TEST BANK FOR Fundamentals of Digital Communication By Upamanyu Madhow

Problem 2.1: Rather than doing the details of the convolution, we simply sketch the shapes of
the waveforms. For a signal s = sc + jss and a filter h = hc + jhs, the convolution
y = s ∗ h = (sc ∗ hc − ss ∗ hs) + j(sc ∗ hs + ss ∗ hc)
For h(t) = smf (t) = s(−t), rough sketches of Re(y), Im(y) and |y| are shown in Figure 1.
Clearly, the maximum occurs at t = 0.
=
c*hc −ss*hs
Re(s* h)
sc*hs ss*hc Im(s* h)
|s* h|
+ =
+
s
Figure 1: The convolution of a signal with its matched filter yields at peak at the origin.
Problem 2.2:
(a) Multiplication in the time domain corresponds to convolution in the frequency domain. The
two sinc functions correspond to boxcars in the frequency domain, convolving which gives that
S(f) has a trapezoidal shape, as shown in Figure 2.
(b) We have
u(t) = s(t) cos(100t) = s(t)
ej100t + e−j100t
2 ↔ U(f) =
S(f − 50) + S(f + 50)
2
The spectrum U(f) is plotted in Figure 2.
1/4
p
sinc(t)
−1/2 1/2
1
1/2
−1 1
sinc(2t)
s(t) = sinc(t) sinc(2t)
S(f)
1/2
1/2 3/2
f f
CONVOLVE
... ...
f
f
U(f)
−51.5 −50 −48.5 48.5 50 51.5
u(t) = s(t) cos(100 t)
Figure 2: Solution for Problem 2.2.
Problem 2.3: The solution is sketched in Figure 3.
(a) We have s(t) = I[−5,5] ∗ I[−5,5]. Since I[−5,5](t) ↔ 10sinc(10f), we have S(f) = 100sinc2(10f).
(b) We have
u(t) = s(t) sin(1000t) = s(t)
ej1000t − e−j100t
2j ↔ U(f) =
S(f − 50) − S(f + 50)
2j
0.1
−5 5 −5 5
S(f) = 100 sinc 2 (10f)
−10 10
t
s(t)
10
1 1
10 sinc(10f)
= t * t
... ...
−0.2 −0.1 0.1 0.2 f
100
... ... 500
−500 f
Im (U(f))
Re(U(f) = 0
50
Figure 3: Solution for Problem 2.3.
Problem 2.4: Part (a) is immediate upon expanding ||s − ar||2.
(b) The minimizing value of a is easily found to be
amin = hs, ri
||r||2
Substituting this value into J(a), we obtain upon simplification that
J(amin) = ||s||2 − hs, ri2
||r||2
The condition J(amin) ≥ 0 is now seen to be equivalent to the Cauchy-Schwartz inequality.
(c) For nonzero s, r, the minimum error J(amin) in approximating s by a multiple of r vanishes if
and only if s is a multiple of r. This is therefore the condition for equality in the Cauchy-Scwartz
inequality. For s = 0 or r = 0, equality clearly holds. Thus, the condition for equality can be
stated in general as: either s is a scalar multiple of r (this includes s = 0 as a special case), or r
is a scalar multiple of s (this includes r = 0 as a special case).
(d) The unit vector in the direction of r is u = r
||r||
. The best approximation of s as a multiple
of r is its projection along u, which is given by
ˆs = hs, uiu = hs,
r
||r||i
r
||r||
and the minimum error is J(amin) = ||s − ˆs||2.
Problem 2.5: We have
y(t) = (x ∗ h)(t) =
Z
As( − t0)h(t − )d = Ahst0 , ht
i
where st0( ) = s( − t0) and ht( ) = h(t − ) are functions of . (Recall that the complex inner
product is defined as hu, vi =
R
uv).
(a) Using the Cauchy-Schwartz inequality, we have
|y(t)| ≤ |A|||st0||||ht
||
2
It is easy to check (simply change variables in the associated integrals) that ||st0 || = ||s|| and
||ht
|| = ||h||. Using the normalization ||h|| = ||s||, we obtain the desired result that |y(t)| ≤ A||s||2.
(b) Equality is attained for t = t0 if ht
= ast0 for t = t0 for some scalar a. Since ||st0 || = ||ht
|| =
||s||, we must have |a| = 1. Thus, we have
h(