Solutions for Problems in Modern Digital and Analog Communication Systems, 5th Edition by Lathi and Ding

Chapter 2
2.1-1 The step function u(t) is a power signal because its energy is innite, i.e., Eu =
∞
−∞
u2
(t) dt = ∞.
2.1-2 Let us denote the signal in question by g(t) and its energy by Eg.
(a),(b) For parts (a) and (b), we write
Eg =
2π
0
sin2
t dt =
1
2
2π
0
dt −
1
2
2π
0
cos 2t dt = π + 0 = π
(c)
Eg =
4π
2π
sin2
t dt =
1
2
4π
2π
dt −
1
2
4π
2π
cos 2t dt = π + 0 = π
(d)
Eg =
2π
0
(2 sin t)
2
dt = 4

1
2
2π
0
dt −
1
2
2π
0
cos 2t dt

= 4[π + 0] = 4π
Sign change and time shift do not aect the signal energy. Doubling the signal quadruples its energy. In the
same way, we can show that the energy of kg(t) is k2
Eg.
2.1-3 Both ϕ(t) and w0(t) are periodic.
The average power of ϕ(t) is Pg = 1
T
T
0
ϕ2
(t) dt = 1
π
π
0
e−t/2
2
dt = 1−e−π
π .
The average power of w0(t) is Pg = 1
T0
T0
o
w2
o(t) dt = 1
T0
T0
0
1 · dt = 1.
2.1-4
(a) Since x(t) is a real signal, Ex =
2
0
x2
(t) dt.
Solving for Fig. S2.1-4(a), we have
Ex =
2
0
(1)2
dt = 2, Ey =
1
0
(1)2
dt +
2
1
(−1)2
dt = 2
Ex+y =
1
0
(2)2
dt = 4, Ex−y =
2
1
(2)2
dt = 4
Therefore, Ex±y = Ex + Ey.
Solving for Fig. S2.1-4(b), we have
Ex =
π
0
(1)2
dt +
2π
π
(−1)2
dt = 2π, Ey =
π/2
0
(1)2
dt +
π
π/2
(−1)2
dt +
3π/2
π
(1)2
dt +
2π
3π/2
(−1)2
dt = 2π
Ex+y =
π/2
0
(2)2
dt +
3π/2
π/2
(0)2
dt +
2π
3π/2
(−2)2
dt = 4π
Ex−y =
π/2
0
(0)2
dt +
π
π/2
(2)2
dt +
3π/2
π/2
(−2)2
dt +
2π
3π/2
(0)2
dt = 4π
Therefore, Ex±y = Ex + Ey.
(b) Ex =
π/4
0
(1)2
dt +
π
π/4
(−1)2
dt = π, Ey =
π
0
(1)2
dt = π
Ex+y =
π/4
0
(2)2
dt +
π
π/4
(0)2
dt = π, Ex−y =
π/4
0
(0)2
dt +
π
π/4
(−2)2
dt = 3π
Therefore, Ex±y 6= Ex + Ey, and Ex̂±ŷ = Ex̂ ± Eŷ are not true in general.
1
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0
2
2 2
0
(a)
-2
(b)
2
0
-2
(c)
0
2
2 2
0
(a)
-2
(b)
2
0
-2
(c)
2
π
3
4
π
4
π
( ) ( )
x t y t
+
2
π
3
4
π
π
π
4
π
( ) ( )
x t y t
−
Fig. S2.1-4
2.1-5
Pg =
1
4
2
−2
(t3
)2
dt = 64/7
(a) P−g(t) =
1
4
2
−2
(−t3
)2
dt = 64/7
(b) P1.5g(t) =
1
4
2
−2
(1.5t3
)2
dt = 144/7
(c) Pg(−t) =
1
4
2
−2
(−t3
)2
dt = 64/7
(d) Pg(1.5t) =
3
8
4/3
−4/3
((1.5t)3
)2
dt = 64/7
Comment: Changing the sign of a signal does not aect its power. Multiplication of a signal by a constant
c increases the power by a factor of c2
. Time-scaling of a signal does not change its power, though the signal
period changes.
2.1-6
Pg =
1
T0
T0
0
C2
cos2
(ω0t + θ) dt =
C2
2T0
T0
0
[1 + cos (2ω0t + 2θ)] dt
=
C2
2T0
T0
0
dt +
T0
0
cos (2ω0t + 2θ) dt
#
=
C2
2T0
[T0 + 0] =
C2
2
2.1-7 If ω1 = ω2, then
g2
(t) = (C1 cos (ω1t + θ1) + C2 cos (ω1t + θ2))
2
= C2
1 cos2
(ω1t + θ1) + C2
2 cos2
(ω1t + θ2) + 2C1C2 cos (ω1t + θ1) cos (ω1t + θ2)
2

