Table Of ContentAnswers to selected exercises for chapter 1
1.1 Apply cos(α + β) = cos α cos β − sinα sin β, then
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,
w
phere C1 = A1 cos φ1 +A2 cos φ2 and C2 = A1 sin φ1 +A2 sin φ2. Put A =
2 2
C1 + C2 and take φ such that cos φ = C1/A and sin φ = C2/A (this is
2 2
possible since (C1/A) +(C2/A) = 1). Now f1(t)+f2(t) = A(cos ωt cos φ−
sin ωt sin φ) = Acos(ωt + φ).
iφ1 iφ2 iωt
1.2 Put c1 = A1e and c2 = A2e , then f1(t) + f2(t) = (c1 + c2)e . Let
iωt
c = c1 + c2, then f1(t) + f2(t) = ce . The signal f1(t) + f2(t) is again a
time-harmonic signal with amplitude | c | and initial phase arg c.
1.5 The power P is given by
Z Z
π/ω 2 π/ω
ω 2 2 A ω
P = A cos (ωt + φ0) dt = (1 + cos(2ωt + 2φ0)) dt
2π −π/ω 4π −π/ω
2
A
= .
2
R
∞ −2t 1
1.6 The energy-content is E = e dt = .
0 2
1.7 The power P is given by
3
X
1 2 1
P = | cos(nπ/2) | = .
2
4
n=0
P
∞ −2n
1.8 The energy-content is E = e , which is a geometric series with
n=0
−2
sum 1/(1 − e ).
1.9 a If u(t) is real, then the integral, and so y(t), is also real.
b Since
˛ Z ˛ Z
˛ ˛
˛ u(τ) dτ ˛ ≤ | 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 integral
R
t
t−1 u(τ − t0) dτ representing the response to u(t − t0).
R
t
d Calculating cos(ωτ) dτ gives the following response: (sin(ωt) −
t−1
sin(ωt − ω))/ω = 2 sin(ω/2) cos(ωt − ω/2)/ω.
R
t
e Calculating sin(ωτ) dτ gives the following response: (− cos(ωt) +
t−1
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) =
−iω/2
2 sin(ω/2)e /ω.
1.11 a The frequency response of the cascade system is H1(ω)H2(ω), since the
iωt iωt iωt
reponse to e is first H1(ω)e and then H1(ω)H2(ω)e .
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).
˛ ˛ √
˛ −2iω ˛
1.12 a The amplitude response is | 1 + i | e = 2.
b The input u[n] = 1 has frequency ω = 0, initial phase 0 and amplitude
iωn iω iωn 0
1. Since e →↦ H(e )e , the response is H(e )1 = 1 + i for all n.
iωn −iωn iωn iω iωn
c Since u[n] = (e + e )/2 we can use e →↦ H(e )e to obtain
iω iωn −iω −iωn
that y[n] = (H(e )e +H(e )e )/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.
2
1.13 a The power is the integral of f (t) over [−π/ | ω | , π/ | ω |], times | ω | /2π.
2
Now cos (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.
2 2
Hence, the power equals (A + 2AB cos(φ0) + B )/2.
R
1 2
b The energy-content is sin (πt) dt = 1/2.
0
2
1.14 The power is the integral of | f(t) | over [−π/ | ω | , π/ | ω |], times | ω | /2π,
2
which in this case equals | c | .
2
1.16 a The amplitude response is |H(ω) | = 1/(1 + ω ). The phase response
is argH(ω) = ω.
iωt iωt
b The input has frequency ω = 1, so it follows from e →↦ H(ω)e that
it i(t+1)
the response is H(1)ie = ie /2.
1.17 a The signal is not periodic since sin(2N) ≠ 0 for all integer N.
iω
iω iω iΦe
b The frequency response H(e ) equals A(e )e , hence, we obtain
iω iω 2 2in −2in
that H(e ) = e /(1 + ω ). The response to u[n] = (e − e )/2i is
2i(n+1) −2i(n+1)
then y[n] = (e − e )/(10i), so y[n] = (sin(2n + 2))/5. The
amplitude is thus 1/5 and the initial phase 2 − π/2.
