Antwoorden - Fourier and Laplace Transforms, Manual Solutions

8/17/2019 Antwoorden - Fourier and Laplace Transforms, Manual Solutions

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Answers to selected exercises for chapter 1

Apply cos(α + β ) = cos α cos β − sin α sin β , then1.1f 1(t) + f 2(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= C 1 cos ωt − C 2 sin ωt,

where C 1 = A1 cos φ1 + A2 cos φ2 and C 2 = A1 sin φ1 + A2 sin φ2. Put A =p C 21 + C

22 and take φ such that cos φ = C 1/A and sin φ = C 2/A (this is

possible since (C 1/A)2+(C 2/A)

2 = 1). Now f 1(t)+ f 2(t) = A(cos ωt cos φ−sin ωt sin φ) = A cos(ωt + φ).

Put c1 = A1eiφ1 and c2 = A2eiφ2 , then f 1(t) + f 2(t) = (c1 + c2)eiωt. Let1.2c = c1 + c2, then f 1(t) + f 2(t) = ce

iωt. The signal f 1(t) + f 2(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 = 1

2.1.6

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 with1.8

sum 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 tt−1

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

R tt−1

cos(ωτ ) dτ gives the following response: (sin(ωt) −sin(ωt − ω))/ω = 2 sin(ω/2) cos(ωt − ω/2)/ω.e Calculating

R tt−1

sin(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 2sin(ω/2)/ω ≥ 0 and −ω/2 + π if 2sin(ω/2)/ω < 0. From

1


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2 Answers to selected exercises for chapter 1

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

a The frequency response of the cascade system is H 1(ω)H 2(ω), since the1.11reponse to eiωt is first H 1(ω)e

iωt and then H 1(ω)H 2(ω)eiωt.

b The amplitude response is | H 1(ω)H 2(ω) | = A1(ω)A2(ω).c The phase response is arg(H 1(ω)H 2(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 → 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 → 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 f 2(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) + B

2)/2.b The energy-content is

R 10

sin2(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 → H (ω)eiωt thatthe response is H (1)ieit = iei(t+1)/2.

a The signal is not periodic since sin(2N ) = 0 for all integer N .1.17b The frequency response H (eiω) equals A(eiω)eiΦe

iω

, 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


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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 +nX

l=n−2

(−1)l = (−1)n−n0 + (−1)n(1 − 1 + 1)

= (−1)n(1 + (−1)n0).


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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 | z 1 | = 4

√ 2, | z 2 | = 4 and arg z 1 = 7π/4, arg z 2 = 2π/3. Hence,2.4

| z 1/z 2 | = | z 1 | / | z 2 | =√

2 and arg(z 1/z 2) = arg(z 1) − arg(z 2) = 13π/12,so z 1/z 2 =

√ 2e13πi/12. Similarly we obtain z 21z

32 = 2048e

3πi/2 and z 21/z 32 =

1

2e3πi/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 + k π/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 z 5 − z 4 + z − 1 as (z − 1)(z 4 + 1) and then solve z 4 = −1 to find2.7the 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 = 2eπi/2 the solutions are z = ln 2 + i(π/2 + 2kπ), where k ∈ Z.2.8Split F (z ) as A/(z − 1

2) + B/(z − 2) and multiply by the denominator of 2.9

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 z 2, A = −5/4 (as in example 2.8).

Trying the first few integers we find the zero z = 1 of the denominator. A2.12 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 denominatorof F (z ) and comparing the coefficients of z 0 = 1, z and z 2 we obtain thatA = 2, B = 0 and C = −1.a Using the chain rule we obtain f (t) = −i(1 + it)−2.2.13Use integration by parts twice and the fact that a primitive of eiω0t is2.14eiω0t/iω0. The given integral then equals 4π(1 − πi)/ω30, since e2πi = 1.Since

˛̨1/(2 − eit) ˛̨ = 1/ ˛̨ 2 − eit ˛̨ and ˛̨ 2 − eit ˛̨ ≥ 2 − ˛̨ eit ˛̨ = 1, the result2.15

follows from˛̨˛R 1

0 u(t) dt

˛̨˛ ≤ R 1

0 | u(t) | dt.

1


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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→∞

˛̨˛̨ 2

n+1 + 1

3n+1 + n + 1

3n + n

2n + 1

˛̨˛̨ = lim

n→∞

2 + 1/2n

3 + (n + 1)/3n1 + n/3n

1 + 1/2n =

2

3.

Determine the radius of convergence as follows:2.19

limn→∞

˛̨˛̨ 2

n+1z 2n+2

(n + 1)2 + 1

n2 + 1

2nz 2n

˛̨˛̨ = lim

n→∞2˛̨

z 2˛̨ 1 + 1/n2

1 + 2/n + 2/n2 = 2

˛̨z 2˛̨

.