Pg = limT0→∞
1
T0
T0
0
(C1 cos (ω1t + θ1) + C2 cos (ω1t + θ2))
2
dt
=
C2
1
2
+
C2
2
2
+ lim
T →∞
2C1C2
1
T0
T0
0
cos (ω1t + θ1) cos (ω1t + θ2) dt
=
C2
1
2
+
C2
2
2
+ lim
T →∞
2C1C2
1
T0
T0
0
1
2

cos (2ω1t + θ1 + θ2) + cos (θ1 − θ2)

dt
=
C2
1
2
+
C2
2
2
+ 0 +
2C1C2
2
cos (θ1 − θ2)
=
C2
1 + C2
1 + 2C1C2 cos (θ1 − θ2)
2
2.1-8
Pg = lim
T →∞
1
T
T/2
−T/2
g(t)g∗
(t) dt
= lim
T →∞
1
T
T/2
−T/2
n
X
k=m
n
X
r=m
DkD∗
rej(ωk−ωr)t
dt
= lim
T →∞
1
T
T/2
−T/2
n
X
k=m
n
X
r=m,r6=k
DkD∗
rej(ωk−ωr)t
dt + lim
T →∞
1
T
T/2
−T/2
n
X
k=m
|Dk|2
dt
The integrals of the cross-product terms (when k 6= r) are nite 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
Pg = lim
T →∞
1
T
T/2
−T/2
n
X
k=m
|Dk|2
dt =
n
X
k=m
|Dk|2
2.1-9
(a) From Eq. (2.5a), the power of a signal of amplitude C is Pg = C2
2 , regardless of phase and frequency;
therefore, Pg = 25/2; the rms value is
p
Pg = 5/
√
2.
(b) From Eq. (2.5b), the power of the sum of two sinusoids of dierent frequencies is the sum of the power
of individual sinusoids, regardless of the phase, C2
1
2 +
C2
2
2 , therefore, Pg = 25/2 + 4/2 = 12.5 + 2 = 14.5;
the rms value is
p
Pg =
√
14.5.
(c) Same as (b), Pg = 25/2 + 4/2 = 12.5 + 2 = 14.5; the rms value is
p
Pg =
√
14.5.
(d) g(t) = 5 sin (55t) sin (πt)=5(cos (55t−πt)−cos (55t+πt))
2
Therefore, Pg = 25/8 + 25/8 = 25/4; the rms value is
p
Pg = 5/2.
(e) Given g(t) = 10 sin (5t) cos (10t) u(t). By denition,
Pg = limT →∞
1
T
T/2
0
100 sin2
(5t) cos2
(10t)dt
= limT →∞
100
T
T/2
0

(1−cos(10t))
2
(1+cos(20t))
2

= 100
4T
T
2 = 25/2.
The rms value is
p
Pg = 5/
√
2.
3

(f) |g(t)|
2
= sin2
(ω0t)
Therefore, Pg = 1/2 = 0.5; the rms value is
p
Pg =
√
0.5.
2.1-10
(a) Pg = 1
4
2
−2
12
dt = 1; the rms value is
p
Pg = 1.
(b) Pg = 1
10π
π
−π
12
dt = 1
5 ; the rms value is
p
Pg = 1/
√
5.
(c) Pg = 1
2π
2π
0
t
2π
2
dt = 1
p
Pg = 1/
√
3.
(d) Pg = 1
π
π/4
−π/4
4t
π
2
dt = 1
p
Pg = 1/
√
6.
(e) Pg = 1
3
1
0
t2
dt = 1
p
Pg = 1/3.
(f) Pg = 1
6
−1
−2
(t + 2)2
dt +
1
−1
dt +
2
1
(t − 2)2
dt

