Chapter 5
5.1
A message signal s (t ) is shown in Figure P5.1. Sketch the FM and PM waveforms
if the carrier frequency is 50 Hz, k f = 30 Hz/V and k p = 10π radians/V.
Figure P5.1
s (t )
1
0
1
2
3
…
t (msec)
−1
s(t), x(t)
Solution:
t (msec)
1
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s(t), x(t)
t (msec)
5.2 An angle-modulated signal is described by
(t )
⎡
θ
⎤
x(t ) = 100 cos ⎢ 2π × 105 t + 0.01cos(4π × 103 t ) ⎥ .
⎥
⎢ φi ( t )
⎣
⎦
a. Write an expression for the instantaneous frequency of x(t ) .
Solution:
The instantaneous frequency of the angle-modulated signal is given by
5
3
1 dφi (t ) 1 d ⎡⎣ 2π × 10 t + 0.01cos(4π × 10 t ) ⎤⎦
fi (t ) =
=
2π dt
2π
dt
5
3
3
= 10 − 0.01× 2 × 10 sin(4π ×10 t )
= 105 − 20sin(4π × 103 t )
By comparing with (5.16), we conclude that ∆f max = 20 Hz.
b. If x(t ) is an FM signal, what is the message signal sn (t ) ?
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Solution:
If x(t ) is an FM signal, the excess phase θ (t ) is given by
t
θ (t ) = 2π ∆f max ∫ sn (t )dt = 0.01cos(4π × 103 t )
−∞
Taking the derivative of both sides yields
2π∆f max sn (t ) = −40π sin(4π ×103 t )
Therefore, the normalized message signal is sn (t ) = − sin(4π × 103 t )
c. If x(t ) is a PM signal, what is the message signal sn (t ) ?
Solution:
If x(t ) is a PM signal, the excess phase θ (t ) is given by
θ (t ) = ∆φmax sn (t ) = 0.01cos(4π ×103 t )
Therefore, the normalized message signal is sn (t ) = cos(4π ×103 t )
5.3 An angle-modulated signal is given by
⎡
⎤
x(t ) = 10 cos ⎢ 2π × 106 t + 2sin(4π × 103 t ) ⎥ .
⎥
⎢⎣
θ (t )
⎦
a. Determine the peak phase deviation.
Solution:
The excess phase θ (t ) is
θ (t ) = ∆φmax sn (t ) = 2sin(4π ×103 t )
Therefore, the peak phase deviation is
∆φmax = 2 radians
b. Determine the peak frequency deviation.
3
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Solution:
6
3
1 dφi (t ) 1 d ⎡⎣ 2π ×10 t + 2sin(4π ×10 t ) ⎤⎦
fi (t ) =
=
2π dt
2π
dt
6
3
3
= 10 + 4 ×10 cos(4π ×10 t )
By comparing with (5.16), we conclude that ∆f max = 4 ×103 Hz.
c. Calculate the average power in the modulated signal x(t ) .
Solution:
The power in the angle-modulated signal x (t ) is
Ac2 102
Px =
=
= 50 W
2
2
c. Is this a frequency- or phase-modulated signal?
Solution:
The angle-modulated signal x(t ) can be interpreted as either a PM or an FM
signal. It is a PM signal with the peak phase deviation ∆φmax = 2 radians and the
normalized message signal sn (t ) = sin(4π × 103 t ) . It is a FM signal with the peak
frequency deviation ∆f max = 4 ×103 and the normalized message
signal sn (t ) = cos(4π × 103 t ) .
5.4 Consider an FM signal generated by modulating a 10 MHz carrier with a 1 kHz
sinusoidal waveform such that peak frequency deviation is 2.5 kHz.
a. Determine the bandwidth of the modulated signal.
Solution:
BT = 2(∆f max + f m )
Substituting f m = 1 kHz and ∆f max = 2.5 kHz , we obtain
BT = 2(2.5 + 1) = 7 kHz
b. If the modulating signal amplitude is doubled, determine the bandwidth of the
modulated signal.
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Solution:
Doubling the amplitude doubles the peak frequency deviation ∆f max . Therefore,
∆f max = 5 kHz
Substituting yields
BT = 2(5 + 1) = 12 kHz
c. Determine the bandwidth of the modulated signal if the modulating signal
frequency is doubled.
Solution:
Now f m = 2 kHz and ∆f max = 2.5 kHz . Substituting
BT = 2(2.5 + 2) = 9 kHz
d. Determine the bandwidth of the modulated signal if both the amplitude and
frequency of the modulating signal are doubled.
Solution:
Now f m = 2 kHz and ∆f max = 5 kHz . Substituting
BT = 2(5 + 2) = 14 kHz
5.5 Let the modulating signal be s (t ) = 10 cos(2π × 300t ) + 25cos(2π × 600t )
a. Write an expression for the FM waveform xFM (t ) when Ac = 100, f c = 5 MHz,
and k f = 200 Hz/V .