This is less than 1 if ˛̨

z 2˛̨


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3

a

2

b

4 5

c

1

d

3

e

32

1 + 2i

2

f

–2

1

g

2 3

2 12

0


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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/2

sin(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/20

g(t)dt

=

Z T T/2

g(t − T )dt +

Z T/20

g(t)dt =

Z 0−T/2

g(τ )dτ +

Z T/20

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

2πZ 0

−π

(−t)e−int dt + 1

2πZ π

0

te−int dt.

As in example 3.2 these integrals can be calculated using integration byparts for n = 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 10

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


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c0 = 1

2Z 1−1

t2

dt + i

2Z 1−1

t dt = 1

3 .

For n = 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 if cn < 0, arg cn = π /2 if cn = iy with y > 0 and arg cn = −π/2 if cn = iywith y


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+ 1

4Z 42

2 sin(nπt/4) dt.

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

bn = 8

n2π2 sin(nπ/2) −

4

nπ cos(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 = 0.

In order to determine the Fourier cosine series we extend the function to3.21

an even function of period 8. We calculate the coefficients an and a0 asfollows (the bn are 0):

a0 = 1

2

Z 40

(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 40

(x2 − 4x) cos(nπx/4) dx

= 1

2

Z 40

x2 cos(nπx/4) dx − 2

Z 40

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

n2 cos(nπx/2).

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

bn = 64((−1)n − 1)

n3π3 for all n ≥ 1.

If f is real and the cn are real, then it follows from (3.13) that bn =3.24 0. 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/20

ei(1−n)ω0t dt − 1

2iT

Z T/20

e−i(1+n)ω0t dt.


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The first integral equals T /2 for n = 1 while for n = 1 it equals i((−1)n

+1)/((1 − n)ω0). The second integral equals T /2 for n = −1 while forn = −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 = 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/20

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

aZ a/2

0

(2bt/a) cos(2nπt/a) dt − 2

aZ 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

a sin(nπt/a) dt

+ 1

a

Z aa/2

(−2bt

a + 2b) sin(nπt/a) dt

= 8b

n2

π2

sin(nπ/2),

where we used integration by parts.


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0 n2 4

a

–2–4 0 n

π

2 4

b

–2–4

π/2


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0 n2 4

a

–2–4 0 n2 4

b

–2–4

π

2

π

2

12

–


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0 n2 4

a

–2–4 0 n2 4

b

–2–4

12

π


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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 = ±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 series P∞

n=11

n2 converges, so

P∞n=−∞ | cn |

2

converges. 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 = 0, cn =

−i

nπ odd 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 of the 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 π0

t dt = π /4, while the Fourier coefficients4.7for n = 0 follow from an integration by parts:

cn = 1

2π

Z π0

te−int dt = (−1)ni

2n +

(−1)n − 1

2n2π .

The Fourier series follows by substituting these cn:

π

4 +

1

2

∞Xn=−∞,n=0

„(−1)ni

n +

(−1)n − 1

n2π

«eint.

12


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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 f 4.9is continuous. We have that f (t) = 2t+1 for−1 < t


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(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 obtain

P∞n=1

1

n4 = π

4

90.

SinceR ba f (t) dt =

R b−T/2 f (t) dt −

R a−T/2 f (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 Since

Z t−π

sin(2n + 1)τ dτ = −cos(2n + 1)t2n + 1

− 12n + 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−π

f (τ ) dτ (theorem 4.9 orbetter 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 cn of f

exist. Since f is continuous , we can apply integrationby parts, as in the proof of theorem 4.10. It then follows that cn = inω0c

n,

where cn are the Fourier coefficients of f . But cn = inω0cn by theorem

4.10, so cn = −n2ω20cn. 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 = 1,−1. Tak-ing positive and negative n in the series together, we obtain the followingFourier series:


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1

2 sin t + 1

π + 2

π

∞Xk=1

1

1 − 4k2 cos 2kt.

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

f (t) = 1

2 cos t −

4

π

∞Xk=1

k

1 − 4k2 sin 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 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


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f (t) = 2

π − 4

π

∞Xn=1

1

4n2 − 1 cos 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 cos2nt = (−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.

Since R π

−π sin2 t dt = π, the result follows.

a Since f 1 is odd it follows that4.31

(f 1 ∗ f 2)(−t) = −1

T

Z T/2−T/2

f 1(t + τ )f 2(τ ) dτ.

Now change the variable from τ to −τ and use that f 2 is odd, then itfollows that (f 1 ∗ f 2)(−t) = (f 1 ∗ f 2)(t).b The convolution product equals

(f ∗ f )(t) = 1

2

Z 1

−1

τ f (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 note

that 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 of

the sawtooth f and applying the convolution theorem gives the Fouriercoefficients of (f ∗ f )(t), namely c20 = 0 and c

2

n = −1/π2n2 (n = 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


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1

2Z 1−1

(f ∗ f )(t) dt =Z 10

(−t2

/2 + t − 1/3) dt = 0.