= 4
p
Pg = 2/3.
2.2-1 If a is real, then both Eg =
∞
−∞
|e−at
|2
· dt = ∞ and Pg = limT →∞
1
T
T/2
−T/2
|e−at
|2
· dt =
limT →∞
1
T
T/2
−T/2
e−2at
· dt = ∞.
If a is purely imaginary, a = iα; then, g(t) = e−iαt
and |g(t)|
2
= 1. Pg = limT →∞
1
T
T/2
−T/2
1 · dt =
limT →∞
1
T T = 1. It is not an energy signal since Eg =
∞
−∞
|g(t)|2
· dt = ∞. Hence it is a power signal.
2.2-2
(a) Pg = limT →∞
1
T
T/2
−T/2
t4
dt = ∞. Hence, it is not a power signal.
(b) Eg =
∞
−∞
t2
dt = ∞. Hence, it is not an energy signal.
2.3-1
g2(t) = g(t − 1) + g1(t − 1), g3(t) = g(t − 1) + g1(t + 1), g4(t) = g(t − 0.5) + g1(t + 0.5)
The signal g5(t) can be obtained by (i) delaying g(t) by 1 second (replace t with t − 1), (ii) then time-
expanding by a factor 2 (replace t with t/2), (iii) then multiplying by 1.5. Thus g5(t) = 1.5g( t
2 − 1).
2.3-2 See Fig. S2.3-2.
2.3-3
(a) See Fig. S2.3-3.
4

0 1 2
t
0
1
(a) xa
(t) = g(-t+1) + g(t-1)
-1 0 1 2
t
0
1
2
(b) xb
(t) = g2
(t) + g3
(t)
0 1 2 3
t
0
1
2.25
(c) xc
(t) = [g5
(t-1)]2
Fig. S2.3-2
(b) Eg =
∞
−∞
|g(t)|2
dt =
15
6
1
6 (t − 12)
2
dt +
24
15

− 1
18 (t − 24)
2
dt = 3.
Based on the properties derived in Prob. 2.3-6, the energies are
(i) Eg(−t) = 3
(ii) Eg(t+2) = 3
(iii) Eg(−3t) = Eg/3 = 1
(iv) Eg(t/3) = 3Eg = 9
(v) Eg(2t+1) = Eg/2 = 1.5
(vi) Eg(2t+2) = Eg/2 = 1.5
2.3-4 Denote g(at) = f(t). Since g(t) is periodic with period T,
g(t) = g(t + T)
g(a t) = g(a t + T) = g

a t +
T
a

f(t) = f

t +
T
a

Therefore, g(at) is periodic with period T/a.
The average power of g(a t) is
Pg(a t) = lim
T →∞
a
T
T/2a
−T/2a
g2
(at) dt = lim
T →∞
a
T
T/2
−T/2
g2
(z)
dz
a
= lim
T →∞
1
T
T/2
−T/2
g2
(z) dz = Pg
Therefore, the average power remains the same.
2.3-5 See Fig. S2.3-5.
5

-24 -15 -6 0
t
-1
0
0.5
(i) g(-t)
0 4 13 22
t
-1
0
0.5
(ii) g(t+2)
-8 -5 -2 0
t
-1
0
0.5
(iii) g(-3t)
0 18 45 72
t
-1
0
0.5
(iv) g(t/3)
0 2.5 7 11.5
t
-1
0
0.5
(v) g(2t+1)
0 2 6.5 11
t
-1
0
0.5
(vi) g(2t+2)
Fig. S2.3-3
6

-2 0 2 4
t
0
1
2
3
4
5
6
g(t-2)
-5.3333 0 2.6667
t
0
1
2
3
4
5
6
g(3t/4)
-0.5 0 1.5 2.5
t
0
1
2
3
4
5
6
g(2t-3)
0 2 4
t
0
1
2
3
4
5
6
g(2-t)
Fig. S2.3-5
7

