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(elec211)[2009](f)midterm~plliu^sol_10300.pdf
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Question 1 (20 points)
(a) (b) (c) (d) (e) (f)
T F T T T T

(g)
Memoryless Linear Time-invariant Causal
F F T F
F T T F

Question 2 (20 points)
(a) (b) (c) (d) (e) (f) (g) (h) (i) (j)
D C C B B B D A A

(a) y[5] = 2.6 (b) y[7] = 5.4 . (1. 0) (1) = 4.4 x[k] 4

x[k] 4 3

3 2

2 1

k


k





0 1 2 3 4 5 0 1 2 3 4 2




21
h[n . k] h[n . k]
0.5

0.5
0.2
0.2
k



k n . 4 n . 1 n . 4 n . 1

j. j n . j. j 2 . j3 . j 4

(c) H (e ) = h[n] e = 2 e + e + 0.5 e + 0.2 e = 2(.1) +1+ 0.5 (.1) + 0.2 =.1.3
n =.

j j n
x[n] is a periodic signal with frequency of . y[n] = H (e ) x[n] =.1.3 e
(d) h(t)2

t

7
. 2

2 3 6
. 2 . 0.5 t 0.5 4sin(/ 2)

(e) s(t) = square pulse =. . S( j) =
0 elsewhere
.

. j 2.5. j6.5
H ( j) = eS( j) . eS( j)
h(t) = s(t . 2.5) . s(t . 6.5) .

4sin(0.5 ) . j 2.5. j6.5
= (e . e )


(f) y(t)

t
2 3 4 6
78
. 4


2

(a) N = 8 (b) = =
o 84
n n 3n 3n
.n .. 3n . 1 j 1 . j 1 j 1 . j
444 4

(c) x[n] = 2 + cos. .+ 2sin. .= 2 + e + e + e . e
. 4 .. 4 . 22 jj

11 . j j
a = 2 a == a == e 2 a = a * = e 2 a = a
o 1 a.13 . 33 kk + 8
2 j
0 1 2 3 4 5 6 7 8 9 10



3 3
j . jj . j

j 044 44
(e) H (e ) = H (e ) = H (e ) = 2 H (e ) = H (e ) = 1
n n 3n 3n
j . j 1 j 1 . j .n .. 3n .
444 4

y[n] = 4 + e + e + e . e = 4 + 2cos. .+ 2sin..jj . 4 .. 4 .
222

(f) Average power of the output = (4) + 2 (1) + 2 (1) = 20

. 8 . 4 . 2 0 2 4 8
2

(b) T = 4 o ==
42
12 2

(c) a = (area of one period) == 1
0 T 4
. 4

1 sin2(k / 2) .2 for k = odd

(d) ak = X ( jko ) = 2 =. (k)
T (k / 2)
.
0 for k = even 0
.

(e) x1(t) is periodic signal . X1( j) = 2 ak (. ko )
k =.

x1(t) is real and even . X1( j) is real and even
X1( j0) = 2 ao = 2 X1( j2o ) = X1( j4o ) = 0
88

X ( j ) = 2 a =(. ) X ( j3 ) = 2 a =(. 3 )
1 o 1 o 1 o 3 o
9
22 22