Prokis Solution Manual

SOLUTIONS MANUAL
Communication Systems Engineering

Second Edition
John G. Proakis
Masoud Salehi
Prepared by Evangelos Zervas

Upper Saddle River, New Jersey 07458

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c 2002 Prentice Hall
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ISBN 0-13-061974-6

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Contents

Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250
Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283

iii

Chapter 2

Problem 2.1
1)

2= ∞ N2
=
x(t) − αiφi(t) dt
−∞ i=1
 
∞N N

x(t) − αiφi(t) x∗(t) − αj∗φj∗(t) dt
−∞ i=1
j=1

∞ N∞ N∞
|x(t)|2dt − αi φi(t)x∗(t)dt − αj∗ φ∗j (t)x(t)dt
=

−∞ i=1 −∞ j=1 −∞

NN ∞

+ αiαj∗ φi(t)φj∗dt

i=1 j=1 −∞

∞ N N∞ N∞ φj∗(t)x(t)dt
|x(t)|2dt + |αi|2 − αi φi(t)x∗(t)dt − αj∗
=

−∞ i=1 i=1 −∞ j=1 −∞

Completing the square in terms of αi we obtain

∞ N∞ 2N ∞ φ∗i (t)x(t)dt 2
2 = |x(t)|2dt − φ∗i (t)x(t)dt + αi −

−∞ i=1 −∞ i=1 −∞

The first two terms are independent of α’s and the last term is always positive. Therefore the

minimum is achieved for ∞

αi = −∞ φi∗(t)x(t)dt

which causes the last term to vanish.

2) With this choice of αi’s ∞ N∞ 2
2= |x(t)|2dt − φ∗i (t)x(t)dt

= −∞ i=1 −∞

∞N

|x(t)|2dt − |αi|2

−∞ i=1

Problem 2.2
1) The signal x1(t) is periodic with period T0 = 2. Thus

1 1 Λ(t)e−j2π n t 1 1
2
x1,n = dt = Λ(t)e−jπntdt
2 −1 2 −1

=1 0 (t + 1)e−jπntdt + 1 1
2 −1 20
(−t + 1)e−jπntdt

1 j te−jπnt 1 e−jπnt 0 j 0
= πn π2n2
+ + e−jπnt
2 −1 2πn
−1

−1 j te−jπnt + 1 e−jπnt 1 j 1
2 πn π2n2
+ e−jπnt
0 2πn
0

1 − 1 (ejπn + e−jπn) = 1 − cos(πn))
π2n2 2π2n2 π2n2 (1

1





































































2jX1(f ) −jX2(f )

1) 2)
e
¡¡ e ee
¡ ee
ee
vv −f0 ¡ f0 e
v ¡¡

3) 2X3(f ) 4) 2X4(f )

ee −f0 ¡ e f0 ¡¡ ee ¡¡
e ¡¡ ee ¡ e ¡

ee −f0 f0
e

X5(f )

5)

−f0 f0

vv
v

−2f0 6) X6(f ) 2f0

ee e ¡¡ ¡¡
e ee ¡ ¡

7) X7(f )

e ¡¡
ee ¡

Problem 2.46
If x(t) is even then X(f ) is a real and even function and therefore −j sgn(f )X(f ) is an imaginary
and odd function. Hence, its inverse Fourier transform xˆ(t) will be odd. If x(t) is odd then X(f )
is imaginary and odd and −j sgn(f )X(f ) is real and even and, therefore, xˆ(t) is even.

Problem 2.47
Using Rayleigh’s theorem of the Fourier transform we have

Ex = ∞|x(t)|2dt = ∞|X(f )|2df

−∞ −∞

and

Exˆ = ∞|xˆ(t)|2dt = ∞| − jsgn(f )X(f )|2df

−∞ −∞

Noting the fact that | − jsgn(f )|2 = 1 except for f = 0, and the fact that X(f ) does not contain

any impulses at the origin we conclude that Ex = Exˆ.

Problem 2.48
Here we use Parseval’s Theorem of the Fourier Transform to obtain

∞x(t)xˆ(t) dt = ∞X(f )[−jsgn(f )X(f )]∗df

−∞ −∞

36

0 +∞ |X(f )|2df
= −j |X(f )|2df + j

−∞ 0

=0

where in the last step we have used the fact that X(f ) is Hermitian and therefore |X(f )|2 is even.

Problem 2.49
We note that C(f ) = M (f ) X(f ). From the assumption on the bandwidth of m(t) and x(t) we
see that C(f ) consists of two separate positive frequency and negative frequency regions that do
not overlap. Let us denote these regions by C+(f ) and C−(f ) respectively. A moment’s thought
shows that

C+(f ) = M (f ) X+(f )

and
C−(f ) = M (f ) X−(f )

To find the Hilbert Transform of c(t) we note that

F[cˆ(t)] = −jsgn(f )C(f )
= −jC+(f ) + jC−(f )
= −jM (f ) X+(f ) + jM (f ) X−(f )
= M (f ) [−jX+(f ) + jX−(f )]
= M (f ) [−jsgn(f )X(f )]
= M (f ) F[xˆ(t)]

Returning to the time domain we obtain

cˆ(t) = m(t)xˆ(t)

Problem 2.50
It is enough to note that

F [xˆˆ(t)] = (−jsgn(f ))2X(f )

and hence

F[xˆˆ(t)] = −X(f )

where we have used the fact that X(f ) does not contain any impulses at the origin.

Problem 2.51
Using the result of Problem 2.49 and noting that the Hilbert transform of cos is sin we have

x(t) cos(2πf0t) = x(t) sin(2πf0t)

Problem 2.52

F [A sin(2πf0t + θ)] = −jsgn(f )A − 1 δ(f + f0)ej2πf θ + 1 − f0)e−j2πf θ
2j 2f0 δ(f 2f0

2j

= A sgn(−f0)δ(f + f0)ej2πf θ − sgn(−f0)δ(f − f0)e−j2πf θ
2 2f0 2f0

= −A δ(f + f0)ej2πf θ + δ(f − f0)e−j2πf θ
2 2f0 2f0

= −AF [cos(2πf0t + θ)]

37

Thus, A sin(2πf0t + θ) = −A cos(2πf0t + θ)

Problem 2.53

Taking the Fourier transform of ej2πf0t we obtain

Thus, F [ej2πf0t] = −jsgn(f )δ(f − f0) = −jsgn(f0)δ(f − f0)
ej2πf0t = F −1[−jsgn(f0)δ(f − f0)] = −jsgn(f0)ej2πf0t

Problem 2.54

F d x(t) = F[x(t) δ (t)] = −jsgn(f )F[x(t) δ (t)]
dt
= −jsgn(f )j2πf X(f ) = 2πf sgn(f )X(f )
= 2π|f |X(f )

Problem 2.55
We need to prove that x (t) = (xˆ(t)) .

F[x (t)] = F[x(t) δ (t)] = −jsgn(f )F[x(t) δ (t)] = −jsgn(f )X(f )j2πf
= F[xˆ(t)]j2πf = F[(xˆ(t)) ]

Taking the inverse Fourier transform of both sides of the previous relation we obtain, x (t) = (xˆ(t))

Problem 2.56

x(t) = sinct cos 2πf0t =⇒ X(f ) = 1 + f0)) + 1 − f0))
Π(f Π(f

2 2

h(t) = sinc2t sin 2πf0t =⇒ H(f ) = −1 Λ(f + f0)) + 1 Λ(f − f0))
2j 2j

The lowpass equivalents are

Xl(f ) = 2u(f + f0)X(f + f0) = Π(f )

Hl(f ) = 2u(f + f0)H(f + f0) 1
Yl(f ) = = Λ(f )
 1 j − 1 < f ≤ 0
1  2j 2
2 Xl(f )Hl(f ) =  1 (f + 1) 1
2j (−f + 1) 0 ≤ f < 2

0 otherwise

Taking the inverse Fourier transform of Yl(f ) we can find the lowpass equivalent response of the
system. Thus,

yl(t) = F −1[Yl(f )]

1 0 (f + 1)ej2πftdf + 1 1
=
2j 1 2j 0 2 (−f + 1)ej2πftdf
2
−

1 1 f ej2πft 1 ej2πf t 0 +1 1 0
2j j2πt 4π2
= + t2 1 2j j2πt ej2πf t 1
2 2
− −

−1 1 f ej2πft + 1 ej2πf t 1 + 1 1 1
2j j2πt 4π2t2 2
ej2πf t 2
0 2j j2πt
0

= j −1 sin πt + 1 (cos πt − 1)
4πt 4π2t2

38

The output of the system y(t) can now be found from y(t) = Re[yl(t)ej2πf0t]. Thus

y(t) = Re (j[− 1 sin πt + 1 (cos πt − 1)])(cos 2πf0t + j sin 2πf0t)
4πt 4π2t2

= 1 (1 − cos πt) + 1 sin πt] sin 2πf0t
[ 4π2t2 4πt

Problem 2.57
1) The spectrum of the output signal y(t) is the product of X(f ) and H(f ). Thus,