Solution:
t
⎡
⎤
xFM (t ) = Ac cos ⎢ 2πf c t + 2π ∆f max ∫ sn (α )dα ⎥
−∞
⎣
⎦
In the present case, max s (t ) = 35 . Therefore,
sn (t ) =
1
[10 cos(2π × 300t ) + 25cos(2π × 600t )]
35
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∆f max = k f max s (t ) = 200 × 35 = 7 × 103
Substituting
⎡
2π × 7 × 103
xFM (t ) = 100 cos ⎢10π × 106 t +
35
⎣
⎤
t
∫ [10 cos(2π × 300α ) + 25cos(2π × 600α )]dα ⎥⎦
−∞
⎡
2π × 7 × 103 ⎡ 10
25
⎤⎤
sin(2π × 300t ) +
sin(2π × 600t ) ⎥ ⎥
= 100 cos ⎢10π × 106 t +
⎢
35
2π × 600
⎣ 2π × 300
⎦⎦
⎣
20
25
⎡
⎤
= 100 cos ⎢10π × 106 t + sin(2π × 300t ) + sin(2π × 600t ) ⎥
3
3
⎣
⎦
b. Determine maximum frequency deviation ∆f max , maximum phase
deviation ∆φmax , and deviation ratio D of the modulated signal.
∆f max = 7 ×103 Hz
The maximum phase deviation ∆φmax is the maximum value of the angle
θ (t ) =
20
25
sin(2π × 300t ) + sin(2π × 600t ) , which equals 15 radians.
3
3
Deviation ratio D =
∆f max 7 ×103 70
=
=
B
600
6
c. Determine bandwidth of the modulated signal.
BT = 2(∆f max + B) = 2(7 ×103 + 600) = 15.2 kHz
5.6 Consider a PM signal generated by modulating a 10 MHz carrier with a 1 kHz
sinusoidal waveform with peak phase deviation of 1 radian.
a. Write an expression for the PM signal and determine its Carson bandwidth.
Solution:
The PM signal is
xPM (t ) = Ac cos ⎡⎣ 2π × 107 t + cos(2π × 103 t ) ⎤⎦
For a sinusoidal modulating signal with β = ∆φmax = 1 kHz , the bandwidth of an
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angle-modulated signal is given by
BT = 2(∆φmax + 1) f m = 2(1 + 1) × 1 = 4 kHz
b. Plot the spectrum of the PM signal.
Solution:
Consider a PM signal with maximum phase deviation β produced by a
sinusoidal modulating signal. That is,
xPM (t ) = Ac cos [ 2πf c t + β cos(2πf mt ) ]
The PM signal can be expressed as
{
xPM (t ) = Ac Re e j 2 πfct e j β cos(2 πfmt )
}
( m)
The function e
is periodic with period 1/ fm, and therefore it can be
expanded in a complex Fourier series:
j β cos 2 πf t
e
j β cos ( 2 πf m t )
∞
∑Ce
=
n =−∞
n
j 2π nf m t
where
Cn = f m ∫
1/ f m
0
e
j β cos ( 2 πf m t ) − j 2π nf m t
e
dt =
1
2π
∫
2π
0
e j ( β cos z − nz ) dz
π⎞
⎛
Substituting cos ( z ) = sin ⎜ z + ⎟ , we obtain
2⎠
⎝
1
Cn =
2π
∫
2π
0
e
⎡
⎤
⎛ π⎞
j ⎢ β sin ⎜ z + ⎟ − nz ⎥
⎝ 2⎠
⎣
⎦
dz
Making change of variable z +
Cn =
1
2π
5π /2
∫π
/2
e
nπ ⎞
⎛
j ⎜ β sin y − ny + ⎟
2 ⎠
⎝
π
2
dy = e
= y yields
jnπ
2
⎧ 1
⎨
⎩ 2π
∫
2π
0
⎫
e j ( β sin y − ny ) dy ⎬ = e
⎭
jnπ
2
Jn ( β )
Since the integrand is periodic function of y with period 2π, the limits of
integral can be changed to any period of length 2π to obtain
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Cn = e
jnπ
2
⎧ 1
⎨
⎩ 2π
∫
2π
0
e
j ( β sin y − ny )
⎫
dy ⎬ = e
⎭
jnπ
2
Jn ( β )
The PM signal can, therefore, be expressed as
nπ ⎞
⎛
∞
j ⎜ n 2 πf m t + ⎟ ⎪
⎫
⎪⎧
j 2 πf c t
2 ⎠
⎝
xPM (t ) = Re ⎨ Ac e
J n ( β )e
⎬
∑
n =−∞
⎪⎩
⎪⎭
∞
nπ ⎤
⎡
= Ac ∑ J n ( β ) cos ⎢ 2π ( f c + nf m ) t +
2 ⎥⎦
⎣
n =−∞
In the present case, the PM signal is given by
xPM (t ) = Ac
∞
∑J
n =−∞
n
nπ ⎤
⎡
(1) cos ⎢ 2π (107 + n ×103 ) t +
2 ⎥⎦
⎣
The spectrum of the PM signal is displayed in the Figure.
