem we then know that the spectrum of f (t) = (e| v | /2e| v | /2)(t) is 1/(1 + 2 )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 that the integrand is even. We then obtain Z 1 1 1 d = f (0) = , 0 (1 + 2 )2 4 which is indeed the case a = b = 1 from exercise 7.22. The Gauss function f (t) = eat belongs to S and so we can apply Poissons p 2 summation formula. Since F () = /ae /4a (see table 3), it follows from (7.23) with T = 1 that X n=2

7.26

ean =

2

X 2 n2 /a p /a e . n=

Replacing a by x the result follows. 7.27 Take f (t) = a/(a2 +t2 ), then F () = ea| | (see table 3); we can then use (7.22) with T = 1 (in example 7.8 the conditions were veried) to obtain

30


X n=

a = a2 + (t + n)2

1+

X n=1

e

2n(a+it)

+

X n=1

! e2n(ait)

.

Here we have also split a sum in terms with n = 0, n > 1 and n < 1, and then changed from n to n in the sum with n < 1. The sums in the righthand side are geometric series with ratio r = e2(a+it) and r = e2(ait) respectively. Note that | r | < 1 since a > 0. Using the formula for the sum of an innite geometric series (example 2.16), then writing the result with a common denominator, and nally multiplying everything out and simplifying, it follows that a X 1 1 e4a . = 2 + (t + n)2 4a e2a (e2it + e2it ) n= a 1+e

Multiplying numerator and denominator by e2a the result follows. 7.28 a To determine the spectrum we write sin t = (eit eit )/2 and calculate the integral dening F () in a direct way: Z Z 1 i(1)t i(1+)t F () = e dt e dt . 2i 0 0 Writing the result with a common denominator and using the fact that ei = ei = 1 gives F () = (1 + ei )/(1 2 ). From theorem 6.10 we know that F () is continuous, so we do not have to calculate F () at the exceptional points = 1. b Apply the fundamental theorem, so (7.9), noting that f (t) is continuous on R. We then obtain Z 1 1 + ei it f (t) = e d. 2 1 2 Split the integral at t = 0 and change from to in the integral over (, 0]. Then Z it 1 e + eit + ei(t) + ei(t) f (t) = d, 2 0 1 2 which leads to the required result. c Take t = /2 in part b and use that f (/2) = 1, thenRthe result follows. d Apply Parsevals identity (7.19) to f and use that 0 sin2 t dt = /2, then it follows that Z 1 | F () |2 d = . 2 2 Since F () can be rewritten as 2ei/2 cos(/2)/(1 2 ) and we have that ei/2 = 1, it follows that | F () |2 = 4 cos2 (/2)/(1 2 )2 . This integrand being even, the result follows. 7.29 a We know from table 3 that p2a (t) 2 sin a/ and e| t | 2/( 2 + 1). From the convolution theorem it then follows that p2a (v)e| v | 4f () = G(). b We now determine g explicitly by calculating the convolution product (use the denition of p2a ): Z t+a Z a (p2a (v) e| v | )(t) = e| t | d = e| u | dua ta


31

where we changed to the variable u = t . Now if a t a, then t a 0 t + a and so Z t+a Z 0 u | v | eu du = 2 2ea cosh t. e du + (p2a (v) e )(t) =ta 0

If t > a, then t a > 0 and so Z t+a eu du = 2et sinh a. (p2a (v) e| v | )(t) =ta

Finally, if t < a, then t + a < 0 and so Z t+a eu du = 2et sinh a. (p2a (v) e| v | )(t) =ta

c The function g from part b is continuous at t = a since limta g(t) = 2ea sinh a = ea (ea ea ) = 1 e2a and g(a) = limta g(t) = 2 2ea cosh a = 2 ea (ea + ea ) = 1 e2a . In the same way it follows that g(t) is continuous at t = a. So g(t) is a piecewise smooth function which moreover is continuous. Also, g(t) is certainly absolutely integrable since e| t | is absolutely integrable over | t | > a. Finally, the Fourier integral exists as improper Riemann integral since G() is even absolutely integrable: | G() | 4/ (1 + 2 ) . We can now apply the duality rule (theorem 7.5) and it then follows that G(t) 2g(), so f (t) g()/2. Since f (t) = f (t) we thus see that F () = ea cosh for | | a and F () = e| | sinh a for | | > a.


