Modern Digital and Analog Communication Systems by B. P. Lathi and Z. Ding 4th Edition Solutions Manual

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1

, Chapter 2


2.1-1 Both ϕ(t) and w0(t) are periodic.
∫T 2 ∫π 2 −π
The average power of ϕ(t) is P g = ϕ (t) dt = e−t/2 dt = .
1 1 1−e
T 0 π 0 π
∫ T0 ∫ T0
The average power of w0 (t) is P g = 1
T0 o
w2o(t) dt = 1
T0 0
1 · dt = 1.


2.1-2

∫2
(a) Since x(t) is a real signal, Ex = 0
x2(t) dt.
Solving for Fig. S2.1-2(a), we have
∫2 2 ∫1 2 ∫2
E x= (1) dt = 2, E y= (1) dt + (−1)2 dt = 2
0 0 1
∫1 2 ∫2 2
Ex+y = 0 (2) dt = 4, E x−y = (2) dt = 4
1

Therefore, Ex±y = Ex + Ey.
Solving for Fig. S2.1-2(b), we have
∫π ∫ 2π ∫ π/2 ∫π ∫ 3π/2 ∫ 2π
E x= (1)2 dt + (−1)2 dt = 2π, Ey = (1)2 dt + (−1)2 dt + (1)2 dt + (−1)2 dt = 2π
0 π 0 π/2 π 3π/2
∫ π/2 ∫ 3π/2 2 ∫ 2π
E x+y = 0
(2)2 dt + π/2
(0) dt + 3π/2(−2)2 dt = 4π
∫ π/2 2 ∫π ∫ 3π/2 ∫ 2π
E x−y = 0
(0) dt + π/2 (2)2 dt + π/2 (−2)2 dt + 3π/2 (0)2 dt = 4π
Therefore, Ex±y = Ex + Ey.



2 2 2
3
2 4 
0 0 0
 
2 4
-2 -2
(a) (b) (c)

x t  y t

2 2 2

3
2 4 
0 0 4 0

2
-2 -2
(a) (b) (c)

x t  y t


Fig. S2.1-2


(b)

∫ π/4 2 ∫π ∫π
Ex= 0
(1) dt + π/4 (−1)2 dt = π, Ey = 0
(1)2 dt = π


2

, ∫ π/4 ∫π ∫ π/4 ∫π
E x+y = 0 (2)2 dt + π/4 (0)2 dt = π, E x−y = 0 (0)2 dt + π/4 (−2)2 dt = 3π Therefore, Ex±y E x + E y, and
Ex̂±ŷ = Ex̂ ± Eŷ are not true in general.


2.1-3
∫ T0
C2
Pg 1 ∫ T0 C2 cos2 (ω0t + θ) dt = [1 + cos (2 ω0t + 2θ)] dt
= T0 0 2T0 0
#
" ∫ T0
2 ∫ T0 C2 2
C + cos (2 + 2 ) = [ + 0] = C
= dt ω 0t θ dt T0
2T 0 0 0 2T 0 2



2.1-4 If ω1 = ω2, then

g2(t) = (C1 cos (ω1t + θ1) + C2 cos (ω1t + θ2))2
= C2 cos2(ω1t + θ1) + C2 cos2(ω1t + θ2) + 2C1C2 cos (ω1t + θ1) cos (ω1t + θ2)
1 2



∫
1 T0
2
Pg = limT0→∞ (C1 cos (ω1t + θ1) + C2 cos (ω1t + θ2)) dt
T0 0
∫ T0
C21 C22 C1 C2 1 ω1t θ1 ω1t θ2 dt
= 2 + 2 + lim T →∞2 T0 0 cos ( + ) cos ( + )
∫ T0
C12
C 2
1 1
= + 2 + lim 2C1C2 cos (2 ω1t + θ1 + θ2 ) + cos (θ1 − θ2 ) dt
C22 C22 T →∞ T0 0 2
2C1C2
= 1
+ 2 +0+ cos (θ1 − θ2)
2 2 2
C 1 + C 1 + 2C1C2 cos (θ1 − θ2)
2 2
=
2



