To obtain the lowpass equivalent signal we have to shift the spectrum positive band of yt to the right by f0... The lowpass filter will cut-off the frequencies above W , where W is the ban
Trang 1Thus,A sin(2πf$ 0t + θ) = −A cos(2πf0t + θ)
Problem 2.53
Taking the Fourier transform of e j2πf# 0t we obtain
F[ e j2πf#0t] =−jsgn(f)δ(f − f0) =−jsgn(f0)δ(f − f0) Thus,
#
e j2πf0t=F −1[−jsgn(f0)δ(f − f0)] =−jsgn(f0)e j2πf0t
Problem 2.54
F
d
dt x(t)
= F[ x(t) δ# (t)] = −jsgn(f)F[x(t) δ (t)]
= −jsgn(f)j2πfX(f) = 2πfsgn(f)X(f)
= 2π |f|X(f)
Problem 2.55
We need to prove thatx# (t) = (ˆ x(t)) .
F[ x# (t)] = F[ x(t) δ# (t)] = −jsgn(f)F[x(t) δ (t)] = −jsgn(f)X(f)j2πf
= F[ˆx(t)]j2πf = F[(ˆx(t)) ] Taking the inverse Fourier transform of both sides of the previous relation we obtain,x# (t) = (ˆ x(t))
Problem 2.56
x(t) = sinct cos 2πf0t =⇒ X(f) = 1
2Π(f + f0)) +
1
2Π(f − f0))
h(t) = sinc2t sin 2πf0t =⇒ H(f) = −1
2j Λ(f + f0)) +
1
2j Λ(f − f0)) The lowpass equivalents are
X l (f ) = 2u(f + f0)X(f + f0) = Π(f )
H l (f ) = 2u(f + f0)H(f + f0) = 1
j Λ(f )
Y l (f ) = 1
2X l (f )H l (f ) =
1
2j (f + 1) −1
2 < f ≤ 0
1
2j(−f + 1) 0 ≤ f < 1
2
Taking the inverse Fourier transform of Y l (f ) we can find the lowpass equivalent response of the
system Thus,
y l (t) = F −1 [Y l (f )]
2j
0
−1(f + 1)e j2πf t df + 1
2j
1
0 (−f + 1)ej2πf t df
2j
1
j2πt f e
j2πf t+ 1
4π2t2e j2πf t
0
−1 2
+ 1
2j
1
j2πt e j2πf t
0
−1 2
−1 2j
1
j2πt f e
j2πf t+ 1
4π2t2e j2πf t
1 2
0
+ 1
2j
1
j2πt e j2πf t
1 2
0
= j − 1
4πt sin πt +
1
4π2t2(cos πt − 1)
Trang 2
The output of the system y(t) can now be found from y(t) = Re[y l (t)e j2πf0t] Thus
y(t) = Re (j[ − 1
4πt sin πt +
1
4π2t2(cos πt − 1)])(cos 2πf0t + j sin 2πf0t)
= [ 1
4π2t2(1− cos πt) + 1
4πt sin πt] sin 2πf0t
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)e j(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,
Y l (f ) = u(f + f0)X(f + f0)A(f0)e j(θ(f0)+f θ (f ) | f =f0)= X l (f )A(f0)e j(θ(f0)+f θ (f ) | f =f0)
2) Taking the inverse Fourier transform of the previous relation, we obtain
y l (t) = F −1
X l (f )A(f0)e jθ(f0)e jf θ (f ) | f =f0
= A(f0)x l (t + 1
2π θ
(f )| f =f0)
With y(t) = Re[y l (t)e j2πf0t ] and x l (t) = V x (t)e jΘx (t) we get
y(t) = Re[y l (t)e j2πf0t]
= Re A(f0)x l (t + 1
2π θ
(f )| f =f0)e jθ(f0)e j2πf0t
= Re A(f0)V x (t + 1
2π θ
(f ) | f =f0)e j2πf0t e jΘx (t+ 2π1 θ (f ) | f =f0)
= A(f0)V x (t − t g ) cos(2πf0t + θ(f0) + Θx (t + 1
2π θ
(f ) | f =f0))
= A(f0)V x (t − t g ) cos(2πf0(t + θ(f0)
2πf0
) + Θx (t + 1
2π θ
(f ) | f =f0))
= A(f0)V x (t − t g ) cos(2πf0(t − t p) + Θx (t + 1
2π θ
(f ) | f =f0)) where
t g =− 1 2π θ
(f ) | f =f0, t p =− 1