8.1

b For t = 0 we have that lima0 Pa (t) = 0, while for t = 0 we have R that lima0 Pa (t) = . Since Pa (t) dt = 1, we see that Pa (t) ts the description of the delta function. c From table 3 it follows that Pa (t) ea| | and lima0 ea| | = 1. Combining this with part b shows that it is reasonable to expect that the spectrum of (t) is 1. a Since (a) C for all S, it follows from (8.10) that (t a) is a mapping from S to C. For c C and S we have that (t a), c = (c)(a) = c (t a), , and for 1 , 2 S we have (t a), 1 + 2 = (1 + 2 )(a) = (t a), 1 + (t a), 2 . So (t a) is a linear mapping from S to C, hence a distribution. b Taking the limit inside the integral in (8.1) gives Z 1 2 sin a lim f (t ) d = f (t) 2 a for any absolutely integrable and piecewise smooth function f (t) on R that is continuous at t. Using (8.3) this can symbolically be written as (take t = a) Z ()f (a ) d = f (a)

8.2

and by changing from to a t we then obtain Z (a t)f (t) dt = f (a).

8.4

Using (a t) = (t a), which by (8.3) is reasonable to expect (see section 8.4 for a proof), this indeed leads to (8.11). R Since 1, = (t) dt C for all S, it follows that 1 is a mapping from S to C. The linearity of this mapping follows from the linearity of integration: for c C and S we have that Z Z 1, c = (c)(t) dt = c (t) dt = c 1, ,

and for 1 , 2 S we have Z Z 1, 1 + 2 = (1 + 2 )(t) dt =

Z 1 (t) dt +

2 (t) dt,

so 1, 1 + 2 = 1, 1 + 1, 2 . This proves that 1 is a linear mapping from S to C, hence a distribution. 8.5 For S there exists a constant M > 0 such that (e.g.) (1+t2 ) | (t) | M for all t R. Hence, Z Z Z 1 | (t) | dt M dt < (t) dt 1 + t2 0 0 0

32


33

8.7

R (the latter integral equals [arctan] = /2). The integral 0 (t) dt thus 0 exists and one can now show that is indeed a distribution precisely as in exercise 8.4 (linearity of integration). R In example 8.4 it was already motivated why the integral | t | (t) dt exists: there exists a constant M > 0 such that (e.g.) (1 + t2 ) | t(t) | M for all t R. Hence, Z Z Z 1 | t(t) | dt M | t | (t) dt dt < . 2 1 + t So | t | , exists and one can now show that | t | is indeed a distribution precisely as in exercise 8.4 (linearity of integration).

8.9

a Z

For the integral over [1, 1] we have Z 1 1 | t |1/2 dt = 2 t1/2 dt = [4t1/2 ]1 = 4, 00 1/2

1

R hence, | t | is integrable over [1, 1]. Since 0 t1/2 dt = 2 limR R does not exist, | t |1/2 is not integrable over R. R b We rst show that | t |1/2 (t) dt exists for S. To do so, we split the integral in an integral over [1, 1] and over | t | 1. For the rst integral we note that | (t) | M1 for some constant M1 > 0. From part a we then get Z 1 Z 1 Z 1 | t |1/2 (t) dt | t |1/2 | (t) | dt M1 | t |1/2 dt < . 1 1 1

For the second integral we use that | t | 1 for | t | 1. Hence, Z Z Z | t |1/2 (t) dt | (t) | dt | (t) | dt. | t |1 | t |1 In example 8.1 itE been shown that the latter integral exists. This shows has D 1/2 that | t | , exists and one can now show that | t |1/2 is indeed a distribution (linearity of integration; see e.g. exercise 8.4). 8.10 a For S there exists a constant M > 0 such that (e.g.) t2 ) | t(t) | M for all t R. Hence, Z Z Z 1 dt < . t(t) dt | t(t) | dt M 1 + t2 As in exercise 8.3 this shows that t denes a distribution. 8.12 a From the linearity for distributions itRfollows immediately that the com plex number 2(0) + i 3 (0) + (1 + i) 0 (((t) (t))dt is assigned. b This denes a distribution if 1, t and t2 are distributions. The rst one is known from example 8.1, the other two from exercise 8.10. From denition 8.15 it thus follows that f (t) denes a distribution as well. a From (8.17) it follows that the complex number (3) (0) is assigned. b This number (3) (0) is meaningfull for all functions that are 3 times continuously dierentiable. a First apply example 8.3, then denition 8.4, and nally integration by parts, then (1 +

