Antwoorden fourier and laplace transforms, manual solutions

Answers to selected exercises for chapter 1
Apply cos(α + β) = cos α cos β − sin α sin β, then1.1
f1(t) + f2(t)
= A1 cos ωt cos φ1 − A1 sin ωt sin φ1 + A2 cos ωt cos φ2 − A2 sin ωt sin φ2
= (A1 cos φ1 + A2 cos φ2) cos ωt − (A1 sin φ1 + A2 sin φ2) sin ωt
= C1 cos ωt − C2 sin ωt,
where C1 = A1 cos φ1 + A2 cos φ2 and C2 = A1 sin φ1 + A2 sin φ2. Put A =p
C2
1 + C2
2 and take φ such that cos φ = C1/A and sin φ = C2/A (this is
possible since (C1/A)2
+(C2/A)2
= 1). Now f1(t)+f2(t) = A(cos ωt cos φ−
sin ωt sin φ) = A cos(ωt + φ).
Put c1 = A1eiφ1
and c2 = A2eiφ2
, then f1(t) + f2(t) = (c1 + c2)eiωt
. Let1.2
c = c1 + c2, then f1(t) + f2(t) = ceiωt
. The signal f1(t) + f2(t) is again a
time-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 =
A2
ω
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
3X
n=0
| cos(nπ/2) |2
= 1
2
.
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.1.9
b Since
˛
˛
˛
˛
Z
u(τ) dτ
˛
˛
˛
˛ ≤
Z
| u(τ) | dτ,
it follows from the boundedness of u(t), so | u(τ) | ≤ 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 ξ = τ − t0 in the integralR t
t−1
u(τ − t0) dτ representing the response to u(t − t0).
d Calculating
R t
t−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 t
t−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 response
is | 2 sin(ω/2)/ω |.
g From the response to cos(ωt) in d it follows that the phase response
is −ω/2 if 2 sin(ω/2)/ω ≥ 0 and −ω/2 + π if 2 sin(ω/2)/ω < 0. From
1

2 Answers to selected exercises for chapter 1
phase and amplitude response the frequency response follows: H(ω) =
2 sin(ω/2)e−iω/2
/ω.
a The frequency response of the cascade system is H1(ω)H2(ω), since the1.11
reponse to eiωt
is ﬁrst H1(ω)eiωt
and then H1(ω)H2(ω)eiωt
.
b The amplitude response is | H1(ω)H2(ω) | = A1(ω)A2(ω).
c The phase response is arg(H1(ω)H2(ω)) = Φ1(ω) + Φ2(ω).
a The amplitude response is | 1 + i |
˛
˛ e−2iω
˛
˛ =
√
2.1.12
b The input u[n] = 1 has frequency ω = 0, initial phase 0 and amplitude
1. 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 obtain
that 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 and
c to obtain y[n] = (1 + i)(1 + cos(4ω(n − 2)))/2.
a The power is the integral of f2
(t) over [−π/ | ω | , π/ | ω |], times | ω | /2π.1.13
Now cos2
(ωt + φ0) integrated over [−π/ | ω | , π/ | ω |] equals π/ | ω | and
cos(ωt) cos(ωt + φ0) integrated over [−π/ | ω | , π/ | ω |] is (π/ | ω |) cos φ0.
Hence, the power equals (A2
+ 2AB cos(φ0) + B2
)/2.
b The energy-content is
R 1
0
sin2
(πt) dt = 1/2.
The power is the integral of | f(t) |2
over [−π/ | ω | , π/ | ω |], times | ω | /2π,1.14
which in this case equals | c |2
.
a The amplitude response is | H(ω) | = 1/(1 + ω2
). The phase response1.16
is arg H(ω) = ω.
b The input has frequency ω = 1, so it follows from eiωt
→ H(ω)eiωt
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.1.17
b The frequency response H(eiω
) equals A(eiω
)eiΦeiω
, hence, we obtain
that H(eiω
) = eiω
/(1 + ω2
). The response to u[n] = (e2in
− e−2in
)/2i is
then y[n] = (e2i(n+1)
− e−2i(n+1)
)/(10i), so y[n] = (sin(2n + 2))/5. The
amplitude is thus 1/5 and the initial phase 2 − π/2.
a If u(t) = 0 for t < 0, then the integral occurring in y(t) is equal to 0 for1.18
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(τ) | ≤ K for some con-
stant 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
y(t) = sin(π(t − t0)) +
Z t
t−1
sin(πτ) dτ = sin(π(t − t0)) − 2(cos πt)/π.
a If u[n] = 0 for n < 0, then y[n] is also equal to 0 for n < 0 whenever1.19
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):
| y[n] | ≤ | u[n − n0] | +
˛
˛
˛
˛
˛
nX
l=n−2
u[l]
˛
˛
˛
˛
˛
≤ K +
nX
l=n−2
| u[l] | ≤ K +
nX
l=n−2
K,

Answers to selected exercises for chapter 1 3
which equals 4K. Hence, the system is stable.
c If u[n] is real, then u[n − n0] is real and also the sum in the 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 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
).

a The absolute values follow from
p
x2 + y2 and are given by
√
2, 2, 3, 22.1
respectively. The arguments follow from standard angles and are given by
3π/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 implies2.2
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 = 7π/4, arg z2 = 2π/3. Hence,2.4
| z1/z2 | = | z1 | / | z2 | =
√
2 and arg(z1/z2) = arg(z1) − arg(z2) = 13π/12,
so z1/z2 =
√
2e13πi/12
. Similarly we obtain z2
1z3
2 = 2048e3πi/2
and z2
1/z3
2 =
1
2
e3πi/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 technique2.6
to solve this binomial equation (as in example 2.3).
b As part a; we now obtain the six solutions 6
√
2(cos(π/9 + kπ/3) +
i sin(π/9 + kπ/3)) where k = 0, 1 . . . , 5.
c By completing the square as in example 2.4 we obtain the two solutions
−1/5 ± 7i/5.
Write z5
− z4
+ z − 1 as (z − 1)(z4
+ 1) and then solve z4
= −1 to find2.7
the roots
√
2(±1 ± i)/2. Combining linear factors with complex conjugate
roots we obtain z5
− z4
+ z − 1 = (z − 1)(z2
+
√
2z + 1)(z2
−
√
2z + 1).
Since 2i = 2eπi/2
the solutions are z = ln 2 + i(π/2 + 2kπ), where k ∈ Z.2.8
Split F(z) as A/(z − 1
2
) + B/(z − 2) and multiply by the denominator of2.9
F(z) to obtain the values A = −1/3 and B = 4/3 (as in example 2.6).
a Split F(z) as A/(z + 1) + B/(z + 1)2
+ C/(z + 3) and multiply by2.11
the denominator of F(z) to obtain the values C = 9/4, B = 1/2 and, by
comparing the coefficient of z2
, A = −5/4 (as in example 2.8).
Trying the first few integers we find the zero z = 1 of the denominator. A2.12
long division gives as denominator (z − 1)(z2
− 2z + 5). We then split F(z)
as A/(z − 1) + (Bz + C)/(z2
− 2z + 5). Multiplying by the denominator
of F(z) and comparing the coefficients of z0
= 1, z and z2
we obtain that
A = 2, B = 0 and C = −1.
a Using the chain rule we obtain f (t) = −i(1 + it)−2
.2.13
Use integration by parts twice and the fact that a primitive of eiω0t
is2.14
eiω0t
/iω0. The given integral then equals 4π(1 − πi)/ω3
0, since e2πi
= 1.
Since
˛
˛ 1/(2 − eit
)
˛
˛ = 1/
˛
˛ 2 − eit
˛
˛ and
˛
˛ 2 − eit
˛
˛ ≥ 2 −
˛
˛ eit
˛
˛ = 1, the result2.15
follows from
˛
˛
˛
R 1
0
u(t) dt
˛
˛
˛ ≤
R 1
0
| u(t) | dt.
1

a Use that | an | = 1/
√
n6 + 1 ≤ 1/n3
and the fact that
P∞
n=1 1/n3
con-2.16
verges (example 2.17).
b Use that | an | ≤ 1/n2
and the fact that
P∞
n=1 1/n2
converges.
c Use that | an | = 1/
˛
˛ nen
eni
˛
˛ = 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
lim
n→∞
˛
˛
˛
˛
n!
(n + 1)!
˛
˛
˛
˛ = lim
n→∞
1
n + 1
= 0.
b The series is convergent; proceed as in part a:
lim
n→∞
˛
˛
˛
˛
2n+1
+ 1
3n+1 + n + 1
3n
+ n
2n + 1
˛
˛
˛
˛ = lim
n→∞
2 + 1/2n
3 + (n + 1)/3n
1 + n/3n
1 + 1/2n
=
2
3
.
Determine the radius of convergence as follows:2.19
lim
n→∞
˛
˛
˛
˛
2n+1
z2n+2
(n + 1)2 + 1
n2
+ 1
2nz2n
˛
˛
˛
˛ = lim
n→∞
2
˛
˛ z2
˛
˛ 1 + 1/n2
1 + 2/n + 2/n2
= 2
˛
˛ z2
˛
˛ .
This is less than 1 if
˛
˛ z2
˛
˛ < 1/2, that is, if | z | <
√
2/2. Hence, the radius
of convergence is
√
2/2.
This is a geometric series with ratio z−i and so it converges for | z − i | < 1;2.20
the sum is (1/(1 − i))(1/(1 − (z − i))), so 1/(2 − z(1 − i)).
b First solving w2
= −1 leads to z2
= 0 or z2
= −2i. The equation2.23
z2
= −2i has solutions −1 + i and 1 − i and z2
= 0 has solution 0 (with
multiplicity 2).
c One has P(z) = z(z4
+ 8z2
+ 16) = z(z2
+ 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)/(z2
−4z +5)+(Cz +D)/(z2
−4z +5)2
and multiply2.25
by the denominator of F(z). Comparing the coeﬃcient of z0
, z1
, z2
and z3
leads to the values A = 0, B = 1, C = −2 and D = 2.
Replace cos t by (eit
+e−it
)/2, then we have to calculate
R 2π
0
(e2it
+1)/2 dt,2.26
which is π.
a Using the ratio test we obtain as limit
√
5/3. This is less than 1 and so2.27
the series converges.
b Since (n + in
)/n2
= (1/n) + (in
/n2
) and the series
P∞
n=1 1/n diverges,
this series is divergent.
The series
P∞
n=0 cn(z2
)n
converges for all z with
˛
˛ z2
˛
˛ < R, so it has radius2.29
of convergence
√
R.
a Determine the radius of convergence as follows:2.30
lim
n→∞
˛
˛
˛
˛
(1 + i)2n+2
zn+1
n + 2
n + 1
(1 + i)2nzn
˛
˛
˛
˛ = lim
n→∞
| z |
n + 1
n + 2
˛
˛ (1 + i)2
˛
˛ = 2 | z | .
This is less than 1 if | z | < 1/2, so the radius of convergence is 1/2.
b Calculate f (z) by termwise diﬀerentiation of the series and multiply
this by z. It then follows that
zf (z) + f(z) =
∞X
n=0
(1 + i)2n
zn
=
∞X
n=0
(2iz)n
.
This is a geometric series with ratio 2iz and so it has sum 1/(1 − 2iz).