PM Spectrum for Ac = 1,
fm(kHz) = 1
& Max phase deviation = 1
1
Magnitude
spectrum
Magnitude
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-10
-8
-6
-4
-2
0
2
Frequency(kHz)
4
6
8
10
f − f c ( kHz )
c. Determine the normalized average power in the PM signal.
Solution:
The power in the PM signal x(t ) is
Px =
Ac2
2
d. Calculate the percentage of the total power at the carrier frequency.
Solution:
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Ac2 2
The average power in the component at the carrier frequency is
J 0 (β )
2
where β = ∆φmax for a PM signal. For β = ∆φmax = 1, J 0 (1) = 0.765 . Therefore,
percentage of the average power at the carrier frequency = 58.5%.
e. Repeat parts (a) and (d), if the amplitude of the modulating signal is doubled.
Solution:
If the amplitude of the modulating signal is doubled, the peak phase deviation is
also doubled. That is, β = ∆φmax = 2 . Therefore, the PM signal becomes
nπ ⎤
⎡
,
xPM (t ) = Ac cos ⎢ 2π × 106 t + 2 cos(2π × 103 t ) +
2 ⎥⎦
⎣
and the bandwidth of the signal is given by
BT = 2(∆φmax + 1) f m = 2(2 + 1) × 1 = 6 kHz
The spectrum of the PM signal is shown below.
PM Spectrum for Ac = 1,
fm(kHz) = 1
& Max phase deviation = 2
1
Magnitude
spectrum
Magnitude
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-10
-8
-6
-4
-2
0
2
Frequency(kHz)
4
6
8
10
f − f c ( kHz )
For β = 2, J 0 (2) = 0.224 . The percentage of the average power at the carrier
frequency = J 02 (2) = ( 0.224 ) = 0.0502 or 5.02% .
2
5.7 A single-tone modulated FM signal is given by
xFM (t ) = 10 cos ⎡⎣ 2π × 106 t + 4sin(4π × 103 t ) ⎤⎦
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a. Plot the spectrum of the FM signal.
Solution:
The FM signal can be expressed as
∞
xFM (t ) = 10 ∑ J n (4) cos ⎡⎣ 2π (106 + 2n × 103 ) t ⎤⎦
n =−∞
FM Spectrum for A c = 1,
fm(kHz) = 2
& beta = 4
0.5
Magnitude
spectrum
Magnitude
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-10
-8
-6
-4
-2
0
2
Frequency(kHz)
4
6
8
10
f − f c ( kHz )
b. Determine Carson’s bandwidth of the FM signal.
Solution:
β = 4, f m = 2 × 103 Hz
BT = 2( β + 1) f m = 2(4 + 1) × 2 ×103 = 20 × 103 Hz
c. Determine the normalized average power of the FM signal.
Solution:
The normalized average power of the FM signal x(t ) is
102
Px =
= 50 W
2
d. Calculate the percentage of the total power at the carrier frequency. How does
the magnitude of power contained in the carrier frequency component changes
if the amplitude of the modulating signal is doubled?
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Solution:
The average power in the component at the carrier frequency is
Ac2 2
J 0 (β )
2
For β = 4, J 0 (4) = −0.3971 . Therefore, percentage of the average power at the
carrier frequency = 15.77%.
If the amplitude of the modulating signal is doubled, β = 8 and J 0 (8) = 0.1717 .
Therefore, percentage of the average power at the carrier frequency = 2.95%. 5.8
The message signal s (t ) modulates a carrier at 1 MHz to produce an anglemodulated signal. s (t ) is periodic waveform with period To as shown in Figure
P5.2. Assuming Am = 1 and To = 1 sec , determine
Figure P5.2
s (t )
…
Am
−To
−
To
2
−
To
4
…
0
− Am
To
4
To
2
t
To
a. Determine the 97% bandwidth of the message signal.
Solution:
The Fourier series coefficients are given by
C0 = 0 because x (t ) is an odd function.
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Cn =
1
x(t )e − j 2π nfot dt
∫
To To
1
=
To
−To /4
∫
−e
− j 2π nf o t
−To /2
− je − j 2π nfot
=
To 2π nf o
1
dt +
To
−To /4
To /4
4t − j 2π nfot
1
e
dt +
∫
T
To
−To / 4 o
je − j 2π nfot
+
To 2π nf o
−T /2
o
To /2
4
+ 2
To
T /4
o
To /2
∫
e − j 2π nfot dt
To /4
To / 4
⎡ jT te − j 2π nfot T 2 e − j 2π nfot ⎤
+ o
⎢ o
2⎥
π
T
nf
2
(To 2π nf o ) ⎥⎦ −To /4
⎢⎣ o
o
⎛ je+ jπ n je+ jπ n /2 ⎞ ⎛ je − jπ n je− jπ n / 2 ⎞
=⎜
−
−
⎟+⎜
⎟
2π n ⎠ ⎝ 2π n
2π n ⎠
⎝ 2π n
⎡ jT 2 − jπ n /2 + jπ n /2
⎤
To2
− jπ n / 2
− jπ n /2
e
e
+e
+
−
⎢ o (e
⎥
)
(
)
2
( 2π n )
⎢⎣ 8π n
⎥⎦
j ⎡
⎛ n ⎞⎤
⎛πn ⎞
⎛πn ⎞
cos (π n ) − cos ⎜
=
⎟ + cos ⎜
⎟ + sinc ⎜ ⎟ ⎥
⎢
πn ⎣
⎝ 2 ⎠⎦
⎝ 2 ⎠
⎝ 2 ⎠
+
=
4
To2
j ⎡
n ⎤
n
( −1) + sinc ⎛⎜ ⎞⎟ ⎥ , n = ±1, ±2,......