2.3-6
E−g =
∞
−∞
[−g(t)]2
dt =
∞
−∞
g2
(t) dt = Eg, Eg(−t) =
∞
−∞
[g(−t)]2
dt =
∞
−∞
g2
(x) dx = Eg
Eg(t−T ) =
∞
−∞
[g(t − T)]2
dt =
∞
−∞
g2
(x) dx = Eg, Eg(at) =
∞
−∞
[g(at)]2
dt =
1
a
∞
−∞
g2
(x) dx = Eg/a
Eg(at−b) =
∞
−∞
[g(at−b)]2
dt =
1
a
∞
−∞
g2
(x) dx = Eg/a, Eg(t/a) =
∞
−∞
[g(t/a)]2
dt = a
∞
−∞
g2
(x) dt = aEg
Eag(t) =
∞
−∞
[ag(t)]2
dt = a2
∞
−∞
g2
(t) dt = a2
Eg
2.4-1 Using the facts that φ(t)δ(t) = φ(0)δ(t) and φ(t)δ(t − T) = φ(T)δ(t − T), we have
(a) tan(3π/4)
π2/8+1 δ(t − π
4 ) = −8
π2+8 δ(t − π
4 )
(b) 1−j
10π δ(ω + π)
(c) −eπ/15
δ(t + π/15)
(d) sin(1.5π)
1−4 δ(t − 1) = 1
3 δ(t − 1)
(e) cos(−3π/2)
1/2 = 0
(f) k2
δ(ω) − 4 sin2
(kπ/2)
π2 δ(ω + π/2) (use L'Hôpital's rule limω→0
sin kω
kω = 1)
2.4-2
(a) −8
π2+8
(b) 1−j
10π
(c) −eπ/15
(d) 1
3
(e) 0
(f) k2
− 4 sin2
(kπ/2)
π2
2.4-3 Letting at = x, we obtain (for a 0)
∞
−∞
φ(t)δ(at) dt =
1
a
∞
−∞
φ
x
a

δ(x) dx =
1
a
φ(0)
Similarly for a 0, we show that this integral is −1
a φ(0). Therefore
∞
−∞
φ(t)δ(at) dt =
1
|a|
φ(0) =
1
|a|
∞
−∞
φ(t)δ(t) dt
δ(at) =
1
|a|
δ(t)
Therefore,
δ(ω) = δ(2πf) =
1
2π
δ(f)
.
8

2.4-4 Using the fact that δ(at) = 1
|a| δ(t) (see Problem 2.4-3), and the equality
b
a
φ(t)δ(t − T) dt = φ(T),
we get
(a) g(−3t + a)
(b) g(t)
(c) ej2ω
(d) 0
(e) e6
/2
(f) 5
(g) 2g(−4)
(h) cos(7π/3)
3
2.5-1
(a) In this case Ex =
1
0
dt = 1, and
c =
1
Ex
1
0
g(t)x(t) dt =
1
1
1
0
t dt = 0.5
(b) Thus, g(t) ≈ 0.5x(t), and the error e(t) = t − 0.5 over (0 ≤ t ≤ 1), and zero outside this interval. Also
Eg and Ee (the energy of the error) are
Eg =
1
0
g2
(t) dt =
1
0
t2
dt = 1/3 and Ee =
1
0
(t − 0.5)2
dt = 1/12
The error (t − 0.5) is orthogonal to x(t) because
1
0
(t − 0.5)(1) dt = 0
Note that Eg = c2
Ex + Ee. To explain these results in terms of vector concepts, we observe from Fig. 2.13
that the error vector e is orthogonal to the component cx. Because of this orthogonality, the length-square
of g [energy of g(t)] is equal to the sum of the square of the lengths of c · x and e [sum of the energies of
cx(t) and e(t)].
2.5-2 In this case Eg =
1
0
g2
(t) dt =
1
0
t2
dt = 1/3, and
c =
1
Eg
1
0
x(t)g(t) dt = 3
1
0
t dt = 1.5
Thus, x(t) ≈ 1.5g(t), and the error e(t) = x(t) − 1.5g(t) = 1 − 1.5t over (0 ≤ t ≤ 1), and zero outside this
interval. Also Ee (the energy of the error) is Ee =
1
0
(1 − 1.5t)2
dt = 1/4.
9

2.5-3
|e|2
= |g|2
+ c2
|x|2
− 2cg · x
To minimize error, set d|e|2
dc = 0:
2c|x|2
− 2g · x = 0
c =
g · x
|x|2
=
g, x
|x|2
2.5-4
(a) In this case Ex =
1
0
sin2
2πt dt = 0.5, and
c =
1
Ex
1
0
g(t)x(t) dt =
1
0.5
1
0
t sin 2πt dt = −1/π
(b) Thus, g(t) ≈ −(1/π)x(t), and the error e(t) = t + (1/π) sin 2πt over (0 ≤ t ≤ 1), and zero outside this
interval.
Also Eg and Ee (the energy of the error) are
Eg =
1
0
g2
(t) dt =
1
0
t2
dt = 1/3
and
Ee =
1
0
[t + (1/π) sin 2πt]2
dt =
1
3
−
1
2π2
The error [t + (1/π) sin 2πt] is orthogonal to x(t) because
1
0
sin 2πt[t + (1/π) sin 2πt] dt = 0
Note that Eg = c2
Ex +Ee. To explain these results in terms of vector concept we observe from Fig. 2.13 that
the error vector e is orthogonal to the component cx. Because of this orthogonality, the square of length of
g [energy of g(t)] is equal to the sum of squares of the lengths cx and f [sum of the energies of cx(t) and
e(t)].
2.5-5
(a) If x(t) and y(t) are orthogonal, then we can show that the energy of x(t) ± y(t) is Ex + Ey.

