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(ELEC211)02fallmidsol.pdf
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Midterm 2002 Suggested Solution
1. (7 pts) Consider two systems with the following input-output relationships:
x[n / 2], n
=
evenSystem 1: y[n]
=
0,
odd
n
=
System 2: y[n] x[2n]
=
(a) Is each of these two systems linear? Is each of them time-invariant?
Ans : System 1 and System 2 ~ linear System 1 and System 2 ~ time-variant
(b) Suppose that these two systems are connected in series as depicted in Fig. P-1. Find the input-output relationship for the overall interconnected system. Is the overall system linear? Is it time-invariant? Notice: You have to consider two cases of series interconnection,
i.e., i) System 1 first
x[n] y[n]
System 1 System 2
Ans : System 1 will insert one zero between two impulses and System 2 will do the compression such that y[n] = x[n] and overall system is LTI.
ii) System 2 first.
x[n]
System 2 System 1
y[n]
Ans : System 2 will compress the signal such that the information for n = odd will be lost and System 2 will insert one zero between two impulses such that
x[n], n y[n]
=
=
even It is linear but time-variant.
0,
n
odd
-1 0 1 2 -1 0 1 2 -3 -2 -1 0 1 2 3 4 5
2. (7 pts) Consider two signals x0(t) and x1(t) shown in Fig. P-2.1 and Fig. P-2.2.
x1(t)
x0(t)
P-2.1
P-2.2 1
1
t
t
-A
A -BB
(a)
Compute the convolution of x0(t) and x1(t): y(t) = x0(t)* x1(t) (you have to show all necessary intermediate steps).
(b)
Sketch y(t) carefully.
Ans :
x1(t) x1(t)
1 1
t t
-BB -BB
x0(t)
1
t
x0(t)
1
t
t -A t + A t -A t + A
For -B < t + A and t -A <-B For -B < t -A and t + A < B -(A + B) < t < A -BA -B < t <-A + B
t + A
y(t) = 2Ay(t) = dt = t + (A + B)
-B
x1(t)
1
t
For t + A <-By(t) = 0
-BB For B < t -Ay(t) = 0
t
t -A t + A
For B < t + A and t -A < B -A + B < t < A + B
. B -
y(t) = dt =-t + (A + B) -(A+B) A-B 0 -A+B A+B
tA
3. (8 pts) Let x[n] be a periodic signal with period N and Fourier series representation
. 2p .
jk..n
. N .
x[n] =.ak e
k =
N
The Fourier series coefficients of each of the following signals can be expressed in terms of ak . Derive the expressions for the following signals:
(a) x[n] -x[n -n0 ] (where 0 < n0 < N )
Ans :
. 2p .. 2p .. 2p .. 2p .
jk .. n jk .. (n -n0) -jk .. n0 jk .. n . N .. N .. N .. N .
x[n] -x[n -n0] =. aK e -. aK e =. ..
ak -ake ..
e
k =<N > k =<N > k =< N >.
.
. 2p .
-jk .. n0
. N .
\ bk = ak (1 -e )
N
(b) x[n] -x[n -] (assume that N is even)
2 Ans :
. 2p .. 2p . N . 2p . N . 2p .
jk .. n jk .. (n -) -jk .. jk .. n
N .
.
. N .. N . 2 . N . 2 . N .
x[n] -x[n -] =. ae -. ae =. .a -ae . e
KK kk2 k =< N > k =< N > k =< N >.
.
-jkp k
\ b = a (1 -e ) = a [1 -(-1) ]
kk k
bk = 2ak fork = odd
bk = 0 for k = even
N
(c) x[n] +x[n + ] ( N is even; and note that this signal is now periodic with period N /2)
2 Ans :
. 2p .. 2p . NL . 4