Y (f ) = H(f )X(f ) = X(f )A(f0)ej(θ(f0)+(f−f0)θ (f)|f=f0 )

y(t) is a narrowband signal centered at frequencies f = ±f0. To obtain the lowpass equivalent
signal we have to shift the spectrum (positive band) of y(t) to the right by f0. Hence,

Yl(f ) = u(f + f0)X(f + f0)A(f0)ej(θ(f0)+fθ (f)|f=f0 ) = Xl(f )A(f0)ej(θ(f0)+fθ (f)|f=f0 )

2) Taking the inverse Fourier transform of the previous relation, we obtain

yl(t) = F −1 Xl(f )A(f0)ejθ(f0)ejfθ (f)|f=f0

1 (f )|f=f0 )
= A(f0)xl(t + θ
2π

With y(t) = Re[yl(t)ej2πf0t] and xl(t) = Vx(t)ejΘx(t) we get

y(t) = Re[yl(t)ej2πf0t]

= Re A(f0)xl(t + 1 (f )|f =f0 )ejθ(f0)ej2πf0t
θ

2π

= Re A(f0)Vx(t + 1 (f )|f =f0 )ej2πf0t ejΘx (t+ 1 θ (f )|f=f0 )
θ 2π

2π

= A(f0)Vx(t − tg ) cos(2πf0t + θ(f0) + Θx(t + 1 (f )|f=f0 ))
θ

2π

= A(f0)Vx(t − tg ) cos(2πf0(t + θ(f0) ) + Θx(t + 1 (f )|f=f0 ))
2πf0 θ

2π

= A(f0)Vx(t − tg ) cos(2πf0(t − tp) + Θx(t + 1 (f )|f=f0 ))
θ

2π

where 1 1 θ(f0) 1 θ(f )
2π 2π f0 2π f
tg = − θ (f )|f=f0 , tp = − = −

f =f0

3) tg can be considered as a time lag of the envelope of the signal, whereas tp is the time

corresponding to a phase delay of 1 θ(f0) .
2π f0

Problem 2.58

1) We can write Hθ(f ) as follows

 f >0
 cos θ − j sin θ f =0
f <0
Hθ(f ) =  0 = cos θ − jsgn(f ) sin θ
cos θ
+ j sin θ

Thus,

hθ (t) = F −1[Hθ(f )] = cos θδ(t) + 1 sin θ
πt

39

2)

1
xθ (t) = x(t) hθ(t) = x(t) (cos θδ(t) + sin θ)
πt

1
= cos θx(t) δ(t) + sin θ x(t)

πt

= cos θx(t) + sin θxˆ(t)

3)

∞∞
|xθ(t)|2dt = | cos θx(t) + sin θxˆ(t)|2dt

−∞ −∞

∞∞

= cos2 θ |x(t)|2dt + sin2 θ |xˆ(t)|2dt

−∞ −∞

∞∞

+ cos θ sin θ x(t)xˆ∗(t)dt + cos θ sin θ x∗(t)xˆ(t)dt

−∞ −∞

But ∞ |x(t)|2dt = ∞ |xˆ(t)|2dt = Ex and ∞ x(t)xˆ∗(t)dt = 0 since x(t) and xˆ(t) are orthogonal.
−∞ −∞ −∞

Thus,

Exθ = Ex(cos2 θ + sin2 θ) = Ex

Problem 2.59
1)

z(t) = x(t) + jxˆ(t) = m(t) cos(2πf0t) − mˆ (t) sin(2πf0t)
+j[m(t)cos(2πf0t) − mˆ (t)sin(2πf0t)

= m(t) cos(2πf0t) − mˆ (t) sin(2πf0t)
+jm(t) sin(2πf0t) + jmˆ (t) cos(2πf0t)

= (m(t) + jmˆ (t))ej2πf0t

The lowpass equivalent signal is given by

xl(t) = z(t)e−j2πf0t = m(t) + jmˆ (t)

2) The Fourier transform of m(t) is Λ(f ). Thus

X(f ) = Λ(f + f0) + Λ(f − f0) − (−jsgn(f )Λ(f ))
2

− 1 δ(f + f0) + 1 δ(f − f0)
2j 2j

= 1 + f0) [1 − sgn(f + f0)] + 1 − f0) [1 + sgn(f − f0)]
Λ(f Λ(f

2 2

. . 1 . .
  d
  . . . . . .
  d
. . . . d

−f0 − 1 −f0 f0 f0 + 1

The bandwidth of x(t) is W = 1.