1/2

8.14

8.15

34


(sgn t) , =

Z0

( (t) (t)) dt = [(t)]0 [(t)] = 2(0), 0

hence, (sgn t) = 2(t). b Since sgn t = 2 (t) 1 (verify this), it follows from the linearity of dierentiation that (sgn t) = 2 (t) = 2(t). Here we used that 1 = 0 and that (t) = (t) (see (8.18)). c Since | t | = sgn t it follows from part a that | t | = (sgn t) = 2(t). 8.16 Since the function | t | from example 8.9 is continuously dierentiable outside t = 0, it follows from the jump formula that | t | = sgn t (at t = 0 there is no jump and outside t = 0 this equality holds for the ordinary derivatives). The function from example 8.10 has a jump of magnitude 1 at 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 ( (t) cos t) = (t) (t) sin t. a The function pa has a jump of magnitude 1 at t = a/2 and of magnitude 1 at t = a/2. Outside t = 0 the ordinary derivative is 0, so it follows from the jump formula that pa (t) = (t + a/2) (t a/2). b The function (t) sin t has no jump at t = 0, for t < 0 the ordinary derivative is 0 and for t > 0 the ordinary derivative is cos t, so it follows from the jump formula that ( (t) sin t) = (t) cos t. a This is entirely analagous to exercises 8.10 and 8.12b. b The function is dierentiable outside t = 1 and the ordinary derivative is 1 for t < 1 and 2t 2 for t > 1. We denote this derivative as the distribution Tf . At t = 1 the jump is 2, so according to the jump formula the derivative is Tf + 2(t 1). From the jump formula it follows that the derivative as distribution is given by a (t)eat + (t), so f (t) af (t) = (t) as distributions. Subsequently apply denition 8.6 and the denition of (t a) in (8.10), then p(t)(t a), = (p)(a) = p(a)(a) = p(a) (t a), . Now use denition 8.5, then p(t)(t a), = p(a)(t a), , which shows that p(t)(t a) = p(a)(t a). 8.23 a The denition becomes: f (t) (t), = (t), f . The product f of two continuously dierentiable functions is continuously dierentiable, so this denition is correct and it gives a mapping from S to C. The linearity follows immediately from the linearity of (t), so f (t) (t) is a distribution. b According to part a we have f (t) (t), = (f ) (0), where we also applied (t) to f . Now apply the product rule for dierentiation and write the result as f (0) (t), + f (0) (t), = f (0) (t) f (0)(t), . c If f (t) = t, then f (0) = 0 and f (0) = 1, so t (t) = (t); if f (t) = t2 then f (0) = 0 and f (0) = 0, so t2 (t) = 0. First apply denition 8.6 and then the denition of pv(1/t) from example 8.5 to obtain Z Z t(t) t pv(1/t), = lim dt = lim (t) dt. 0 | t | 0 | t | t Since S is certainly integrable over R, the limit exists and it will be R equal to (t) dt. Hence, t pv(1/t) = 1. 8.26 Let T be an even distribution, then T (t) = T (t) (deniton 8.8), so

8.17

8.18

8.19 8.22

8.25


35

T (t), (t) = T (t), (t) for all S, where we used denition 8.7. Similarly for odd T . 8.27 a From the denition of sgn t in example 8.3 it follows that sgn t, (t) = sgn t, (t) for all S. This shows that sgn t is odd according to exercise 8.26. Similarly for pv(1/t) (change from t to t in the integrals dening pv(1/t)). b From the denition of | t | in example 8.4 it follows that | t | , (t) = | t | , (t) for all S (change from t to t in the integral dening | t |). This shows that | t | is even according to exercise 8.26. a Applying (8.12) to f (t) gives Z 0 Z Tf , = 2t(t) dt + t2 (t) dt 0

8.29

and in e.g. exercises 8.10, 8.12b and 8.18a we have seen that such integrals are well-dened for S. This gives a mapping from S to C and the linearity of this mapping follows precisely as in e.g. exercise 8.3 or 8.4. Hence, f indeed denes a distribution Tf . b Apply the jump formula (8.21): outside t = 0 the function f is continuously dierentiable with derivative f (t) = 2t for t > 0 and f (t) = 2 for t < 0. Note that f again denes a distribution Tf . At t = 0 the function has no jump, hence (8.21) implies that Tf = Tf . c Again we have that the function f is continuously dierentiable outside t = 0 and f (t) = 2 for t > 0 and f (t) = 0 for t < 0. Let Tf be the distribution dened by f . At t = 0 the function f has a jump f (0+) f (0) = 0 2 = 2, and according to (8.21) (applied to Tf and using that Tf = Tf and so Tf = Tf ) we have that Tf = Tf = Tf + (f (0+) f (0))(t) = Tf 2(t). The second derivative of f considered as distribution is the same as the second derivative of f outside t = 0, minus the distribution 2(t) at t = 0. 8.30 a Since (t) can be dened for all twice continuously dierentiable functions, the product f (t) (t) can also be dened for all twice continuously dierentiable functions f (t) by f (t) (t), (t) = (t), f (t)(t) . This is because it follows from the product rule that the product f (t)(t) is again twice continuously dierentiable. b From part b and the denition of the second derivative of a distribution (formula (8.17) for k = 2) we obtain f (t) (t), (t) = (t), (f (t)(t)) . Since (f (t)(t)) = f (t)(t) + 2f (t) (t) + f (t) (t) we thus obtain that f (t) (t), (t) = f (0)(0) + 2f (0) (0) + f (0) (0), which equals f (0)(t) 2f (0) (t) + f (0) (t), (t) ( S). This proves the identity. c Apply part b to the function f (t) = t2 and use that f (0) = f (0) = 0 and f (0) = 2, then t2 (t) = 2(t). Next apply b to f (t) = t3 and use that f (0) = f (0) = f (0) = 0, then it follows that t3 (t) = 0. d According to denition 8.7 we have that (at), (t) = | a |1 (t), (a1 t) . Now put (t) = (a1 t), then the right-hand side equals | a |1 (0). Next we use the chain rule twice to obtain that (0) = a2 (0). Hence (at), (t) = | a |1 a2 , .