2π
θ(f0)
f0
=− 1 2π
θ(f ) f
f =f0
3) t g can be considered as a time lag of the envelope of the signal, whereas t p is the time corresponding to a phase delay of 2π1 θ(f f0)
0
Problem 2.58
1) We can write H θ (f ) as follows
H θ (f ) =
cos θ − j sin θ f > 0
cos θ + j sin θ f < 0
= cos θ − jsgn(f) sin θ
Thus,
h θ (t) = F −1 [H
θ (f )] = cos θδ(t) + 1
πt sin θ
Trang 3x θ (t) = x(t) h θ (t) = x(t) (cos θδ(t) + 1
πt sin θ)
= cos θx(t) δ(t) + sin θ 1
πt x(t)
= cos θx(t) + sin θ ˆ x(t)
3)
∞
−∞ |x θ (t) |2dt =
∞
−∞ | cos θx(t) + sin θˆx(t)|2dt
= cos2θ
∞
−∞ |x(t)|2dt + sin2θ
∞
−∞ |ˆx(t)|2dt + cos θ sin θ
∞
−∞ x(t)ˆ x
∗ (t)dt + cos θ sin θ ∞
−∞ x
∗ (t)ˆ x(t)dt
But∞
−∞ |x(t)|2dt =∞
−∞ |ˆx(t)|2dt = E xand∞
−∞ x(t)ˆ x ∗ (t)dt = 0 since x(t) and ˆ x(t) are orthogonal.
Thus,
E x θ = E x(cos2θ + sin2θ) = E x
Problem 2.59
1)
z(t) = x(t) + j ˆ x(t) = m(t) cos(2πf0t) − ˆ m(t) sin(2πf0t)
+j[m(t) cos(2πf# 0t) − ˆ m(t) sin(2πf# 0t)
= m(t) cos(2πf0t) − ˆ m(t) sin(2πf0t) +jm(t) sin(2πf0t) + j ˆ m(t) cos(2πf0t)
= (m(t) + j ˆ m(t))e j2πf0t
The lowpass equivalent signal is given by
x l (t) = z(t)e −j2πf0t = m(t) + j ˆ m(t)
2) The Fourier transform of m(t) is Λ(f ) Thus
X(f ) = Λ(f + f0) + Λ(f − f0)
2 − (−jsgn(f)Λ(f))
− 1 2j δ(f + f0) +
1
2j δ(f − f0)
2Λ(f + f0) [1− sgn(f + f0)] + 1
2Λ(f − f0) [1 + sgn(f − f0)]
@
1
The bandwidth of x(t) is W = 1.
Trang 4z(t) = x(t) + j ˆ x(t) = m(t) cos(2πf0t) + ˆ m(t) sin(2πf0t)
+j[m(t) cos(2πf# 0t) + ˆ m(t) sin(2πf# 0t)
= m(t) cos(2πf0t) + ˆ m(t) sin(2πf0t) +jm(t) sin(2πf0t) − j ˆ m(t) cos(2πf0t)
= (m(t) − j ˆ m(t))e j2πf0t
The lowpass equivalent signal is given by
x l (t) = z(t)e −j2πf0t = m(t) − j ˆ m(t) The Fourier transform of x(t) is
X(f ) = Λ(f + f0) + Λ(f − f0)
2 − (jsgn(f)Λ(f))
− 1 2j δ(f + f0) +
1
2j δ(f − f0)
2Λ(f + f0) [1 + sgn(f + f0)] +
1
2Λ(f − f0) [1− sgn(f − f0)]
.
@
@
@
1
−f0+ 1
Trang 5Chapter 3
Problem 3.1
The modulated signal is
u(t) = m(t)c(t) = Am(t) cos(2π4 × 103t)
= A 2 cos(2π200
π t) + 4 sin(2π
250
π t +
π
3)
cos(2π4 × 103t)
= A cos(2π(4 × 103+200
π )t) + A cos(2π(4 × 103−200
π )t) +2A sin(2π(4 × 103+250
π )t +
π
3)− 2A sin(2π(4 × 103−250
π )t − π
3) Taking the Fourier transform of the previous relation, we obtain
U (f ) = A δ(f − 200
π ) + δ(f +
200
π ) +
2
j e
j π3δ(f −250
π )− 2
j e
−j π
3δ(f + 250
π )
1
2[δ(f − 4 × 103) + δ(f + 4 × 103)]
2 δ(f − 4 × 103−200
π ) + δ(f − 4 × 103+200
π ) +2e −j π
6δ(f − 4 × 103−250
π ) + 2e
j π6δ(f − 4 × 103+250
π ) +δ(f + 4 × 103−200
π ) + δ(f + 4 × 103+200
π ) +2e −j π
6δ(f + 4 × 103− 250
π ) + 2e
j π6δ(f + 4 × 103+ 250
π )
The next figure depicts the magnitude and the phase of the spectrum U (f ).