3
a
2
b
4 5
c
1
d
3
e
3
2
1 + 2i
2
f
–2
1
g
2 3
2 1
2
0

A trigonometric polynomial can be written as3.2
f(t) =
a0
2
+
kX
m=1
(am cos(mω0t) + bm sin(mω0t)).
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
the integral
R 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/2
0
g(t)dt
=
Z T
T/2
g(t − T)dt +
Z T/2
0
g(t)dt =
Z 0
−T/2
g(τ)dτ +
Z T/2
0
g(t)dt
=
Z T/2
−T/2
g(t)dt.
Multiplying by 2/T gives an.
From a sketch of the periodic function with period 2π given by f(t) = | t |3.6
for t ∈ (−π, π) we obtain
cn =
1
2π
Z 0
−π
(−t)e−int
dt +
1
2π
Z π
0
te−int
dt.
As in example 3.2 these integrals can be calculated using integration by
parts for n = 0. Calculating c0 separately (again as in example 3.2) we
obtain
c0 =
π
2
, cn =
(−1)n
− 1
n2π
Substituting these values of cn in (3.10) we obtain the Fourier series. One
can also write this as a Fourier cosine series:
π
2
−
4
π
∞X
k=0
cos((2k + 1)t)
(2k + 1)2
.
From the description of the function we obtain that3.7
cn =
1
2
Z 1
0
e−(1+inπ)t
dt.
This integral can be evaluated immediately and leads to
cn =
inπ − 1
2(n2π2 + 1)
`
(−1)n
e−1
− 1
´
.
The Fourier series follows from (3.10) by substituting cn.
The Fourier coeﬃcients are calculated by splitting the integrals into a real3.9
and an imaginary part. For c0 this becomes:
3

c0 =
1
2
Z 1
−1
t2
dt +
i
2
Z 1
−1
t dt =
1
3
.
For n = 0 we have that
cn =
1
2
Z 1
−1
t2
e−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 the
results and simplifying somewhat we obtain the Fourier coefficients (and
thus the Fourier series):
cn =
(−1)n
(2 − nπ)
n2π2
.
From the values of the coefficients cn calculated earlier in exercises 3.6, 3.73.10
and 3.9, one can immediately obtain the amplitude spectrum | cn | and the
phase 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 = iy
with y < 0). This results in three figures that are given separately on the
website.
a By substituting a = T/4 in (3.14) it follows that3.11
cn =
sin(nπ/4)
nπ
for n = 0, c0 =
1
4
.
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.
By substituting a = T/2 in (3.15) it follows that3.12
c0 =
1
2
, cn = 0 for n = 0 even, cn =
2
n2π2
for n odd.
We have that f(t) = 2p2,4(t) − q1,4(t) and so the Fourier coefficients follow3.14
by linearity from table 1:
c0 = 3/4, cn = (2nπ sin(nπ/2) − 4 sin2
(nπ/4))/(n2
π2
) for n = 0.
Note that f(t) can be obtained from the sawtooth z(t) by multiplying the3.15
shifted version z(t − T/2) by the factor T/2 and then adding T/2, that
is, f(t) = T
2
z(t − T
2
) + T
2
. Now use the Fourier coefficients of z(t) (table 1
e.g.) and the properties from table 2 to obtain that
c0 =
T
2
, cn =
iT
2πn
for all n = 0.
Shifts over a period T (use the shift property and the fact that e−2πin
= 13.17
for all n).
In order to determine the Fourier sine series we extend the function to an3.19
odd function of period 8. We calculate the coefficients bn as follows (the
an are 0):
bn =
1
4
Z −2
−4
(−2 sin(nπt/4)) dt +
1
4
Z 2
−2
t sin(nπt/4) dt

+
1
4
Z 4
2
2 sin(nπt/4) dt.
The second integral can be calculated by an integration by parts and one
then obtains that
bn =
8
n2π2
sin(nπ/2) −
4
nπ
cos(nπ),
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 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 as
follows (the bn are 0):
a0 =
1
2
Z 4
0
(x2
− 4x) dx = −
16
3
,
while for n ≥ 1 we have
an =
1
4
Z 0
−4
(x2
+ 4x) cos(nπx/4) dx +
1
4
Z 4
0
(x2
− 4x) cos(nπx/4) dx
=
1
2
Z 4
0
x2
cos(nπx/4) dx − 2
Z 4
0
x cos(nπx/4) dx.
The first integral can be calculated by applying integration by parts twice;
the second integral can be calculated by integration by parts. Combining
the 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
∞X
n=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). The
result 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 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(ω0t) = (eiω0t
− e−iω0t
)/2i we have3.25
cn =
1
2iT
Z T/2
0
ei(1−n)ω0t
dt −
1
2iT
Z T/2
0
e−i(1+n)ω0t
dt.

The first integral equals T/2 for n = 1 while for n = 1 it equals i((−1)n
+
1)/((1 − n)ω0). The second integral equals T/2 for n = −1 while for
n = −1 it equals i((−1)n+1
− 1)/((1 + n)ω0). The Fourier coefficients are
thus 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 can3.27
thus calculate the coefficients an and a0 as follows (the bn are 0):
a0 =
2
a
Z a/2
0
2bt/a dt −
2
a
Z 0
−a/2
2bt/a dt = b.
while for n ≥ 0 we obtain from an integration by parts that
an =
2
a
Z a/2
0
(2bt/a) cos(2nπt/a) dt −
2
a
Z 0
−a/2
(2bt/a) cos(2nπt/a) dt
=
2b((−1)n
− 1)
n2π2
,
which gives the Fourier cosine series. It can also be determined using the
result of exercise 3.6 by applying a multiplication and a scaling.
The odd extension has period 2a and the coefficients bn are given by (the
an 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 a
a/2
(
−2bt
a
+ 2b) sin(nπt/a) dt
=
8b
n2π2
sin(nπ/2),
where we used integration by parts.

0 n2 4
a
–2–4 0 n
π
2 4
b
–2–4
π/2

0 n2 4
a
–2–4 0 n2 4
b
–2–4
π
2
π
2
1
2
–

0 n2 4
a
–2–4 0 n2 4
b
–2–4
1
2
π

a The periodic block function from section 3.4.1 is a continuous function4.1
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 coefficients 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
T2
+
8
T2ω2
0
∞X
n=1
1
n2
since sin2
(nω0a/2) ≤ 1. The series
P∞
n=1
1
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 and4.4
use 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.4.6
Using 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 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
+
2
π
∞X
k=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 the
series converges to f(π/2) = 1. Since sin((2k + 1)π/2) = (−1)k
, formula
(4.11) follows:
∞X
k=0
(−1)k
2k + 1
=
π
4
.
a We have that c0 = (2π)−1
R π
0
t dt = π/4, while the Fourier coefficients4.7
for n = 0 follow from an integration by parts:
cn =
1
2π
Z π
0
te−int
dt =
(−1)n
i
2n
+
(−1)n
− 1
2n2π
.
The Fourier series follows by substituting these cn:
π
4
+
1
2
∞X
n=−∞,n=0
„
(−1)n
i
n
+
(−1)n
− 1
n2π
«
eint
.
12

b From the fundamental theorem it follows that the series will converge
to 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 π
4
to the other side of the
=-sign, then multiply by 2, and finally split the sum into a sum from n = 1
to ∞ and a sum from n = −1 to −∞, then it follows that
π
2
= 2
∞X
n=1
(−1)n
− 1
n2π
(−1)n
(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:
π2
8
=
∞X
k=1
1
(2k − 1)2
.
a From f(0+) = 0 = f(0−) and f(1−) = 0 = f((−1)+) it follows that f4.9
is continuous. We have that f (t) = 2t+1 for −1 < t < 0 and f (t) = −2t+1
for 0 < t < 1. Calculating the defining 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 defined for −1 < t ≤
1 by g(t) = t and h(t) = t2
for −1 < t ≤ 0 and h(t) = −t2
for 0 < t ≤ 1.
Since g is a sawtooth, the Fourier coefficients are cn = (−1)n
i/πn (see
section 3.4.3). The function h is the odd extension of −t2
on (0, 1] and
its Fourier coefficients have been determined in the first example of section
3.6. By linearity one obtains the Fourier coefficients of f. In terms of the
an and bn they become an = 0 and bn = 4(1 − (−1)n
)/π3
n3
. Hence, they
decrease as 1/n3
.
c Use e.g. the fundamental theorem for odd functions to obtain
f(t) =
8
π3
∞X
k=0
sin(2k + 1)πt
(2k + 1)3
.
Now substitute t = 1/2 and use that f(1/2) = 1/4 and sin((2k + 1)π/2) =
(−1)n
to obtain the required result.
Use (3.8) to write the right-hand side of (4.14) as a2
0/4 + 1
2
P∞
n=1(a2
n + b2
n).4.10
a The Fourier coefficients of f and g are (see table 1 or section 3.4.1),4.12
respectively, fn = (sin na)/nπ for n = 0 and f0 = a/π and gn = (sin nb)/nπ
for n = 0 and g0 = b/π. Substitute into Parseval (4.13) and calculate the
integral (1/π)
R a/2
−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
(nπ/2) = 1 for n odd and 0 for n even, then (4.10) follows.
a The Fourier coefficients are (see table 1 or section 3.4.2 and use that4.13
sin2
(nπ/2) = 1 for n odd and 0 for n even): cn = 2/n2
π2
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
3
=
1
4
+
8
π2
∞X
k=1
1
(2k − 1)4