This is in agreement with the result in part c since c20 = 0. Since thiscoefficient 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 isalso odd, since f is even and for 0 ≤ t ≤ 1 it equalsZ 0

−1

(−τ 2/2 − τ − 1/3) dτ +

Z t0

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


20/87


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 of the 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 = −

1π

R π0

u(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 thaty(t) = y−4e

−4it + y−1e−it + y0 + y1e

it + y4e4it

= 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

104 cos 4t +

30

104 sin 4t +

1

8.

We have that5.6

H (ω) = 1

−ω2 + ω20.

18


21/87


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) = πq π,2π(t) the Fouriercoefficients of u follow immediately from table 1:

u0 = π

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

2

(2k + 1)2π2.

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

0

un, 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/20

x sin(nπx/L) dx + 2

L

Z LL/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 = 0 the characteristicequation s2 − c = 0 has two distinct roots ±s1. The general solution isthen X (x) = αes1x + βe−s1x, so X (x) = s1αe

s1x − s1βe−s1x. The first

boundary condition X (0) = 0 gives s1(α − β ) = 0 , s o 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 X n(x) = cos((2n +1)πx/2L) (n = 0, 1, 2, 3, . . .). Since T n(t) remains as in the textbook (forother parameters), we have thus found the fundamental solutions

un(x, t) = e−(2n+1)2π2kt/4L2 cos((2n + 1)πx/2L).

Superposition gives

u(x, t) =∞Xn=0

Ane−(2n+1)2π2kt/4L2 cos((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).


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

7e−25π2kt/4L2 cos(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 substitute5.12

the 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 = 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 = 0 the characteristicequation s2 − c = 0 has two distinct roots ±s1. The general solutionis then X (x) = αes1x + β e−s1x, so X (x) = s1αe

s1x − s1βe−s1x. The

boundary condition X (0) = 0 gives s1(α − β ) = 0 , s o β = α. Fromthe boundary condition X (π) = 0 we obtain s1α(e

s1π − 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 X n(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 T n(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.


23/87


We have A0 = (2/π)R π0 kxdx = kπ and An = (2/π)

R π0 kx cos(nx) dx for

n = 0, which can be calculated by an integration by parts: An = 0 for neven (n = 0) and An = −4k/n

2π for n odd. The solution is thus

u(x, t) = kπ

2 −

4k

π

∞Xn=0

1

(2n + 1)2 cos((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 = (e

iω0t − 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 sinω0t − 2 sin(ω0(t − 2)) +sin(ω0(t − 4)).c

A signal with period 1 has Fourier series of the formP∞

n=−∞ une

2πint

.The response is

P∞

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

all 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/30

| y(t) |2 dt =∞X

n=−∞

| 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 follows

that 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/20

a sin(nπx/L) dx = 2a(1 − cos(nπ/2))/nπ, which givesthe (formal) solution

u(x, t) = 2a

π

∞Xn=1

1

n(1 − cos(nπ/2))e−n

2π2kt/L2 sin(nπx/L).


24/87


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 b oundary 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.


25/87


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 limit

does 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

R ω/(a2 + ω2) dω = 12 ln(a

2 + ω2), so this improper integral,which is the Fourier integral for t = 0, does not exits (limA→∞ ln(a

2 + A2)does not exist e.g.).c We have lima→0 g(t) = lima→0 (t)e

−at = (t), while for ω = 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 10

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 10

t cos ωtdt, which can be calculated for ω = 0 using an integ-ration by parts. The result is:

G(ω) = 2sin ω

ω +

2(cos ω − 1)ω2

.

For ω = 0 we have that G(0) = 2R 10 t dt = 1. Since limω→0 sin ω/ω = 1 andlimω→0(cos ω − 1)/ω2 = − 12 (use e.g. De l’Hôpital’s rule), we obtain that

limω→0 G(ω) = G(0), so G is continuous.

a Calculating the integral we have that6.5

F (ω) = 2icos(aω/2)− 1

ω for ω = 0, F (0) = 0.

b Using Taylor or De l’Hôpital 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


26/87


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

(F (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

(F (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 ωtdt.

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) = −2iω1 + ω2

.

The spectrum is given by R a/2

−a/2 te−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 (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


27/87


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

R t0

1 dτ = t remains.) Since (t)t is not absolutelyintegrable, the function ( ∗ )(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 12 and 1 for | τ | < 12 , we have6.29

( p1 ∗ p3)(t) =Z 1/2−1/2

p3(t− τ ) dτ.


28/87


Here p3(t − τ ) = 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 ; i f t


29/87


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.7.2Take 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

u du = lim

A

→∞Z AC

0

sin v

v dv =

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 and

R −2−∞ F (ω) dω both exist. For the former integral this can be shown as

follows (the other integral can be treated similarly):˛̨̨˛Z ∞2

F (ω) dω

˛̨̨˛ ≤

Z ∞2

2

ω2 − 1 dω

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 − 1 eiωt dω

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

f (t) = 1

π Z ∞

0

cos(π − t)ω − cos(π + t)ω1

−ω2

dω.

