Exam (elaborations) TEST BANK FOR Fundamentals of Digital Communication By Upamanyu Madhow (Solution Manual)
Solutions to Chapter 2
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Fundamentals of Digital Communication
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(t0 − ) = as( − t0) (|a| = 1)
for all . (We would get a = 1 if we insisted that y(t0) = A||s||2 rather than |y(t0)| = |A|||s||2.)
Setting a = 1, therefore, we have h(t) = s(−t).
f
p
2
Im(X p (f))
1/ 2
−22 −20 20 22
21 23
−23 −21
f
Re(X (f))
Figure 4: Passband spectrum for Problem 2.6(a).
1
Re(X(f))
0 2 f
2
Im(X(f))
1 3 f
Figure 5: Complex baseband spectrum for Problem 2.6(b).
Problem 2.6 (a) The real and imaginary parts of Xp(f) are sketched in Figure 4.
(b) Passing √2xp(t) cos 20t through a lowpass filter yields xc(t), the I component with respect
to fc = 20. In this case, it is easiest to find the Fourier transform (see Figure 5), and then the
time domain expression, for the complex baseband signal x(t), and then take the real part. We
see that
Re(X(f)) ↔ 4sinc(2t)ej2t
Im(X(f)) ↔ sinc2(t)ej4t
so that
X(f) = Re(X(f)) + jIm(X(f)) ↔ x(t) = 4sinc(2t)ej2t + jsinc2(t)ej4t
3
We can now read off
xc(t) = Re(x(t)) = 4sinc(2t) cos(2t) − sinc2(t) sin(4t)
−101
p
2
−99 99 101 f
V (f)
Figure 6: Passband spectrum for Problem 2.7(a).
+1
V(f)/ 2
−j2p t
−1 1
f
2
1 1
−1 1 −1
1
−1
f f
= +
d/df
−1
f
+1
A(f) B(f)
C(f)
−1 1
Figure 7: Computing the complex baseband representation in Problem 2.7(c).
2
V(f)/ 2
~
−j2p t
f
2
−2 0 f
d/df
1
−2
U(f)
Figure 8: Computing the complex baseband representation in Problem 2.7(d).
Problem 2.7 (a) The passband spectrum Vp(f) is sketched in Figure 6, where we use the fact
that V (f) = V (−f), since v(t) is real-valued.
(b) We have v(t) = v(−t) if V (f) = V (−f), which is clearly true here.
(c) The complex baseband spectrum V (f), with respect to f0 = 100, is sketched in Figure 7. In
order to compute v(t) with minimal work, we break up V (f)/√2 into two components, A(f) and
B(f). Clearly, a(t) = 2sinc(2t). To find b(t), we minimize work further by differentiating in the
frequency domain to obtain C(f) = d/df B(f). We have c(t) = 2 cos 2t−2sinc(2t) = −j2tb(t),
which implies that
v(t)/√2 = a(t) + b(t) = 2sinc(2t) +
cos 2t − sinc(2t)
−jt
4
We therefore obtain that the I and Q components are given by
vc(t) = Re(v(t)) = 2√2sinc(2t)
vs(t) = Im(v(t)) = √2
cos 2pit − sinc(2t)
t
Note that we could have done this even more easily by directly differentiating V (f). We illustrate
this in the next part.
(d) Denoting the complex baseband representation with respect to f0 = 101 as ˜v, we have that
vp(t) = Re
√2˜v(t)ej202t
= Re
√2v(t)ej200t
so that
˜v(t) = v(t)e−j2t
Substituting into (c) and reading off real and imaginary parts gives us the desired result. Alternatively,
let us proceed from scratch, as shown in Figure 8. The complex baseband representation
˜ V (f) is shown. To minimize computations, we differentiate in the frequency domain to get U(f),
as shown. Clearly,
u(t) = 2e−j4t + 2sinc(2t)e−j2t = −j2t˜v(t)/√2
We can now simplify to read off ˜v(t) and the corresponding I and Q components.
Problem 2.8 (a) For a frequency reference fc = 50, clearly the complex envelope for up is given
by u(t) = sinc(2t). For vp, we write
vp(t) = √2sinc(t) sin(101t +
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