40

3)

z(t) = x(t) + jxˆ(t) = m(t) cos(2πf0t) + mˆ (t) sin(2πf0t)

+j[m(t)cos(2πf0t) + mˆ (t)sin(2πf0t)
= m(t) cos(2πf0t) + mˆ (t) sin(2πf0t)

+jm(t) sin(2πf0t) − jmˆ (t) cos(2πf0t)
= (m(t) − jmˆ (t))ej2πf0t

The lowpass equivalent signal is given by
xl(t) = z(t)e−j2πf0t = m(t) − jmˆ (t)

The Fourier transform of x(t) is

X(f ) = Λ(f + f0) + Λ(f − f0) − (jsgn(f )Λ(f ))
2

−1 δ(f + f0) + 1 δ(f − f0)
2j 2j

= 1 + f0) [1 + sgn(f + f0)] + 1 − f0) [1 − sgn(f − f0)]
Λ(f Λ(f

2 2

.. d..... dd 1    ..... ..
. .
. ... . . ... .
. .

−f0 −f0 + 1 f0 − 1 f0

41

Chapter 3

Problem 3.1
The modulated signal is

u(t) = m(t)c(t) = Am(t) cos(2π4 × 103t)

= A 200 250 π cos(2π4 × 103t)
2 cos(2π t) + 4 sin(2π t + )
π π3

= A cos(2π(4 × 103 + 200 + A cos(2π(4 × 103 − 200
)t) )t)
ππ

+2A sin(2π(4 × 103 + 250 + π ) − 2A sin(2π(4 × 103 − 250 − π )
)t )t
π3 π3

Taking the Fourier transform of the previous relation, we obtain

U (f ) = A δ(f − 200 ) + δ(f + 200 ) + 2 ej π δ(f − 250 ) − 2 e−j π δ(f + 250 )
3 3
π πj πj π

1 − 4 × 103) + δ(f + 4 × 103)]
[δ(f
2

= A δ(f − 4 × 103 − 200 + δ(f − 4 × 103 + 200
) )
2π π

+2e−j π δ(f − 4 × 103 − 250 + 2ej π δ(f − 4 × 103 + 250
6 ) 6 )
ππ

+δ(f + 4 × 103 − 200 + δ(f + 4 × 103 + 200
) )
ππ

+2e−j π δ(f + 4 × 103 − 250 + 2ej π δ(f + 4 × 103 + 250
6 ) 6 )
ππ

The next figure depicts the magnitude and the phase of the spectrum U (f ).

T T |U (f )| A ......................
TT

. . . . . . . . . . . . . . . . . . . . . . .A/2 TT
TT

−fc − 2π5−0 fc − 200 −fc + 2π00−fc + 250 fc − 250 fc − 200 fc + 200 fc + 250
π π π π π π

U (f )
π
s s
. . . . . . . . . . . . . . . . . . . . . . .6. . . . . .

s . . . . . . . .−. .π6. . . . . . . . . . . . . . . . . . . s

To find the power content of the modulated signal we write u2(t) as

u2(t) = A2 cos2(2π(4 × 103 + 200 + A2 cos2(2π(4 × 103 − 200
)t) )t)
ππ

+4A2 sin2(2π(4 × 103 + 250 )t + π ) + 4A2 sin2(2π(4 × 103 − 250 )t − π )
π3 π3

+terms of cosine and sine functions in the first power

Hence,

P = lim T u2(t)dt = A2 A2 4A2 4A2 = 5A2
2 ++ +

T →∞ − T 22 2 2
2

42

Problem 3.2

u(t) = m(t)c(t) = A(sinc(t) + sinc2(t)) cos(2πfct)
Taking the Fourier transform of both sides, we obtain

U (f ) = A (δ(f − fc) + δ(f + fc))
[Π(f ) + Λ(f )]

2

= A [Π(f − fc) + Λ(f − fc) + Π(f + fc) + Λ(f + fc)]
2

Π(f − fc) = 0 for |f − fc| < 1 , whereas Λ(f − fc) = 0 for |f − fc| < 1. Hence, the bandwidth of
2

the bandpass filter is 2.

