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(ELEC211)[2010](sum)final~kytangab^_46745.pdf
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Problem 1 (Total = 20 points)
2. 2. j10.t . j10.tj 20.t . j 20.tj30.t . j30.t
a. .o .. . 10. x(t) . 2 e . 2 e . e . e . e . e
T 0.2
b.
X ( j.)
4

2


.

. 30
. 20.10 0 10 20
30
5 . j10.t . j10.t 4
c. y(t) ... . (2 e . 2 e ) . cos(10.t)
.15 . 3
d.
Hp ( j.) 20


.

. 40. 25. 15 0 15 25 40 55
4 5(2) 80 40
e. z(t) . (20) cos(10.t) . (20) cos(30.t) . cos(10.t) . cos(30.t)
3 15 33
f.
Hd (ej. ) 20


.

. 2. 1.25. 0.75 0 0.75 1.25 2
2. 1
a. Sampling frequency = 20 .. . 20. T .. 0.1STs s10
b. Y1(j...
....

....

..
..........................
c.

Y1(j... ..
....
.... ..

..........................
d. Synchronous detection is not needed. Demodulation becomes simpler. Just use a diode, capacitor and a resistor.

2. 1 j60.t 1 . j60.t
f. i. .. . 60. cos 60.t . e . e
oT 22 1
a1 . a.1 . ak = 0 for other ks 2
11 1
ii. b0 . b1 . b.1 . b2 . b.2 . 0 b3.. 2 . 3.

11
c0 . c . c..
. 1 14
g.
Z(j... ... .... ...
.... ..


......................................... ..................
h. Local oscillator frequency = 20 rad/sec
11
a. H (s) .. . 2 . Re{s} ..1
s . 2 s .1
b.
No. h(t) is non-zero for t is negative. Its ROC shows that its not a right-sided function.

c.
No. This is because h(t) dt will go to infinity. Its ROC does not contain j.-axis.


..
..
11
d. H (s) . . H ( j.) .
s2 . 2s .101 ( j.)2 . 2( j.) .101 . 2 .
22 . 4 (101) j.
e. Poles . ..1 . j10
2

10
.
. 1 0
. 10
f.
Yes. This is because poles contain imaginary part which will result in an impulse response with complex exponential term to give oscillation.

g.
.1= 10

h.
. = 1 Y (s)1

i.
. s2Y (s) . X (s) . 2sY (s) .101 Y (s)


X (s) s2 . 2s .101

1 . K .
K 2 . 4 (9)
j. H (s) . 2 Poles.
s . Ks . 92 For a causal and stable system, all the poles must be on the left hand side. . K > 0 Multiple Choice
(a) (b) (c) (d) (e) (f) (g) (h) (i) (j)
C A C C B B D D B C

Problem 5 (Total = 10 points)
True / False
(a) (b) (c) (d) (e) (f)
F F T F T F

(g) (h) (i) (j)
F T F T