⎢
πn ⎣
⎝ 2 ⎠⎦
Using the procedure illustrated in Example 2.27, the 97% bandwidth of the
10
signal is calculated as B = .
To
b. Calculate the bandwidth of an FM signal with k f = 100 Hz/V .
Solution:
The peak frequency deviation, ∆f max , of the FM signal is given from (5.15) by
∆f max = k f max s (t ) = 100 × 1 = 100 Hz
BT = 2(∆f max + B) = 2(100 + 10) = 220 Hz
c. Determine the bandwidth of an PM signal with k p = π / 2 .
Solution:
The peak phase deviation, ∆φmax , of the PM signal is given from (5.9) by
∆φmax = k p max s (t ) = π / 2
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BT = 2(∆φmax + 1) B = 2(π / 2 + 1)10 = 51.41 Hz
d. Calculate the bandwidth of an FM signal with k f = 1000 Hz/V .
Solution:
∆f max = k f max s (t ) = 1000 × 1 = 1000 Hz
BT = 2(∆f max + B) = 2(1000 + 10) = 2020 Hz
e. How do answers to parts (a) − (c) change for the case where
Am = 2 and To = 1 msec .
Solution:
For To = 1 msec , the message bandwidth B = 10 kHz
For Am = 2 , ∆f max = 200 , BT = 2(∆f max + B) = 2(200 + 10000) = 20.4 kHz
For Am = 2 , ∆φmax = π and BT = 2(∆φmax + 1) B = 2(π + 1)10000 = 82.83 kHz
f. How do answers to parts (a) − (c) change for the case where
Am = 1 and To = 0.5 msec .
Solution:
For To = 0.5 msec , the message bandwidth B = 20 kHz
For Am = 1 , ∆f max = 100 , BT = 2(∆f max + B) = 2(100 + 20000) = 40.2 kHz
For Am = 1 , ∆φmax = π / 2 , BT = 2(∆φmax + 1) B = 2(π / 2 + 1)20000 = 102.83 kHz
5.9
A 1 MHz carrier is frequency-modulated by a single tone of frequency 2 kHz,
resulting in the peak frequency deviation of 10 kHz.
a. What is the bandwidth occupied by the modulated signal? Plot the spectrum of
the FM signal (only sidebands in the Carson’s bandwidth).
Solution:
∆f max = 10 kHz, B = f m = 2 kHz
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BT = 2(∆f max + B) = 2(10 + 2) = 24 kHz
β=
∆f max 10
=
=5
2
fm
The spectrum of the FM signal contains 2(β +1) = 12 sidebands within the
Carson’s bandwidth of 24 kHz.
FM Spectrum for Ac = 1,
fm(kHz) = 2
& beta = 5
0.5
Magnitude
spectrum
Magnitude
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-10
-5
0
Frequency(kHz)
5
10
f − f c ( kHz )
b. If the amplitude of the modulating sinusoidal signal is increased by a factor of 3
and its frequency is decreased to 1 kHz, how is the bandwidth of the modulated
signal modified? Plot the spectrum of the FM signal.
Solution:
If the amplitude of the modulating sinusoidal signal is increased by a factor of
3, the peak FM deviation is increased by a factor of 3 to ∆f max = 30 kHz .
∆f max = 30 kHz, B = f m = 1 kHz
BT = 2(∆f max + B) = 2(30 + 1) = 62 kHz
β=
∆f max 30
=
= 30
1
fm
The spectrum of the FM signal contains 2(β +1) = 62 sidebands within the
Carson’s bandwidth of 62 kHz.
14
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FM Spectrum for Ac = 1,
fm(kHz) = 1
& beta = 30
0.5
Magnitude
spectrum
Magnitude
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-30
-20
-10
0
Frequency(kHz)
10
20
30
f − f c ( kHz )
c. Repeat part (b) with frequency of the modulating sinusoidal signal increased to
3 kHz.
Solution:
∆f max = 30 kHz, B = f m = 3kHz
BT = 2(∆f max + B) = 2(30 + 3) = 66 kHz
∆f
30
β = max =
= 10
3
fm
The spectrum of the FM signal contains 2(10 + 1) = 22 sidebands within the
Carson’s bandwidth of 66 kHz.