1.18 a If u(t) = 0 for t < 0, then the integral occurring in y(t) is equal to 0 for
t < 0. For t0 ≥ 0 the expression u(t − t0) is also causal. Hence, the system
is causal for t0 ≥ 0.
b It follows from the boundedness of u(t), so | u(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
Z
t
y(t) = sin(π(t − t0)) + sin(πτ) dτ = sin(π(t − t0)) − 2(cos πt)/π.
t−1
1.19 a If u[n] = 0 for n < 0, then y[n] is also equal to 0 for n < 0 whenever
n0 ≥ 0. Hence, the system is causal for n0 ≥ 0.
b It follows from the boundedness of u[n], so | u[n] | ≤ K for some constant
K and all n, that y[n] is also bounded (use the triangle inequality):
˛ ˛
˛ n ˛ n n
X X X
˛ ˛
| y[n] | ≤ | u[n − n0] | + ˛ u[l] ˛ ≤ K + | u[l] | ≤ K + K,
˛ ˛
l=n−2 l=n−2 l=n−2
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.
n
d The response to u[n] = cos πn = (−1) is
n
X
n−n0 l n−n0 n
y[n] = (−1) + (−1) = (−1) + (−1) (1 − 1 + 1)
l=n−2
n n0
= (−1) (1 + (−1) ).
Answers to selected exercises for chapter 2
p √
2 2
2.1 a The absolute values follow from x + y and are given by 2, 2, 3, 2
respectively. The arguments follow from standard angles and are given by
3π/4, π/2, π, 4π/3 respectively.
√ √
πi/4
b Calculating modulus and argument gives 2+2i = 2 2e , − 3+i =
5πi/6 3πi/2
2e and −3i = 3e .
2.2 In the proof of theorem 2.1 it was shown that | Re z | ≤ | z |, which implies
that − | z | ≤ ± |Re z | ≤ | z |. Hence,
2
| z ± w | = (z ± w)(z ± w) = zz ± zw ± wz + ww
2 2
= | z | ± 2Re(zw) + |w |
2 2 2
≥ | z | − 2 | z | | w | + |w | = (| z | − |w |) .
2 2
This shows that | z ± w | ≥ (| z | − |w |) .
√
2.4 We have | z1 | = 4 2, | z2 | = 4 and arg z1 = 7π/4, arg z2 = 2π/3. Hence,
√
| z1/z2 | = | z1 | / | z2 | = 2 and arg(z1/z2) = arg(z1) − arg(z2) = 13π/12,
√
13πi/12 2 3 3πi/2 2 3
so z1/z2 = 2e . Similarly we obtain z1z2 = 2048e and z1/z2 =
1 3πi/2
e .
2
2.5 The solutions are given in a separate figure on the website.
2.6 a The four solutions ±1± i are obtained by using the standard technique
to solve this binomial equation (as in example 2.3).
√
6
b As part a; we now obtain the six solutions 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.
5 4 4 4
2.7 Write z − z + z − 1 as (z − 1)(z + 1) and then solve z = −1 to find
√
the roots 2(±1 ± i)/2. Combining linear factors with complex conjugate
√ √
5 4 2 2
roots we obtain z − z + z − 1 = (z − 1)(z + 2z + 1)(z − 2z + 1).
πi/2
2.8 Since 2i = 2e the solutions are z = ln 2 + i(π/2 + 2kπ), where k ∈ Z.
1
2.9 Split F (z) as A/(z − ) + B/(z − 2) and multiply by the denominator of
2
F (z) to obtain the values A = −1/3 and B = 4/3 (as in example 2.6).
2
2.11 a Split F (z) as A/(z + 1) + B/(z + 1) + C/(z + 3) and multiply by
the denominator of F (z) to obtain the values C = 9/4, B = 1/2 and, by
2
comparing the coefficient of z , A = −5/4 (as in example 2.8).
