Download EE 3015 Midterm 1 Solutions and more Exams Signals and Systems in PDF only on Docsity!
University of Minnesota
EE 3015: Signals& Systems
Midterm 1 Solutions
Feb. 24, 2017
- Concept Testing [18 points total]
a) [5 points] Consider the system described by the input-output relationship
y[n] = cos
2 ฯ
n
x[n].
Determine if this system is linear. Justify your answer.
Solution: Define H(x[n]) = cos
2 ฯ
5
n
x[n].
We consider two signals, x 1 [n] and x 2 [n] , with corresponding outputs, y 1 [n] = cos
2 ฯ
5
n
x 1 [n]
and y 2 [n] = cos
2 ฯ
5
n
x 2 [n], respectively.
We input a linear combination of these two signals, as ax 1 [n] + bx 2 [n] for some constant a, b:
H(ax 1 [n] + bx 2 [n]) = cos
2 ฯ
n
(ax 1 [n] + bx 2 [n])
= a cos
2 ฯ
n
x 1 [n] + b cos
2 ฯ
n
x 2 [n]
= ay 1 [n] + by 2 [n]
Thus the system is linear.
b) [6 points] Consider the system in (a) described by the input-output relationship
y[n] = cos
2 ฯ
n
x[n].
Determine if this system is time invariant. Justify your answer.
Solution: Define H(x[n]) = cos
2 ฯ
5
n
x[n].
For a fixed n 0 , we input a time-shifted signal as
H(x[n โ n 0 ]) = cos
2 ฯ
n
x[n โ n 0 ]
We now check the time-shifted output
y[n โ n 0 ] = cos
2 ฯ
(n โ n 0 )
x[n โ n 0 ]
These two expressions are not equivalent for all choices of n 0 , thus the system is not time-
invariant.
c) [2 points] The impulse response of an LTI system is the output for which input signal?
Solution: To the unit impulse, or ฮด(t) in continuous-time or ฮด[n] in discrete-time.
d) [5 points] Determine whether the continuous-time LTI system with the impulse response
h(t) = e
2 t u(โ 1 โ t)
is stable. Justify your answer.
Solution: Check if h(t) is absolutely integrable:
โ
โโ
|h(t)|dt =
โ 1
โโ
e
2 t dt
e
2 t
โ 1
t=โโ
e
โ 2 โ 0
e
โ 2
- Convolution [16 points]
a) [10 points] Impulse response of a continuous-time LTI system is given by h(t) =
2 e
โ 2 t u(t). Find the output signal y(t) when the input signal x(t) is
x(t) =
1 , 0 < t < 1
0 , otherwise
Solution: We need to calculate
y(t) =
โ
โโ
h(ฯ )x(t โ ฯ )dฯ.
We can sketch h(ฯ ) and x(t โ ฯ ) as functions of ฯ for a given t.
If t < 0 , then there is no overlap between h(ฯ ) and x(t โ ฯ ), i.e. y(t) = 0.
If 0 < t < 1 , then there is partial overlap between h(ฯ ) and x(t โ ฯ ) for 0 < ฯ < t, thus
y(t) =
t
0
2 e
โ 2 ฯ dฯ = โe
โ 2 ฯ
t
ฯ =
= 1 โ e
โ 2 t .
If t > 1 , then there is overlap between h(ฯ ) and x(t โ ฯ ) for t โ 1 < ฯ < t, thus
y(t) =
t
tโ 1
2 e
โ 2 ฯ dฯ = โe
โ 2 ฯ
t
ฯ =tโ 1
= e
โ2(tโ1) โ e
โ 2 t = (1 โ e
โ 2 )e
โ2(tโ1) .
Thus, we have
y(t) =
0 , t < 0
1 โ e
โ 2 t , 0 < t < 1
(1 โ e
โ 2 )e
โ2(tโ1) , t > 1
b) [6 points] Roughly sketch h(t), x(t) (if you have not done so already) and y(t).
Solution:
