Problems And Solutions In Communication Systems
SOLUTION MANUAL
,Solutions to Chapter 2 Problems
Introduction to Communication Systems, by Upamanyu Madhow
Problem 2.1 (a) We do this in two ways. The first is to directly show that y = x ∗ h for some
h, which shows that the system is LTI with impulse response h. We can rewrite
∫ ∞ ∫ ∞
y(t) = e−(t−u)I{u≤t}x(u)du = e−(t−u)I{t−u≥0}x(u)du
−∞ −∞
We see that this is in the convolution form
∫ ∞
y(t) = h(t — u)x(u)du
−∞
where h(t) = e−tIt≥0.
The second approach is to show that the system is LTI and then feed in an impulse to find
the impulse response. Linearity of y in x is clear, hence let us check for time invariance. Let
x1(t) = x(t — t0) be a delayed version of the input. The corresponding output is given by
∫ t ∫ t
y1(t) = eu−tx1(u)du = eu−tx(u — t0)du
−∞ −∞
Making the change of variables v = u — t0, we obtain that
∫ t−t0 ∫ t−t0
y1(t) = ev+t0 −tx(v)dv = ev−(t−t0)x(v)dv
−∞ −∞
Comparing with the original expression for y(t), it is clear that we have simply replaced t by
t — t0. That is, y1(t) = y(t —
t0) and the system is time invariant. We now find the impulse
response by setting the input to an impulse:
∫t
h(t) = eu−tδ(u)du
−∞
The impulse at time zero falls into the integration interval only if t ≥ 0, in which case we select
the value of the integrant at u = 0. We therefore obtain h(t) = e−tI{t≥0} as before.
(b) It is easy to see that the Fourier transform of the impulse response is H(f ) = j2πf1 +1 , with
magnitude |H(f )| = √ 21 2 . The plot is omitted, but it is clear that this is a sloppy lowpass
4π f +1
filter.
(c) We compute the energy in the frequency domain using Parseval’s identity. The input x(t) =
2sinc(2t) x X(f ) = I[−1,1](f ), and the output Y (f ) = H(f )X(f ) = 1
j2πf +1 I[−1,1] (f ). The energy
is given by
∫ ∞ ∫ 1 1
Ey = |Y (f )|2df = 2 2
df
−∞ −1 4π f + 1
Making the standard substitution 2πf = tan θ, so that 2πdf = sec2 θ dθ and 4π2f 2 + 1 =
tan2 θ + 1 = sec2 θ, we obtain (the limits of the transformed integral are ± tan−1(2π) = ±1.413)
∫ 1.413
1
Ey = dθ = 0.4498
−1.413 2π
,
SOLUTION MANUAL
,Solutions to Chapter 2 Problems
Introduction to Communication Systems, by Upamanyu Madhow
Problem 2.1 (a) We do this in two ways. The first is to directly show that y = x ∗ h for some
h, which shows that the system is LTI with impulse response h. We can rewrite
∫ ∞ ∫ ∞
y(t) = e−(t−u)I{u≤t}x(u)du = e−(t−u)I{t−u≥0}x(u)du
−∞ −∞
We see that this is in the convolution form
∫ ∞
y(t) = h(t — u)x(u)du
−∞
where h(t) = e−tIt≥0.
The second approach is to show that the system is LTI and then feed in an impulse to find
the impulse response. Linearity of y in x is clear, hence let us check for time invariance. Let
x1(t) = x(t — t0) be a delayed version of the input. The corresponding output is given by
∫ t ∫ t
y1(t) = eu−tx1(u)du = eu−tx(u — t0)du
−∞ −∞
Making the change of variables v = u — t0, we obtain that
∫ t−t0 ∫ t−t0
y1(t) = ev+t0 −tx(v)dv = ev−(t−t0)x(v)dv
−∞ −∞
Comparing with the original expression for y(t), it is clear that we have simply replaced t by
t — t0. That is, y1(t) = y(t —
t0) and the system is time invariant. We now find the impulse
response by setting the input to an impulse:
∫t
h(t) = eu−tδ(u)du
−∞
The impulse at time zero falls into the integration interval only if t ≥ 0, in which case we select
the value of the integrant at u = 0. We therefore obtain h(t) = e−tI{t≥0} as before.
(b) It is easy to see that the Fourier transform of the impulse response is H(f ) = j2πf1 +1 , with
magnitude |H(f )| = √ 21 2 . The plot is omitted, but it is clear that this is a sloppy lowpass
4π f +1
filter.
(c) We compute the energy in the frequency domain using Parseval’s identity. The input x(t) =
2sinc(2t) x X(f ) = I[−1,1](f ), and the output Y (f ) = H(f )X(f ) = 1
j2πf +1 I[−1,1] (f ). The energy
is given by
∫ ∞ ∫ 1 1
Ey = |Y (f )|2df = 2 2
df
−∞ −1 4π f + 1
Making the standard substitution 2πf = tan θ, so that 2πdf = sec2 θ dθ and 4π2f 2 + 1 =
tan2 θ + 1 = sec2 θ, we obtain (the limits of the transformed integral are ± tan−1(2π) = ±1.413)
∫ 1.413
1
Ey = dθ = 0.4498
−1.413 2π
,