Problem 3.3
The following figure shows the modulated signals for A = 1 and f0 = 10. As it is observed
both signals have the same envelope but there is a phase reversal at t = 1 for the second signal
Am2(t) cos(2πf0t) (right plot). This discontinuity is shown clearly in the next figure where we
plotted Am2(t) cos(2πf0t) with f0 = 3.

1 1
0.8 0.8
0.6 0.6
0.4 0.4
0.2 0.2

0 0
-0.2 -0.2
-0.4 -0.4
-0.6 -0.6
-0.8 -0.8

-10 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -10 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

0.8
0.6
0.4
0.2

0
-0.2
-0.4
-0.6
-0.8

-10 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

Problem 3.4
y(t) = x(t) + 1 x2(t)
2

43

= 1 m2(t) + cos2(2πfct) + 2m(t) cos(2πfct)
m(t) + cos(2πfct) + 2

= m(t) + cos(2πfct) + 1 m2(t) + 1 + 1 cos(2π2fct) + m(t) cos(2πfct)
2 4 4

Taking the Fourier transform of the previous, we obtain

Y (f ) = 1 M (f ) + 1 (M (f − fc) + M (f + fc))
M (f ) + M (f ) 2

2

1 + 1 (δ(f − fc) + δ(f + fc)) + 1 (δ(f − 2fc) + δ(f + 2fc))
+ δ(f ) 2 8

4

The next figure depicts the spectrum Y (f )

1/2

1/4 1/8

-2fc -fc -2W 2W fc 2fc

Problem 3.5

u(t) = m(t) · c(t)
= 100(2 cos(2π2000t) + 5 cos(2π3000t)) cos(2πfct)

Thus,

U (f ) = 100 δ(f − 2000) + δ(f + 2000) + 5 (δ(f − 3000) + δ(f + 3000))
2 2

[δ(f − 50000) + δ(f + 50000)]

= 50 δ(f − 52000) + δ(f − 48000) + 5 − 53000) + 5 δ(f − 47000)
δ(f
22

55
+δ(f + 52000) + δ(f + 48000) + δ(f + 53000) + δ(f + 47000)

22

A plot of the spectrum of the modulated signal is given in the next figure

. ..... .. ..... ..... .. .... 125 T T
T T

. . . . . . . . . . . . . . . . . . . . .5.0. . . . . . . . . . . . . . . . . . .
TT TT

-53 -52 -48 -47 0 47 48 52 53 KHz

Problem 3.6
The mixed signal y(t) is given by

y(t) = u(t) · xL(t) = Am(t) cos(2πfct) cos(2πfct + θ)
A

= 2 m(t) [cos(2π2fct + θ) + cos(θ)]

44

The lowpass filter will cut-off the frequencies above W , where W is the bandwidth of the message
signal m(t). Thus, the output of the lowpass filter is

A
z(t) = m(t) cos(θ)

2

If the power of m(t) is PM , then the power of the output signal z(t) is Pout = PM A2 cos2(θ). The
4
A2
power of the modulated signal u(t) = Am(t) cos(2πfct) is PU = 2 PM . Hence,

Pout = 1 cos2(θ)
PU 2

A plot of Pout for 0≤θ≤π is given in the next figure.
PU

0.5

0.45

0.4

0.35

0.3

0.25

0.2

0.15

0.1

0.05

00 0.5 1 1.5 2 2.5 3 3.5
Theta (rad)

Problem 3.7
1) The spectrum of u(t) is

U (f ) = 20 [δ(f − fc) + δ(f + fc)]
2

2 [δ(f − fc − 1500) + δ(f − fc + 1500)
+

4

+δ(f + fc − 1500) + δ(f + fc + 1500)]

10 [δ(f − fc − 3000) + δ(f − fc + 3000)
+

4

+δ(f + fc − 3000) + δ(f + fc + 3000)]

The next figure depicts the spectrum of u(t).

......................................
T 10 T

T T . 1. ./.2. . . . . . . . . . . . . . . . . . . . . . . .
T T TT

.5./.2. . . . . . . . . . . . . . . . . . . .
TT

-1030-1015-1000 -985 -970 0 970 985 1000 10151030
X 100 Hz

2) The square of the modulated signal is
u2(t) = 400 cos2(2πfct) + cos2(2π(fc − 1500)t) + cos2(2π(fc + 1500)t)
+25 cos2(2π(fc − 3000)t) + 25 cos2(2π(fc + 3000)t)
+ terms that are multiples of cosines

45

If we integrate u2(t) from − T to T , normalize the integral by 1 and take the limit as T → ∞,
2 2 T

then all the terms involving cosines tend to zero, whereas the squares of the cosines give a value of

1 . Hence, the power content at the frequency fc = 105 Hz is Pfc = 400 = 200, the power content
2 2

at the frequency Pfc+1500 is the same as the power content at the frequency Pfc−1500 and equal to
1 25
2 , whereas Pfc+3000 = Pfc−3000 = 2 .