(again we split the sum in a part from n = 1 to ∞ and from n = −1 to
−∞). Take all constants together and multiply by π2
/8, then the required
result follows.
b Since
S =
∞X
n=1
1
n4
=
∞X
k=1
1
(2k)4
+
∞X
k=0
1
(2k + 1)4
,
it follows from part a that
S =
1
16
∞X
k=1
1
k4
+
π4
96
=
1
16
S +
π4
96
.
Solving for S we obtain
P∞
n=1
1
n4 = π4
90
.
Since
R b
a
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 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 and4.16
c−n = (an + ibn)/2 (n ∈ N).
a The Fourier series is given by4.17
4
π
∞X
n=0
sin(2n + 1)t
2n + 1
.
b Since
Z t
−π
sin(2n + 1)τ dτ = −
cos(2n + 1)t
2n + 1
−
1
2n + 1
,
the integrated series becomes
−
4
π
∞X
n=0
1
(2n + 1)2
−
4
π
∞X
n=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 or
better still, exercise 4.16). Calculating this integral we obtain the function
g(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 a
Fourier 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 Fourier4.20
coefficients 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 = inω0cn,
where cn are the Fourier coefficients of f . But cn = inω0cn by theorem
4.10, so cn = −n2
ω2
0cn. Now apply the Riemann-Lebesgue lemma to cn,
then it follows that limn→±∞ n2
cn = 0.
a The Fourier coefficients have been determined in exercise 3.25: c1 =4.22
1/(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 following
Fourier series:

1
2
sin t +
1
π
+
2
π
∞X
k=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 differentiating
its Fourier series for t = nπ:
f (t) =
1
2
cos t −
4
π
∞X
k=1
k
1 − 4k2
sin 2kt.
At t = nπ the differentiated series converges to (f (t+) + f (t−))/2, which
equals 1/2 for t = 0, while it equals −1/2 for t = π. Hence, the differen-
tiated series is a periodic function with period 2π which is given by 0 for
−π < t < 0, 1
2
for t = 0, cos t for 0 < t < π, −1
2
for t = π.
Write down the expression for Si(−x) and change from the variable t to4.25
−t, then it follows that Si(−x) = −Si(−x).
a From the definition of Si(x) it follows that Si (x) = sin x/x. So Si (x) =4.26
0 if sin x/x = 0. For x > 0 we thus have Si (x) = 0 for x = kπ with k ∈ N.
A candidate for the first maximum is thus x = π. Since sin x/x > 0 for
0 < x < π and sin x/x < 0 for π < x < 2π, it follows that Si(x) indeed has
its first maximum at x = π.
b The value at the first maximum is Si(π). Since Si(π) = 1.852 . . . and
π/2 = 1.570 . . ., the overshoot is 0.281 . . .. The jump of f at x = 0 is
π = 3.141 . . ., so the overshoot is 8.95 . . .%, so about 9%.
a The function f is continuous for t = (2k + 1)π (k ∈ Z) and it then4.28
converges to f(t), which is 2t/π for 0 ≤ | t | < π/2, 1 for π/2 ≤ t < π and −1
for −π < t ≤ −π/2. For t = (2k+1)π it converges to (f(t+)+f(t−))/2 = 0.
b Since f is odd we have an = 0 for all n. The bn can be found using an
integration by parts:
bn =
4
π2
Z π/2
0
t sin nt dt +
2
π
Z π
π/2
sin nt dt =
4
n2π2
sin(nπ/2) −
2(−1)n
nπ
.
Since sin(nπ/2) = 0 if n even and (−1)k
if n = 2k + 1, the Fourier series is
−
2
π
∞X
n=1
(−1)n
n
sin nt +
4
π2
∞X
n=0
(−1)n
(2n + 1)2
sin(2n + 1)t.
Substituting t = 0 and t = π it is easy to verify the fundamental theorem
for these values.
c We cannot differentiate the series; the resulting series is divergent be-
cause limn→∞(−1)n
cos nt = 0. Note that theorem 4.10 doesn’t apply since
f is not continuous.
d We can integrate the series since theorem 4.9 can be applied (note that
c0 = 0). If we put g(t) =
R t
−π
f(τ) dτ, then g is even, periodic with period
2π and given by (t2
/π)−(3π/4) for 0 ≤ t < π/2 and by t−π for π/2 ≤ t ≤ π.
Use table 1 to obtain the Fourier coefficients and then apply Parseval, that4.29
is, (4.13). Calculating the integral in Parseval’s identity will then give the
first result; choosing a = π/2 gives the second result.
a The Fourier series has been determined in the last example of section4.30
3.6. Since f is continuous (and piecewise smooth), the Fourier series con-
verges to f(t) for all t ∈ R:

f(t) =
2
π
−
4
π
∞X
n=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 Fourier
series; since f(π/2) = 1 and cos 2nt = (−1)n
for all n, the second result
follows.
c One should recognize the squares of the Fourier coefficients here. Hence
we have to apply Parseval’s identity (4.14), or the alternative form given
in exercise 4.10. This leads to
1
2π
Z π
−π
sin2
t dt =
4
π2
+
1
2π2
∞X
n=1
16
(4n2 − 1)2
.
Since
R π
−π
sin2
t dt = π, the result follows.
a Since f1 is odd it follows that4.31
(f1 ∗ f2)(−t) = −
1
T
Z T/2
−T/2
f1(t + τ)f2(τ) dτ.
Now change the variable from τ to −τ and use that f2 is odd, then it
follows that (f1 ∗ f2)(−t) = (f1 ∗ f2)(t).
b The convolution product equals
(f ∗ f)(t) =
1
2
Z 1
−1
τf(t − τ) dτ.
Since f is odd, part a implies that f ∗ f is even. It is also periodic with
period 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 and
0 ≤ t ≤ 1 we see that t − 1 ≤ t − τ ≤ t + 1. From 0 ≤ t ≤ 1 it follows
that −1 ≤ t − 1 ≤ 0, and so close to τ = 1 the function f(t − τ) is given
by t − τ. Since 1 ≤ t + 1 ≤ 2, the function f(t − τ) is given by t − τ − 2
close to τ = −1. Hence, we have to split the integral precisely at the point
where t − τ gets larger than 1, because precisely then the function changes
from t − τ to t − τ − 2. But t − τ ≥ 1 precisely when τ ≤ t − 1, and so we
have 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 result
is (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 Fourier
coefficients of (f ∗ f)(t), namely c2
0 = 0 and c2
n = −1/π2
n2
(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 < 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 differentiable. So theorem 4.10 implies that the differentiated
series represents the function (f ∗ f) (t) on [−1, 1], except at t = 0. At
t = 0 the differentiated series converges to ((f ∗f) (0+)+(f ∗f) (0−))/2 =
(1 − 1)/2 = 0.
f The zeroth Fourier coefficient of f ∗ f is given by

1
2
Z 1
−1
(f ∗ f)(t) dt =
Z 1
0
(−t2
/2 + t − 1/3) dt = 0.
This is in agreement with the result in part c since c2
0 = 0. Since this
coeﬃcient is 0, we can apply theorem 4.9. The function represented by the
integrated series is given by the (periodic) function
R t
−1
(f ∗ f)(τ) dτ. It is
also odd, since f is even and for 0 ≤ t ≤ 1 it equals
Z 0
−1
(−τ2
/2 − τ − 1/3) dτ +
Z t
0
(−τ2
/2 + τ − 1/3) dτ = −t(t − 1)(t − 2)/6.

For a stable LTC-system the real parts of the zeroes of the characteristic5.1
polynomial are negative. Fundamental solutions of the homogeneous equa-
tions 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 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+1
odd 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 cannot5.3
be described by a differential equation (5.3).
b Since H(nω0) = H(n) = 0 for | n | ≥ 4 (because 4 > π), we only need to
consider the Fourier coefficients of y with | n | ≤ 3. From Parseval it then
follows that P =
P3
n=−3 | yn |2
with yn as calculated in exercise 5.2. This
sum is equal to P = 1
4
+ 20
9π2 .
Note that u has period π and that the integral to be calculated is thus the5.4
zeroth Fourier coefficient of y. Since y0 = H(0ω0)u0 = H(0)u0 and H(0) =
−1 (see example 5.6 for H(ω)), it follows that y0 = −u0 = − 1
π
R π
0
u(t) dt =
− 2
π
.
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 are
ω = ±1.
b Write u(t) = e−4it
/4 − e−it
/2i + 1/2 + eit
/2i + e4it
/4. It thus follows
that the Fourier coefficients unequal to 0 are given by u−4 = u4 = 1/4,
u−1 = −1/2i, u1 = 1/2i and u0 = 1/2. Since yn = H(nω0)un = H(n)un
and H(1) = H(−1) = 0 we thus obtain that
y(t) = y−4e−4it
+ y−1e−it
+ y0 + y1eit
+ y4e4it
=
15
12 + 8i
·
1
4
e−4it
+
1
4
·
1
2
+
15
12 − 8i
·
1
4
e4it
.
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 + ω2
0
.
18

Since | ω0 | is not an integer, there are no homogeneous solutions having
period 2π, while u does have period 2π. There is thus a uniquely determined
periodic solution y corresponding to u. Since u(t) = πqπ,2π(t) the Fourier
coefficients 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-5.7
tion (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 coefficients
An =
2
L
Z L/2
0
x sin(nπx/L) dx +
2
L
Z L
L/2
(L − x) sin(nπx/L) dx,
which can be calculated using an integration by parts. The result is: An =
(4L/n2
π2
) sin(nπ/2) (which is 0 for n even). We thus obtain the (formal)
solution
u(x, t) =
4L
π2
∞X
n=0
(−1)n
(2n + 1)2
e−(2n+1)2
π2
kt/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 characteristic
equation s2
− c = 0 has two distinct roots ±s1. The general solution is
then X(x) = αes1x
+ βe−s1x
, so X (x) = s1αes1x
− s1βe−s1x
. The first
boundary condition X (0) = 0 gives s1(α − β) = 0, so β = α. Next
we obtain from the second boundary condition X(L) = 0 the equation
α(es1L
+ e−s1L
) = 0. For α = 0 we get the trivial solution. So we must
have es1L
+ e−s1L
= 0, implying that e2s1L
= −1. From this it follows
that s1 = i(2n + 1)π/2L. This gives us eigenfunctions Xn(x) = cos((2n +
1)πx/2L) (n = 0, 1, 2, 3, . . .). Since Tn(t) remains as in the textbook (for
other parameters), we have thus found the fundamental solutions
un(x, t) = e−(2n+1)2
π2
kt/4L2
cos((2n + 1)πx/2L).
Superposition gives
u(x, t) =
∞X
n=0
Ane−(2n+1)2
π2
kt/4L2
cos((2n + 1)πx/2L).
Substituting t = 0 (and using the remaining initial condition) leads to
u(x, 0) =
∞X
n=0
An cos((2n + 1)πx/2L) = 7 cos(5πx/2L).

