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THE UNIVERSITY OF HONG KONG
Bachelor of Engineering
Department of Electrical & Electronic Engineering
ELEC3241 Signal and linear systems
Online Examination
Date: 12 December 2019
Answer ALL questions.
Time: _2:30pm-4:30pm_
All questions carry equal marks
Upload ONE file (size not more than 10 Mb) for submission.
Use of Electronic Calculators:
"Only approved calculators as announced by the Examinations Secretary can be used
in this examination. It is candidates' responsibility to ensure that their calculator
operates satisfactorily, and candidates must record the name and type of the calculator
used on the front page of the examination script."
EEE/ ELEC3241/2019 Dec page 1 of 6
ELEC3241, Dec. 2019
Problem 1. (30 marks) Basic Properties of Signals and Systems
(a) (6 marks) Determine whether the following signals are periodic or not. For the periodic
signal(s), please give the period:
i) (3 marks) x(t) = jeJ 10t.
ii) (3 marks)x[n] = sin(2mr).
(b) (24 marks) Consider a continuous-time system where the input is x(t) and the output is y(t).
The relationship between them is given by y(t) = x(sin(t)).
i) (12 marks) Show this system is a linear system by the definition of linearity.
ii) (12 marks) Show this system is a non-causal system. (Hint: Recall the definition of
causality, you may try some particular values oft to find counter examples.)
Problem 2. (35 marks) Time-Domain Analysis of Systems
(a) (10 marks) x(t) = u(t-3) -u(t-5) and h(t) = Lk=-l 5(t-k). Determine y(t) = x(t) * h(t).
(b) (25 marks) An LTID system is specified by the following equation:
2y[n + 1] - 3y[n] + y[n - 1] = 4x[n + 1] - 3x[n].
i) [8 marks] Draw a schematic diagram for the LTID system in terms of delay compo­
nents "D".
ii) [12 marks] Determine the impulse response h[n] of the LTID system, when n � 2.
iii) [5 marks] Determine the impulse response h[n] for n ?: 0.
(Hint: You can do it by either the following two ways. {1) Direct calculation. (2)
Calculate more values of n {e.g., n = 3,4,5, .. .) and discover the laws.)
Problem 3. (35 marks) Fourier Series and Fourier Transform
(a) (7 marks) The Fourier series for the triangular wave x(t) = ltl, -1r � t � 1r, x(t + 21r) = x(t),
is
ao � . 1r 4� 1 x(t) = - + � (ancosnt + bnsmnt) = - - - � ( k )2 cos(2k - l)t. 2 2 1r 2 - 1
n=O k=l
Please find the value of the sum:
I: (a;+ b;)
n=l
page 2 of 6
..
ELEC3241, Dec. 2019
(b) (8 marks) Find the Fourier transform of x(t) = e-2tu(t - 3).
(c) (20 marks) The output of a causal LTI system is related to the input x(t) by the differential
equation
d
��
t) + 2y(t) = x(t)
(i) (10 marks) Determine the frequency response H(w) = Y(w)/X(w).
(ii) (10 marks) If x(t) = e-tu(t), determine Y(w), the Fourier transform of the output.
page 3 of 6
ELEC3241, Dec. 2019
No.
1
2
3
4
5
6
7
Sa
Sb
Operation
***
Time differentiation
*** Appendix A: Laplace transform table ***
x(t)
�(!)
u(t)
tu(t)
eA.1u(t)
te,..1 u(t)
cos ·bt,u(t)
sin bt u(t)
X(s)
1
s2
n!
5n+I
1
s->..
(s - >..)2
n!
s
s2 +b2
b
s2 +b2
Appendix B: Laplace time differentiation property ***
x(t)
dx
dt
d2x
dt2
d3x
dt3
page 4 of 6
X(s)
sX(s) - x(O-)
ELEC3241, Dec. 2019
*** Appendix C: Z transform table ***
No.
1
2
3
4
5
6
7
8
9
x[n)
S[n -k]
u[n]
nu[n]
n2u[n]
n3u[n]
ynu[n]
yn-lu[n - 1]
nynu[n]
n2y"u[n]
Xlz]
,-k
z --
z-1
z
(z -1)�
z(z + 1)
( z -1)3
z(z2 + 4z + 1)
(z -1)4
z
z-y
1
z-y
yz
(z -y)2
yz(z + y)
(z -y)J
10
n(n - l)(n -2) · · · (n -m + 1)
ynu[n]
z
ymm! (z-y)m+I
*** Appendix D: Fourier series table ***
Series Form
'Iiigonometlic
f (t) = ao + Lan cos nwot + bn sin naJu(
•=I
Compact trisunomet.rlc
f(t) =Co+ L c. cos(nwnr + e.)
-�1
Exponential
/(t) = L D.ei"""''
11=-oc
Coefficient Computatiorn
ao = _!__ r I c,) dt
To lr0
a. = 2.1 f(t) e-0snTo To
b. = � f f(t) sai nwot dt
To }70
Co =ao
c. = y'a. 2 + b.2
o. = _!__ r f(l)e-i•"'D• dt
To Jr0
page 5 of 6
Conversion Formulas
ao =Co= Do
On - jb. = c.eil, = 2D.
Cu = Do
c. = 21Dnl n ;:,:: 1
e. = LD.
ELEC3241, Dec. 2019
*** Appendix E: Fourier transform table ***
No.
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
x(t)
e0'u(-t)
o(t)
1
COS Wot
sin wot
u(t)
sgnt
cos wot u(t)
sin w0t u(t)
e-01 sin wot u(t)
e-a, cos wot u(t)
rect ( �)
w . ( -smc _Wt)
JT
� (�)
W sinc
2 (Wt)
21r 2
X(w)
I
a+ jw
1
a - jw
2a
a2 + w2
1
(a+ jw)2
n!
(a+ jw)n+l
I
2d(w)
2rro(w - Wo)
rr[o(w - Wo) + o(w + Wo)]
jrr[o(w + WI)) - o(w - Wo)]
I
rro(w) + -.
2
JW
jw
7r jw -
2
[o(w - Wo) + o(w + Wo)] + 2 2 Wo - W
JT Wo
--;-[O(W - Wo) - O(W + Wo)] + 2 w2 2J Wo-
Wo
(a+ jw)2 + w5
a+ jw
r sinc (�
r
)
rect (:W)
r . 2 (wr)
2
smc
4
*** END OF PAPER ***
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