2.12 Trying the first few integers we find the zero z = 1 of the denominator. A
2
long division gives as denominator (z −1)(z −2z +5). We then split F (z)
2
as A/(z − 1) + (Bz + C)/(z − 2z + 5). Multiplying by the denominator
0 2
of F (z) and comparing the coefficients of z = 1, z and z we obtain that
A = 2, B = 0 and C = −1.
′ −2
2.13 a Using the chain rule we obtain f (t) = −i(1 + it) .
iω0t
2.14 Use integration by parts twice and the fact that a primitive of e is
iω0t 3 2πi
e /iω0. The given integral then equals 4π(1 − πi)/ω0, since e = 1.
˛ ˛ ˛ ˛ ˛ ˛ ˛ ˛
2.15 Since ˛ 1/(2 − eit) ˛ = 1/ ˛ 2 − eit ˛ and ˛ 2 − eit ˛ ≥ 2 − ˛ eit ˛ = 1, the result
˛ ˛
˛ R 1 ˛ R 1
follows from ˛ u(t) dt ˛ ≤ | u(t) | dt.
0 0
1
2 Answers to selected exercises for chapter 2
√ P
6 3 ∞ 3
2.16 a Use that | an | = 1/ n + 1 ≤ 1/n and the fact that
n=1 1/n con-
verges (example 2.17).
P
2 ∞ 2
b Use that | an | ≤ 1/n
˛and the˛ fact that n=1 1/n converges.
˛ n ni ˛ n n
c Use that | an | = 1/ ne e = 1/(ne ) ≤ 1/e and the fact that
P
∞ n
1/e converges since it is a geometric series with ratio 1/e.
n=1
2.17 a Use the ratio test to conclude that the series is convergent:
˛ ˛
˛ ˛
n! 1
˛ ˛
lim = lim = 0.
˛ ˛
n→∞ (n + 1)! n→∞ n + 1
b The series is convergent; proceed as in part a:
˛ ˛
n+1 n n n
˛ ˛
2 + 1 3 + n 2 + 1/2 1 + n/3 2
˛ ˛
lim = lim = .
˛ n+1 n ˛ n n
n→∞ 3 + n + 1 2 + 1 n→∞ 3 + (n + 1)/3 1 + 1/2 3
2.19 Determine the radius of convergence as follows:
˛ ˛
n+1 2n+2 2 2
˛ ˛ ˛ ˛ ˛ ˛
2 z n + 1 2 1 + 1/n 2
˛ ˛ ˛ ˛ ˛ ˛
lim = lim 2 z = 2 z .
˛ 2 n 2n ˛ 2
n→∞ (n + 1) + 1 2 z n→∞ 1 + 2/n + 2/n
˛ ˛ √
˛ 2 ˛
This is less than 1 if z < 1/2, that is, if | z | < 2/2. Hence, the radius
√
of convergence is 2/2.
2.20 This is a geometric series with ratio z−i and so it converges for | z − i | < 1;
the sum is (1/(1 − i))(1/(1 − (z − i))), so 1/(2 − z(1 − i)).
2 2 2
2.23 b First solving w = −1 leads to z = 0 or z = −2i. The equation
2 2
z = −2i has solutions −1 + i and 1 − i and z = 0 has solution 0 (with
multiplicity 2).
4 2 2 2 2 2
c One has P(z) = z(z + 8z + 16) = z(z + 4) = z(z − 2i) (z + 2i) , so
0 is a simple zero and ±2i are two zeroes of multiplicity 2.
2 2 2
2.25 Split F (z) as (Az+B)/(z −4z+5)+(Cz+D)/(z −4z+5) and multiply
0 1 2 3
by the denominator of F (z). Comparing the coefficient of z , z , z and z
leads to the values A = 0, B = 1, C = −2 and D = 2.
R
it −it 2π 2it
2.26 Replace cos t by (e +e )/2, then we have to calculate (e +1)/2 dt,
0
which is π.
√
2.27 a Using the ratio test we obtain as limit 5/3. This is less than 1 and so
the series converges.
P
n 2 n 2 ∞
b Since (n + i )/n = (1/n) + (i /n ) and the series 1/n diverges,
n=1
this series is divergent.