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

2

π

Z ∞0

1 − cos aωω

sin ωtdω = 1

2(f (t+) + f (t−)).

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

f (t) = 2

π

Z ∞0

1 − cos aωω

sin ωtdω.

27


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b At t = a the function is discontinuous, so we have convergence to12

(f (a+) + f (a−)) = 12

.

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

The spectrum of pa(t) is 2sin(aω/2)/ω (table 3). From duality it then7.11follows that the spectrum of sin(at/2)/t is πpa(ω) at the points where

pa(t) is continuous; at ω = ±a/2 we should take the value π/2. (We canapply duality since the Fourier integral exists as improper integral; this isexercise 7.5b).

The spectrum of q a(t) is F (ω) = 4 sin2(aω/2)/(aω2) (see table 3). From du-7.12

ality it then follows that the spectrum of sin2(at/2)/t2 is (aπ/2)q a(ω). (Wecan apply duality since q a is continuous, piecewise smooth, and absolutelyintegrable and since its Fourier integral exists as improper integral; thislatter fact follows immediately if we use that F (ω) is even and continuousand that e.g. F (ω) ≤ 1/ω2 for ω ≥ 1).The function 1/(a + iω) is not integrable on R (see exercise 6.2), so duality7.14cannot be applied.

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

constants). For example:p

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

This is an important exercise: it teaches to recognize useful properties.7.16a Complete the square, then one can apply a shift in time: f (t) = 1/(1 +(t − 1)2). Since the spectrum of 1/(1 + t2) is πe−|ω |, the result follows:πe−iωe−|ω |.b Here we have a shift from t to t − 3; from the spectrum of sin 2πt/t theresult follows: πe−3iω p4π(ω) with value 12 at ω = ±2π.c We now have 1/(t2 − 4t + 7), multiplied by a sine function. The sinefunction is easy to deal with using exercise 6.9 (a variant of the modulationtheorem). As in a we complete the square and note that 1/(3 + (t − 2)2)has spectrum F (ω) = πe−2iωe−

√ 3|ω |/

√ 3. Hence, the result is now (F (ω −

4) − F (ω + 4))/2i.d We use that 3πq 6(ω) is the spectrum of sin

2

(3t)/t2

and apply a shift intime from t to t − 1, then the result is 3πe−iωq 6(ω).Again, this is an important exercise: it teaches to recognize useful proper-7.17ties for the inverse transform.a We immediately use table 3 to obtain that 1/(4 + ω2) is the spectrumof f (t) = e−2| t |/4.b Apply a shift in the frequency domain to the spectrum πp2a(ω) of sin(at)/t, then it follows that f (t) = (eiω0t + e−iω0t) sin(at)/(πt), so f (t) =2cos(ω0t) sin(at)/(πt).c As in part b it follows that f (t) = 3e9it/(π(t2 + 9)).

From the convolution theorem it follows that F (P a ∗ P b)(ω) = (F P a)(ω) ·7.19


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(F P b)(ω) = e−(a+b)

|ω|, where we also used table 3. But also (F P a+b)(ω) =

e−(a+b)|ω |, and since F is one-to-one (theorem 7.4) it then follows thatP a+b = P a ∗ P b.a Use the result of exercise 6.14 (G(ω) = −2iω/(1+ ω2)), the fundamental7.21theorem and the fact that the spectrum is odd to change from

R ∞−∞ to

R ∞0

.It then follows that (use x instead of ω)Z ∞0

x sin xt

1 + x2 dt =

π

2e−t.

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

R ∞−∞ | g(t) |2 dt =

R 0−∞ e2t dt + R ∞0 e−2t dt, which is 1. In R ∞−∞ | G(ω) |

2 dω we can use the

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

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

R ∞0

e−(a+b)t dt = 2/(a + b). The spectra of f and g are2a/(a2 + ω2) and 2b/(b2 + ω2) (table 3).

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

R ∞−∞ sin

4 t/t4 dt is the Fourier transform of sin4 x/x4 cal-

culated at ω = 0, henceR ∞−∞ sin

4 x/x4 dx = (π/2)(q 2 ∗ q 2)(0). Using thatq 2 is an even function we obtain that

(q 2

∗q 2)(0) = Z

∞

−∞q 2(t)q 2(

−t) dt = 2Z

2

0

(1

−t/2)2 dt.

This integral equals 4/3 and soR ∞−∞ sin

4 x/x4 dx = 2π/3.

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

1

π

Z ∞0

1

(1 + ω2)2 dω = f (0) =

1

4,

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

The Gauss function f (t) = e−at2

belongs to S and so we can apply Poisson’s7.26∗summation formula. Since F (ω) =

p π/ae−ω

2/4a (see table 3), it followsfrom (7.23) with T = 1 that

∞Xn=−∞

e−an2

=p

π/a

∞Xn=−∞

e−π2n2/a.