9.1

Let S. From theorem 6.12 it follows that the spectrum belongs to S. Since T is a distribution, we then have that T, C, and so FT, = T, C as well. So F T is a mapping from S to C. The linearity of F T follows from the linearity of T and F ; we will only give the necessary steps for FT, c , since the rule for FT, 1 + 2 follows similarly. FT, c = T, F (c) = T, c = c T, = c FT, .

9.3

a Use table 5 to obtain that (t 4) e4i . b Again use table 5 to obtain that e3it 2( 3). c First write the sine function as combination of exponentials, so sin at = (eiat eiat )/2i. From linearity and table 5 it then follows that sin at i(( a) ( + a)). d First determine the spectrum of pv(1/t) and 4 cos 2t = 2e2it + 2e2it using table 5 and then (again) apply linearity to obtain the spectrum 4(( 2) + ( + 2)) + 2sgn . a From example 9.1 (or table 5) we obtain the result e5it /2. b See example 9.2: 2 cos 2t. c The spectrum of pv(1/t) is isgn (table 5). Note that 2 cos = ei + ei and that the spectrum of (t a) is eia (table 5). Hence the answer is i 1 pv(1/t) + (t 1) + (t + 1). Let T be an even distribution with spectrum U . We have to show that U () = U (), so U, (t) = U, (t) for all S (see exercise 8.26). But U, (t) = T, F (t) and from table 4 we know that (F (t))() = () if is the spectrum of . Since T is even, we have that T, () = T, () . From these observations it follows that U, (t) = T, () = U, (t) , which shows that U is even. Similarly for odd T . It is obvious that (t) = (1 + sgn t)/2 by looking at the cases t > 0 and t < 0. Since 2() is the spectrum of 1 and 2ipv(1/) is the spectrum of (t), it follows that (t) has spectrum () ipv(1/). a Let S have spectrum . From denition 9.1 and the action of it follows that F , = , = , = (0). From the dierentiation rule in the frequency domain (table 4) with k = 2 we see that () = F ((it)2 (t))(), and hence 2 is proven as follows: Z Z F , = F (t2 (t))(0) = t2 (t) dt = 2 () d,

9.4

9.5

9.7

9.8

so F , = , for all S, proving the required result. Parts b and c can be proven using similar steps. 9.9 a Subsequently apply denitions 9.1 and 8.7: FT (at), = T (at), = | a |1 T, (a1 ) ( S having spectrum ). From table 4 we see that (a1 ) = | a | (F (at))(), so it follows that FT (at), = T, F (at) = U, (at) , where we again used denition 9.1 in the nal step. Now again apply denition 8.7, then FT (at), = | a |1 U (a1 ), , which proves T (at) | a |1 U (a1 ).