2.1-5
∫ 2
1
Pg = (t3 )2 dt = 64/7
4 −2

(a) P−g ∫2 −t 3 2 dt = 64 /7
1
4
= −2 ( )
∫
(b) P2g 1 2 t3 2 dt = 4(64/7) = 256 /7
= 4 −2(2 )
∫
1 2
(c) Pcg = (ct3 )2 dt = 64c2/ 7
4 −2

Changing the sign of a signal does not affect its power. Multiplication of a signal by a constant c increases the
power by a factor of c2.


2.1-6 Let us denote the signal in question by g(t) and its energy by Eg.


(a),(b) For parts (a) and (b), we write

∫ 2π
1 ∫ 2π 1 ∫ 2π
Eg = sin2 t dt = dt − cos 2t dt = π + 0 = π
0 2 0 2 0


3

, (c)
∫ 4π
1 ∫ 4π 1 ∫ 4π
Eg = sin2 t dt = dt — cos 2t dt = π + 0 = π
2π 2 2π 2 2π

(d) ∫
∫ 2π
1 2π
1 ∫ 2π
Eg = (2 sin t) dt = 4 2
2
0 dt − cos 2t dt = 4[π + 0] = 4π
0 2 0

Sign change and time shift do not affect the signal energy. Doubling the signal quadruples its energy. In the same
way, we can show that the energy of kg(t) is k2Eg.


2.1-7
∫
lim 1
T/2
Pg = g(t)g∗(t) dt
T →∞ T −T/2
∫ n Σ
Σ n
lim 1
T/2
= Dk D∗ r ej(ωk −ωr )t dt
T →∞ T −T/2 k=m r=m

∫ Σ ∫
Σ
n n
Dk D∗ r ej(ωk −ωr )t dt + lim 1 Σ
n
lim 1
T/2 T/2
= |Dk| 2 dt
T →∞ T −T /2 k=m r=m,r/=k T →∞ T −T/2 k=m

The integrals of the cross-product terms (when k /= r) are finite because the integrands (functions to be integrated)
are periodic signals (made up of sinusoids). These terms, when divided by T → ∞ , yield zero. The remaining terms
(k = r) yield
∫ T/2 Σ n Σ
n
Pg = lim 1 |Dk|2 dt = |Dk| 2
T →∞ T −T/2
k=m k=m




2.1-8

2
(a) From Eq. (2.5a), the power of a signal of amplitude C is P = C , regardless of phase and frequency; therefore,
√
g
√ 2
Pg = 100/2 = 50; the rms value is Pg = 5 2.


(b) From Eq. (2.5b), the power of the sum ofC2twoC2 sinusoids of different frequencies is the sum of the power of
individual sinusoids, regardless of the phase, + , therefore, P = 100/2 + 256/2 = 50 + 128 = 178; the rms
√ √ 1
2
2
g
value is P g = 178. 2



(c) g(t) = (10 + 2 sin (3t)) cos (10t)=10 cos (10t) + 2 sin (3t) cos (10t) = 10 cos (10t) + sin (13t) − cos (7t)
√ √
Therefore, Pg = 100/2 + 1/2 + 1/2 = 50 + 0.5 + 0.5 = 51; the rms value is Pg = 51.


(d) g(t) = 10 cos (5t) cos (10t)= 10(cos (15t)+cos
2
(5t))
= 5 cos (15t) + 5 cos (5t)
√
Therefore, Pg = 25/2 + 25/2 = 25; the rms value is Pg = 5.


(e) g(t) = 10 sin (5t) cos (10t)=5 (cos (15t) − cos (5t)) = 5 cos (15t) − 5 cos (5t)
√
Therefore, Pg = 25/2 + 25/2 = 25; the rms value is Pg = 5.

2
(f) |g(t)| = cos2(ω0t)
√ √
Therefore, Pg = 1/2 = 0.5; the rms value is Pg = 0.5


4