Replacing a by π x the result follows.

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

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


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∞Xn=−∞

a

a2 + (t + n)2 = π

1 +

∞Xn=1

e−2πn(a+it)

+

∞Xn=1

e−2πn(a

−it)!

.

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

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

a

π

∞Xn=−∞

1

a2 + (t + n)2 =

1 − e−4πa1 + e−4πa − e−2πa(e2πit + e−2πit) .

Multiplying numerator and denominator by e2πa the result follows.

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

F (ω) = 1

2i

„Z π0

ei(1−ω)t dt −Z π0

e−i(1+ω)t dt«

.

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

f (t) = 1

2πZ ∞

−∞

1 + e−iωπ

1 − ω2 eiωt dω.

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

f (t) = 1

2π

Z ∞0

eiωt + e−iωt + eiω(t−π) + e−iω(t−π)

1 − ω2 dω,

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

R π0

sin2 t dt = π/2,then it follows that

1

2π

Z ∞−∞

| F (ω) |2 dω = π2

.

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

˛̨̨e−iωπ/2

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

integrand being even, the result follows.

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

( p2a(v) ∗ e−| v |)(t) =Z a−a

e−| t−τ | dτ =Z t+at−a

e−|u | du


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

( p2a(v) ∗ e−| v |)(t) =Z 0t−a

eu du +

Z t+a0

e−u du = 2 − 2e−a cosh t.

If t > a, then t − a > 0 and so

( p2a(v) ∗ e−| v |)(t) =Z t+at−a

e−u du = 2e−t sinh a.

Finally, if t < −a, then t + a a. Finally, the Fourier integral ex-ists as improper Riemann integral since G(ω) is even absolutely integrable:| G(ω) | ≤ 4/ ˛̨ω(1 + ω2) ˛̨. We can now apply the duality rule (theorem7.5) and it then follows that G(−t) ↔ 2πg(ω), so f (−t) ↔ πg(ω)/2. Sincef (−t) = f (t) we thus see that F (ω) = π − πe−a cosh ω for | ω | ≤ a andF (ω) = πe−|ω | sinh a for | ω | > a.


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

R ∞−∞

P a(t) dt = 1, we see that P a(t) fits thedescription of the delta function.c From table 3 it follows that P a(t) ↔ e−a|ω | and lima↓0 e−a|ω | = 1.Combining this with part b shows that it is reasonable to expect that thespectrum of δ (t) is 1.

a Since φ(a) ∈ C for all φ ∈ S , it follows from (8.10) that δ (t − a) is a8.2mapping from S to C. For c ∈ C and φ ∈ S we have thatδ (t − a), cφ = (cφ)(a) = c δ (t − a), φ ,and for φ1, φ2 ∈ S we haveδ (t − a), φ1 + φ2 = (φ1 + φ2)(a) = δ (t − a), φ1 + δ (t − a), φ2 .So δ (t − a) is a linear mapping from S to C, hence a distribution.b Taking the limit inside the integral in (8.1) givesZ ∞−∞

„ 1

2π lima→∞

2sin aω

ω

«f (t − ω) dω = f (t)

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

Z ∞

−∞

δ (ω)f (a

−ω) dω = f (a)

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

δ (a − t)f (t) dt = f (a).

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

Since 1, φ = R ∞−∞

φ(t) dt ∈ C for all φ ∈ S , it follows that 1 is a mapping8.4from S to C. The linearity of this mapping follows from the linearity of integration: for c ∈ C and φ ∈ S we have that

1, cφ =Z ∞−∞

(cφ)(t) dt = c

Z ∞−∞

φ(t) dt = c 1, φ ,

and for φ1, φ2 ∈ S we have

1, φ1 + φ2 =Z ∞−∞

(φ1 + φ2)(t) dt =

Z ∞−∞

φ1(t) dt +

Z ∞−∞

φ2(t) dt,

so 1, φ1 + φ2 = 1, φ1 + 1, φ2. This proves that 1 is a linear mappingfrom S to C, hence a distribution.For φ ∈ S there exists a constant M > 0 such that (e.g.) (1+t2) | φ(t) | ≤ M 8.5for all t ∈ R. Hence,˛̨˛̨ Z ∞

0

φ(t) dt

˛̨˛̨ ≤

Z ∞0

| φ(t) | dt ≤ M Z ∞0

1

1 + t2 dt


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(the latter integral equals [arctan]∞0 = π /2). The integral

R ∞0 φ(t) dt thus

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

In example 8.4 it was already motivated why the integralR ∞−∞

| t | φ(t) dt8.7exists: there exists a constant M > 0 such that (e.g.) (1 + t2) | tφ(t) | ≤ M for all t ∈ R. Hence,˛̨˛̨ Z ∞