2

36


37

b We have that (4t+3) is the distribution (t+3) scaled by 4. According to the shift rule in the time domain (see table 6) it follows from (t) 1 that (t+3) e3i . From part a it then follows that (4t+3) 41 U (/4) with U () = e3i . Hence, (4t + 3) 41 e3i/4 . (This can also be solved by considering (4t + 3) as the distribution (4t) shifted over 3/4.) 9.11 From table 5 it follows that (t) i. Using (9.12) we then obtain that it (t) (i) = i, so t (t) 1. Exercise 8.23c gives: t (t) = (t) and since (t) 1 we indeed get t (t) 1 again. Similarly we get t (t) 2i using (9.12) or using exercise 8.30b: t (t) = 2 (t). From iT = 1 we may not conclude that T = 1/(i) since there exist distributions S = 0 such that S = 0 (e.g. ()). The linearity follows as in denition 8.6. The main point is that one has to show that eiat (t) S whenever S. So we have to show that for any m, n Z+ there exists an M > 0 such that tn (eiat (t))(m) < M . From the product rule for dierentiation it follows that (eiat (t))(m) is a sum of terms of the form ceiat (k) (t) (k Z+ ). It is now sucient to show that tn eiat (k) (t) < M for some M > 0 and all k, n Z+ . But since iat e = 1 this means that we have to show that tn (k) (t) < M for some M > 0 and all k, n Z+ , which indeed holds precisely because S. 9.15 From denition9.1 and the denition of eiat T (see exercise 9.13) it follows iat that F e T, = eiat T, = T, eiat ( S having spectrum ). According to the shift property in the frequency domain (table 4) we have that eiat (t) = F ((+a))(t) (note that for convenience weve interchanged the role of the variables and t). Hence, F eiat T, = T, F (( + a))(t) = U, ( + a) = U ( a), , where we used denition 9.2 in the last step. So we indeed have eiat T U ( a). a Use table 5 for (t) and apply a shift in the time domain, then it follows that (t 1) ei (() ipv(1/)). b Use table 5 for (t) and apply a shift in the frequency domain, then it follows that eiat (t) ( a) ipv(1/( a)). c We have (t) () ipv(1/) and if we now write the cosine as a combination of exponentials, then we can use a shift in the frequency domain (as in part b) to obtain that (t) cos at 1 (( a) ipv(1/( 2 a)) + ( + a) ipv(1/( + a))). d Use that 1 2() and (t) i (table 5), so 3i 6i() and (apply a shift) (t 4) e4i i; the sum of these gives the answer. e First note that (t)sgn t = (t) and the spectrum of this is known; furthermore we have that t3 2i3 (3) () (table 5), so the result is 2 2 (3) () + () ipv(1/).1 a Use table 5 for the sign function and apply a shift: 2 ieit sgn t. 1 b Write sin t as a combination of exponentials and apply a shift to 2 isgn t, then we obtain the result 1 (sgn(t + 3) sgn(t 3)). 4 c Apply reciprocity to (t), then we obtain ((t) ipv(1/t))/2 (). Now (t) = (t) and pv(1/t) = pv(1/t), hence, the result is: 1 i 1 pv(1/t) + 1 (t). 2 2 d Apply the scaling property (table 6) to 1 2() to obtain 1 6(3). Next we apply a shift in the frequency domain (table 6), which

9.12 9.13

9.16

9.17

38


results in e2it/3 6(3 2). Using the dierentiation rule in the frequency domain (table 6) we obtain from 1/2 () that (it)2 /2 (). From linearity it then follows that (31 e2it/3 t2 )/2 (3 2) + (). 9.19 9.20

We know that T = T , so | t | = | t | = sgn t (by example 8.9). According to denition 9.3 we have that T (t) (t a), = T ( ), (t a), (t + ) . Since (t a), (t + ) = (a + ), the function (t a), (t + ) belongs to S, so T (t) (t a) exists and T (t) (t a), = T ( ), (a + ) = T ( a), ( ) (the last step uses denition 9.2). This proves that T (t)(ta) = T (ta).

9.21

Use exercise 9.20 with T (t) = (t b). The convolution theorem leads to the obvious eib eia = ei(a+b) . a Use table 5: (t 3) e3i . b Since (t + 4) e4i (as in part a) and cos t = (eit + eit )/2 we apply a shift in the frequency domain: cos t(t + 4) (e4i(1) + e4i(+1) )/2. c From table 5 we have (t) ()ipv(1/). Apply the dierentiation rule in the time domain (with n = 2), then we obtain t2 (t) () + ipv(1/) . d Apply the dierentiation rule in the time domain to the result obtained in exercise 9.16c, then it follows that (2 (t) cos t) i(( 1) + ( + 1) ipv(1/( 1)) ipv(1/( + 1))). e Since (t) 1 it follows from rst the scaling property and then a shift in the time domain that (7(t 1/7)) ei/7 7. Finally apply the dierentiation rule in the time domain to obtain the result: ((7t 1)) iei/7 7. f This is a convergent Fourier series and so we can determine the spectrum term-by-term. Since 1 2() and e(2k+1)it 2( (2k + 1)) we obtain the following result: 2 () 4 X k=

9.24

(2k + 1)2 ( (2k + 1)).