s
s
s s
6
6 6
6
|U(f)|
U (f )
−fc− 250 π−fc−200π −fc+ 200 π −fc+250π fc− 250 π fc−200
π fc+ 200 π fc+ 250 π
− π
6
π
6
A/2 A
To find the power content of the modulated signal we write u2(t) as
u2(t) = A2cos2(2π(4 × 103+200
π )t) + A
2cos2(2π(4 × 103−200
π )t) +4A2sin2(2π(4 × 103+250
π )t +
π
3) + 4A
2sin2(2π(4 × 103−250
π )t − π
3) +terms of cosine and sine functions in the first power
Hence,
P = lim
T →∞
T
2
− T
2
u2(t)dt = A
2
2 +
A2
2 +
4A2
2 +
4A2
2 = 5A
2
Trang 6Problem 3.2
u(t) = m(t)c(t) = A(sinc(t) + sinc2(t)) cos(2πf c t)
Taking the Fourier transform of both sides, we obtain
U (f ) = A
2 [Π(f ) + Λ(f )] (δ(f − f c ) + δ(f + f c))
2 [Π(f − f c ) + Λ(f − f c ) + Π(f + f c ) + Λ(f + f c)]
Π(f − f c) = 0 for |f − f c | < 1
2, whereas Λ(f − f c)= 0 for |f − f c | < 1 Hence, the bandwidth of
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
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -1
-0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8
0 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
2x
2(t)
Trang 7= m(t) + cos(2πf c t) +1
2
m2(t) + cos2(2πf c t) + 2m(t) cos(2πf c t)
= m(t) + cos(2πf c t) +1
2m
2(t) + 1
4+
1
4cos(2π2f c t) + m(t) cos(2πf c t) Taking the Fourier transform of the previous, we obtain
Y (f ) = M (f ) +1
2M (f ) M (f ) +
1
2(M (f − f c ) + M (f + f c)) +1
4δ(f ) +
1
2(δ(f − f c ) + δ(f + f c)) +1
8(δ(f − 2f c ) + δ(f + 2f c))
The next figure depicts the spectrum Y (f )
1/4
1/8 1/2
Problem 3.5
u(t) = m(t) · c(t)
= 100(2 cos(2π2000t) + 5 cos(2π3000t)) cos(2πf c t)
Thus,
U (f ) = 100
2 δ(f − 2000) + δ(f + 2000) + 5
2(δ(f − 3000) + δ(f + 3000))
[δ(f − 50000) + δ(f + 50000)]
= 50 δ(f − 52000) + δ(f − 48000) +5
2δ(f − 53000) +5
2δ(f − 47000) +δ(f + 52000) + δ(f + 48000) +5
2δ(f + 53000) +
5
2δ(f + 47000)
A plot of the spectrum of the modulated signal is given in the next figure
6
6
6 6
125 50 0
Problem 3.6
The mixed signal y(t) is given by
y(t) = u(t) · x L (t) = Am(t) cos(2πf c t) cos(2πf c t + θ)
2m(t) [cos(2π2f c t + θ) + cos(θ)]
Trang 8The 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
z(t) = A
2m(t) cos(θ)
If the power of m(t) is P M , then the power of the output signal z(t) is Pout = P M A
2
4 cos2(θ) The power of the modulated signal u(t) = Am(t) cos(2πf c t) is P U = A22P M Hence,
Pout
P U
= 1
2cos
2(θ)
A plot of Pout
P U for 0≤ θ ≤ π is given in the next figure.
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0 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
2 [δ(f − f c ) + δ(f + f c)]
+2
4[δ(f − f c − 1500) + δ(f − f c+ 1500)
+δ(f + f c − 1500) + δ(f + f c+ 1500)]
+10
4 [δ(f − f c − 3000) + δ(f − f c+ 3000)
+δ(f + f c − 3000) + δ(f + f c+ 3000)]
The next figure depicts the spectrum of u(t).