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) =
7e−25π2
kt/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 as
in 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 textbook
and lead to the solution
u(x, t) =
∞X
n=1
An cos(nπat/L) sin(nπx/L).
Substituting t = 0 (and using the remaining initial condition) gives
u(x, 0) =
∞X
n=1
An sin(nπx/L) = 0.05 sin(4πx/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(4πat/L) sin(4πx/L).
Separation of variables leads to X (x) − cX(x) = 0 for 0 < x < π, X (0) =5.15
X (π) = 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) = αes1x
+ βe−s1x
, so X (x) = s1αes1x
− s1βe−s1x
. The
boundary condition X (0) = 0 gives s1(α − β) = 0, so β = α. From
the boundary condition X (π) = 0 we obtain s1α(es1π
− e−s1π
) = 0. For
α = 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 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 < π.

We have A0 = (2/π)
R π
0
kx dx = kπ and An = (2/π)
R π
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
π for n odd. The solution is thus
u(x, t) =
kπ
2
−
4k
π
∞X
n=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 response5.16
y(t) to a real signal u(t) is real: since u−n = un we also have y−n = yn.
b Since we can write sin ω0t = (eiω0t
− e−iω0t
)/2i, the response is equal
to (H(ω0)eiω0t
− H(−ω0)e−iω0t
)/2i, which is ((1 − e−2iω0
)2
eiω0t
− (1 −
e2iω0
)2
e−iω0t
)/2i. This can be rewritten as sin ω0t − 2 sin(ω0(t − 2)) +
sin(ω0(t − 4)).
c A signal with period 1 has Fourier series of the form
P∞
n=−∞ une2πint
.
The response 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) = 05.18
and has zeroes s = −1 and s = ±2i. The zeroes on the imaginary axis
correspond to periodic eigenfrequencies with period π and so the response
to 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 power
is given by
P =
3
2π
Z 2π/3
0
| y(t) |2
dt =
∞X
n=−∞
| yn |2
=
∞X
n=−∞
| H(nω0)un |2
.
We have that
H(ω) =
1 + iω
4 − ω2 + iω(4 − ω2)
.
Now use that only u3 = u−3 = 1
2
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) =5.19
u0 in (5.10). The solution is thus given by (5.16), where now the An are the
Fourier 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/nπ
for n odd. This gives
u(x, t) =
4u0
π
∞X
n=0
1
(2n + 1)2
e−(2n+1)2
π2
kt/L2
sin((2n + 1)πx/L).
Substituting x = L/2 in the x-derivative and using the fact that cos((2n +
1)π/2) = 0 for all n leads to ux(L/2, t) = 0.
a As in the previous exercise the solution is given by (5.16). The An are5.20
given by (2/L)
R L/2
0
a sin(nπx/L) dx = 2a(1 − cos(nπ/2))/nπ, which gives
the (formal) solution
u(x, t) =
2a
π
∞X
n=1
1
n
(1 − cos(nπ/2))e−n2
π2
kt/L2
sin(nπx/L).

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 ﬁrst two terms of the series to obtain the
equation u(20, t) ≈ 63.662e−0.0000308t
= 36 (terms of the series tend to 0
very rapidly, so two terms suﬃce). We then obtain 18509 seconds, which
is approximately 5 hours.

We have to calculate (the improper integral)
R ∞
−∞
e−iωt
dt. Proceed as in6.1
eaxample 6.1, but we now have to determine limB→∞ e−iωB
. This 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−αR
e−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 = 1
2
ln(a2
+ ω2
), so this improper integral,
which is the Fourier integral for t = 0, does not exits (limA→∞ ln(a2
+ A2
)
does not exist e.g.).
c We have lima→0 g(t) = lima→0 (t)e−at
= (t), while for ω = 0 we have
that lima→0 G(ω) = −i/ω.
To calculate the spectrum we split the integral at t = 0:6.4
G(ω) =
Z 1
0
te−iωt
dt −
Z 0
−1
te−iωt
dt.
Changing from the variable t to −t in the second integral we obtain that
G(ω) = 2
R 1
0
t cos ωt dt, which can be calculated for ω = 0 using an integ-
ration by parts. The result is:
G(ω) =
2 sin ω
ω
+
2(cos ω − 1)
ω2
.
For ω = 0 we have that G(0) = 2
R 1
0
t dt = 1. Since limω→0 sin ω/ω = 1 and
limω→0(cos ω − 1)/ω2
= −1
2
(use e.g. De l’Hˆopital’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) = 2i
cos(aω/2) − 1
ω
for ω = 0, F(0) = 0.
b Using Taylor or De l’Hˆopital 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-6.9
lows 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 and
apply part a (and use the fact that sin(πω ± π) = − sin(ωπ)), then
F(ω) =
2i sin(πω)
ω2 − 1
.
23

Use section 6.3.3 (or exercise 6.2) and the modulation theorem 6.17, and6.10
write 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 6.9a
that
(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 ωt dt.
a We have F(−ω) = F(ω) and F(ω) is even, so F(ω) = F(ω), and thus6.13
F(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 techniques6.14
as in example 6.3.3 (or use (6.20) and twice an integration by parts):
F(ω) =
−2iω
1 + 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 in
example 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) 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 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-6.19
quency 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
)−1
6.20
e.g. is not bounded.
Use the fact that limx→∞ xa
e−x
= 0 for all a ∈ R and change to the variable6.21
x = at2
in tk
/eat2
(separate the cases t ≥ 0 and t < 0). Then part a follows
and, hence, part b also follows since we have a finite sum of these terms.
Apply the product rule repeatedly to get an expression in terms of the de-6.22
rivatives of f and g (this involves the binomial coefficients 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.

We have that ( ∗ )(t) =
R ∞
0
(t − τ) dτ. Now treat the cases t > 0 and6.23
t ≤ 0 separately, then it follows that ( ∗ )(t) = (t)t. (If t ≤ 0, then
t − τ < 0 for τ > 0 and so (t − τ) = 0; if t > 0 then (t − τ) = 0 for
τ > t and the integral
R t
0
1 dτ = t remains.) Since (t)t is not absolutely
integrable, 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 < 0 this is 0. For t ≥ 0 it equals
R t
0
f(τ)g(t − τ) dτ.
a We use the definition of convolution and then split the integral at τ = 0:6.26
(e−| v |
∗ e−| v |
)(t) =
Z ∞
0
e−τ
e−| t−τ |
dτ +
Z 0
−∞
eτ
e−| t−τ |
dτ.
First we take t ≥ 0. Then − | t − τ | = τ −t for τ < 0. Furthermore we have
for τ > t that − | t − τ | = t − τ and for 0 ≤ τ < t that − | t − τ | = τ − t.
Hence,
(e−| v |
∗ e−| v |
)(t) =
Z t
0
e−t
dτ +
Z ∞
t
et−2τ
dτ +
Z 0
−∞
e2τ−t
dτ.
A straightforward calculation of these integrals gives (1 + t)e−t
.
Next we take t < 0. Then − | t − τ | = t−τ for τ > 0. Furthermore we have
for τ < t that − | t − τ | = τ − t and for t ≤ τ < 0 that − | t − τ | = t − τ.
Hence,
(e−| v |
∗ e−| v |
)(t) =
Z ∞
0
et−2τ
dτ +
Z 0
t
et
dτ +
Z t
−∞
e2τ−t
dτ.
A straightforward calculation of these integrals gives (1 − t)et
.
b Use the result from section 6.3.3 and the convolution theorem to obtain
the spectrum (2(1 + ω2
)−1
)2
= 4/(1 + ω2
)2
.
c Since (1 + | t |)e−| t |
= e−| t |
+ | t | e−| t |
and the spectrum of e−| t |
is
2(1 + ω2
)−1
, we only need to determine the spectrum of f(t) = | t | e−| t |
.
But f(t) = tg(t) with g(t) the function from exercise 6.14, whose spec-
trum we’ve already determined: G(ω) = −2iω(1 + ω2
)−1
. Apply theorem
6.8 (differentiation rule in the frequency domain): the spectrum of f(t)
is −G (ω)/i. Calculating this and taking the results together we obtain
4/(1 + ω2
)2
, in agreement with part b.
a From the differentiation rule in the frequency domain we obtain that6.28
the spectrum of tg(t) is iG (ω) = −iω
√
2πe−ω2
/2
. Since (Ftg(t))(0) = 0,
we may apply the integration rule to obtain that F1(ω) = −
√
2πe−ω2
/2
.
b Apply the differentiation rule in the frequency domain with n = 2, then
F2(ω) =
√
2π(1 − ω2
)e−ω2
/2
.
c Since f3(t) = f2(t − 1), it follows from the shift property that F3(ω) =
e−iω
F2(ω).
d From part a and exercise 6.9 it follows that F4(ω) = (−
√
2πe−(ω−4)2
/2
+
√
2πe−(ω+4)2
/2
)/2i.
e Use the scaling property from table 4 with c = 4, then F5(ω) =
G(ω/4)/4.
b Since p1(τ) = 0 for | τ | > 1
2
and 1 for | τ | < 1
2
, we have6.29
(p1 ∗ p3)(t) =
Z 1/2
−1/2
p3(t − τ) dτ.

Here p3(t − τ) = 0 only if t − 3/2 ≤ τ ≤ t + 3/2. Moreover, we have
that −1/2 ≤ τ ≤ 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
(p1 ∗ p3)(t) = 0; if −1 ≤ t ≤ 1, then (p1 ∗ p3)(t) =
R 1/2
−1/2
1 dτ = 1; if
1 < t ≤ 2, then (p1 ∗ p3)(t) =
R 1/2
t−3/2
1 dτ = 2 − t; ﬁnally, if −2 ≤ t < −1,
then (p1 ∗ p3)(t) =
R t+3/2
−1/2
1 dτ = 2 + t.
c Apply the convolution theorem to T(t) = (p1 ∗p3)(t), then the spectrum
of T(t) follows: 4 sin(ω/2) sin(3ω/2)/ω2
.

