This document contains an exam for ECES 302 with 4 problems. It specifies the exam conditions such as being closed book/notes and lasting 50 minutes. Problem 1 involves drawing and labeling signals x1(t) and x2(t). Problem 2 involves determining properties of signals including periodicity and even parts. Problem 3 evaluates whether a given system is linear, time invariant, and BIBO stable. Problem 4 determines the output signal y2(t) given input x2(t) for a linear time invariant system and evaluates causality.

1. ECES 302 - Spring 2014
April 25, 2015
Name:
Section: 060 061 062
Exam # 1 Solutions
This examination is closed book and closed notes. You may use a calculator with basic functionality. You may
NOT use a graphing calculator or any device capable of advanced operations such as symbolic integration.
You must write your name clearly at the top of every page.
The examination will last 50 minutes. There are four questions worth 100 points in total. Answer as many ques-
tions as you can in the time allowed.
You must show all of your work to receive full or partial credit for each problem.
Please write clearly and box or circle your

2. nal answer.

3. ECES 302 - Spring 2014
April 25, 2015
Name:
Section: 060 061 062
Problem 1 (15 points)
Consider the continuous time signal x1(t) = u(t + 1) u(t 1) + 2(t 2).
a) (5 points) Draw and clearly label x1(t).
b) (10 points) Let x2(t) = 2x1( 1
3 t + 2). Draw and clearly label x2(t).
Solution:
a) x1(t) is shown below:
b)
x2(t) = 2x1( 1
u(( 1
3 t + 2) = 2
3 t + 2) + 1) u(( 1
3 t + 2) 1) + 2(( 1
3 t + 2) 2)
= 2u( 1
3 t + 3) 2u( 1
3 t + 1) + 4( 1
3 t)
= 2u( 1
3 t + 3) 2u( 1
3 t + 1) + 4(t)
x2(t) is shown below:

4. ECES 302 - Spring 2014
April 25, 2015
Name:
Section: 060 061 062
Problem 2 (30 points)
Consider the following discrete time signals x1[n], x2[n] and x3[n] de

5. ned below:
a) (10 points) Is the signal x1[n] = ej( 8
5 n2
3 ) periodic? If yes, what is its fundamental period? If no, explain
why it is not periodic.
b) (10 points) What is the even part of x2[n] = ej 4
3 n? Express your

6. nal answer in terms of sine and cosine.
c) (10 points) What is the total energy of the signal x3[n] =
2
3
n[n 1]? In other words, what is E1?
Solution:
a) We can see that x1[n] is a discrete time complex exponential of the form ej(!n+) where ! = 8
5 . We can also
see that ! is a rational multiple of 2, therefore x1[n] is periodic.
For periodic discrete time complex exponentials, we can express the frequency ! as
! =
m
N
2 (1)
where N is the period of the complex exponential and m is an integer. For this particular complex exponential,
we can determine the fundamental frequency as follows:
!0 =
m
N
2 =
8
5
=
4
5
2
Therefore, the period is N = 5:
b) Evfx2[n]g = 1
2
x2[n] + x2[n]
= 1
2
ej 4
3 n
3 n + ej 4
= cos( 4
3 n):
Alternatively, we can use Euler's formula to write ej 4
3 n = cos( 4
3 n)+j sin( 4
3 n). We use the fact that cos() = cos()
and sin() = sin() to write
Evfx2[n]g =
1
2
x2[n] + x2[n]
=
1
2
cos( 4
3 n) + j sin( 4
3 n)
+
cos(4
3 n) + j sin(4
3 n)
=
1
2
cos( 4
3 n) + j sin( 4
3 n) + cos( 4
3 n) j sin( 4
3 n)
= cos( 4
3 n):
c) E1 =
1X
n=1

10. x3[n]
2 =
1X
n=1

13. 2
3
n

16. [n 1]
2
=
1X
n=1
2
3
2n
2
[n 1]
=
1X
n=1
4
9
n
[n 1] since
[n 1]
2 = [n 1][n 1] = [(1) 1][n 1] = (1)[n 1] = [n 1]
=
1X
n=1
4
9
(1)
[n 1] =
4
9
1X
n=1
[n 1] = 4=9 since
1X
n=1
[n 1] = 1.

17. ECES 302 - Spring 2014
April 25, 2015
Name:
Section: 060 061 062
Problem 3 (32 points)
Consider the system y(t) = 1
2x(t) + 2.
a) (12 points) Is this system linear? You must justify your answer.
b) (12 points) Is this system time invariant? You must justify your answer.
c) (8 points) Is this system BIBO stable? You must justify your answer.
Solution:
a) Let x1(t) ! y1(t) = 1
2x1(t) + 2 and x2(t) ! y2(t) = 1
2x2(t) + 2. Now de

18. ne x3(t) = ax1(t) + bx2(t). We can
now express y3(t) as
y3(t) = 1
2x3(t) + 2 = 1
2
ax1(t) + bx2(t)
+ 2 = a1
2x1(t) + b 1
2x2(t) + 2
6= ay1(t) + by2(t) = a
1
2x1(t) + 2
+ b
1
2x2(t) + 2
As a result, the system is not linear.
b) Let x1(t) ! y1(t) = 1
2x1(t) + 2 and de

19. ne x2(t) = x1(t t0). We can express y2(t) as
y2(t) = 1
2x2(t) + 2 = 1
2x1(t t0) + 2: (2)
Additionally, we can express y1(t t0) as
y1(t t0) = 1
2x1(t t0) + 2: (3)
Since y2(t) = y1(t t0), the system is time invariant.
c) Let B be a positive number such that jBj 1 and jx(t)j B. Using this along with the triangle inequality
ja + bj jaj + jbj we can write
jy(t)j =

21. 1
2x(t) + 2

23. 1
2 jx(t)j + j2j 1
2B + 2 1: (4)
Therefore, the system is BIBO stable.

24. ECES 302 - Spring 2014
April 25, 2015
Name:
Section: 060 061 062
Problem 4 (23 points)
Assume that we know that a particular system is linear and time invariant. If the input to the system is x1[n]
then the output of the system is y1[n], where x1[n] and y1[n] are shown below.
Now consider the input signal x2[n] shown below.
a) (6 points) Express x2[n] in terms of x1[n].
b) (10 points) If the input to the system is x2[n] what is the output signal y2[n]? You may draw your answer.
c) (7 points) Is this system causal? You must justify your answer.
Solution:
a) By visually inspecting the plot of x2[n] we can see that x2[n] = 2x1[n] + x1[n 2]:
b) Since the system is linear and time invariant we know x2[n] = 2x1[n] + x1[n 2] ! y2[n] = 2y1[n] + y1[n 2].
As a result, y2[n] is the signal shown below.
c) For a system to be causal, the output at any time must depend only on past and current values of the input.
We can see y1[1]6= 0 even though x1[n] = 0 for n 1. As a result, this system is not causal.