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                            935_Ch1Answ.pdf
936_Ch2Answ.pdf
955_fig.02.06.pdf
937_Ch3Answ.pdf
956_fig.03.20.pdf
957_fig.03.21.pdf
958_fig.03.23.pdf
938_Ch4Answ.pdf
939_Ch5Answ.pdf
940_Ch6Answ.pdf
941_Ch7Answ.pdf
942_Ch8Answ.pdf
943_Ch9Answ.pdf
944_Ch10Answ.pdf
945_Ch11Answ.pdf
946_Ch12Answ.pdf
947_Ch13Answ.pdf
948_Ch14Answ.pdf
949_Ch15Answ.pdf
950_Ch16Answ.pdf
951_Ch17Answ.pdf
952_Ch18Answ.pdf
953_Ch19Answ.pdf
                        
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Page 1

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
C21 + C

2
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 = A1e
iφ1 and c2 = A2e

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

c = c1 + c2, then f1(t) + f2(t) = ce
iωt. 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



Z π/ω
−π/ω

A
2
cos

2
(ωt + φ0) dt =

A2ω



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

Page 2

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)e

iωt and then H1(ω)H2(ω)e
iωt.

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 7→ H(eiω)eiωn, the response is H(e0)1 = 1 + i for all n.
c Since u[n] = (eiωn + e−iωn)/2 we can use eiωn 7→ H(eiω)eiωn to 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) + B

2)/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 7→ H(ω)eiωt that
the response is H(1)ieit = iei(t+1)/2.

a The signal is not periodic since sin(2N) 6= 0 for all integer N .1.17
b The frequency response H(eiω) equals A(eiω)eiΦe



, 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,

Page 43

Answers to selected exercises for chapter 10 41

b The function u has a Fourier series with terms cne
int (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)c1e

it +H(−1)c−1e−it = 12−
2


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)e
t ↔ 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 − ω20

ω2 − i


2ω0ω − ω20
.

b Write the cosine as a combination of exponentials, then it follows from
eiωt 7→ 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 − ω20).

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

Page 44

42 Answers to selected exercises for chapter 10

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) = e
isx

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)e
isx

ds.

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

u(x, 0) =
1

1 + x2
=

Z ∞
−∞

F (s)e
isx

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)

e
isx

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).

Page 86

84 Answers to selected exercises for chapter 19

1]. For the z-transforms we have (use the shift rule in the n-domain)
H(z) = A(z) − A(z)/z = A(z)(z − 1)/z. From table 13 it follows that
(1/2)n�[n] ↔ z/(z−1/2), n(1/2)n�[n] ↔ (z/2)/(z−1/2)2,

`
n
2

´
(1/2)n�[n] ↔

(z/4)/(z − 1/2)3 and since n2 = 2
`

n
2

´
+ n it then follows that

A(z) =
z/2

(z − 1/2)3
+

z/2

(z − 1/2)2
=

1

2

z(z + 1/2)

(z − 1/2)3
.

This means that the transfer function H(z) equals

H(z) =
1

2

(z − 1)(z + 1/2)
(z − 1/2)3

.

b The impulse response is causal, so the system is causal.
c From einω 7→ H(eiω)einω it follows that

y[n] =
1

2

(eiω − 1)(eiω + 1/2)
(eiω − 1/2)3

e
inω

.

a From the difference equation we obtain that (1− z−1/2)Y (z) = (z−1 +19.22
z−2)U(z) and hence

H(z) =
z−1 + z−2

1− z−1/2
=

z + 1

z2 − z/2
.

Since �[n] ↔ z/(z − 1) = U(z) we have

a[n] ↔ A(z) = H(z)U(z) =
z + 1

(z − 1/2)(z − 1)
.

A partial fraction expansion of A(z)/z gives

A(z)

z
=

2

z
+

4

z − 1


6

z − 1/2
.

Multiply this by z and use table 13 to obtain that a[n] = 2δ[n] + (4 −
6(1/2)n)�[n].
b The (rational) transfer function H(z) has poles at z = 0 and z = 1/2,
which lie inside the unit circle, so the system is stable.

a From the difference equation we obtain that (6 − 5z−1 + z−2)Y (z) =19.23
(6− 6z−2)U(z) and hence

H(z) =
6− 6z−2

6− 5z−1 + z−2
=

6(z2 − 1)
6z2 − 5z + 1

.

A partial fraction expansion of H(z)/z (note that the denominator equals
z(2z − 1)(3z − 1)) gives

H(z)

z
= −

6

z
+

16

z − 1/3


9

z − 1/2
.

Multiply this by z and use table 13 to obtain that h[n] = −6δ[n] +
(16(1/3)n − 9(1/2)n)�[n].
b The frequency response H(eiω) is obtained from the transfer function
H(z) by substituting z = eiω, so H(eiω) = 6(e2iω − 1)/(6e2iω − 5eiω + 1).
c From (19.10) we know that Y (eiω) = H(eiω)U(eiω). Since y[n] is
identically 0, we know that Y (eiω) = 0 for all frequencies ω. Since H(eiω) 6=
0 for e2iω 6= 1 we have U(eiω) = 0 for e2iω 6= 1. The frequencies ω = 0
or ω = π may still occur in the input. These frequencies correspond to

Page 87

Answers to selected exercises for chapter 19 85

the time-harmonic signals ei0n = 1 and eiπn = (−1)n respectively. Hence,
the input consists of a linear combination of 1 and (−1)n, which means
that the solution equals u[n] = A + B(−1)n, where A and B are complex
constants.

a The frequency response can be written as H(eiω) = 1+2 cos ω+cos 2ω =19.24
1 + eiω + e−iω + e2iω/2 + e−2iω/2. Since H(eiω) =

P∞
n=−∞ h[n]e

−inω it
follows that h[n] = δ[n] + δ[n + 1] + δ[n− 1] + δ[n + 2]/2 + δ[n− 2]/2.
b If we write U(eiω) = 1+sin ω+sin 2ω as complex exponentials, as in part
a, then it follows that u[0] = 1, u[1] = u[2] = i/2, u[−1] = u[−2] = −i/2
and so u[n] = δ[n]+ i(δ[n−1]+ δ[n−2]− δ[n+1]− δ[n+2])/2. We have to
calculate E =

P∞
n=−∞ | y[n] |

2
. In this case it is easiest to do this directly

(and so not using Parseval). From the expression for u[n] it follows (by
linearity and time-invariance) that y[n] = h[n] + i(h[n − 1] + h[n − 2] −
h[n +1]−h[n +2])/2. Substituting h[n] from part a we get y[n] as a linear
combination of δ[n− 4], δ[n− 3], . . . , δ[n + 3], δ[n + 4]. The coefficients in
this combination are not hard to determine. In fact, they are −i/4, −3i/4,
(1/2)− i, 1−3i/4, 1, 1+3i/4, (1/2)+ i, 3i/4, i/4. Their contribution to E
is 1/16, 9/16, 20/16, 25/16, 1, 25/16, 20/16, 9/16, 1/16. The sum of these
contributions is 126/16, hence, E = 7 7

8
.

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