−∞

| t | φ(t) dt˛̨˛̨ ≤

Z ∞−∞

| tφ(t) | dt ≤ M Z ∞−∞

1

1 + t2 dt 0. From part awe then get˛̨̨˛Z 1−1

| t |−1/2 φ(t) dt˛̨̨˛ ≤

Z 1−1

| t |−1/2 | φ(t) | dt ≤ M 1Z 1−1

| t |−1/2 dt 0 such that (e.g.) (1 +8.10t2) | tφ(t) | ≤ M for all t ∈ R. Hence,˛̨˛̨ Z ∞

−∞

tφ(t) dt

˛̨˛̨ ≤

Z ∞−∞

| tφ(t) | dt ≤ M Z ∞−∞

1

1 + t2 dt


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˙(sgn t)

, φ¸

=Z ∞0

(φ

(−t) − φ

(t)) dt = [φ(t)]0−∞ − [φ(t)]

∞0 = 2φ(0),

hence, (sgn t) = 2δ (t).b Since sgn t = 2(t) − 1 (verify this), it follows from the linearity of differentiation that (sgn t) = 2(t) = 2δ (t). Here we used that 1 = 0 andthat (t) = δ (t) (see (8.18)).c Since | t | = sgn t it follows from part a that | t | = (sgn t) = 2δ (t).Since the function | t | from example 8.9 is continuously differentiable out-8.16side t = 0, it follows from the jump formula that | t | = sgn t (at t = 0there is no jump and outside t = 0 this equality holds for the ordinaryderivatives). The function from example 8.10 has a jump of magnitude 1at t = 0, while for t < 0 the derivative is 0 and for t > 0 the derivative is− sin t. Hence, the jump formula implies that ((t)cos t) = δ (t) − (t)sin t.a The function pa has a jump of magnitude 1 at t = −a/2 and of mag-8.17nitude −1 at t = a/2. Outside t = 0 the ordinary derivative is 0, so itfollows from the jump formula that pa(t)

= δ (t + a/2) − δ (t − a/2).b The function (t)sin t has no jump at t = 0, for t < 0 the ordinaryderivative is 0 and for t > 0 the ordinary derivative is cos t, so it followsfrom the jump formula that ((t)sin t) = (t)cos t.

a This is entirely analagous to exercises 8.10 and 8.12b.8.18b The function is differentiable outside t = 1 and the ordinary derivativeis 1 for t < 1 and 2t − 2 for t > 1. We denote this derivative as thedistribution T f . At t = 1 the jump is 2, so according to the jump formulathe derivative is T f + 2δ (t − 1).From the jump formula it follows that the derivative as distribution is given8.19

by a(t)eat

+ δ (t), so f

(t) − af (t) = δ (t) as distributions.Subsequently apply definition 8.6 and the definition of δ (t − a) in (8.10),8.22then

p(t)δ (t − a), φ = ( pφ)(a) = p(a)φ(a) = p(a) δ (t − a), φ .Now use definition 8.5, then p(t)δ (t − a), φ = p(a)δ (t − a), φ, whichshows that p(t)δ (t − a) = p(a)δ (t − a).a The definition becomes: f (t)δ (t), φ = δ (t), f φ. The product f φ of 8.23two continuously differentiable functions is continuously differentiable, sothis definition is correct and it gives a mapping from S to C. The linearityfollows immediately from the linearity of δ (t), so f (t)δ (t) is a distribution.b According to part a we have f (t)δ (t), φ = −(f φ)(0), where we alsoapplied δ (t) to f φ. Now apply the product rule for differentiation and writethe result as −f (0) δ (t), φ + f (0) δ (t), φ = f (0)δ (t) − f (0)δ (t), φ.c If f (t) = t, then f (0) = 0 and f (0) = 1, so tδ (t) = −δ (t); if f (t) = t2then f (0) = 0 and f (0) = 0, so t2δ (t) = 0.

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

t · pv(1/t), φ = limα↓0

Z | t |≥α

tφ(t)

t dt = lim

α↓0

Z | t |≥α

φ(t) dt.

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

R ∞−∞

φ(t) dt. Hence, t · pv(1/t) = 1.Let T be an even distribution, then T (−t) = T (t) (definiton 8.8), so8.26


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

a From the definition of sgnt in example 8.3 it follows that sgn t, φ(t) =8.27−sgn t, φ(−t) for all φ ∈ S . This shows that sgnt is odd according toexercise 8.26. Similarly for pv(1/t) (change from t to −t in the integralsdefining pv(1/t)).b From the definition of | t | in example 8.4 it follows that | t | , φ(t) =| t | , φ(−t) for all φ ∈ S (change from t to −t in the integral defining | t |).This shows that | t | is even according to exercise 8.26.a Applying (8.12) to f (t) gives8.29