FM Spectrum for Ac = 1,
fm(kHz) = 3
& beta = 10
0.5
Magnitude
spectrum
Magnitude
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-30
-20
-10
0
10
Frequency(kHz)
20
30
f − f c ( kHz )
5.10 Consider the parallel RLC tuned circuit in Figure P5.3 used as a slope detector.
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Figure P5.3 Parallel resonant circuit
R
I ( jω )
X ( jω )
L
C
V ( jω )
a. Show that the transfer function of the tuned circuit is given by
H ( jω ) =
jωωo / Q
V ( jω )
= 2
Comment:
X ( jω ) ωo − ω 2 + jωωo / Q
where
ωo =
Q=
1
1
, ω3dB =
RC
LC
ωo
C
=R
L
ω3dB
Solution:
⎛ 1
⎞
+ jωC ⎟
I = V ( jω ) ⎜
⎝ jω L
⎠
⎛ 1
⎞
+ jωC ⎟ + V ( jω )
X ( jω ) = RI + V ( jω ) = RV ( jω ) ⎜
⎝ jω L
⎠
⎛ R
⎞
= V ( jω ) ⎜
+ RjωC + 1⎟
⎝ jω L
⎠
The transfer function H ( jω ) of the parallel RLC tuned circuit can be expressed
as
H ( jω ) =
V ( jω )
1/ R
=
X ( jω ) ⎛ 1
1⎞
⎜ jω L + jωC + R ⎟
⎝
⎠
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Substituting ωo =
1
C
, we obtain
,Q = R
L
LC
jωωo2 L
R
=
=
H ( jω ) =
2
jω L ⎞ ⎛ ω
⎛
jω L ⎞ ⎛ 2
2
jωωo
ωo − ω 2 +
⎜ 1 − ω LC +
⎟ ⎜1 − 2 +
⎟
⎜
R ⎠ ⎝ ωo
⎝
R ⎠ ⎝
R
jω L
R
=
jω L
R
jωωo
R
L
C
⎛ 2
jωωo
2
⎜ ωo − ω +
R
⎝
L⎞
⎟
C⎠
=
L⎞
⎟
C⎠
jωωo / Q
ω − ω 2 + jωωo / Q
2
o
b. Determine appropriate values for R, L and C for ωo = 2π ×106 and Q = 20 .
Solution:
We choose C = 0.01 µ F and L = 0.1 mH so that
f o = 106 =
1
−3
0.1× 10 × 0.01× 10−6
Next we calculate R using Q = 20 .
20 = R
C
0.01× 10−6
⇒ 400 = R 2
= R 210−4
−3
L
0.1× 10
R = 400 × 104 = 2000 ohms
c. Plot the magnitude frequency response of the parallel RLC tuned circuit from
940 kHz to 1 MHz. Select the discriminator center frequency, discriminator
constant K FD , and permissible peak frequency deviation for the input signal.
Solution:
The frequency response H ( jω ) of the tuned circuit from 940 kHz to 1 MHz is
plotted in the Figure. We choose the discriminator center frequency as 970 kHz.
The discriminator constant K FD is now calculated from the figure as
K FD =
0.7778 − 0.5222
0.2556
=
= 12.78 µ V/Hz
6
( 0.98 − 0.96 ) ×10 0.02 ×106
17
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The determination of permissible peak frequency deviation is subject to the
amount of acceptable distortion in the demodulated signal. From the amplitude
vs frequency characteristic above, we observe that ∆f max = 20 kHz is a good
choice.
Frequency Response of Parallel Resonant Circuit
1
0.9
|H(j
ω)|
Magnitude
0.8
0.7
0.6
0.5
0.4
940
950
960
970
Frequency(kHz)
980
990
1000
5.11 Delay-line FM discriminator Consider the FM signal
xFM (t ) = Ac cos[2πf c t + θ (t )]
t
where θ (t ) = 2π ∆f max ∫ sn (t )dt . We use the FM demodulation scheme illustrated in
−∞
Figure P5.4. The input FM signal is passed through a delay line that produces a
delay of π / 2 radians at the carrier frequency f c . That is, 2π f c td = π / 2 . The
output of delay-line is subtracted from the incoming FM signal, and the resulting
difference signal is envelope detected. Let
y (t ) = xFM (t ) − xFM (t − td )
Assume θ (t ) − θ (t − td ) 1 .