9.25

a From table 5 we know that eit 2( 1) and similarly for eit . Hence, i 1 sin t ( 1) ( + 1). b Apply the dierentiation rule in the time domain (with n = 2) to (t) 1, then (t) 2 . c From table 5 we obtain that (t + 1 )/4 ei/2 /4. 2 d From table 5 (and linearity) we obtain that ((t + 1) (t 1))/2i (ei ei )/2i, which is sin . Now apply dierentiation in the time domain (with n = 3), then we obtain that ( (3) (t + 1) (3) (t 1))/2 3 sin . e From exercise 9.4c and a shift in the frequency domain it follows that e4it ((t + 1) + (t 1))/2 cos( 4). a In exercise 9.25b it was shown that (t) 2 . Applying a shift in the frequency domain leads to eit (t) ( 1)2 . b From the dierentiation rule in the time domain and table 5 it follows as in exercise 9.25b that (t) i and (t) 2 , so (t) + 2i (t) + (t) 2 2 + 1.

9.26


39

c Since ( 1)2 = 2 2 + 1, the results in part a and b should be the same. Using exercise 8.30b with f (t) = eit we indeed obtain that eit (t) = (t) 2i (t) (t). 9.27 a From exercise 9.25b it follows that (t) 2 . The convolution theorem then implies that T (t) 2 U where U is the spectrum of T . This also follows by applying the dierentiation rule in the time domain to T , which equals T (t) by (9.21). b As noted in part a we have that | t | = | t | . In exercise 8.15c it was shown that | t | = 2, so we indeed get | t | = 2. Now let V be the spectrum of | t |. Since 2 and 1 it then follows as in part a from the convolution theorem that 2 V = 2.


10.1

a When the system is causal, then the response to the causal signal (t) is again causal, so h(t) is causal. On the other hand, if h(t) is causal, then Rt it follows that y(t) = (u h)(t) = h(t )u( ) d and if we now have a causal input u, then the integral will be 0 for t < 0 and so y(t) is causal as well, proving that the system is causal. b When the system is real, then the response to the real signal (t) is again real, so h(t) is real. On the other hand, if h(t) is real, then it follows from the integral for y(t) = (u h)(t) that if u(t) is real, then y(t) is also real, proving that the system is real. a If we substitute u(t) = (t) then it follows that h(t) = (t1)+ (t)e2t , so h(t) is causal and real and according to exercise 10.1 the system is then causal and real. b We substitute u(t) = (t), then it follows for t 0 that a(t) = (t Rt 1 1) + 0 e2(t ) d = (t 1) + 2 (1 e2t ), while for t < 0 the integral is 0 and so a(t) = (t 1). This result can be written for all t as a(t) = (t 1) + 1 (t)(1 e2t ). 2 We can express p2 (t) as p2 (t) = (t + 1) (t 1). Since a(t) is (by denition) the response to (t) and we have a linear system, the response to p2 (t) = (t + 1) (t 1) is a(t + 1) a(t 1). a We dierentiate a(t) in distribution sense, which results in h(t) = (t) (t)e3t (2 sin 2t + 3 cos 2t), since a(t) has a jump at t = 0 of magnitude 1 and we can dierentiate in ordinary sense outside t = 0. b We use theorem 10.1, which implies that we may ignore the delta com3t ponent and only have to R that (t)e (2 sin 2t + 3 cos 2t) is absolutely show R integrable. Since both 0 R e3t sin 2t dt and 0 e3t cos 2t dt exist (e.g., both are smaller than 0 e3t dt), this is indeed the case and hence the system is stable. a The impulse response is h1 h2 . b If the input for the rst system is bounded, then the output is bounded since the system is stable. This output is then used as input for the second system, which is again stable. So the output of the second system is again bounded. This means that the cascade system itself is stable: the response to a bounded input is bounded. a The spectrum of h is H() = 1+1/(1+i)2 since 1 and tet (t) 1/(1 + i)2 . b Since eit H()eit it follows that the response is given by eit (1 + 1/(1 + i)2 ). a We need to determine the inverse Fourier transform of the function 1 H() = cos /( 2 + 1). First note that 2 e| t | 1/( 2 + 1). Writing the cosine as a combination of exponentials we obtain from the shift rule that 1 h(t) = 2 (e| t+1 | + e| t1 | ). b It now suces to use the time-invariance of the system: the reponse to (t) is h(t), so the response to (t 1) is h(t 1). a We need to determine the inverse Fourier transform of the function from gure 10.3, which is H() = qc (). This is h(t) = 2 sin2 (c t/2)/(c t2 ). 40

