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(ELEC211)02fallmid.pdf
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Midterm (for ELEC211), 24th of October, 2002
Remarks: (1) This is a 2-hour, closed-book exam. (2) Do all four problems. (3) Total marks = 30
(4) Please show all the steps clearly.
1. (7 pts) Consider two systems with the following input-output relationships:
.x[n / 2], n = evenSystem 1: y[n] =. 0, n = odd
System 2: y[n] = x[2n]
(a)
Is each of these two systems linear? Is each of them time-invariant?
(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 and
x[n]
System 1 System 2
y[n]
Fig. P-1.
ii) System 2 first.
x[n]
System 2 System 1
y[n]
2. (7 pts) Consider two signals x0(t) and x1(t) shown in Fig. P-2.1 and Fig. P-2.2.
x0(t) x1(t)
P-2.1 P-2.2
1 1
t t
-AA -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.
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 )
N
(b)
x[n] -x[n -] (assume that N is even)
2
N
(c) x[n] +x[n + ] (assume that N is even; and note that this signal is now periodic
2
with period N /2)
4. (8 pts) Consider the following three LTI systems with impulse responses: h1(t) = u(t),
h2(t) =-2d(t) + 5e -2tu(t), and
h3(t) = 2te -tu(t).
(a)
Compute the response of each system to input x(t) = cost .
(Hints: use Fourier transform in each case.) Comment on the results.
(b)
Find the impulse response of another LTI system with the same response to x(t) = cost .
(c)
What conclusion can you draw on the use of cost to specify an LTI system?
Given equations : jkw0 t 1 -jkw0 t 2p
CTFS x(t) = . ae a = x(t) e dt w =
kk 0
k =- TT T
jkw n 1 -jkw n 2p
00
DTFS x[n] =. ae a =. x[n] e w =
kk 0 k =<N > Nn =<N > N
1
jw t -jw t
CTFT x(t) = X ( jw) edw X ( jw) = x(t) e dt
2p .- .-
Convolution Integral y(t) = .- x (t ) h (t -t ) dt
Convolution Sum y[n] = . x [k] h [n -k]
k =-