Figure P5.4
xFM (t ) Delay t = 1
d
y (t )
−
4 fc
Envelope
detector
yD (t )
+
18
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a. Show that the envelope of y (t ) is given by
⎡ π θ (t ) − θ (t − td ) ⎤
v(t ) = 2 Ac sin ⎢ +
⎥
2
⎣4
⎦
Solution:
y (t ) = xFM (t ) − xFM (t − td )
= Ac cos [ 2πf c t + θ (t ) ] − Ac cos [ 2πf c (t − td ) + θ (t − td )]
θ (t ) + θ (t − td )
θ (t ) − θ (t − td ) ⎤
⎡
⎤
⎡
− π ⎥ 2 Ac sin ⎢ πf c td +
y (t ) = sin ⎢ 2πf ct − πf ctd +
⎥
2
2
⎣
⎣
⎦ ⎦
Hig-frequency term
Low-frequency envelope
The first term on the right-hand side is modulating signal dependent carrier
term. The second-term is modulating signal dependent low-frequency
envelope. Since θ (t ) − θ (t − td ) 1 , the argument of the second term exhibits
π
. This implies that envelope always remains positive
4
to assure proper detection.
small variations around
The envelope detector tracks the envelope of y (t ) which is given by
θ (t ) − θ (t − td ) ⎤
⎡
v(t ) = 2 Ac sin ⎢ πf c td +
⎥
2
⎣
⎦
⎡ π θ (t ) − θ (t − td ) ⎤
= 2 Ac sin ⎢ +
⎥
2
⎣4
⎦
b. Show that the envelope detector output can be expressed as
yD (t ) =
1
2
⎡ 1 dθ ⎤ 1
⎢⎣1 + 2 dt td ⎥⎦ = 2 [1 + π ∆f max td sn (t ) ]
Solution:
⎡ π θ (t ) − θ (t − td ) ⎤
v(t ) = 2 Ac sin ⎢ +
⎥
2
⎣4
⎦
⎡ ⎛π⎞
⎛ θ (t ) − θ (t − td ) ⎞
⎛ π ⎞ ⎛ θ (t ) − θ (t − td ) ⎞ ⎤
= 2 Ac ⎢sin ⎜ ⎟ cos ⎜
⎟ + cos ⎜ ⎟ sin ⎜
⎟⎥
2
2
⎝4⎠ ⎝
⎝
⎠
⎠⎦
⎣ ⎝4⎠
⎡ ⎛ θ (t ) − θ (t − td ) ⎞
⎛ θ (t ) − θ (t − td ) ⎞ ⎤
= 2 Ac ⎢ cos ⎜
⎟ + sin ⎜
⎟⎥
2
2
⎠
⎝
⎠⎦
⎣ ⎝
19
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Since θ (t ) − θ (t − td ) 1 , the output of the envelope detector can be
approximated as
⎡ θ (t ) − θ (t − td ) ⎤
yD (t ) = 2 Ac ⎢1 +
⎥
2
⎣
⎦
⎡ 1 dθ (t ) ⎤
td ⎥
= 2 Ac ⎢1 +
⎣ 2 dt
⎦
Now
dθ (t )
=
dt
t
d ⎡ 2π ∆f max ∫ sn (t )dt ⎤
⎢⎣
⎥⎦
−∞
= 2π ∆f max sn (t )
dt
Substituting
⎡ θ (t ) − θ (t − td ) ⎤
yD (t ) = 2 Ac ⎢1 +
⎥
2
⎣
⎦
(*)
= 2 Ac [1 + π ∆f max td sn (t ) ]
c. Calculate the output yD (t ) for a single-tone modulated FM signal
xFM (t ) = Ac cos[2πf c t + β sin 2πf mt ] . Assume f c f m so that f mtd 1 .
Solution:
For a single-tone modulating signal
t
θ (t ) = 2π ∆f max ∫ sn (t )dt = β sin 2πf mt
−∞
Taking the derivative of both sides
2π ∆f max sn (t ) = β 2π f m cos ( 2πf mt ) ⇒ sn (t ) = cos ( 2πf mt )
Substituting into equation (*), we obtain
yD (t ) = 2 Ac [1 + π ∆f max td sn (t ) ] = 2 Ac ⎡⎣1 + π ∆f max td cos ( 2π f mt ) ⎤⎦
5.12
A superheterodyne FM receiver operates in the frequency range of 88-108 MHz.
Assuming that an IF of 10.7 MHz is selected, determine the range of variation of
20
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the local oscillator frequency f LO . Does the range of image frequencies fall
outside of the 88-108 MHz band?
Solution:
We will assume high-side injection. So the LO frequency is f LO = f c + f IF . The
range of LO frequencies is obtained as [88 + 10.7 , 108 + 10.7 ] = [98.7 , 118.7 ] .
The image frequencies are located at fimage = f c + 2 f IF . So the range of image
frequencies is obtained as [88 + 21.4 , 108 + 21.4] = [109.4 , 129.4] . Image
frequencies fall outside the signal band.
5.13
A first-order PLL has phase detector characteristic shown in Figure P5.5. Assume
that the phase detector output voltage swing is ±1.5 V. Determine
a. Phase detector gain constant
Solution:
Phase detector gain constant K PD =
3
π
V/radians
b. Hold-in range of the PLL.
Solution:
The magnitude of the hold-in range ∆ωH is calculated by finding the frequency
offset of the input that causes a phase error of ±π/2 radians. The phase detector
output voltage swing is ±1.5 V for this range of phase error variation. We can
obtain the PLL hold-in range by multiplying the change in phase detector voltage
by KVCO F (0) . Since F (0) = 1 for a first-order PLL, ∆ωH is given by
∆ωH = 3KVCO
Figure P5.5
υe (t )
…
−π
0
π
2
π
2π
…
Phase error θ e
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5.14 A first-order PLL is operating in phase lock when a frequency step ∆ω is applied.