T f , φ =Z 0−∞

2tφ(t) dt +

Z ∞0

t2φ(t) dt

and in e.g. exercises 8.10, 8.12b and 8.18a we have seen that such integralsare well-defined for φ ∈ S . This gives a mapping from S to C and thelinearity of this mapping follows precisely as in e.g. exercise 8.3 or 8.4.Hence, f indeed defines a distribution T f .b Apply the jump formula (8.21): outside t = 0 the function f is continu-ously differentiable with derivative f (t) = 2t for t > 0 and f (t) = 2 fort 0 and f (t) = 0 for t < 0. Let T f

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

f = T

f ) we have that

T f = T f = T f + (f

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

a Since δ (t) can be defined for all twice continuously differentiable func-8.30tions, the product f (t)δ (t) can also be defined for all twice continuouslydifferentiable functions f (t) by f (t)δ (t), φ(t) = δ (t), f (t)φ(t). This isbecause it follows from the product rule that the product f (t)φ(t) is againtwice continuously differentiable.b From part b and the definition of the second derivative of a distribution(formula (8.17) for k = 2) we obtain f (t)δ (t), φ(t) = δ (t), (f (t)φ(t)).Since (f (t)φ(t)) = f (t)φ(t) + 2f (t)φ(t) + f (t)φ(t) we thus obtainthat

f (t)δ (t), φ(t)

= f (0)φ(0) + 2f (0)φ(0) + f (0)φ(0), which equals

f (0)δ (t) − 2f (0)δ (t) + f (0)δ (t), φ(t) (φ ∈ S ). This proves the iden-tity.c Apply part b to the function f (t) = t2 and use that f (0) = f (0) = 0and f (0) = 2, then t2δ (t) = 2δ (t). Next apply b to f (t) = t3 and usethat f (0) = f (0) = f (0) = 0, then it follows that t3δ (t) = 0.d According to definition 8.7 we have that˙

δ (at), φ(t)¸

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


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

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

a Use table 5 to obtain that δ (t − 4) ↔ e−4iω.9.3b Again use table 5 to obtain that e3it ↔ 2πδ (ω − 3).

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

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

a From example 9.1 (or table 5) we obtain the result e−5it/2π.9.4b See example 9.2: 2 cos2t.c The spectrum of pv(1/t) is −πisgn ω (table 5). Note that 2 cos ω =eiω + e−iω and that the spectrum of δ (t − a) is e−iaω (table 5). Hence theanswer is iπ−1pv(1/t) + δ (t − 1) + δ (t + 1).

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

It is obvious that (t) = (1 + sgn t)/2 by looking at the cases t > 0 and9.7t


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

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

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

The linearity follows as in definition 8.6. The main point is that one has9.13

to show that eiatφ(t) ∈ S whenever φ ∈ S . So we have to show that forany m, n ∈ Z+ there exists an M > 0 such that

˛̨˛ tn(eiatφ(t))(m)

˛̨˛ < M .

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

that˛̨˛ tneiatφ(k)(t)

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

eiat ̨̨

= 1 this means that we have to show that˛̨˛ tnφ(k)(t)

˛̨˛ < M for some

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

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

˙F eiatT, φ

¸ =

˙eiatT, Φ

¸ =

˙T, eiatΦ

¸ (φ ∈ S having spectrum Φ). Ac-

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

role of the variables ω and t). Hence,˙

F eiat

T, φ¸

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

a Use table 5 for (t) and apply a shift in the time domain, then it follows9.16that (t − 1) ↔ e−iω(πδ (ω) − ipv(1/ω)).b Use table 5 for (t) and apply a shift in the frequency domain, then itfollows that eiat(t) ↔ πδ (ω − a) − ipv(1/(ω − a)).c We have (t) ↔ πδ (ω) − ipv(1/ω) and if we now write the cosine asa combination of exponentials, then we can use a shift in the frequencydomain (as in part b) to obtain that (t)cos at ↔ 12 (πδ (ω − a) − ipv(1/(ω −a)) + πδ (ω + a) − ipv(1/(ω + a))).d Use that 1 ↔ 2πδ (ω) and δ (t) ↔ iω (table 5), so 3i ↔ 6iπδ (ω) and(apply a shift) δ (t − 4) ↔ e−4iωiω; the sum of these gives the answer.e

First note that (t)sgn t = (t) and the spectrum of this is known;furthermore we have that t3 ↔ 2πi3δ (3)(ω) (table 5), so the result is2π2δ (3)(ω) + πδ (ω) − ipv(1/ω).

