A Foundation in Digital Communication (2nd Edition, 2017) – Solutions Manual – Lapidoth

ALL 29 CHAPTERS COVERED




SOLUTIONS MANUAL

, Solutions to Exercises of Chapter 2

Exercise 2.1 (problem0109m)
Let x , Re(z) and y , Im(z). By Equation (2.3) in the book, the desired integral is defined as
Z ∞ Z ∞  Z ∞ 
−zt −xt −xt
e dt = e cos(−yt) dt + i e sin(−yt) dt
0 0
Z ∞  Z ∞0 
−xt −xt
= e cos(yt) dt − i e sin(yt) dt .
0 0

In the following, we consider each of the two integrals separately. We find that:
Z ∞
e−xt ∞
e−xt cos(yt) dt = 2 (−x cos(yt) + y sin(yt))
0 x + y2 t=0
1
= 2 x,
x + y2
and
∞
e−xt ∞
Z
e−xt sin(yt) dt = (−x sin(yt) − y cos(yt))
0 x2 + y 2 t=0
1
= 2 y.
x + y2
Thus, for Re(z) > 0 both of these expressions are finite. Combining these two expressions, we
obtain:
Z ∞
1 z∗ 1
e−zt dt = 2 2
(x − iy) = 2
= .
0 x +y |z| z



Exercise 2.2 (problem0103m)
For any two complex numbers w, z:

|w + z|2 = |w|2 + |z|2 + 2 Re(wz ∗ ) (2.1)

and
(|w| + |z|)2 = |w|2 + |z|2 + 2|w||z|. (2.2)
Since the real part of a complex number is never larger than its absolute value:

Re(wz ∗ ) ≤ |wz ∗ | = |w||z|. (2.3)

Combining this with (2.1) and (2.2), the Triangle-Inequality for Complex Numbers (2.11) follows.
Moreoever, we observe that equality holds only if (2.3) is satisfied with equality, which is the case
when (wz ∗ ) is real, i.e., when there exists a real number α so that w = αz.


Exercise 2.3 (problem0104m)
Two complex numbers are identical, whenever the real and imaginary parts coincide. Thus, for
two complex numbers z and w:
 
w=z ⇐⇒ Re(w) = Re(z) and Im(w) = Im(z) . (2.4)


c Amos Lapidoth, 2009 1

,Notice that for every complex number v the imaginary part satisfies

Im(v) = Re(−iv). (2.5)

By combining (2.4) and (2.5), it follows that w = z holds whenever the two conditions

Re(1w) = Re(1z) and Re(−iw) = Re(−iz)

are satisfied. This concludes the proof.


Exercise 2.4 (problem0105m)
By the Triangle Inequality for Complex Numbers (see Equation (2.11) in the book), for every t ∈ R

|u(t) − w(t)| = |(u(t) − v(t)) + (v(t) − w(t))| (2.6)
≤ |u(t) − v(t)| + |v(t) − w(t)| . (2.7)

Since (2.6) holds for all t ∈ R, by integrating over the real line the following inequality results:
Z ∞ Z ∞ Z ∞
|u(t) − w(t)| dt ≤ |u(t) − v(t)| dt + |v(t) − w(t)| dt,
−∞ −∞ −∞

where the integrals exist because u, v, and w are integrable. Indeed, e.g.,
Z ∞ Z ∞ Z ∞
|u(t) − w(t)| dt ≤ |u(t)| dt + |w(t)| dt < ∞
−∞ −∞ −∞

where the first inequality follows again by the Triangle Inequality for Complex Numbers and the
second inequality by the integrability of u and w.


Exercise 2.5 (problem0106m)
The function t 7→ I{t = 17}e−i2πf t is indistinguishable from the all-zero function, since these two
functions differ only on a set of Lebesgue measure 0, namely on the single point t = 17. Thus, the
integral under consideration equals 0.


Exercise 2.6 (problem0111m)
Let N ⊂ R be a set of Lebesgue measure zero. Thus, by Definition 2.5.1 in the book, for every
ǫ > 0 there exists a sequence of intervals [a1 , b1 ], [a2 , b2 ], . . . so that

• the union of these intervals covers N , and

• the sum of the lengths of the intervals is smaller than ǫ.

Notice that the sequence of intervals [a1 , b1 ], [a2 , b2 ], . . . , obviously also covers every subset of N .
Therefore, every subset of N trivially satisfies Definition 2.5.1 in the book, and thus is also a set
of Lebesgue measure 0.




c Amos Lapidoth, 2009 2

, Exercise 2.7 (problem0113m)




c Amos Lapidoth, 2009 3