School of Engineering Science, Simon Fraser University
ENSC327 Communication Systems
Assignment 3-Solutions
1.
(SSB Lower Sideband) A SSB signal is generated by modulating a 270-kHz carrier by the signal m(t)
= cos (2020πt) + 8sin(2020πt). The amplitude of the carrier is Ac = 50. Determine the (time domain)
expression for the lower sideband of the SSB signal.
SOLUTION:
𝐴𝐴𝑐𝑐
𝑆𝑆𝐿𝐿𝐿𝐿𝐿𝐿 (𝑡𝑡) =
[𝑚𝑚(𝑡𝑡)𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋𝜋𝜋𝑐𝑐𝑡𝑡) + 𝑚𝑚
� (𝑡𝑡)𝑠𝑠𝑠𝑠𝑠𝑠(2𝜋𝜋𝜋𝜋𝑐𝑐𝑡𝑡)]
2
𝑚𝑚(𝑡𝑡) = 𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋1010𝑡𝑡) + 8𝑠𝑠𝑠𝑠𝑠𝑠(2𝜋𝜋1010𝑡𝑡)
We know that:
ℋ
𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋𝑓𝑓0 𝑡𝑡) → 𝑠𝑠𝑠𝑠𝑠𝑠(2𝜋𝜋𝑓𝑓0 𝑡𝑡)
ℋ
𝑠𝑠𝑠𝑠𝑠𝑠(2𝜋𝜋𝑓𝑓0 𝑡𝑡) → −𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋𝑓𝑓0 𝑡𝑡)
Therefore:
𝑆𝑆𝐿𝐿𝐿𝐿𝐿𝐿 (𝑡𝑡) =
Using:
ℋ: Hilbert transform
50
[ [𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋1010𝑡𝑡) + 8𝑠𝑠𝑠𝑠𝑠𝑠(2𝜋𝜋1010𝑡𝑡)] 𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋 270 × 103 𝑡𝑡) +
2
[𝑠𝑠𝑠𝑠𝑠𝑠(2𝜋𝜋1010𝑡𝑡) − 8𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋1010𝑡𝑡)] 𝑠𝑠𝑠𝑠𝑠𝑠(2𝜋𝜋 270 × 103 𝑡𝑡)]
We can write:
SLSB(t) = 25 [ cos(2π (268,990) t) - 8sin(2π (268,990) t)]
2.
(Complex Envelope) Consider the signal x(t) = 2W sinc(2Wt) cos(2πf 0t) .
a) Find the pre-envelope (analytic signal)-positive frequencies of 𝑥𝑥(𝑡𝑡) in frequency domain 𝑋𝑋𝑝𝑝 (𝑓𝑓).
b) Find the complex envelope of 𝑥𝑥(𝑡𝑡) in frequency domain, 𝑋𝑋�(f).
c) Find the complex envelope in time domain, 𝑥𝑥�(t).
d) Sketch 𝑋𝑋𝑝𝑝 (𝑓𝑓) and 𝑋𝑋�(f).
SOLUTION:
(d)
3. Angle Modulation An angle modulated signal has the form
𝑥𝑥𝑐𝑐 (𝑡𝑡) = 50 𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋𝑓𝑓𝑐𝑐 𝑡𝑡 + 2𝑠𝑠𝑠𝑠𝑠𝑠1000𝜋𝜋𝜋𝜋)
where fc = 5 MHz. Assume the amplitude of the single tone message to be equal to 1.
(a) Determine the average transmitted power.
(b) Determine the maximum phase deviation.
(c) Determine the maximum frequency deviation.
(d) Find the message signal with unit amplitude when the signal is a FM signal.
(e) Find the message signal with unit amplitude when the signal is a PM signal.