|x(t) ± y(t)|2
dt =
∞
−∞
|x(t)|2
dt +
∞
−∞
|y(t)|2
dt ±
∞
−∞
x(t)y∗
(t) dt ±
∞
−∞
x∗
(t)y(t) dt
=
∞
−∞
|x(t)|2
dt +
∞
−∞
|y(t)|2
dt
The last result follows from the fact that because of orthogonality, the two integrals of the cross products
x(t)y∗
(t) and x∗
(t)y(t) are zero. Thus the energy of x(t) + y(t) is equal to that of x(t) − y(t) if x(t) and y(t)
are orthogonal.
10

(b) We can use a similar argument to show that the energy of c1x(t)+c2y(t) is equal to that of c1x(t)−c2y(t)
if x(t) and y(t) are orthogonal. This energy is given by |c1|2
Ex + |c2|2
Ey.
(c) If z(t) = x(t) ± y(t), then it follows from part (a) in the preceding derivation that
Ez = Ex + Ey ± (Exy + Eyx)
2.6-1 We shall use Eq. (2.51) to compute ρn for each of the four cases. Let us rst compute the energies
of all the signals:
Ex =
1
0
sin2
2πt dt = 0.5
In the same way, we nd Eg1 = Eg2 = Eg3 = Eg4 = 0.5.
From Eq. (2.51), the correlation coecients for four cases are found as follows:
(1) 1
√
(0.5)(0.5)
1
0
sin 2πt sin 4πt dt = 0
(2) 1
√
(0.5)(0.5)

(sin 2πt) (− sin 2πt) dt = −1
(3) 1
√
(0.5)(0.5)
1
0
0.707 sin 2πt dt = 0
(4) 1
√
(0.5)(0.5)
h 0.5
0
0.707 sin 2πt dt −
1
0.5
0.707 sin 2πt dt
i
= 2.828/π
Signals x(t) and g2(t) provide the maximum protection against noise.
2.6-2 Eg1 =
∞
−∞
g2
1(t)dt =
2
0
1dt = 2 and Eg2 =
∞
−∞
g2
2(t)dt =
∞
0
e−t
= 1.
Therefore, ρ = 1
√
2·1
2
0
e−0.5t
dt =
√
2 1 − e−1

.
2.6-3 Given g(t) = exp(−2t) cos(πt)u(t), the autocorrelation function is
ψg(τ) =
∞
−∞
g(t)g(t + τ)dt
=
∞
−∞
exp(−2t) cos(πt)u(t) exp(−2(t + τ)) cos(π(t + τ))u(t + τ)dt
If τ ≥ 0,
ψg(τ) =
∞
0
exp(−2t) cos(πt) exp(−2(t + τ)) cos(π(t + τ))dt
=
∞
0
exp(−4t − 2τ)

1
2
cos(2πt + πτ) +
1
2
cos(πτ)

dt
=
1
2
∞
0
exp(−4(t +
τ
2
)) · cos(2π(t +
τ
2
)) dt +
1
2
exp(−2τ) cos(πτ)
∞
0
exp(−4t) dt
=
1
8
exp(−2τ) cos(πτ) +
1
8π2 + 32
exp(−2τ)(4 cos(πτ) − 2π sin(πτ)), τ ≥ 0.
Because of symmetry,
ψg(τ) =
1
8
exp(−2|τ|)

π2
+ 8
π2 + 4
cos(πτ) −
2π
π2 + 4
sin(π|τ|)).