10.2

10.3

10.5

10.6

10.7

10.8

10.9


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b The function u has a Fourier series with terms cn eint (note that 0 = 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 c1 . These can easily be calculated 1 from the dening integrals: c0 = 2 and c1 = c1 = 1/. Hence, the 2 response follows: y(t) = H(0)c0 +H(1)c1 eit +H(1)c1 eit = 1 3 cos t. 2 10.11 a Put s = i and apply partial fraction expansion to the system function (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) 2 1 4 1 2 1 4 1 =1 =1 . (s 1)(s + 2) 3s1 3s+2 3 i 1 3 i + 2 Now (t) 1 and (t)e2t 1/(i + 2) (table 3, no. 7) and from time 1 reversal (scaling with a = 1 from table 4, no. 5) it follows for i1 = t t 1 1 2 4 that (t)e i1 . Hence, h(t) = (t) + 3 (t)e 3 (t)e2t . i()+1 b The impulse reponse h(t) is not causal, so the system is not causal. c The modulus of H() is 1, so it is an all-pass system and from Parseval it then follows that the energy-content of the input is equal to the energycontent of the output (if necessary, see the textbook, just above example 10.7). 10.13 a From the dierential equation we immediately obtain the frequency response: H() =2 2 0 . 2 2 i 20 0

b Write the cosine as a combination of exponentials, then it follows from eit H()eit that y(t) = (H(0 )ei0 t + H(0 )ei0 t )/2. However H(0 ) = 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 we obtain that (t) pv(1/i) + (). Write the cosine as a combination of exponentials, then it follows from the shift rule that the spectrum of u(t) = cos(0 t) (t) is given by U () = pv(1/(2i( 0 )) + pv(1/(2i( + 0 )) + (/2)( 0 ) + (/2)( + 0 ). To determine Y () = H()U () we use that H()( 0 ) = H(0 )( 0 ) = 0. Hence, writing 2 everything with a common denominator, Y () = /i( 2 i 20 0 ). Put s = i and apply partial fraction expansion obtain that 2Y () = to (1 + i)/(s + 0 (1 i)/ 2) + (1 i)/(s + 0 (1 + i)/ 2). The inverse Fourier transform of this equals e0 t/ 2 (cos(0 t/ 2) sin(0 t/ 2)) (t). 10.14 a From the dierential equation we immediately obtain the frequency response: H() = 1 + /2 1 i + 1 + = + , 2i + 2 2i +

where we also used a long division. Using the tables it then follows that h(t) = 1 ((t) + (1 + /2) (t)et/2 ). 2 b We have to interpret the dierential equation the other way around, so with input and output interchanged. This means that the system function is now 1/H(), so 2i + 2+ =2 i + 1 + i + 1 +

42


where we also used a long division. Using the tables it then follows that h1 (t) = 2(t) (2 + ) (t)et(1+) . c Note that the spectrum of (h h1 )(t) is the function H() (1/H()), which is 1. Since (t) 1, it follows that (h h1 )(t) = (t). 10.16 We use separation of variables, so we substitute u(x, y) = X(x)Y (y) into uyy + uxx = 0. This gives for some arbitrary constant c (the separation constant) that X + cX = 0, Y cY = 0. In order to satisfy the linear homogeneous condition as well, X(x)Y (y) has to be bounded, and this implies that both X(x) and Y (y) have to be bounded functions. Solving the dierential equations we obtain from the boundedness condition that X(x) = if c = 0 and ei cx , ei cx if c > 0. Similarly Y (y) = 1 if c = 0 and 1 e cy , e cy if c > 0. But e cy is not bounded for y > 0 so Y (y) = e cy for c 0. We put c = s2 , then it follows that the class of functions satisfying the dierential equation and being bounded, can be described by X(x)Y (y) = eisx e| s |y , where s R.

By superposition we now try a solution u(x, y) of the form Z u(x, y) = e| s |y F (s)eisx ds.

If we substitute y = 0 in this integral representation, then we obtain that Z 1 u(x, 0) = = F (s)eisx ds. 1 + x2 Since 1/(1 + 2 ) this means that Z 1 1 | t | it = e e dt. 1 + 2 2 1 | t | e 2

The (formal) solution is thus given by Z Z 1 u(x, y) = 2 e| s |(1+y) eisx ds = e| s |(1+y) cos(sx) ds. 0

10.18

We substitute u(t) = (t),then Z t Z h(t) = e(t ) ( ) d = ( ( + 1 t) ( t))e(t ) ( ) d,t1 t

a

so h(t) = ( (t) (t 1))e . Applying table 3, no. 7 and a shift in the time domain gives H() = (1 e(1+i) )/(1 + i). b The impulse response is causal, so the system is causal. c It is straightforward to verify that the impulse response is absolutely integrable, so the system is stable. d The response y(t) to the block function p2 (t) is equal to the convolution of h(t) with p2 (t), which equals Z 1 Z t+1 y(t) = h(t ) d = h( ) d.1 t1

This is 0 for t < 1 or for t > 2. For 1 < t < 0 it equals 1 e(t+1) , for 0 t < 1 it equals 1 e1 , and for 1 t < 2 it equals e(t1) e1 . 10.19 a The impulse response is the derivative in the sense of distributions of a(t) = et (t), which is (t) et (t).