From the spectra calculated in exerices 6.2 to 6.5 it follows immediately7.1
that the limits for ω → ±∞ 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 limω→±∞ 1/(a + iω) = 0.
Use table 3 with a = 2A and substitute ω = s − t.7.2
Take C > 0, then it follows by ﬁrst changing from the variable Au to v and7.3
then applying (7.3) that
lim
A→∞
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+7.4
ω2
). The limit of A → ∞ of the integrals over [−A, A] of these parts gives
limA→∞ 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). The7.6
function 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 fundamental
theorem 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 it
is 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 the
right-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 and
that 2 sin πω sin ωt = cos(π − t)ω − cos(π + t)ω, then
f(t) =
1
π
Z ∞
0
cos(π − t)ω − cos(π + t)ω
1 − ω2
dω.
a In exercise 6.15b it was shown that Fs(ω) = (1 − cos aω)/ω. This7.8
exercise used the odd extension to R. So f(t) is odd and using (7.12) we
thus obtain
2
π
Z ∞
0
1 − cos aω
ω
sin ωt dω =
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 aω
ω
sin ωt dω.
27

b At t = a the function is discontinuous, so we have convergence to
1
2
(f(a+) + f(a−)) = 1
2
.
If we take g(t) = 0 in theorem 7.4, then G(ω) = 0 and so we get the7.10
statement: if F(ω) = 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(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 spectrum
of (f − g)(t); but (F − G)(ω) = F(ω) − G(ω) = 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(aω/2)/ω (table 3). From duality it then7.11
follows 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 can
apply duality since the Fourier integral exists as improper integral; this is
exercise 7.5b).
The spectrum of qa(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)qa(ω). (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(ω) is even and continuous
and 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 duality7.14
cannot 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.7.16
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 πe−| ω |
, the result follows:
πe−iω
e−| ω |
.
b Here we have a shift from t to t − 3; from the spectrum of sin 2πt/t the
result follows: πe−3iω
p4π(ω) with value 1
2
at ω = ±2π.
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 complete the square and note that 1/(3 + (t − 2)2
)
has spectrum F(ω) = πe−2iω
e−
√
3| ω |
/
√
3. Hence, the result is now (F(ω −
4) − F(ω + 4))/2i.
d We use that 3πq6(ω) is the spectrum of sin2
(3t)/t2
and apply a shift in
time from t to t − 1, then the result is 3πe−iω
q6(ω).
Again, this is an important exercise: it teaches to recognize useful proper-7.17
ties for the inverse transform.
a We immediately use table 3 to obtain that 1/(4 + ω2
) is the spectrum
of 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) =
2 cos(ω0t) sin(at)/(πt).
c As in part b it follows that f(t) = 3e9it
/(π(t2
+ 9)).
From the convolution theorem it follows that F(Pa ∗ Pb)(ω) = (FPa)(ω) ·7.19

(FPb)(ω) = e−(a+b)| ω |
, where we also used table 3. But also (FPa+b)(ω) =
e−(a+b)| ω |
, and since F is one-to-one (theorem 7.4) it then follows that
Pa+b = Pa ∗ Pb.
a Use the result of exercise 6.14 (G(ω) = −2iω/(1+ω2
)), the fundamental7.21
theorem 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
2
e−t
.
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.
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 are
2a/(a2
+ ω2
) and 2b/(b2
+ ω2
) (table 3).
a Since sin4
t/t4
is the square of sin2
t/t2
and (F sin2
t/t2
)(ω) = πq2(ω)7.23
(table 3), it follows from the convolution theorem in the frequency domain
that (F sin4
t/t4
)(ω) = (π/2)(q2 ∗ q2)(ω).
b The integral
R ∞
−∞
sin4
t/t4
dt is the Fourier transform of sin4
x/x4
cal-
culated at ω = 0, hence
R ∞
−∞
sin4
x/x4
dx = (π/2)(q2 ∗ q2)(0). Using that
q2 is an even function we obtain that
(q2 ∗ q2)(0) =
Z ∞
−∞
q2(t)q2(−t) dt = 2
Z 2
0
(1 − t/2)2
dt.
This integral equals 4/3 and so
R ∞
−∞
sin4
x/x4
dx = 2π/3.
From table 3 we know that e−| t |
/2 ↔ 1/(1 + ω2
). By the convolution7.24
theorem we then know that the spectrum of f(t) = (e−| v |
/2∗e−| v |
/2)(t) is
1/(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 that
the 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−ω2
/4a
(see table 3), it follows
from (7.23) with T = 1 that
∞X
n=−∞
e−an2
=
p
π/a
∞X
n=−∞
e−π2
n2
/a
.
Replacing a by πx the result follows.
Take f(t) = a/(a2
+t2
), then F(ω) = πe−a| ω |
(see table 3); we can then use7.27∗
(7.22) with T = 1 (in example 7.8 the conditions were veriﬁed) to obtain

∞X
n=−∞
a
a2 + (t + n)2
= π 1 +
∞X
n=1
e−2πn(a+it)
+
∞X
n=1
e−2πn(a−it)
!
.
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 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 the
sum of an infinite geometric series (example 2.16), then writing the result
with a common denominator, and finally multiplying everything out and
simplifying, it follows that
a
π
∞X
n=−∞
1
a2 + (t + n)2
=
1 − e−4πa
1 + e−4πa − e−2πa(e2πit + e−2πit)
.
Multiplying numerator and denominator by e2πa
the result follows.
a To determine the spectrum we write sin t = (eit
− e−it
)/2 and calculate7.28
the integral defining F(ω) in a direct way:
F(ω) =
1
2i
„Z π
0
ei(1−ω)t
dt −
Z π
0
e−i(1+ω)t
dt
«
.
Writing the result with a common denominator and using the fact that
eπi
= e−πi
= −1 gives F(ω) = (1 + e−iωπ
)/(1 − ω2
). From theorem 6.10
we know that F(ω) is continuous, so we do not have to calculate F(ω) at
the exceptional points ω = ±1.
b Apply the fundamental theorem, so (7.9), noting that f(t) is continuous
on R. We then obtain
f(t) =
1
2π
Z ∞
−∞
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 π
0
sin2
t dt = π/2,
then it follows that
1
2π
Z ∞
−∞
| F(ω) |2
dω =
π
2
.
Since F(ω) can be rewritten as 2e−iωπ/2
cos(ωπ/2)/(1 − ω2
) and we have
that
˛
˛
˛ e−iωπ/2
˛
˛
˛ = 1, it follows that | F(ω) |2
= 4 cos2
(ωπ/2)/(1−ω2
)2
. This
integrand being even, the result follows.
a We know from table 3 that p2a(t) ↔ 2 sin aω/ω and e−| t |
↔ 2/(ω2
+1).7.29
From the convolution theorem it then follows that p2a(v)∗e−| v |
↔ 4f(ω) =
G(ω).
b We now determine g explicitly by calculating the convolution product
(use the definition of p2a):
(p2a(v) ∗ e−| v |
)(t) =
Z a
−a
e−| t−τ |
dτ =
Z t+a
t−a
e−| u |
du

where we changed to the variable u = t − τ. Now if −a ≤ t ≤ a, then
t − a ≤ 0 ≤ t + a and so
(p2a(v) ∗ e−| v |
)(t) =
Z 0
t−a
eu
du +
Z t+a
0
e−u
du = 2 − 2e−a
cosh t.
If t > a, then t − a > 0 and so
(p2a(v) ∗ e−| v |
)(t) =
Z t+a
t−a
e−u
du = 2e−t
sinh a.
Finally, if t < −a, then t + a < 0 and so
(p2a(v) ∗ e−| v |
)(t) =
Z t+a
t−a
eu
du = 2et
sinh a.
c The function g from part b is continuous at t = a since limt↓a g(t) =
2e−a
sinh a = e−a
(ea
− e−a
) = 1 − e−2a
and g(a) = limt↑a g(t) = 2 −
2e−a
cosh a = 2−e−a
(ea
+e−a
) = 1−e−2a
. In the same way it follows 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 ex-
ists as improper Riemann integral since G(ω) is even absolutely integrable:
| G(ω) | ≤ 4/
˛
˛ ω(1 + ω2
)
˛
˛. We can now apply the duality rule (theorem
7.5) and it then follows that G(−t) ↔ 2πg(ω), so f(−t) ↔ πg(ω)/2. Since
f(−t) = f(t) we thus see that F(ω) = π − πe−a
cosh ω for | ω | ≤ a and
F(ω) = πe−| ω |
sinh a for | ω | > a.

b For t = 0 we have that lima↓0 Pa(t) = 0, while for t = 0 we have8.1
that lima↓0 Pa(t) = ∞. Since
R ∞
−∞
Pa(t) dt = 1, we see that Pa(t) ﬁts the
description of the delta function.
c From table 3 it follows that Pa(t) ↔ e−a| ω |
and lima↓0 e−a| ω |
= 1.
Combining this with part b shows that it is reasonable to expect that the
spectrum of δ(t) is 1.
a Since φ(a) ∈ C for all φ ∈ S, it follows from (8.10) that δ(t − a) is a8.2
mapping from S to C. For c ∈ C and φ ∈ S we have that
δ(t − a), cφ = (cφ)(a) = c δ(t − a), φ ,
and for φ1, φ2 ∈ S we have
δ(t − a), φ1 + φ2 = (φ1 + φ2)(a) = δ(t − a), φ1 + δ(t − a), φ2 .
So δ(t − a) is a linear mapping from S to C, hence a distribution.
b Taking the limit inside the integral in (8.1) gives
Z ∞
−∞
„
1
2π
lim
a→∞
2 sin aω
ω
«
f(t − ω) dω = f(t)
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 ∞
−∞
δ(ω)f(a − ω) dω = f(a)
and by changing from ω to a − t we then obtain
Z ∞
−∞
δ(a − t)f(t) dt = f(a).
Using δ(a−t) = δ(t−a), which by (8.3) is reasonable to expect (see section
8.4 for a proof), this indeed leads to (8.11).
Since 1, φ =
R ∞
−∞
φ(t) dt ∈ C for all φ ∈ S, it follows that 1 is a mapping8.4
from S to C. The linearity of this mapping follows from the linearity 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 mapping
from S to C, hence a distribution.
For φ ∈ S there exists a constant M > 0 such that (e.g.) (1+t2
) | φ(t) | ≤ M8.5
for all t ∈ R. Hence,
˛
˛
˛
˛
Z ∞
0
φ(t) dt
˛
˛
˛
˛ ≤
Z ∞
0
| φ(t) | dt ≤ M
Z ∞
0
1
1 + t2
dt < ∞
32