3)

u(t) = (20 + 2 cos(2π1500t) + 10 cos(2π3000t)) cos(2πfct)
11

= 20(1 + 10 cos(2π1500t) + 2 cos(2π3000t)) cos(2πfct)

This is the form of a conventional AM signal with message signal

11
m(t) = cos(2π1500t) + cos(2π3000t)

10 2
= cos2(2π1500t) + 1 cos(2π1500t) − 1

10 2

The minimum of g(z) = z2 + 1 z − 1 is achieved for z = − 1 and it is min(g(z)) = − 201 . Since
10 2 20 400
− 1 − 201
z = 20 is in the range of cos(2π1500t), we conclude that the minimum value of m(t) is 400 .

Hence, the modulation index is

α = − 201
400

4)

u(t) = 20 cos(2πfct) + cos(2π(fc − 1500)t) + cos(2π(fc − 1500)t)
= 5 cos(2π(fc − 3000)t) + 5 cos(2π(fc + 3000)t)

The power in the sidebands is

1 1 25 25
Psidebands = 2 + 2 + 2 + 2 = 26

The total power is Ptotal = Pcarrier + Psidebands = 200 + 26 = 226. The ratio of the sidebands power

to the total power is

Psidebands = 26
Ptotal 226

Problem 3.8
1)

u(t) = m(t)c(t)

= 100(cos(2π1000t) + 2 cos(2π2000t)) cos(2πfct)

= 100 cos(2π1000t) cos(2πfct) + 200 cos(2π2000t) cos(2πfct)

= 100 [cos(2π(fc + 1000)t) + cos(2π(fc − 1000)t)]
2

200 [cos(2π(fc + 2000)t) + cos(2π(fc − 2000)t)]
2

Thus, the upper sideband (USB) signal is

uu(t) = 50 cos(2π(fc + 1000)t) + 100 cos(2π(fc + 2000)t)

46

2) Taking the Fourier transform of both sides, we obtain

Uu(f ) = 25 (δ(f − (fc + 1000)) + δ(f + (fc + 1000)))
+50 (δ(f − (fc + 2000)) + δ(f + (fc + 2000)))

A plot of Uu(f ) is given in the next figure.
. . . . . . . . . . . . . . . . . . . . . . . .5.0. . . . . . . . . . . . . . . . . . . . .
TT
. . . . . . . . . . . . . . . . .2.5. . . . . . . . . . . . . . .
TT

-1002 -1001 0 1001 1002 KHz

Problem 3.9

If we let Tp Tp
4 4
x(t) = −Π t + +Π t −

Tp Tp

2 2

then using the results of Problem 2.23, we obtain

∞

v(t) = m(t)s(t) = m(t) x(t − nTp)

n=−∞

1 ∞ n )ej2π n t
= m(t) Tp
X(
Tp n=−∞ Tp

where

n = F −Π t + Tp +Π t − Tp
X( ) 4 4

Tp Tp Tp f = n
Tp
2 2

= Tp sinc(f Tp ) e−j 2πf Tp − ej2πf Tp
4 4
22
f = n
Tp

= Tp sinc( n )(−2j) sin(n π )
22 2

Hence, the Fourier transform of v(t) is

1 ∞ sinc( n )(−2j) π n
V (f ) = sin(n )
2 n=−∞ 2 )M (f −
2 Tp

The bandpass filter will cut-off all the frequencies except the ones centered at 1 , that is for n = ±1.
Tp
Thus, the output spectrum is

U (f ) = sinc( 1 )(−j)M (f − 1 1 1
) + sinc( )jM (f + )
2 Tp 2 Tp

= − 2 jM (f − 12 1
) + jM (f + )
π Tp π Tp

= 4 M (f ) 1 δ(f − 1 ) − 1 δ(f + 1 )
π 2j Tp 2j Tp

Taking the inverse Fourier transform of the previous expression, we obtain

41
u(t) = m(t) sin(2π t)

π Tp

47