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b The frequency response is H() = i/(1 + i). (Apply e.g. the dierentiation rule to h(t) = a (t).) c We have Y () = H()U () (using obvious notations), so Y () = i/(1 + i)2 = 1/(1 + i) 1/(1 + i)2 . The response is the inverse Fourier transform of Y (): y(t) = (1 t)et (t). 10.21 a The frequency response H() is the triangle function qc (). The inverse Fourier transform follows from table 3: h(t) = 2 sin2 (c t/2)/(c t2 ). b Since a (t) = h(t) and h(t) 0, the function a(t) is a monotone increasing function. The frequency response follows immediately from the dierential equation: H() = 1 + i . 2 + 2i + 2

10.22

Applying partial fraction expansion (use s = i) we obtain H() = 1 1 + . 2(i + 1 i) 2(i + 1 + i)

The inverse Fourier transform is then h(t) = (et cos t) (t). Integrating this over (, t] gives the step response a(t) = 1 (1 + et (sin t cos t)) (t). 2 b We have Y () = H()U () (using obvious notations), so Y () = 1/( 2 + 2i + 2). Applying partial fraction expansion we obtain Y () = 1 1 . 2i(i + 1 i) 2i(i + 1 + i)

10.23

The response is the inverse Fourier transform of Y (), which gives as repsonse the function y(t) = (et sin t) (t). a Since | i 1 i | = | i + 1 i | and eit0 = 1 we have | H() | = 1 and so L is an all-pass system. b First write (i 1 i)/(i + 1 i) as 1 2/(i + 1 i) and then use the inverse Fourier transform (the tables) to obtain (t) 2e(1i)t (t) 1 2/(i + 1 i). From the shift property in the time domain we obtain h(t) = (t t0 ) 2e(1i)(tt0 ) (t t0 ). c Ignoring the delta function it is easy to verify that h(t) is absolutely integrable ( ei(tt0 ) = 1), hence the system is stable. d Write u(t) = 1 + eit + eit and use that eit H()eit , then y(t) = H(0)+H(1)eit +H(1)eit , which equals ieit0 eit +(4i3)eit0 eit /5.

10.24

We use separation of variables, so we substitute u(x, y) = X(x)Y (y) into uxx 2uy = 0. This gives for some arbitrary constant c (the separation constant) that X +cX = 0, 2Y +cY = 0. Here X(x) and Y (y) have to be bounded functions. Solving the dierential equations, obtain from the we boundedness condition that X(x) = 1 if c = 0 and ei cx , ei cx if c > 0. For c 0 we obtain for Y (y) the solution ecy/2 . We put c = s2 with s R. It then follows that the class of functions satisfying the dierential equation and being bounded, can be described by X(x)Y (y) = eisx es2

y/2

,

where s R.

By superposition we now try a solution u(x, y) of the form Z 2 u(x, y) = es y/2 F (s)eisx ds.

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If we substitute y = 0 in this integral representation, then we obtain that Z u(x, 0) = xex (x) = F (s)eisx ds. 1 Since te (t) 1/(1+i)2 this means that F (s) = 2(1+is)2 . The (formal) solution is thus given by Z 2 1 1 u(x, y) = es y/2 eisx ds 2 (1 + is)2 Z (1 s2 ) cos sx + 2s sin sx s2 y/2 1 = e ds. 2 (1 + s2 )2 t


11.1 11.2

In parts a, b and c the domain is C and the range is C as well. In part d the 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 the imaginary part is y. b Expanding z 3 = (x + iy)3 we see that the real part is x3 3xy 2 and that the imaginary part is 3x2 y y 3 . 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 + y 2 ) and the imaginary part is (2y + 3x + 9)/((x + 3)2 + y 2 ). To see this, write z = x + iy, then z + 3 = x + 3 + iy and so f (z) = 3y 2x 6 + i(2y + 3x + 9) (3i 2)(x + 3 iy) = . (x + 3 + iy)(x + 3 iy) (x + 3)2 + y 2

11.3

Apply denition 11.3 and expand the squares; several of the exponentials cancel and only 1/2+1/2 remains, so sin2 z +cos2 z = 1. Similarly it follows by substitution that 2 sin z cos z = sin 2z. a From denition 11.3 it follows that sin(iy) = (ey ey )/2i = i sinh y and cos(iy) = (ey + ey )/2 = cosh y. b Write z = x + iy; from exercise 11.4 with z = x and w = iy it follows that sin(x + iy) = sin x cos(iy) + cos x sin(iy). Now apply part a, then sin(x + iy)

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