6 6
6 6
6 6
6 6
-1030-1015-1000 -985 -970 970 985
X 100 Hz
1000 1015 1030 0
5/2 1/2 10
2) The square of the modulated signal is
u2(t) = 400 cos2(2πf c t) + cos2(2π(f c − 1500)t) + cos2(2π(f c + 1500)t)
+25 cos2(2π(f c − 3000)t) + 25 cos2(2π(f c + 3000)t)
+ terms that are multiples of cosines
Trang 9If we integrate u2(t) from − T
2 to T2, normalize the integral by T1 and take the limit as T → ∞,
then all the terms involving cosines tend to zero, whereas the squares of the cosines give a value of
1
2 Hence, the power content at the frequency f c = 105 Hz is P f c = 4002 = 200, the power content
at the frequency P fc+1500 is the same as the power content at the frequency P fc−1500 and equal to
1
2, whereas P fc+3000 = P fc−3000 = 252
3)
u(t) = (20 + 2 cos(2π1500t) + 10 cos(2π3000t)) cos(2πf c t)
= 20(1 + 1
10cos(2π1500t) +
1
2cos(2π3000t)) cos(2πf c t) This is the form of a conventional AM signal with message signal
m(t) = 1
10cos(2π1500t) +
1
2cos(2π3000t)
= cos2(2π1500t) + 1
10cos(2π1500t) −1
2
The minimum of g(z) = z2+ 101z −1
2 is achieved for z = −1
20 and it is min(g(z)) = −201
400 Since
z = −1
20 is in the range of cos(2π1500t), we conclude that the minimum value of m(t) is −201
400 Hence, the modulation index is
α = −201
400
4)
u(t) = 20 cos(2πf c t) + cos(2π(f c − 1500)t) + cos(2π(f c − 1500)t)
= 5 cos(2π(f c − 3000)t) + 5 cos(2π(f c + 3000)t)
The power in the sidebands is
Psidebands= 1
2 +
1
2 +
25
2 +
25
2 = 26
The total power is Ptotal= Pcarrier+ Psidebands= 200 + 26 = 226 The ratio of the sidebands power
to the total power is
Psidebands
Ptotal
= 26 226
Problem 3.8
1)
u(t) = m(t)c(t)
= 100(cos(2π1000t) + 2 cos(2π2000t)) cos(2πf c t)
= 100 cos(2π1000t) cos(2πf c t) + 200 cos(2π2000t) cos(2πf c t)
= 100
2 [cos(2π(f c + 1000)t) + cos(2π(f c − 1000)t)]
200
2 [cos(2π(f c + 2000)t) + cos(2π(f c − 2000)t)]
Thus, the upper sideband (USB) signal is
u u (t) = 50 cos(2π(f c + 1000)t) + 100 cos(2π(f c + 2000)t)
Trang 102) Taking the Fourier transform of both sides, we obtain
U u (f ) = 25 (δ(f − (f c + 1000)) + δ(f + (f c+ 1000)))
+50 (δ(f − (f c + 2000)) + δ(f + (f c+ 2000)))
A plot of U u (f ) is given in the next figure.
6 6
50 25
Problem 3.9
If we let
x(t) = −Π
t + Tp4 Tp
2
+ Π
t − Tp
4
Tp
2
then using the results of Problem 2.23, we obtain
v(t) = m(t)s(t) = m(t)
∞
n= −∞
x(t − nT p)
= m(t) 1
T p
∞
n= −∞
X( n
T p )e
j2π n
Tp t
where
X( n
T p) = F
−Π
t + Tp4
T p
2
+ Π
t − Tp
4
T p
2
f = n Tp
= T p
2 sinc(f
T p
2 ) e
−j2πf Tp
4 − e j2πf Tp4
f = n Tp
= T p
2 sinc(
n
2)(−2j) sin(n π
2)
Hence, the Fourier transform of v(t) is
V (f ) = 1
2
∞
n= −∞
sinc(n
2)(−2j) sin(n π
2)M (f − n
T p
)
The bandpass filter will cut-off all the frequencies except the ones centered at Tp1 , that is for n = ±1.
Thus, the output spectrum is
U (f ) = sinc(1
2)(−j)M(f − 1
T p) + sinc(
1
2)jM (f +
1
T p)
= −2
π jM (f − 1
T p
) + 2
π jM (f +
1
T p
)
π M (f )
1
2j δ(f − 1
T p)− 1 2j δ(f +
1
T p)
Taking the inverse Fourier transform of the previous expression, we obtain
u(t) = 4
π m(t) sin(2π
1
T p t)
Trang 11which has the form of a DSB-SC AM signal, with c(t) = π4 sin(2π Tp1 t) being the carrier signal.