11

2.7-1
(a) g1 = (2, −1), g2 = (−1, 2), g3 = (0, −1), g4 = (1, 2), g5 = (2, 1), and g6 = (3, 0).
(b) Signal pairs (g3, g6), (g1, g4) and (g2, g5) are orthogonal. We can verify this analytically.
g3, g6 = (0 × 3) + (−2 × 0) = 0,
g1, g4 = (2 × 1) + (−1 × 2) = 0,
g2, g5 = (−1 × 2) + (2 × 1) = 0
We can show that the corresponding signal pairs are also orthogonal.
∞
−∞
g3(t)g6(t) dt =
∞
−∞
[−x2(t)][3x1(t)] dt = 0
∞
−∞
g1(t)g4(t) dt =
∞
−∞
[2x1(t) − x2(t)][x1(t) + 2x2(t)] dt = 0
∞
−∞
g2(t)g5(t) dt =
∞
−∞
[−x1(t) + 2x2(t)][2x1(t) + x2(t)] dt = 0
In deriving these results, we used the fact that
∞
−∞
x2
1 dt =
∞
−∞
x2
2(t) dt = 1 and
∞
−∞
x1(t)x2(t) dt = 0.
(c) Because
∞
−∞
x2
1 dt =
∞
−∞
x2
2(t) dt = 1 and
∞
−∞
x1(t)x2(t) dt = 0, signal energy Eg = c2
1 + c2
2 for
g(t) = c1x1(t) + c2x2(t). Therefore, Eg1 = 5, Eg2 = 5, Eg3 = 4, Eg4 = 5, Eg5 = 5, Eg6 = 9.
2.7-2
(a) Since x(t) and e(t) are mutually orthogonal, we can choose these two functions as the basis functions.
After normalizing, we obtain ψ1(t) = x(t)/
√
Ex = x(t) = 1, 0 ≤ t ≤ 1 and ψ2(t) = e(t)/
√
Ee = 2
√
3 e(t) =
2
√
3(t − 0.5), 0 ≤ t ≤ 1.
(b) Through orthogonal decomposition, we know that g(t) = 0.5 x(t) + e(t) = 0.5 ψ1(t) + 1
2
√
3
ψ2(t) and
x(t) = ψ1(t) + 0 ψ2(t). Thus, the vector representations are g = [1
2
1
2
√
3
] and x = [1 0], respectively.
(c) Similarly, this time ψ1(t) = x(t)/
√
Ex =
√
2 x(t) =
√
2 sin(2πt), 0 ≤ t ≤ 1 and ψ2(t) = e(t)/
√
Ee =
1
√
1/3−1/(2π2)
(t + sin(2πt)/π), 0 ≤ t ≤ 1.
g(t) = − 1
π x(t) + e(t) = − 1
π
√
2
ψ1(t) +
q
1
3 − 1
2π2 ψ2(t) and x(t) = 1
√
2
ψ1(t) + 0 ψ2(t). Thus, the vector
representations are g = [− 1
π
√
2
q
1
3 − 1
2π2 ] and x = [ 1
√
2
0], respectively.
2.7-3
12

(a) We can choose the normalized x(t), g1(t), g3(t) as the rst three orthonormal bases for the set of signals,
since all the correlations between these signals are zeros. Therefore,
φ1(t) =
√
2x(t)
φ2(t) =
√
2g1(t)
φ3(t) =
√
2g3(t).
Signal g2(t) is the negative of signal x(t). Therefore, g2(t) = − 1
√
2
φ1(t).
To represent g4(t), we need an additional basis function φ4(t). We can represent g4(t) as g4(t) = c1φ1(t) +
c2φ2(t) + c3φ3(t) + c4φ4(t), where,
c1 =
1
0
g4(t)φ∗
1(t) dt =
√
2
1
0
g4(t)x(t) dt =
2
π
c2 =
1
0
g4(t)φ∗
2(t) dt =
√
2
1
0
g4(t)g1(t) dt =
√
2
0.5
0
0.707 sin 4πt dt −
1
0.5
0.707 sin 4πt dt