(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 in
exercise 8.4 (linearity of integration).
In example 8.4 it was already motivated why the integral
R ∞
−∞
| t | φ(t) dt8.7
exists: 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 < ∞.
So | t | , φ exists and one can now show that | t | is indeed a distribution
precisely as in exercise 8.4 (linearity of integration).
a For the integral over [−1, 1] we have8.9
Z 1
−1
| t |−1/2
dt = 2
Z 1
0
t−1/2
dt = [4t1/2
]1
0 = 4,
hence, | t |−1/2
is integrable over [−1, 1]. Since
R ∞
0
t−1/2
dt = 2 limR→∞
√
R
does not exist, | t |−1/2
is not integrable over R.
b We first show that
R ∞
−∞
| t |−1/2
φ(t) dt exists for φ ∈ S. To do so, we
split the integral in an integral over [−1, 1] and over | t | ≥ 1. For the first
integral we note that | φ(t) | ≤ M1 for some constant M1 > 0. From part a
we then get
˛
˛
˛
˛
Z 1
−1
| t |−1/2
φ(t) dt
˛
˛
˛
˛ ≤
Z 1
−1
| t |−1/2
| φ(t) | dt ≤ M1
Z 1
−1
| t |−1/2
dt < ∞.
For the second integral we use that | t |−1/2
≤ 1 for | t | ≥ 1. Hence,
˛
˛
˛
˛
˛
Z
| t |≥1
| t |−1/2
φ(t) dt
˛
˛
˛
˛
˛
≤
Z
| t |≥1
| φ(t) | dt ≤
Z ∞
−∞
| φ(t) | dt.
In example 8.1 it has been shown that the latter integral exists. This shows
that
D
| t |−1/2
, φ
E
exists and one can now show that | t |−1/2
is indeed a
distribution (linearity of integration; see e.g. exercise 8.4).
a For φ ∈ S there exists a constant M > 0 such that (e.g.) (1 +8.10
t2
) | tφ(t) | ≤ M for all t ∈ R. Hence,
˛
˛
˛
˛
Z ∞
−∞
tφ(t) dt
˛
˛
˛
˛ ≤
Z ∞
−∞
| tφ(t) | dt ≤ M
Z ∞
−∞
1
1 + t2
dt < ∞.
As in exercise 8.3 this shows that t defines a distribution.
a From the linearity for distributions it follows immediately that the com-8.12
plex number 2φ(0) + i
√
3φ (0) + (1 + i)
R ∞
0
((φ(t) − φ(−t))dt is assigned.
b This defines a distribution if 1, t and t2
are distributions. The first
one is known from example 8.1, the other two from exercise 8.10. From
definition 8.15 it thus follows that f(t) defines a distribution as well.
a From (8.17) it follows that the complex number −φ(3)
(0) is assigned.8.14
b This number −φ(3)
(0) is meaningfull for all functions that are 3 times
continuously differentiable.
a First apply example 8.3, then definition 8.4, and finally integration by8.15
parts, then

˙
(sgn t) , φ
¸
=
Z ∞
0
(φ (−t) − φ (t)) dt = [φ(t)]0
−∞ − [φ(t)]∞
0 = 2φ(0),
hence, (sgn t) = 2δ(t).
b Since sgn t = 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 and
that (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-8.16
side 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) = δ(t) − (t) sin t.
a The function pa has a jump of magnitude 1 at t = −a/2 and of mag-8.17
nitude −1 at t = a/2. Outside t = 0 the ordinary derivative is 0, so it
follows 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 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.8.18
b The function is differentiable 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 + 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),8.22
then
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 , which
shows that p(t)δ(t − a) = p(a)δ(t − a).
a The definition becomes: f(t)δ (t), φ = δ (t), fφ . The product fφ of8.23
two continuously differentiable functions is continuously differentiable, so
this definition is correct and it gives a mapping from S to C. The linearity
follows immediately from the linearity of δ (t), so f(t)δ (t) is a distribution.
b According to part a we have f(t)δ (t), φ = −(fφ) (0), where we also
applied δ (t) to fφ. Now apply the product rule for differentiation and write
the result as −f (0) δ(t), φ + f(0) δ (t), φ = f(0)δ (t) − f (0)δ(t), φ .
c If f(t) = t, then f(0) = 0 and f (0) = 1, so tδ (t) = −δ(t); if f(t) = t2
then f(0) = 0 and f (0) = 0, so t2
δ (t) = 0.
First apply definition 8.6 and then the definition of pv(1/t) from example8.25
8.5 to obtain
t · pv(1/t), φ = lim
α↓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 be
equal to
R ∞
−∞
φ(t) dt. Hence, t · pv(1/t) = 1.
Let T be an even distribution, then T(−t) = T(t) (definiton 8.8), so8.26

T(t), φ(t) = T(t), φ(−t) for all φ ∈ S, where we used definition 8.7.
Similarly for odd T.
a From the definition of sgn t in example 8.3 it follows that sgn t, φ(t) =8.27
− sgn t, φ(−t) for all φ ∈ S. This shows that sgn t is odd according to
exercise 8.26. Similarly for pv(1/t) (change from t to −t in the integrals
defining pv(1/t)).
b From the definition of | t | in example 8.4 it follows that | t | , φ(t) =
| t | , φ(−t) for all φ ∈ S (change from t to −t in the integral defining | t |).
This shows that | t | is even according to exercise 8.26.
a Applying (8.12) to f(t) gives8.29
Tf , φ =
Z 0
−∞
2tφ(t) dt +
Z ∞
0
t2
φ(t) dt
and in e.g. exercises 8.10, 8.12b and 8.18a we have seen that such integrals
are well-defined for φ ∈ 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 defines a distribution Tf .
b Apply the jump formula (8.21): outside t = 0 the function f is continu-
ously differentiable with derivative f (t) = 2t for t > 0 and f (t) = 2 for
t < 0. Note that f again defines 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 differentiable out-
side t = 0 and f (t) = 2 for t > 0 and f (t) = 0 for t < 0. Let Tf
be the distribution defined by f . At t = 0 the function f has a 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−))δ(t) = Tf − 2δ(t).
The second derivative of f considered as distribution is the same as the
second 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-8.30
tions, the product f(t)δ (t) can also be defined for all twice continuously
differentiable functions f(t) by f(t)δ (t), φ(t) = δ (t), f(t)φ(t) . This is
because it follows from the product rule that the product f(t)φ(t) is again
twice 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 obtain
that 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) = 0
and f (0) = 2, then t2
δ (t) = 2δ(t). Next apply b to f(t) = t3
and use
that f(0) = f (0) = f (0) = 0, then it follows that t3
δ (t) = 0.
d According to definition 8.7 we have that
˙
δ (at), φ(t)
¸
= | a |−1 ˙
δ (t), φ(a−1
t)
¸
.
Now put ψ(t) = φ(a−1
t), then the right-hand side equals | a |−1
ψ (0).
Next we use the chain rule twice to obtain that ψ (0) = a−2
φ (0). Hence
δ (at), φ(t) = | a |−1
a−2
δ , φ .

Let φ ∈ S. From theorem 6.12 it follows that the spectrum Φ belongs9.1
to S. Since T is a distribution, we then have that T, Φ ∈ C, and so
FT, φ = T, Φ ∈ C as well. So FT is a mapping from S to C. The
linearity of FT follows from the linearity of T and F; we will only give
the necessary steps for FT, cφ , since the rule for FT, φ1 + φ2 follows
similarly.
FT, cφ = T, F(cφ) = T, cΦ = c T, Φ = c FT, φ .
a Use table 5 to obtain that δ(t − 4) ↔ e−4iω
.9.3
b 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 spectrum
4π(δ(ω − 2) + δ(ω + 2)) + 2πsgn ω.
a From example 9.1 (or table 5) we obtain the result e−5it
/2π.9.4
b See example 9.2: 2 cos 2t.
c The spectrum of pv(1/t) is −πisgn ω (table 5). Note that 2 cos ω =
eiω
+ e−iω
and that the spectrum of δ(t − a) is e−iaω
(table 5). Hence the
answer is iπ−1
pv(1/t) + δ(t − 1) + δ(t + 1).
Let T be an even distribution with spectrum U. We have to show that9.5
U(−ω) = U(ω), so U, φ(t) = U, φ(−t) for all φ ∈ S (see exercise
8.26). But U, φ(−t) = T, Fφ(−t) and from table 4 we know that
(Fφ(−t))(ω) = Φ(−ω) if Φ is the spectrum of φ. Since T is even, we
have that T, Φ(−ω) = T, Φ(ω) . From these observations it follows that
U, φ(−t) = T, Φ(ω) = U, φ(t) , which shows that U is even. Similarly
for odd T.
It is obvious that (t) = (1 + sgn t)/2 by looking at the cases t > 0 and9.7
t < 0. Since 2πδ(ω) is the spectrum of 1 and −2ipv(1/ω) is the spectrum
of (t), it follows that (t) has spectrum πδ(ω) − ipv(1/ω).
a Let φ ∈ S have spectrum Φ. From definition 9.1 and the action of δ it9.8
follows that Fδ , φ = δ , Φ = δ, Φ = Φ (0). From the differentiation
rule in the frequency domain (table 4) with k = 2 we see that Φ (ω) =
F((−it)2
φ(t))(ω), and hence δ ↔ −ω2
is proven as follows:
˙
Fδ , φ
¸
= F(−t2
φ(t))(0) = −
Z ∞
−∞
t2
φ(t) dt = −
Z ∞
−∞
ω2
φ(ω) dω,
so Fδ , φ =
˙
−ω2
, φ
¸
for all φ ∈ S, proving the required result.
Parts b and c can be proven using similar steps.
a Subsequently apply definitions 9.1 and 8.7: FT(at), φ = T(at), Φ =9.9
| a |−1 ˙
T, Φ(a−1
ω)
¸
(φ ∈ S having spectrum Φ). From table 4 we see that
Φ(a−1
ω) = | a | (Fφ(at))(ω), so it follows that FT(at), φ = T, Fφ(at) =
U, φ(at) , where we again used definition 9.1 in the final step. Now again
apply definition 8.7, then FT(at), φ = | a |−1 ˙
U(a−1
ω), φ
¸
, which proves
T(at) ↔ | a |−1
U(a−1
ω).
36

