Fourier and Laplace Transforms (2003) – Answers to Exercises – Beerends

All 19 Chapters & Figures Covered

, Answers to selected exercises for chapter 1




1.1 Apply cos(α + β) = cos α cos β − sin α 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,
where C1 = A1 cos φ1 + A2 cos φ2 and C2 = A1 sin φ1 + A2 sin φ2 . Put A =
p
C12 + C22 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).
1.2 Put c1 = A1 eiφ1 and c2 = A2 eiφ2 , then f1 (t) + f2 (t) = (c1 + c2 )eiωt . Let
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.
1.5 The power P is given by
Z π/ω
A2 ω π/ω
Z
ω
P = A2 cos2 (ωt + φ0 ) dt = (1 + cos(2ωt + 2φ0 )) dt
2π −π/ω 4π −π/ω
A2
= .
2

e−2t dt = 12 .
R∞
1.6 The energy-content is E = 0

1.7 The power P is given by
3
1X
P = | cos(nπ/2) |2 = 12 .
4 n=0

P∞
1.8 The energy-content is E = n=0 e−2n , which is a geometric series with
sum 1/(1 − e−2 ).
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
Rt follows from the substitution ξ = τ − t0 in the integral
t−1
u(τ − t 0 ) dτ representing the response to u(t − t0 ).
Rt
d Calculating t−1 cos(ωτ ) dτ gives the following response: (sin(ωt) −
sin(ωt − ω))/ω = R t2 sin(ω/2) cos(ωt − ω/2)/ω.
e Calculating 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.
1.11 a The frequency response of the cascade system is H1 (ω)H2 (ω), since the
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).
˛ ˛ √
1.12 a The amplitude response is | 1 + i | ˛ e−2iω ˛ = 2.
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.
1.13 a The power is the integral of f 2 (t) over [−π/ | ω | , π/ | ω |], times | ω | /2π.
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 .
2 2
Hence, the power equals (A R 1 + 2AB cos(φ0 ) + B )/2.
b The energy-content is 0 sin2 (πt) dt = 1/2.

1.14 The power is the integral of | f (t) |2 over [−π/ | ω | , π/ | ω |], times | ω | /2π,
which in this case equals | c |2 .
1.16 a The amplitude response is | H(ω) | = 1/(1 + ω 2 ). The phase response
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.
1.17 a The signal is not periodic since sin(2N ) 6= 0 for all integer N .
iω
b The frequency response H(eiω ) equals A(eiω )eiΦe , hence, we obtain
that H(e ) = e /(1 + ω ). The response to u[n] = (e2in − e−2in )/2i is
iω iω 2

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.
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
˛ X ˛ n
X Xn
| 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.
d The response to u[n] = cos πn = (−1)n is
n
X
y[n] = (−1)n−n0 + (−1)l = (−1)n−n0 + (−1)n (1 − 1 + 1)
l=n−2
= (−1)n (1 + (−1)n0 ).