P ˛ ˛
∞ 2 n
˛ 2 ˛
2.29 The series
n=0 cn(z ) converges for all z with z < R, so it has radius
√
of convergence R.
2.30 a Determine the radius of convergence as follows:
˛ ˛
2n+2 n+1
˛ ˛ ˛ ˛
(1 + i) z n + 1 n + 1 2
˛ ˛ ˛ ˛
lim = lim | z | (1 + i) = 2 | z | .
˛ 2n n ˛
n→∞ n + 2 (1 + i) z n→∞ n + 2
This is less than 1 if | z | < 1/2, so the radius of convergence is 1/2.
′
b Calculate f (z) by termwise differentiation of the series and multiply
this by z. It then follows that
∞ ∞
X X
′ 2n n n
zf (z) + f(z) = (1 + i) z = (2iz) .
n=0 n=0
This is a geometric series with ratio 2iz and so it has sum 1/(1 − 2iz).
a b c
3 2 4 5
d e f
3
2
1 + 2i
1 0 3 –2 2
g
1 2 3
1
2
2
Answers to selected exercises for chapter 3
3.2 A trigonometric polynomial can be written as
k
X
a0
f(t) = + (am cos(mω0t) + bm sin(mω0t)).
2
m=1
Now substitute this for f(t) in the right-hand side of (3.4) and use the
fact that all the integrals in the resulting expression are zero, except for
R
T/2
the integral
−T/2 sin(mω0t) sin(nω0t) dt with m = n, which equals T/2.
Hence, one obtains bn.
3.4 The function g(t) = f(t) cos(nω0t) has period T , so
Z Z Z
T T T/2
g(t)dt = g(t)dt + g(t)dt
0 T/2 0
Z Z Z Z
T T/2 0 T/2
= g(t − T)dt + g(t)dt = g(τ)dτ + g(t)dt
T/2 0 −T/2 0
Z
T/2
= g(t)dt.
−T/2
Multiplying by 2/T gives an.
3.6 From a sketch of the periodic function with period 2π given by f(t) = | t |
for t ∈ (−π, π) we obtain
Z Z
0 π
1 −int 1 −int
cn = (−t)e dt + te dt.
2π −π 2π 0
As in example 3.2 these integrals can be calculated using integration by
parts for n ≠ 0. Calculating c0 separately (again as in example 3.2) we
obtain
n
π (−1) − 1
c0 = , cn =
2
2 n π
Substituting these values of cn in (3.10) we obtain the Fourier series. One
can also write this as a Fourier cosine series:
∞
X
π 4 cos((2k + 1)t)
− .
2
2 π (2k + 1)
k=0
3.7 From the description of the function we obtain that
Z
1
1 −(1+inπ)t
cn = e dt.
2 0
This integral can be evaluated immediately and leads to
inπ − 1 ` n −1 ´
cn = (−1) e − 1 .
2 2
2(n π + 1)
The Fourier series follows from (3.10) by substituting cn.
3.9 The Fourier coefficients are calculated by splitting the integrals into a real
and an imaginary part. For c0 this becomes:
3
4 Answers to selected exercises for chapter 3
Z Z
1 1
1 2 i 1
c0 = t dt + t dt = .
2 2 3
−1 −1
For n ≠ 0 we have that
Z Z
1 1
1 2 −inπt i −inπt
cn = t e dt + te dt.
2 2
−1 −1
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):
n
(−1) (2 − nπ)
cn = .
2 2
n π
3.10 From the values of the coefficients cn calculated earlier in exercises 3.6, 3.7
and 3.9, one can immediately obtain the amplitude spectrum | cn | and the
phase spectrum arg cn (note e.g. that arg cn = 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.
3.11 a By substituting a = T/4 in (3.14) it follows that
sin(nπ/4) 1
cn = for n ≠ 0, c0 = .
nπ 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.
3.12 By substituting a = T/2 in (3.15) it follows that
1 2
c0 = , cn = 0 for n ≠ 0 even, cn = for n odd.