b We have that δ(4t+3) is the distribution δ(t+3) scaled by 4. According
to the shift rule in the time domain (see table 6) it follows from δ(t) ↔ 1
that δ(t+3) ↔ e3iω
. From part a it then follows that δ(4t+3) ↔ 4−1
U(ω/4)
with U(ω) = e3iω
. Hence, δ(4t + 3) ↔ 4−1
e3iω/4
. (This can also be solved
by 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 get
tδ (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 exist9.12
distributions S = 0 such that ωS = 0 (e.g. δ(ω)).
The linearity follows as in definition 8.6. The main point is that one has9.13
to show that eiat
φ(t) ∈ S whenever φ ∈ S. So we have to show that for
any m, n ∈ Z+
there exists an M > 0 such that
˛
˛
˛ tn
(eiat
φ(t))(m)
˛
˛
˛ < M.
From the product rule for differentiation it follows that (eiat
φ(t))(m)
is a
sum of terms of the form ceiat
φ(k)
(t) (k ∈ Z+
). It is now sufficient to show
that
˛
˛
˛ tn
eiat
φ(k)
(t)
˛
˛
˛ < M for some M > 0 and all k, n ∈ Z+
. But since
˛
˛ eiat
˛
˛ = 1 this means that we have to show that
˛
˛
˛ tn
φ(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 eiat
T (see exercise 9.13) it follows9.15
that
˙
Feiat
T, φ
¸
=
˙
eiat
T, Φ
¸
=
˙
T, eiat
Φ
¸
(φ ∈ S having spectrum Φ). Ac-
cording to the shift property in the frequency domain (table 4) we have that
eiat
Φ(t) = F(φ(ω+a))(t) (note that for convenience we’ve interchanged the
role of the variables ω and t). Hence,
˙
Feiat
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 eiat
T ↔ U(ω − a).
a Use table 5 for (t) and apply a shift in the time domain, then it follows9.16
that (t − 1) ↔ e−iω
(πδ(ω) − ipv(1/ω)).
b Use table 5 for (t) and apply a shift in the frequency domain, then it
follows 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 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
2
(πδ(ω −a)−ipv(1/(ω −
a)) + πδ(ω + a) − ipv(1/(ω + a))).
d Use that 1 ↔ 2πδ(ω) and δ (t) ↔ iω (table 5), so 3i ↔ 6iπδ(ω) and
(apply a shift) δ (t − 4) ↔ e−4iω
iω; 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
↔ 2πi3
δ(3)
(ω) (table 5), so the result is
2π2
δ(3)
(ω) + πδ(ω) − ipv(1/ω).
a Use table 5 for the sign function and apply a shift: 1
2
ieit
sgn t.9.17
b Write sin t as a combination of exponentials and apply a shift to 1
2
isgn t,
then we obtain the result 1
4
(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:
1
2
iπ−1
pv(1/t) + 1
2
δ(t).
d Apply the scaling property (table 6) to 1 ↔ 2πδ(ω) to obtain 1 ↔
6πδ(3ω). Next we apply a shift in the frequency domain (table 6), which

results in e2it/3
↔ 6πδ(3ω − 2). Using the differentiation rule in the fre-
quency domain (table 6) we obtain from 1/2π ↔ δ(ω) that (−it)2
/2π ↔
δ (ω). From linearity it then follows that (3−1
e2it/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−ibω
e−iaω
= e−i(a+b)ω
.
a Use table 5: δ(t − 3) ↔ e−3iω
.9.24
b Since δ(t + 4) ↔ e4iω
(as in part a) and cos t = (eit
+ e−it
)/2 we apply
a 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 differentiation
rule 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 obtained
in 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 a
shift in the time domain that δ(7(t − 1/7)) ↔ e−iω/7
7. Finally apply the
differentiation rule in the time domain to obtain the result: (δ(7t − 1)) ↔
iωe−iω/7
7.
f This is a convergent Fourier series and so we can determine the spectrum
term-by-term. Since 1 ↔ 2πδ(ω) and e(2k+1)it
↔ 2πδ(ω − (2k + 1)) we
obtain the following result:
π2
δ(ω) − 4
∞X
k=−∞
(2k + 1)−2
δ(ω − (2k + 1)).
a From table 5 we know that eit
↔ 2πδ(ω − 1) and similarly for e−it
.9.25
Hence, iπ−1
sin t ↔ δ(ω − 1) − δ(ω + 1).
b Apply the differentiation rule in the time domain (with n = 2) to δ(t) ↔
1, then −δ (t) ↔ ω2
.
c From table 5 we obtain that δ(t + 1
2
)/4 ↔ eiω/2
/4.
d From table 5 (and linearity) we obtain that (δ(t + 1) − δ(t − 1))/2i ↔
(eiω
−e−iω
)/2i, which is sin ω. Now apply differentiation in the time domain
(with n = 3), then we obtain that (δ(3)
(t + 1) − δ(3)
(t − 1))/2 ↔ ω3
sin ω.
e From exercise 9.4c and a shift in the frequency domain it follows that
e4it
(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 in9.26
the frequency domain leads to −eit
δ (t) ↔ (ω − 1)2
.
b From the differentiation rule in the time domain and table 5 it follows
as in exercise 9.25b that δ (t) ↔ iω and δ (t) ↔ −ω2
, so −δ (t)+2iδ (t)+
δ(t) ↔ ω2
− 2ω + 1.

c Since (ω − 1)2
= ω2
− 2ω + 1, the results in part a and b should be
the same. Using exercise 8.30b with f(t) = eit
we indeed obtain that
eit
δ (t) = δ (t) − 2iδ (t) − δ(t).
a From exercise 9.25b it follows that δ (t) ↔ −ω2
. The convolution9.27
theorem then implies that T ∗ δ (t) ↔ −ω2
U where U is the spectrum of
T. This also follows by applying the diﬀerentiation rule in the time domain
to 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 it
was shown that | t | = 2δ, so we indeed get δ ∗ | t | = 2δ. Now let V be
the spectrum of | t |. Since δ ↔ −ω2
and δ ↔ 1 it then follows as in part
a from the convolution theorem that ω2
V = −2.

a When the system is causal, then the response to the causal signal δ(t)10.1
is again causal, so h(t) is causal. On the other hand, if h(t) is causal, then
it follows that y(t) = (u ∗ h)(t) =
R t
−∞
h(t − τ)u(τ) dτ 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 δ(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) = δ(t) then it follows that h(t) = δ(t−1)+ (t)e−2t
,10.2
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 −
1) +
R t
0
e−2(t−τ)
dτ = (t − 1) + 1
2
(1 − e−2t
), 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
2
(t)(1 − e−2t
).
We can express p2(t) as p2(t) = (t + 1) − (t − 1). Since a(t) is (by10.3
definition) 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 differentiate a(t) in distribution sense, which results in h(t) =10.5
δ(t) − (t)e−3t
(2 sin 2t + 3 cos 2t), since a(t) has a jump at t = 0 of mag-
nitude 1 and we can differentiate in ordinary sense outside t = 0.
b We use theorem 10.1, which implies that we may ignore the delta com-
ponent and only have to show that (t)e−3t
(2 sin 2t + 3 cos 2t) is absolutely
integrable. Since both
R ∞
0
˛
˛ e−3t
sin 2t
˛
˛ dt and
R ∞
0
˛
˛ e−3t
cos 2t
˛
˛ dt exist
(e.g., both are smaller than
R ∞
0
e−3t
dt), this is indeed the case and hence
the system is stable.
a The impulse response is h1 ∗ h2.10.6
b If the input for the first 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(ω) = 1+1/(1+iω)2
since δ ↔ 1 and te−t
(t) ↔10.7
1/(1 + iω)2
.
b Since eiωt
→ H(ω)eiωt
it follows that the response is given by eiωt
(1 +
1/(1 + iω)2
).
a We need to determine the inverse Fourier transform of the function10.8
H(ω) = cos ω/(ω2
+ 1). First note that 1
2
e−| t |
↔ 1/(ω2
+ 1). Writing the
cosine as a combination of exponentials we obtain from the shift rule that
h(t) = 1
2
(e−| t+1 |
+ e−| t−1 |
).
b It now suffices to use the time-invariance of the system: the reponse to
δ(t) is h(t), so the response to δ(t − 1) is h(t − 1).
a We need to determine the inverse Fourier transform of the function from10.9
figure 10.3, which is H(ω) = qωc (ω). This is h(t) = 2 sin2
(ωct/2)/(πωct2
).
40

b The function u has a Fourier series with terms cneint
(note that ω0 = 1).
But the response to eint
is H(n)eint
and H(n) = 0 for n > 1 and n < −1.
So we only have to determine c0, c1 and c−1. These can easily be calculated
from the defining integrals: c0 = 1
2
and c−1 = c1 = −1/π. Hence, the
response follows: y(t) = H(0)c0 +H(1)c1eit
+H(−1)c−1e−it
= 1
2
− 2
3π
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 time
reversal (scaling with a = −1 from table 4, no. 5) it follows for 1
iω−1
=
−1
i(−ω)+1
that − (−t)et
↔ 1
iω−1
. Hence, h(t) = δ(t) + 2
3
(−t)et
− 4
3
(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 Parseval
it 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 example
10.7).
a From the differential equation we immediately obtain the frequency10.13
response:
H(ω) =
ω2
− ω2
0
ω2 − i
√
2ω0ω − ω2
0
.
b Write the cosine as a combination of exponentials, then it follows from
eiωt
→ H(ω)eiωt
that y(t) = (H(ω0)eiω0t
+ 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 we
obtain that (t) ↔ pv(1/iω) + πδ(ω). Write the cosine as a combination
of 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, writing
everything with a common denominator, Y (ω) = ω/i(ω2
− i
√
2ω0ω − ω2
0).
Put s = iω and apply partial fraction expansion to obtain that 2Y (ω) =
(1 + i)/(s + ω0(1 − i)/
√
2) + (1 − i)/(s + ω0(1 + i)/
√
2). The inverse Fourier
transform of this equals e−ω0t/
√
2
(cos(ω0t/
√
2) − sin(ω0t/
√
2)) (t).
a From the differential equation we immediately obtain the frequency10.14
response:
H(ω) =
iω + 1 + α
2iω + α
=
1
2
+
1 + α/2
2iω + α
,
where we also used a long division. Using the tables it then follows that
h(t) = 1
2
(δ(t) + (1 + α/2) (t)e−tα/2
).
b We have to interpret the differential equation ‘the other way around’, so
with input and output interchanged. This means that the system function
is now 1/H(ω), so
2iω + α
iω + 1 + α
= 2 −
2 + α
iω + 1 + α

