School of Engineering Science, Simon Fraser University
ENSC327 Communication Systems – Spring 2020
Assignment 1
Important Notes:
1) Please upload ONE PDF containing the solutions on Canvas only. Please write clearly; if it is not clear and
can't be marked, it will not be marked. No other format for submission (hard copy, email, etc…) is accepted.
2) Please do not share your solutions with your friends. Any solutions found to be the same will be marked
zero. The goal is to get everyone to do the work themselves so that each individual benefits by testing their
understanding with these questions.
1. (Fourier Transform Practice): The magnitude and the phase of the Fourier transform of a signal are
plotted in figure 1. Find the signal in time domain.
|X (f) |
6
4
-40
𝜋𝜋
2
-10
10
40
f
40
f
<X (f)
-40
-10
10
𝜋𝜋
-2
Figure 1
SOLUTION: X(f) = 4 δ(f +10) + 4 δ(f -10)
+ 6 δ(f+40) 𝑒𝑒 𝑗𝑗𝑗𝑗/2 +6 δ(f-40) 𝑒𝑒 −𝑗𝑗𝑗𝑗/2
1
1
= 8[ δ (f + 10) + δ (f - 10) ]
2
1
2
1
+ 12 [ 𝑗𝑗 δ (f +40) - 𝑗𝑗 δ (f - 40)]
2
2
x(t) = 8 cos 2π 10t + 12 sin 2π 40t
2. (Fourier Transform Properties) Find the Fourier transform of the signal 𝑦𝑦(𝑡𝑡) =
8 𝑥𝑥(𝑡𝑡) sin(2𝜋𝜋 𝑓𝑓1 𝑡𝑡) sin(2𝜋𝜋 𝑓𝑓2 𝑡𝑡) in terms of the Fourier transform of x(t).
1
SOLUTION: We know if ℱ { x (t) } = X (f), then ℱ { x (t) cos(2 𝜋𝜋 𝑓𝑓0 𝑡𝑡) } = [X (f - 𝑓𝑓0 ) + X (f + 𝑓𝑓0 ) ]
1
We also have: sinA sinB = [cos (A − B) − cos (A + B)]
2
2
Applying the trig identity to y(t): y(t) = 4x(t) cos (2𝜋𝜋 [𝑓𝑓1 − 𝑓𝑓2 ] t ) - 4x(t) cos (2𝜋𝜋 [𝑓𝑓1 + 𝑓𝑓2 ] t)
Then Fourier Transform:
Y(f) = 2 [X (f – [𝒇𝒇𝟏𝟏 − 𝒇𝒇𝟐𝟐 ]) + X (f + [𝒇𝒇𝟏𝟏 − 𝒇𝒇𝟐𝟐 ]) - X (f - [𝒇𝒇𝟏𝟏 + 𝒇𝒇𝟐𝟐 ]) - X (f + [𝐟𝐟𝟏𝟏 + 𝒇𝒇𝟐𝟐 ])]
3. (Fourier Transform Properties) Show that sinc(t) * sinc(t) = sinc(t) where * is the convolution
operator.
SOLUTION: We know that : ℱ { sinc(t) } = rect (f)
Where:
rect(f) = {1
0
|f| <= 0.5
Otherwise
we also have : rect (f) . rect (f) = rect (f)
ℱ −1 {rect(f) . rect(f)} = ℱ −1 rect (f)
Therefore, sinc (t) * sinc (t) = sinc(t).