= 0
c3 =
1
0
g4(t)φ∗
3(t) dt =
√
2
1
0
g4(t)g3(t) dt =
√
2 (0.707)
1
0
g4(t) dt = 0
Therefore,
c4φ4(t) = g4(t) − c1φ1(t) = g4(t) −
2
√
2
π
x(t)
The energy of the signal is
Ec4φ4 =
1
0

g4(t) −
2
√
2
π
x(t)
#2
dt =
1
0

g2
4(t) +
8
π2
x2
(t)2
−
4
√
2
π
g4(t)x(t)
#
dt
=
1
2
+
4
π2
−
4
√
2
π
·
√
2
π
=
1
2
+
4
π2
−
8
π2
=
1
2
−
4
π2
= 0.0946
Therefore,
φ4(t) =
g4(t) − 2
√
2
π x(t)
p
Ec4φ4
=
g4(t) − 2
√
2
π x(t)
0.3076
(b)
x1(t) = [ 1
√
2
0 0 0 ]T
g1(t) = [ 0 1
√
2
0 0 ]T
g2(t) = [ − 1
√
2
0 0 0 ]T
g3(t) = [ 0 0 1
√
2
0 ]T
g4(t) = [ 2
π 0 0 0.3076 ]T
2.8-1
(a) T0 = 4, ω0 = 2π
T0
= π
2 . Because of even symmetry, all sine terms are zero.
13

0 1 2 3 4 5 6 7 8 9 10
-0.5
0
0.5
1
1.5
C
n
0 1 2 3 4 5 6 7 8 9 10
-0.5
0
0.5
C
n
0 2 4 6 8 10
0
0.5
C
n
0 2 4 6 8 10
0
1
2
∠θ
n
0 2 4 6 8 10
-0.5
0
0.5
C
n
0 5 10 15 20
-2
0
2
∠θ
n
0 2 4 6 8 10
0
0.1
0.2
C
n
0 2 4 6 8 10
-2
-1
0
1
∠θ
n
0 1 2 3 4 5 6 7 8 9 10
0
0.5
1
C
n
(a)
(b)
(c)
(d)
(e)
(f)
Fig. S2.8-1
g(t) = a0 +
∞
X
n=1
an cos
nπ
2
t

a0 = 0 (by inspection of its lack of dc)
an =
4
4
1
0
cos
nπ
2
t

dt −
2
1
cos
nπ
2
t

dt

=
4
nπ
sin
nπ
2
Therefore, the Fourier series for g(t) is
g(t) =
4
π

cos
πt
2
−
1
3
cos
3πt
2
+
1
5
cos
5πt
2
−
1
7
cos
7πt
2
+ · · ·

Here bn = 0, and we allow Cn to take negative values. Figure S2.8-1(a) shows the plot of Cn which is
real-valued.
14

(b) T0 = 10π , ω0 = 2π
T0
= 1
5 . Because of even symmetry, all the sine terms are zero.
g(t) = a0 +
∞
X
n=1
an cos
n
5
t

+ bn sin
n
5
t

a0 =
1
5
(by inspection)
an =
2
10π
π
−π
cos
n
5
t

dt
=
1
5π

5
n

sin
n
5
t
π
−π
=
2
πn
sin
nπ
5

bn =
2
10π
π
−π
sin
n
5
t

dt
= 0 (integrand is an odd function of t)
Here bn = 0, and we allow Cn to take negative values. Note that Cn = an for n = 0, 1, 2, 3, . . .. Fig.
S2.8-1(b) shows the plot of Cn which is real-valued.
(c) T0 = 2π, ω0 = 1, and
g(t) = a0 +
∞
X
n=1
an cos nt + bn sin nt
with
a0 = 0.5 (by inspection of the dc or average)
an =
1
π
2π
0
t
2π
cos nt dt = 0, and bn =
1
π
2π
0
t
2π
sin nt dt = −
1
πn
and
g(t) = 0.5 −
1
π

sin t +
1
2
sin 2t +
1
3
sin 3t +
1
4
sin 4t + · · ·

= 0.5 +
1
π

cos

t +
π
2

+
1
2
cos

2t +
π
2

+
1
3
cos

3t +
π
2

+ · · ·

The cosine terms vanish because when 0.5 (the dc component) is subtracted from g(t), the remaining function
has odd symmetry. Hence, the Fourier series would contain dc and sine terms only. Figure S2.8-1(c) shows
the plots of |Cn| and θn. Note that Cn is purely imaginary.
(d) T0 = π, ω0 = 2 and g(t) = 4
π t.
a0 = 0 (by inspection)
an = 0 (n 0) (because of odd symmetry)
bn =
4
π
π/4
0
4
π
t sin 2nt dt =
2
πn

2
πn
sin
πn
2
− cos
πn
2

15

Solutions for Problems in Modern Digital and Analog Communication Systems, 5th Edition by Lathi and Ding

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