where we also used a long division. Using the tables it then follows that
h1(t) = 2δ(t) − (2 + α) (t)e−t(1+α)
.
c Note that the spectrum of (h ∗ h1)(t) is the function H(ω) · (1/H(ω)),
which is 1. Since δ(t) ↔ 1, it follows that (h ∗ h1)(t) = δ(t).
We use separation of variables, so we substitute u(x, y) = X(x)Y (y) into10.16
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 diﬀerential equations we obtain from the boundedness condition that
X(x) = 1 if c = 0 and ei
√
cx
, e−i
√
cx
if c > 0. Similarly Y (y) = 1 if c = 0 and
e
√
cy
, e−
√
cy
if c > 0. But e
√
cy
is not bounded for y > 0 so Y (y) = e−
√
cy
for
c ≥ 0. We put c = s2
, then it follows that the class of functions satisfying
the diﬀerential 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
u(x, y) =
Z ∞
−∞
e−| s |y
F(s)eisx
ds.
If we substitute y = 0 in this integral representation, then we obtain that
u(x, 0) =
1
1 + x2
=
Z ∞
−∞
F(s)eisx
ds.
Since 1
2
e−| t |
↔ 1/(1 + ω2
) this means that
1
1 + ω2
=
1
2
Z ∞
−∞
e−| t |
e−iωt
dt.
The (formal) solution is thus given by
u(x, y) = 1
2
Z ∞
−∞
e−| s |(1+y)
eisx
ds =
Z ∞
0
e−| s |(1+y)
cos(sx) ds.
a We substitute u(t) = δ(t),then10.18
h(t) =
Z t
t−1
e−(t−τ)
δ(τ) dτ =
Z ∞
−∞
( (τ + 1 − t) − (τ − t))e−(t−τ)
δ(τ) dτ,
so h(t) = ( (t) − (t − 1))e−t
. Applying table 3, no. 7 and a shift in the
time domain gives H(ω) = (1 − e−(1+iω)
)/(1 + iω).
b The impulse response is causal, so the system is causal.
c It is straightforward to verify that the impulse response is 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
y(t) =
Z 1
−1
h(t − τ) dτ =
Z t+1
t−1
h(τ) dτ.
This is 0 for t < −1 or for t > 2. For −1 < t < 0 it equals 1 − e−(t+1)
, for
0 ≤ t < 1 it equals 1 − e−1
, and for 1 ≤ t < 2 it equals e−(t−1)
− e−1
.
a The impulse response is the derivative in the sense of distributions of10.19
a(t) = e−t
(t), which is δ(t) − e−t
(t).

b The frequency response is H(ω) = iω/(1 + iω). (Apply e.g. the differ-
entiation rule to h(t) = a (t).)
c We have Y (ω) = H(ω)U(ω) (using obvious notations), so Y (ω) =
iω/(1 + iω)2
= 1/(1 + iω) − 1/(1 + iω)2
. The response is the inverse
Fourier transform of Y (ω): y(t) = (1 − t)e−t
(t).
a The frequency response H(ω) is the triangle function qωc(ω). The in-10.21
verse Fourier transform follows from table 3: h(t) = 2 sin2
(ωct/2)/(πωct2
).
b Since a (t) = h(t) and h(t) ≥ 0, the function a(t) is a monotone increas-
ing function.
The frequency response follows immediately from the differential equation:10.22
H(ω) =
1 + iω
−ω2 + 2iω + 2
.
Applying partial fraction expansion (use s = iω) we obtain
H(ω) =
1
2(iω + 1 − i)
+
1
2(iω + 1 + i)
.
The inverse Fourier transform is then h(t) = (e−t
cos t) (t). Integrating
this over (−∞, t] gives the step response a(t) = 1
2
(1+e−t
(sin t−cos t)) (t).
b We have Y (ω) = H(ω)U(ω) (using obvious notations), so Y (ω) =
1/(−ω2
+ 2iω + 2). Applying partial fraction expansion we obtain
Y (ω) =
1
2i(iω + 1 − i)
−
1
2i(iω + 1 + i)
.
The response is the inverse Fourier transform of Y (ω), which gives as
repsonse the function y(t) = (e−t
sin t) (t).
a Since | iω − 1 − i | = | iω + 1 − i | and
˛
˛ e−iωt0
˛
˛ = 1 we have | H(ω) | = 110.23
and so L is an all-pass system.
b First write (iω − 1 − i)/(iω + 1 − i) as 1 − 2/(iω + 1 − i) and then use
the inverse Fourier transform (the tables) to obtain δ(t) − 2e−(1−i)t
(t) ↔
1 − 2/(iω + 1 − i). From the shift property in the time domain we obtain
h(t) = δ(t − t0) − 2e−(1−i)(t−t0)
(t − t0).
c Ignoring the delta function it is easy to verify that h(t) is absolutely
integrable (
˛
˛
˛ ei(t−t0)
˛
˛
˛ = 1), hence the system is stable.
d Write u(t) = 1 + eit
+ e−it
and use that eiωt
→ H(ω)eiωt
, then y(t) =
H(0)+H(1)eit
+H(−1)e−it
, which equals −i−e−it0
eit
+(4i−3)eit0
e−it
/5.
We use separation of variables, so we substitute u(x, y) = X(x)Y (y) into10.24
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 differential equations, we obtain from the
boundedness condition that X(x) = 1 if c = 0 and ei
√
cx
, e−i
√
cx
if c > 0.
For c ≥ 0 we obtain for Y (y) the solution e−cy/2
. We put c = s2
with
s ∈ R. It then follows that the class of functions satisfying the differential
equation and being bounded, can be described by
X(x)Y (y) = eisx
e−s2
y/2
, where s ∈ R.
By superposition we now try a solution u(x, y) of the form
u(x, y) =
Z ∞
−∞
e−s2
y/2
F(s)eisx
ds.

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

In parts a, b and c the domain is C and the range is C as well. In part d11.1
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 the11.2
imaginary part is −y.
b Expanding z3
= (x + iy)3
we see that the real part is x3
− 3xy2
and
that the imaginary part is 3x2
y − y3
.
c The real part is x − 4 and the imaginary part is −y − 1.
d The real part is (3y − 2x − 6)/((x + 3)2
+ y2
) and the imaginary part
is (2y + 3x + 9)/((x + 3)2
+ y2
). To see this, write z = x + iy, then
z + 3 = x + 3 + iy and so
f(z) =
(3i − 2)(x + 3 − iy)
(x + 3 + iy)(x + 3 − iy)
=
3y − 2x − 6 + i(2y + 3x + 9)
(x + 3)2 + y2
.
Apply definition 11.3 and expand the squares; several of the exponentials11.3
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 definition 11.3 it follows that sin(iy) = (e−y
− ey
)/2i = i sinh y11.5
and cos(iy) = (e−y
+ 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) = sin x cosh y + i cos x sinh y, so the real part is sin x cosh y and
the imaginary part is cos x sinh y.
The proofs can be copied from the real case; this is a straightforward mat-11.6
ter. The same applies to exercise 11.7.
This rational function is continuous for all z ∈ C for which the denominator11.8
is unequal to 0. But the denominator is 0 for z = 1, z = −i or z = 2i. So
g(z) is continuous on G = C {1, −i, 2i}.
The proof can be copied from the real case; this is a straightforward matter:11.9
lim
w→z
f(w) − f(z)
w − z
= lim
w→z
w2
− z2
w − z
= lim
w→z
(w + z) = 2z.
a The derivative is 4(z − 1)3
; the function is differentiable on C, so it is11.11
analytic on C.
b The derivative is 1 − 1/z2
; since the function is not differentiable at
z = 0, it is analytic on C {0}.
c The derivative is ((z3
+ 1)(2z − 3) − 3z2
(z2
− 3z + 2))/(z3
+ 1)2
; the
function is not differentiable when z3
= −1. Solving this equation one
obtains that it is analytic on C {−1, 1
2
± 1
2
i
√
3}.
d The derivative is 2zez2
; the function is analytic on C.
Using definition 11.3 and the chain rule it follows that (cos z) = (ieiz
−11.12
ie−iz
)/2 = (−eiz
+ e−iz
)/2i = − sin z.
The step
√
wz =
√
w
√
z cannot be applied for non-real numbers (so e.g.11.13
for w = −1, z = −1).
The real and imaginary part are u(x, y) = x2
− y2
and v(x, y) = 2xy11.14∗
45

(example 11.4). So ∂u/∂x = 2x, which equals ∂v/∂y, and ∂u/∂y = −2y
which equals −∂v/∂x. Hence, the Cauchy-Riemann equations are satisﬁed
on the whole of R2
.
a This is not true; take e.g. z = 2i, then cos 2i = (e−2
+ e2
)/2 > 3.11.16
b Use deﬁnition 11.3; expand the exponentials in the resulting expression
for cos z cos w − sin z sin w.
c Write z = x + iy and use part b, then cos(x + iy) = cos x cos iy −
sin x sin iy. But cos iy = cosh y and sin iy = i sinh y (see exercise 11.5a), so
cos(x+iy) = cos x cosh y −i sin x sinh y, which gives the real and imaginary
parts.
d Since ez
is analytic on C, it follows from theorem 11.5 that cos z =
(eiz
+ e−iz
)/2 is also analytic on C.
a This function is analytic on C {1}. From the quotient rule it follows11.17
that the derivative is given by (2z3
− 3z2
− 1)/(z − 1)2
.
b This function is analytic when z4
= −16, so on C {±
√
2 ± i
√
2}. The
derivative is given by −40z3
/(z4
+ 16)11
.
c This function is analytic when z2
= −3, so on C {±i
√
3}. The deriv-
ative is given by ez
(z2
− 2z + 3)/(z2
+ 3)2
(use the quotient rule).
d This function is analytic on C since both ez
and sin w are analytic on
C. The derivative is given by ez
cos ez
(use the chain rule).

Antwoorden fourier and laplace transforms, manual solutions

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