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(ELEC211)[2011](sum)midterm~4341^_80252.pdf
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a) s(t) . cos . e . e . a . a.. ak. 0 for k .1 and 1
222 112 b)
..
./2 0 /2
12 . j . 2 j .
4 *4
c) H1( j. / 2) .. ... eH1(. j. / 2) . H ( j. / 2) .. e . j
22
1 j.t /2 1 . j.t /2 2 ..t ..
Output = . eH ( j. / 2) . eH (. j. / 2) . cos. ..
22 .. 24 .
.0.5 k . 0 sin k.oT1 sin k (2. / 4) (1) sin(k. / 2) ..0 k ..2, . 4, . 6, . 8
d) bk .. ...
k. k. k. 1/ . ,1/5. k ..1, . 5
. ..1/3. , .1/7. k ..3, . 7
.
e) System 2 will block all harmonics and only allow the dc term to pass through. 12 j02 1
H1( j0) .. H 2( j0) . 1 g(t) . H ( j0)H ( j0) be . (1) (0.5) (1) .
1 20
. /2 . .. f) y(t) is real and even so that the FS is also real and even. By the multiplication property:
.
c .. ab
k lk . l
l ...
1 . 1 . 1 . 1 . 1
c . ab . ab . ... ...
0 .11 1 .1
2 ... 2 ... . k .1 0 k+1
11111 1 . 1 . 1
c . c . ab . ab ... ... c . c . ab . ...
23. 2 . 3. 2 . 2 . 4
k .1 0 k+1
2
2005 2006 2007 2008 2009 2010 2011 2012
n .1 . 2011 . n . 2012
c) This is NOT a causal system because h(t) is not equal to zero for all t < 0.
. t . (t . 5)
d) y1(t) . x1(t)* h(t) . h(t) . 0.5h(t . 5) . e (u(t .1) . u(t .1)) . 0.5 e (u(t . 4) . u(t . 6)) e) The unit step response of an LTI system is the integral of the impulse response. Hence,
. t .t
y2(t) . 0 t ..1 y2(t) .. te dt . e . e .1 . t .1
.1 . 1
. t .1
y2(t) . e dt . e . et .1
.1
f)
e . e .1
t
. 1 0 1
j.
. (t . 1)
eg) e . eu(t .1) . e .
1. j.
. j. j.. j.
. 1 . (t . 1) .1 .1
e ee
h) e . eu(t .1) . e . H ( j.) . e .. e .
1. j. 1. j. 1 . j.
j . j . jj1 jt 1 . jt e .1 ee .1 e
i) cost . e . eH ( j) . e .. e . H (. j) . e .. e .
22 1 . j 1. j 1 . j 1 . j
j . j . jj
1 jt . e .1 e . 1 . jt . e .1 e .
y(t) . e .e .. e . .. e .e .. e ..
2 . 1 . j 1. j . 2 . 1. j 1. j .
3
2 point for CORRECT answer, C 1 point for INCORRECT answer, and 0 point for BLANK.
Remarks : Only