Assume that the loop gain K = 500π . Assume K PD = 0.8V/radians
a. Determine the steady-state phase error, in degrees, for
∆ω = 100π , 200π , and 800π .
Solution:
For a first-order PLL, the steady-state phase-error due to frequency-step ∆ω is
given by
θ e (∞ ) =
∆ω
K
∆ω
θ e (∞ )
100π
200π
800π
0.2
0.4
1.6
b. Plot the control voltage in each case. What is the steady-state control voltage
after the initial transients have died?
Solution:
The control voltage υe (t ) is given from (5.110) as
υe (t ) =
K PD ∆ω
1 − e− K t
K
(
)
1.4
1.2
1
Freq. offset = 100*pi
Freq. offset = 200*pi
Freq. offset = 800*pi
ve(t)
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
t(sec)
3
3.5
4
4.5
5
-3
x 10
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∆ω
υ e (∞ )
100π
200π
800π
0.16
0.32
1.28
c. What is the 3-dB bandwidth of the loop in each case? Comment on the tradeoff
between the noise performance of the first-order PLL versus its steady-state
phase error in part(a).
Solution:
The 3-dB bandwidth of the PLL is K = 500π (1/sec). The shortcoming of the
first-order PLL is that steady-state phase error θ e (∞) is inversely proportional
to the loop gain K. Larger the loop gain, smaller the steady-state phase error
θ e (∞) . However, the larger loop gain implies larger loop bandwidth and more
output noise.
5.15 Consider two-pole, second-order PLL with loop filter F ( s ) =
1
.
1 + sτ 1
a. Derive closed-loop transfer function of the PLL in the standard form.
Solution:
The open-loop transfer function G ( s ) for the loop is given by
G ( s) =
K
s (1 + sτ 1 )
We can now write the closed-loop transfer function H ( s ) as
H (s) =
K
⎛
⎞
K
s (1 + sτ 1 ) ⎜⎜1 +
⎟⎟
⎝ s (1 + sτ 1 ) ⎠
=
K / τ1
s
s 2 + + K / τ1
τ1
We can express H ( s ) in the standard form as
H (s) =
ωn2
s 2 + 2sζωn + ωn2
where
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K
ωn =
ζ =
τ1
1
2 Kτ 1
b. Determine and compare the 3-dB bandwidth and noise equivalent bandwidth BN
of the PLL.
Solution:
H ( jω ) =
ωn2
ωn2 − ω 2 + 2ζ jωnω
H ( jω ) =
ωn4
2
(ω
2
n
−ω2
) + ( 2ζω ω )
2
2
n
Now
H ( jω3dB ) =
ωn4
2
(ω
2
n
2
− ω3dB
) + ( 2ζω ω )
2
n
2
=
3dB
1
2
Solving for ω3dB, we have
⎡
⎣
(
⎡
2
⎢1 − 2ς +
⎣
(1 − 2ς )
ω3dB = ωn ⎢1 − 2ς 2 +
B3dB
ω
= n
2π
∞
1 − 2ς 2
2
∞
BN = ∫ H ( f ) df = ∫
0
(ω
2
1/2
⎤
+ 1⎥
⎦
1/2
⎤
+ 1⎥
⎦
ωn4
2
0
)
2
2
n
− ω 2 ) + 4ζ 2ωn2ω 2
2
df
∞
ωn4
df
4
2 2 2
2
4
0 ( 2π f ) + 8π ωn f ( 2ζ − 1) + ωn
=∫
=
ωn
8ζ
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B3dB ωn
=
2π
BN
⎡
2
⎢1 − 2ς +
⎣
(1 − 2ς )
8ζ
=
2π
⎡
2
⎢1 − 2ς +
⎣
(1 − 2ς )
2
2
8ζ
It can be shown that
2π
B3dB < BN .
2
2
8ζ
⎤
+ 1⎥ ×
ωn
⎦
1/2
1/2
⎤
+ 1⎥
⎦
⎡
2
⎢1 − 2ζ +
⎣
(1 − 2ζ )
2
2
1/2
⎤
+ 1⎥
⎦
< 1 for ζ > 0. Therefore,
5.16 A frequency step of ∆ω is applied to the second-order PLL with loop active PI
loop filter analyzed in Section 5.5.4.
a. Derive an expression for the control voltage applied to the VCO. What is the
steady-state control voltage after the initial transients have died?