2 2
2 n π
3.14 We have that f(t) = 2p2,4(t) − q1,4(t) and so the Fourier coefficients follow
by linearity from table 1:
2 2 2
c0 = 3/4, cn = (2nπ sin(nπ/2) − 4 sin (nπ/4))/(n π ) for n ≠ 0.
3.15 Note that f(t) can be obtained from the sawtooth z(t) by multiplying the
shifted version z(t − T/2) by the factor T/2 and then adding T/2, that
T T T
is, f(t) = z(t − ) + . Now use the Fourier coefficients of z(t) (table 1
2 2 2
e.g.) and the properties from table 2 to obtain that
T iT
c0 = , cn = for all n ≠ 0.
2 2πn
−2πin
3.17 Shifts over a period T (use the shift property and the fact that e = 1
for all n).
3.19 In order to determine the Fourier sine series we extend the function to an
odd function of period 8. We calculate the coefficients bn as follows (the
an are 0):
Z Z
−2 2
1 1
bn = (−2 sin(nπt/4)) dt + t sin(nπt/4) dt
4 4
−4 −2
Answers to selected exercises for chapter 3 5
Z
4
1
+ 2 sin(nπt/4) dt.
4
2
The second integral can be calculated by an integration by parts and one
then obtains that
8 4
bn = sin(nπ/2) − cos(nπ),
2 2
n π 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
8
a0 = 3, an = (cos(nπ/2) − 1) for all n ≠ 0.
2 2
n π
3.21 In order to determine the Fourier cosine series we extend the function to
an even function of period 8. We calculate the coefficients an and a0 as
follows (the bn are 0):
Z
4
1 2 16
a0 = (x − 4x) dx = − ,
2 3
0
while for n ≥ 1 we have
Z Z
0 4
1 2 1 2
an = (x + 4x) cos(nπx/4) dx + (x − 4x) cos(nπx/4) dx
4 4
−4 0
Z Z
4 4
1 2
= x cos(nπx/4) dx − 2 x cos(nπx/4) dx.
2
0 0
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
n n n
64(−1) 32((−1) − 1) 32((−1) + 1)
an = − = ,
2 2 2 2 2 2
n π n π n π
which also gives the Fourier cosine series. One can write this series as
∞
X
8 16 1
− + cos(nπx/2).
2 2
3 π n
n=1
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
n
64((−1) − 1)
bn = for all n ≥ 1.
3 3
n π
3.24 If f is real and the cn are real, then it follows from (3.13) that bn =
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.
iω0t −iω0t
3.25 Since sin(ω0t) = (e − e )/2i we have
Z Z
T/2 T/2
1 i(1−n)ω
0t 1 −i(1+n)ω0t
cn = e dt − e dt.
2iT 2iT
0 0
6 Answers to selected exercises for chapter 3
n
The first integral equals T/2 for n = 1 while for n ≠ 1 it equals i((−1) +
1)/((1 − n)ω0). The second integral equals T/2 for n = −1 while for
n+1
n ≠ −1 it equals i((−1) − 1)/((1 + n)ω0). The Fourier coefficients are
n 2
thus c1 = 1/(4i), c−1 = −1/(4i) and ((−1) +1)/(2(1−n )π) for n ≠ 1,−1;
the Fourier series follows immediately from this.
3.27 b The even extension has period 2a, but it has period a as well. We can
thus calculate the coefficients an and a0 as follows (the bn are 0):
Z Z
a/2 0
2 2
a0 = 2bt/a dt − 2bt/a dt = b.
a a
0 −a/2
while for n ≥ 0 we obtain from an integration by parts that
Z Z
a/2 0
2 2
an = (2bt/a) cos(2nπt/a) dt − (2bt/a) cos(2nπt/a) dt
a a
0 −a/2
n
2b((−1) − 1)
= ,
2 2
n π
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):
Z Z
−a/2 a/2
1 −2bt 1 2bt
bn = ( − 2b) sin(nπt/a) dt + sin(nπt/a) dt
a a a a
−a −a/2
Z
a
1 −2bt
+ ( + 2b) sin(nπt/a) dt
a a
a/2
8b
= sin(nπ/2),
2 2
n π
where we used integration by parts.