Problem 3.10
Assume that s(t) is a periodic signal with period T p , i.e s(t) =
n x(t − nT p) Then
v(t) = m(t)s(t) = m(t)
∞
n= −∞
x(t − nT p)
= m(t) 1
T p
∞
n= −∞
X( n
T p )e
j2π n
Tp t
T p
∞
n=−∞
X( n
T p )m(t)e
j2π n
Tp t
where X( Tp n) =F[x(t)]| f = n
Tp The Fourier transform of v(t) is
V (f ) = 1
T p F
∞
n= −∞
X( n
T p )m(t)e j2π Tp n t
T p
∞
n= −∞
X( n
T p
)M (f − n
T p
)
The bandpass filter will cut-off all the frequency components except the ones centered at f c =± 1
T p Hence, the spectrum at the output of the BPF is
U (f ) = 1
T p
X( 1
T p
)M (f − 1
T p
) + 1
T p
X( − 1
T p
)M (f + 1
T p
)
In the time domain the output of the BPF is given by
u(t) = 1
T p X(
1
T p )m(t)e
j2π 1
Tp t+ 1
T p X
∗( 1
T p )m(t)e
−j2π 1
Tp t
T p m(t)
X( 1
T p
)e j2π Tp1 t + X ∗( 1
T p )e −j2π 1
Tp t
T p
2Re(X( 1
Tp
))m(t) cos(2π 1
Tp
t)
As it is observed u(t) has the form a modulated DSB-SC signal The amplitude of the modulating signal is A c = T1
p2Re(X(T1
p)) and the carrier frequency f c = T1
p
Problem 3.11
1) The spectrum of the modulated signal Am(t) cos(2πf c t) is
V (f ) = A
2[M (f − f c ) + M (f + f c)]
The spectrum of the signal at the output of the highpass filter is
U (f ) = A
2[M (f + f c )u −1(−f − f c ) + M (f − f c )u −1 (f − f c)]
Multiplying the output of the HPF with A cos(2π(f c + W )t) results in the signal z(t) with spectrum
Z(f ) = A
2[M (f + f c )u −1(−f − f c ) + M (f − f c )u −1 (f − f c)]
A
2[δ(f − (f c + W )) + δ(f + f c + W )]
Trang 12= A
2
4 (M (f + f c − f c − W )u −1(−f + f c + W − f c)
+M (f + f c − f c + W )u −1 (f + f c + W − f c)
+M (f − 2f c − W )u −1 (f − 2f c − W ) +M (f + 2f c + W )u −1(−f − 2f c − W ))
2
4 (M (f − W )u −1(−f + W ) + M(f + W )u −1 (f + W ) +M (f − 2f c − W )u −1 (f − 2f c − W ) + M(f + 2f c + W )u −1(−f − 2f c − W ))
The LPF will cut-off the double frequency components, leaving the spectrum
Y (f ) = A
2
4 [M (f − W )u −1(−f + W ) + M(f + W )u −1 (f + W )]
The next figure depicts Y (f ) for M (f ) as shown in Fig P-5.12.
Y(f)
2) As it is observed from the spectrum Y (f ), the system shifts the positive frequency components
to the negative frequency axis and the negative frequency components to the positive frequency
axis If we transmit the signal y(t) through the system, then we will get a scaled version of the original spectrum M (f ).
Problem 3.12
The modulated signal can be written as
u(t) = m(t) cos(2πf c t + φ)
= m(t) cos(2πf c t) cos(φ) − m(t) sin(2πf c t) sin(φ)
= u c (t) cos(2πf c t) − u s (t) sin(2πf c t) where we identify u c (t) = m(t) cos(φ) as the in-phase component and u s (t) = m(t) sin(φ) as the
quadrature component The envelope of the bandpass signal is
V u (t) =
%
u2
c (t) + u2(t) =
%
m2(t) cos2(φ) + m2(t) sin2(φ)
=
%
m2(t) = |m(t)|
Hence, the envelope is proportional to the absolute value of the message signal
Problem 3.13
1) The modulated signal is
u(t) = 100[1 + m(t)] cos(2π8 × 105t)
= 100 cos(2π8 × 105t) + 100 sin(2π103t) cos(2π8 × 105t)
+500 cos(2π2 × 103t) cos(2π8 × 105t)
= 100 cos(2π8 × 105t) + 50[sin(2π(103+ 8× 105)t) − sin(2π(8 × 105− 103)t)] +250[cos(2π(2 × 103+ 8× 105)t) + cos(2π(8 × 105− 2 × 103)t)]