Solution:
The closed-loop transfer function H ( s ) the second-order PLL with loop active
PI loop filter is given by
H (s) =
Θout ( s )
2sζωn + ωn2
= 2
Θin ( s ) s + 2 sζωn + ωn2
The control voltage υe (t ) in s-domain is now obtained as
Ve ( s ) =
(
)
sΘin 2 sζωn + ωn2
sΘout ( s )
=
K VCO
K VCO s 2 + 2 sζωn + ωn2
(
)
For a step change ∆f in frequency, Θin ( s ) =
Ve ( s ) =
(
2π∆f 2 sζωn + ωn2
(
)
sK VCO s + 2 sζωn + ωn2
2
∆ω
. Substituting yields
s2
)
The control voltage υe (t ) is now obtained by taking the inverse Laplace transform as
υe (t ) =
2π∆f
K VCO
⎛
⎞
ς e−ςωnt
⎜1 − e −ςωnt cos( 1 − ς 2 ωnt ) +
sin( 1 − ς 2 ωnt ) ⎟ u (t ), ς < 1
⎜
⎟
1− ς 2
⎝
⎠
25
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Therefore, the steady-state control voltage after the initial transients have died is
given by
lim υe (t ) =
t →∞
2π∆f
K VCO
The above result can also be derived by applying the final value theorem of the
Laplace transform.
b. Calculate the peak phase error for the step frequency change ∆ω .
Solution:
The phase error θ e (t ) , for the step frequency change ∆ω , is given from (5.130)
as
θ e (t ) =
∆ω e −ζωnt ⎛ 1
⎜
sin
ωn ⎜⎝ 1 − ζ 2
(
⎞
1 − ζ 2 ωn t ⎟ , ζ < 1
⎟
⎠
)
The peak phase error θ p is obtained by setting
θp =
∆ω
ωn
e
(
dθ e (t )
= 0 and solving to give
dt
)
− tan −1 κ / κ
where
κ = 1− ζ 2 / ζ 5.17 Consider the second-order PLL with an active PI loop filter. A single tone
modulated FM signal with maximum frequency deviation ∆f max is applied to the
PLL, where the frequency of the modulating signal is f m .
a. Calculate the magnitude of the phase error.
Solution:
For the second-order active PI loop, the phase error in the s-domain is given
from (5.124) as
26
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Θin ( s ) s 2
Θe ( s ) = 2
s + 2 sζωn + ωn2
The magnitude of the phase error in the frequency domain can be expressed as
Θe ( jω ) =
Θin ( jω ) ω 2
(ω
2
n
− ω2
)
2
+ 4ζ 2ωn2ω 2
For an FM signal with maximum frequency deviation ∆f max caused by a
sinusoidal tone of frequency ωm , the maximum phase deviation from (5.26) is
β=
∆f max ∆ωmax
=
ωm
fm
The magnitude of the PLL phase error in tracking the FM signal with the phase
deviation β is now obtained as
Θe ( jω ) =
βω 2
(ω
2
n
− ω2
)
2
+ 4ζ 2ωn2ω 2
b. For fixed ∆f max , calculate the value of modulating signal frequency f m for
which the peak phase error occurs. What is the corresponding value of the peak
phase error?
Solution:
The phase error at ω = ωm = 2π f m is given by
Θe ( jωm ) =
βωm2
(ω
2
n
−ω
)
2 2
m
+ 4ζ 2ωn2ωm2
For fixed ∆f max , we calculate
=
d Θe ( jωm )
d ωm
occurs for ωm = ωn . Substituting yields
β
2
2
⎛ ωn2
⎞
2 ωn
ζ
−
1
+
4
⎜ 2
⎟
ωm2
⎝ ωm ⎠
= 0 . We find that the maximum error
27
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⎛ ∆ωmax ⎞ 1
⎟
⎝ ωn ⎠ 2ζ
ε p = max Θe ( jωm ) = ⎜
f
m
c. Calculate the parameters of PLL ( K ,τ 1 ,τ 2 ) for a FM signal with
∆f max = 75 kHz and f m = 15 kHz . The PLL uses the phase detector with
characteristic in Figure P5.5. Assume ζ = 0.707 , and τ 1 = τ 2 .
Solution:
The PLL must demodulate the signal with uniform gain and stay within the
linear portion of the phase detector characteristic. To assure uniform gain, we
choose
ωn > ωm = 15 kHz
In order that the PLL operates within the linear portion of the phase detector
characteristic, the peak phase error ε p < π / 2 . That is,
⎛ ∆ωmax ⎞ 1 π
<
⎟
⎝ ωn ⎠ 2ζ 2
εp =⎜
or
75
π
<
fn 2 2
or
fn >
2 × 75
π
= 33.76 kHz
Therefore, we choose f n = 34 kHz . Now from Table 5.4, we obtain parameters
of the second-order PLL with an active PI loop filter as follows:
ωn =
ζ =
K
τ1
τ 2ωn
2
and ζ =
⇒τ2 =
Also, ωn × ζ =
2ζ
ωn
τ2
2
=
K
τ1
2
= 6.62 µ sec
2π × 34 × 103
τ2 K
. Assuming τ 2 = τ 1 ,
2 τ1
K = 2 × ωn × ζ = 2 × 0.707 × 2π × 34 × 103 = 3.02 × 105 (1/ sec)
28
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