Assignment 3 Solutions

ENGG 2310-B: Principles of Communication Systems
2014–15 First Term
Assignment 3 Solutions
Instructor: Wing-Kin Ma
5:30pm, November 11, 2014
Problem 1 (30%) Sketch the PM and FM waves produced by the sawtooth wave shown in
Figure 1. You may use the following parameters: fc = 2Hz, T0 = 1s, kf = 1, kp = π/2. Also, you
are allowed to use software like MATLAB to plot the waveforms, rather than sketching.
m(t)
1
T0
2T0
3T0
t
4T0
−1
Figure 1: Problem 1.
Solution:
t
T0
2T0
3T0
4T0
t
T0
2T0
3T0
4T0
Problem 2 Suppose that the message signal is a sinusoidal wave; i.e., m(t) = Am sin(2πfm t). The
amplitude and frequency of the modulating signal is Am = 10 volts and fm = 8 kHz, respectively.
(a) (10%) Use Carson’s rule to give an estimate of the transmission bandwidth of the FM signal,
when the frequency sensitivity is kf = 10 kHz/volt and the carrier frequency is 450 MHz.
(b) (10%) Repeat Problem 2.(a) when the message signal is a sum of two sinusoidal signals,
i.e., m(t) = Am,1 sin(2πfm,1 t) + Am,2 sin(2πfm,2 t), with Am,1 = 10 volts, fm,1 = 8 kHz,
Am,1 = 20 volts, fm,2 = 20 kHz.
(c) (15%) Repeat Problem 2.(a) when the modulation scheme is changed to PM, with phase
sensitivity kp = 3 rad./volt.
1
Solution: Recall that Carson’s rule estimates the FM bandwidth by the formula BT = 2∆f + 2W ,
where ∆f = kf maxt |m(t)|.
(a) For the message signal in this problem, the message bandwidth is W = fm = 8 kHz. Also,
since |m(t)| = Am |sin(2πfm t)| ≤ Am , the maximum signal amplitude is maxt |m(t)| = Am .
It follows that
∆f = kf Am = 10 kHz/volt × 10 volts = 100 kHz,
and that
BT = 2 × (100 kHz + 8 kHz) = 216 kHz.
(b) When the message signal is a sum of two sinusoidal waves, the bandwidth equals
W = max{fm,1 , fm,2 } = 20 kHz.
Also, since
|m(t)| ≤ Am,1 |sin(2πfm,1 t)| + Am,1 |sin(2πfm,2 t)| ≤ Am,1 + Am,2 ,
we may use Am,1 + Am,2 = 30 volts as a conservative estimate of maxt |m(t)|. The resulting
FM transmission bandwidth estimate by Carson’s rule is therefore
BT = 2(10 kHz/volt × 30 volts + 20 kHz) = 640 kHz.
(c) Let m′ (t) =
dm(t)
. The PM signal can be rewritten as
dt
Z
kp t
′
m (τ )dτ
s(t) = cos 2πfc t + 2π
2π −∞
From the above expression, it is seen that the PM signal is equivalent to an FM signal
k
with message signal m′ (t) and frequency sensitivity 2πp . Hence, we can estimate the PM
transmission bandwidth by applying Carson’s formula on the above equivalent FM expression.
With the aforementioned idea in mind, we consider
m′ (t) =
dm(t)
= Am · (2πfm ) · cos(2πfm t).
dt
It is seen that W = fm , and maxt |m′ (t)| = 2πfm Am . It follows that
kp
BT = 2
× 2πfm Am + fm
2π
= 2(kp Am fm + fm )
= 2(3 × 10 × 8 + 8) = 496 kHz.
Problem 3
Thirty different message signals, each with a bandwidth of 3 kHz, are to be
multiplexed and transmitted. Determine the minimum transmission bandwidth required for each
method if the multiplexing/modulation method used is
2
(a) (10%) FDM, DSB-SC.
(b) (10%) FDM, SSB.
(b) (15%) TDM, PAM, with the system specification that the pulse is rectangular and the pulse
width is half of the interval between successive pulses.
Solution:
(a) The transmission bandwidth for each DSB-SC-modulated message signal is BT = 2×3 kHz =
6 kHz. To have FDM of 30 such message signals, we would at least need a total transmission
bandwidth of 30 × 6 kHz = 180 kHz.
(b) For SSB modulation, the transmission bandwidth of each message signal is BT = 3 kHz.
Therefore, like Problem 3.(a), the total transmission bandwidth is 30 × 3 kHz = 90 kHz.
(c) By the sampling theorem, the sampling rate of each message should at least be 2 × 3 kHz =
6 kHz in order to guarantee perfect reconstruction. This means that to send one PAM signal,
we need to transmit one pulse every 1/6000 seconds. This further implies that to send 30
PAM signals by TDM, we need to transmit 30 times faster, i.e., to transmit one pulse every
1/(30 × 6000) seconds. Since the pulse has a width that is half of the interval between
successive pulses, the pulse width is 1/(2 × 30 × 6000) seconds. It follows that TDM PAM
transmission bandwidth is (approximately)
2 × 30 × 6000 = 360 kHz.
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