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(ELEC211)2007_f_Midterm_soln.pdf
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Problem 1 (30 points)
.2t
a) Let h(t) = eu(t) be the impulse response of an LTI system. Determine the frequency response H ( j ) of this
system.

. j6t
b) Determine the output y(t) of the above system when the input is x(t) = e
.3t
c) Determine the output y(t) of the above system when the input is x(t) = eu(t)
d) Express this LTI system in the form of a linear differential equation (LDE).

.3t
e) Assume that x(t) = eu(t) is again the input to the system described by the LDE in part d). It is given, however, that
the initial condition is such that the output y1(t) has an initial value of 1; that is, y1(0) = 1.
Is the output y1(t) the same as the output y(t) as determine in part c)? Briefly explain your answer.
j+ 4

f) For the causal and stable LTI system with the frequency response H ( j ) = , determine | H ( j ) | and
6 . 2 + 5 j
H ( j ) when = 0 and when .

(a).
. 2tF 1
By use of the table given h(t) = e u(t). H(j) =
2 + j
(b).
. j6t
x(t) = e is an eign function
. j6t

. output = eH(j)|=.6
1 . j6t
=e
2 . 6j
(c). F (e). No.
. 3t 1
x(t) = e u(t). X(j) =
In part (c) we assumed an LTI system. For LDE system
3 + j
to be an LTI system, the system must be assumed to be
Y(j) = X(j)H(j)
at initial rest. However, the system is not at initial rest 1 1 in this case.
Y(j) = ()( )
3 + j 2 + j j+ 4
. 1 1 (f). H(j) =
Y(j) =+ 2
6 . + 5j
3 + j 2 + j
as = 0,
.2t .3t

. y(t) = e u(t) . e u(t) 42
H(j)|= 0 ==
631
(d). H(j) =. | H(j) | =2/3, H ( j ) =0
2 + j
as ,
Y(j) = X(j)H(j)
1 lim H(j) = 0 . | H(j) | =0,

Y(j) = X(j)
2 + j
X(j) = (2 + j)Y(j)

as >6 and dy(t) .1 5
.1
.+ 2y(t) = x(t) H(j) = tan () . {+ tan ( 2)}dt 46 . = /2- = -/2
Problem 2 (30 points)
a) (25 points) x[n] is a periodic discrete-time signal with period N=4.
2 2 2

. .
1
1


n

. - 3 - 2 -1 0 1 2 3 4..
i. Find a0.
ii. Without finding the individual ak , determine the value of a + a + a + a .
1

0123
iii. Note that x[n] is real and even. It is known that a1 =. . Without using the analytic equation, determine the
4
value of determine the value a.1.

iv.
Continuing from part iii., determine the value of a3.

v.
Using results from i. to iv. or otherwise, determine the value of a2.


b) (5 points) The signal in part a) is the input to an LTI system with unit sample response h[n]. The output y[n] is a constant as plotted below (i.e., y[n]=1): y[n]

1
.

.




. - 3 - 2 -1 0 1 2 3 n
2

. jn
4
What is the value of h[n]e ? Justify your answer.
n=. 2 2
(a). . j 4 n . jn
(b). h[n]e = 0 where h[n]e4 = H(e jko)|
k=11 . jkon n=. n=.
(i). ak = x[n]e
N
n =< N>
jkon jk
and y[n] = ake H(e o)
1
a= x[n] k =<N>
0
N