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

b Write sin t as a combination of exponentials and apply a shift to 12 isgn t,then we obtain the result 14 (sgn(t + 3) − sgn(t − 3)).c Apply reciprocity to (t), then we obtain (πδ (−t) − ipv(−1/t))/2π ↔(ω). Now δ (−t) = δ (t) and pv(−1/t) = −pv(1/t), hence, the result is:12 iπ

−1pv(1/t) + 12δ (t).d Apply the scaling property (table 6) to 1 ↔ 2πδ (ω) to obtain 1 ↔6πδ (3ω). Next we apply a shift in the frequency domain (table 6), which


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results in e2it/3

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

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

According to definition 9.3 we have that9.20∗

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

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

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

(the last step uses definition 9.2). This proves that T (t)∗δ (t−a) = T (t−a).Use exercise 9.20 with T (t) = δ (t − b). The convolution theorem leads to9.21∗

the obvious e−ibωe−iaω = e−i(a+b)ω.

a Use table 5: δ (t − 3) ↔ e−3iω.9.24b Since δ (t + 4) ↔ e4iω (as in part a) and cos t = (eit + e−it)/2 we applya shift in the frequency domain: cos tδ (t + 4) ↔ (e4i(ω−1) + e4i(ω+1))/2.c From table 5 we have (t) ↔ πδ (ω)−ipv(1/ω). Apply the differentiationrule in the time domain (with n = 2), then we obtain t2(t) ↔ −πδ (ω) +ipv(1/ω).d Apply the differentiation rule in the time domain to the result obtainedin exercise 9.16c, then it follows that (2(t)cos t) ↔ iω(πδ (ω − 1) + πδ (ω +1) − ipv(1/(ω − 1)) − ipv(1/(ω + 1))).e Since δ (t) ↔ 1 it follows from first the scaling property and then ashift in the time domain that δ (7(t − 1/7)) ↔ e−iω/77. Finally apply thedifferentiation rule in the time domain to obtain the result: (δ (7t − 1)) ↔iωe−iω/77.f This is a convergent Fourier series and so we can determine the spectrumterm-by-term. Since 1 ↔ 2πδ (ω) and e(2k+1)it ↔ 2πδ (ω − (2k + 1)) weobtain the following result:

π2δ (ω) − 4∞X

k=−∞

(2k + 1)−2δ (ω − (2k + 1)).

a From table 5 we know that eit ↔ 2πδ (ω − 1) and similarly for e−it.9.25Hence, iπ−1 sin t ↔ δ (ω − 1) − δ (ω + 1).b Apply the differentiation rule in the time domain (with n = 2) to δ (t) ↔1, then −δ (t) ↔ ω2.c From table 5 we obtain that δ (t + 12 )/4 ↔ e

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

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


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c Since (ω − 1)2

= ω2

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

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


42/87


a When the system is causal, then the response to the causal signal δ (t)10.1is again causal, so h(t) is causal. On the other hand, if h(t) is causal, thenit follows that y(t) = (u ∗ h)(t) = R t−∞ h(t − τ )u(τ ) dτ and if we now havea causal input u, then the integral will be 0 for t


43/87


b The function u has a Fourier series with terms cneint

(note that ω0 = 1).But the response to eint is H (n)eint and H (n) = 0 for n > 1 and n < −1.So we only have to determine c0, c1 and c−1. These can easily be calculatedfrom the defining integrals: c0 =

12 and c−1 = c1 = −1/π. Hence, the

response follows: y(t) = H (0)c0+H (1)c1eit +H (−1)c−1e−it = 12− 23π cos t.

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

(s + 1)(s − 2)(s − 1)(s + 2) = 1 −

2

3

1

s − 1 − 4

3

1

s + 2 = 1 − 2

3

1

iω − 1 − 4

3

1

iω + 2.

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

−1 from table 4, no. 5) it follows for 1iω

−1 =

−1i(−ω)+1 that −(−t)et ↔ 1iω−1 . Hence, h(t) = δ (t) + 23 (−t)et − 43 (t)e−2t.b The impulse reponse h(t) is not causal, so the system is not causal.c The modulus of H (ω) is 1, so it is an all-pass system and from Parsevalit then follows that the energy-content of the input is equal to the energy-content of the output (if necessary, see the textbook, just above example10.7).

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

H (ω) = ω2 − ω20

ω2 − i√ 2ω0ω − ω20.

b Write the cosine as a combination of exponentials, then it follows fromeiωt

→ H (ω)eiωt that y(t) = (H (ω0)e

iω0t + H (−

ω0)e−iω0t)/2. However

H (±ω0) = 0, so y(t) = 0 for all t.c Note that we cannot use the method from part b. Instead we use(10.6) to determine the spectrum of the response y(t). From table 5 weobtain that (t) ↔ pv(1/iω) + πδ (ω). Write the cosine as a combinationof exponentials, then it follows from the shift rule that the spectrum of u(t) = cos(ω0t)(t) is given by U (ω) = pv(1/(2i(ω − ω0)) + pv(1/(2i(ω +ω0)) + (π/2)δ (ω − ω0) + (π/2)δ (ω + ω0). To determine Y (ω) = H (ω)U (ω)we use that H (ω)δ (ω ± ω0) = H (±ω0)δ (ω ± ω0) = 0. Hence, writingeverything with a common denominator, Y (ω) = ω/i(ω2 − i√ 2ω0ω − ω20).Put s = iω and apply partial fraction expansion to obtain that 2Y (ω) =(1 + i)/(s +

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Antwoorden - Fourier and Laplace Transforms, Manual Solutions

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