4. (AM Modulator Signal Parameter Estimation) The signal 𝑚𝑚(𝑡𝑡) = cos(2𝜋𝜋 𝑓𝑓𝑚𝑚 𝑡𝑡) is applied to an AM
modulator and the output y(t) is depicted in the figure 2. We know that y(t) is the in the following form:
𝑦𝑦(𝑡𝑡) = 𝐴𝐴𝑐𝑐 [1 + 𝑘𝑘𝑎𝑎 𝑚𝑚(𝑡𝑡)] cos(2𝜋𝜋 𝑓𝑓𝑐𝑐 𝑡𝑡)
Find 𝐴𝐴𝑐𝑐 , 𝑘𝑘𝑎𝑎 , 𝑓𝑓𝑚𝑚 , 𝑓𝑓𝑐𝑐
Figure 2
SOLUTION: Based on the figure, we have:
Max: Ac [1 + Ka] = 3.4
Min: Ac [1 – Ka] = 0.6
Therefore, Max/Min = (1 + Ka) / (1 – Ka) = 3.4/0.6
0.6+0.6K = 3.4-3.4K
4K = 3.4-0.6=2.8
Using simple algebra, Ka = 2.8/4 = 7/10 or 0.7 and Ac = 3.4/(1+0.7) = 2
1/fc = 0.025 ms fc = 1/0.025= 40 KHz
1/fm = 0.25 ms
fm = 1/0.25 ~= 4 KHz
5. (AM Power efficiency) The following periodic message signal m(t) with period T is applied to an AM
modulator. Find out its power efficiency when amplitude sensitivity 𝑘𝑘𝑎𝑎 = 0.3, 0.5, and 1, respectively.
2
…
…
T/2
0
t
T
-2
SOLUTION: Power efficiency = 𝑘𝑘𝑎𝑎2 𝑃𝑃𝑚𝑚 /(1 + 𝑘𝑘𝑎𝑎2 𝑃𝑃𝑚𝑚 )
1
𝑇𝑇
2
𝑇𝑇/2
Pm = ∫0 𝑚𝑚(𝑡𝑡)2 𝑑𝑑𝑑𝑑 = ∫0
𝑇𝑇
𝒌𝒌𝒂𝒂
0.3
0.5
1.0
𝑇𝑇
2
𝑇𝑇/2 2×2
𝑚𝑚(𝑡𝑡)2 𝑑𝑑𝑑𝑑 = ∫0
𝑇𝑇
(
𝑇𝑇
4
𝑡𝑡)2 𝑑𝑑𝑑𝑑 = = 1.33
3
Power Efficiency
0.1069
0.2495
0.5708
6. (AM Signal Summary) An AM modulator has output:
𝑠𝑠(𝑡𝑡) = 4 𝑐𝑐𝑐𝑐𝑐𝑐(𝜋𝜋20𝑡𝑡) + 𝑐𝑐𝑐𝑐𝑐𝑐(𝜋𝜋30𝑡𝑡) + 𝑐𝑐𝑐𝑐𝑐𝑐(𝜋𝜋10𝑡𝑡)
Knowing that the message signal is of the form m(t) = A cos(2 𝜋𝜋 𝑓𝑓0 𝑡𝑡)
(a) Find the carrier frequency, message frequency, and modulation factor.
(b) Find the power efficiency.
SOLUTION:
𝛼𝛼+𝛽𝛽
𝛼𝛼−𝛽𝛽
(a) Using: cos 𝛼𝛼 + cos 𝛽𝛽 = 2 cos(
) cos
2
2
𝑠𝑠(𝑡𝑡) = 4 𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋10𝑡𝑡) + 𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋15𝑡𝑡) + 𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋5𝑡𝑡)
15 − 5
15 + 5
𝑡𝑡� 𝑐𝑐𝑐𝑐𝑐𝑐 �2𝜋𝜋
𝑡𝑡��
= 4 𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋10𝑡𝑡) + 2 �𝑐𝑐𝑐𝑐𝑐𝑐 �2𝜋𝜋
2
2
= 4 𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋10𝑡𝑡) + 2 𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋10𝑡𝑡) 𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋5𝑡𝑡)
= 4[1 + 0.5 𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋5𝑡𝑡)] 𝑐𝑐𝑐𝑐𝑐𝑐(2𝜋𝜋10𝑡𝑡)
Therefore, carrier frequency, message frequency, and modulation factor are 10 Hz, 5 Hz and 0.5,
respectively.
𝜇𝜇 2
(b) Power Efficiency = 2+𝜇𝜇2 ,
(0.5)2
⇒ Power Efficiency = 2+(0.5)2 = 0.11 = 11%