(CEAB indicators: 1.1: Mathematical Knowledge, 1.4: Discipline‐specific Knowledge, 2.3: Problem Solution)
Solution:
a) Since angle modulation is essentially a sinusoidal signal with constant amplitude, we have
b) ΔΦ𝑀𝑀𝑀𝑀𝑀𝑀 = max|2 ∗ sin(1000πt)| = 2
𝑃𝑃 =
𝐴𝐴2𝑐𝑐 502
=
= 1250
2
2
c) The instantaneous frequency is
𝑓𝑓𝑖𝑖 = 𝑓𝑓𝑐𝑐 +
1 𝑑𝑑
2
Φ(𝑡𝑡) = 𝑓𝑓𝑐𝑐 +
cos(1000𝜋𝜋𝜋𝜋) 1000𝜋𝜋 = 𝑓𝑓𝑐𝑐 + 1000cos (1000𝜋𝜋𝜋𝜋)
2𝜋𝜋 𝑑𝑑𝑑𝑑
2𝜋𝜋
Hence, the maximum freq. deviation is
Δ𝑓𝑓𝑚𝑚𝑚𝑚𝑚𝑚 = max |𝑓𝑓𝑖𝑖 − 𝑓𝑓𝑐𝑐 | = 1000
Parts (d and e) The angle modulated signal can be interpreted both as a PM and an FM signal. It is a PM signal
with phase deviation constant kp = 2 and message signal m(t) = sin (1000πt) or an FM signal with kf = 1000 and
m(t) = cos(1000πt)
4. An FM signal with carrier frequency 𝑓𝑓𝑐𝑐 = 100 𝐾𝐾𝐾𝐾𝐾𝐾 is given by:
𝑠𝑠(𝑡𝑡) = 10𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋𝑓𝑓𝑐𝑐 𝑡𝑡 + 5𝑠𝑠𝑠𝑠𝑠𝑠3000𝜋𝜋𝜋𝜋 + 10𝑠𝑠𝑠𝑠𝑠𝑠2000𝜋𝜋𝜋𝜋)
a) Find the frequency deviation ∆f
b) Find the deviation ratio 𝐷𝐷 =
∆𝑓𝑓
𝑊𝑊
c) Estimate the bandwidth of the FM signal using Carson’s rule 𝐵𝐵𝐵𝐵 = 2(∆𝑓𝑓 + 𝑊𝑊), where W is the bandwidth
of the message.
Solution:
1 𝑑𝑑
[2𝜋𝜋𝑓𝑓𝑐𝑐 𝑡𝑡 + 5 sin 3000𝜋𝜋𝜋𝜋 + 10 sin 2000𝜋𝜋𝜋𝜋]
a) Instantaneous frequency =
2𝜋𝜋 𝑑𝑑𝑑𝑑
= 𝑓𝑓𝑐𝑐 + 7500 cos 3000𝜋𝜋𝜋𝜋 + 10000 cos 2000𝜋𝜋𝜋𝜋
∆𝑓𝑓 = max |𝑓𝑓𝑖𝑖 − 𝑓𝑓𝑐𝑐 | =
1 𝑑𝑑𝜙𝜙(𝑡𝑡)
�
�
= 7500 + 10000 = 𝟏𝟏𝟏𝟏, 𝟓𝟓𝟓𝟓𝟓𝟓 𝑯𝑯𝑯𝑯
2𝜋𝜋 𝑑𝑑𝑑𝑑 𝑚𝑚𝑚𝑚𝑚𝑚
b) Note that since the message has 2 tones, we have:
3000𝜋𝜋 2000𝜋𝜋
� = 1500 Hz
Message Bandwidth = 𝑚𝑚𝑚𝑚𝑚𝑚 �
,
2𝜋𝜋
Frequency Deviation = 𝐷𝐷 =
∆𝑓𝑓
𝑊𝑊
=
2𝜋𝜋
17500
1500
= 𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔
C) Carson’s Rule: 𝐵𝐵 = 2(∆𝑓𝑓 + 𝑊𝑊) = 𝟑𝟑𝟑𝟑 𝑲𝑲𝑲𝑲𝑲𝑲