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UNIVERSITY OF CALIFORNIA, SAN DIEGO
Electrical and Computer Engineering Department ECE 101 - Winter 2009 Linear Systems Fundamentals
MIDTERM EXAM
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Student ID Number
Problem Weight Score 1 25 2 25 3 25 4 25
Please do not begin until told. Write your name on all pages. Show your work. Use back of previous page and scratch sheets as needed. Tables 3.1, 3.2, 4.1, 4.2, 5.1, and 5.2 from the textbook are at the back of the exam. Good luck!
Problem 1 (25 points) For the following questions, be sure to provide sufficient justification for your answers.
(1A) Let x[n] = cos^2
[π 4
n
]
Is this function periodic? If so, what is the period N?
Solution: The period is 4, as can be seen by either inserting sequential values into x[n]
x[0] = cos^2
[π 4
]
x[1] = cos^2
[π 4
]
x[2] = cos^2
[π 4
]
x[3] = cos^2
[π 4
]
x[4] = cos^2
[π 4
]
or via trigonometry
cos^2
[π 4
(n + 4)
]
1 + 2 cos
[π 2
(n + 4)
])
1 + 2 cos
[π 2
n
])
= cos^2
[π 4
n
]
Problem 1 (cont)
(1C) You are given the discrete LTI system x[n] → y[n], with impulse response h[n]. If y[n] = δ[n] and x[n] = u[n], calculate h[n]. Is the system causal? Is the system stable?
Solution: One way to solve this is to remember that δ[n] = u[n] − u[n − 1] and y[n] = x[n] ∗ h[n]. Thus,
u[n] − u[n − 1] = u[n] ∗ h[n]
By the sifting property of delta functions we must have
h[n] = δ[n] − δ[n − 1]
The system is causal since h[n] = 0 for n < 0. It is stable since
∑^ ∞
k=−∞
|h[n]| = 2 < ∞
Problem 1 (cont)
(1D) The discrete function x[n] is real and odd, has period N = 3, and has the Fourier series coefficient a 7 = j. Determine the impulse response function h[n]. Calculate x[n].
Solution: Since x[n] has period N = 3, it follows that ak is also periodic with period 3. Thus, a 1 = a 7. x[n] is real and odd, so it also follows that a 1 = a∗− 1 and that a 0 = 0. Finally,
x[n] =
∑^1
k=− 1
akejk(2π/N^ )n^ = (−j)e−j^2 πn/^3 + (j)ej^2 πn/^3 = −2 sin
2 π 3 n
Problem 2 (cont)
(2B) Indicate which of the following properties are satisfied by the system (or not): Linear, Time-invariant, stable, causal. Justify your answers.
Solution: Linear - yes: Define xA[n] = C 1 x 1 [n] + C 2 x 2 [n]. Then yA[n] = (n + 3)xA[n + 1] − 2 u[n − 1]xA[n − 1]. Next, define yB[n] = C 1 y 1 [n] + C 2 y 2 [n], where y 1 [n] = (n + 3)x 1 [n + 1] − 2 u[n − 1]x 1 [n − 1] and y 2 [n] = (n + 3)x 2 [n + 1] − 2 u[n − 1]x 2 [n − 1]. Since yA[n] = yB[n], then by the principle of superposition the system is linear.
Time Invariant - no: Let x 1 [n] = x[n − n 0 ]. Then y 1 [n] = (n + 3)x 1 [n + 1] − 2 u[n − 1]x 1 [n − 1]. But, y 2 [n] = y[n − n 0 ] yields y 2 [n] = (n − n 0 − 3)x 1 [n + 1] − 2 u[n − n 0 − 1]x 1 [n − 1]. Thus, y 1 [n] 6 = y 2 [n] and the system is not time invariant.
Stable - no: Stability requires that y[n] be bounded for all n assuming x[n] is likewise bounded. For this problem, there is no way to stop (n + 3)x[n + 1] from going to infinity as n goes to infinity.
Causal - no: y[n] depends on x[n + 1]; i.e., on future values of x, violating causality.
Note that one cannot use tests on h[n] to determine the system properties since the system is not LTI.
Problem 3 (25 points) Assume that you are given an LTI discrete system x[n] → y[n], with impulse response h[n], and with the following properties:
- When x[n] = (1/2)nu[n], then y[n] = 0 for n ≥ 2 and n < 0.
- h[1] = 0 and H(ejπ/^2 ) = 1
Determine the impulse response function h[n]. Hint: Start by calculating H(ejω) = Y (ejω)/X(ejω).
Solution: First, from Table 5.2, the transform of x[n] is
X(ejω) =
1 − (1/2)e−jω
We also know that y[n] = 0 for n < 0 or n > 1. Thus
Y (ejω) =
∑^ ∞
n=−∞
y[n]e−iωn^ = y[0] + y[1]e−jω
The frequency response is therefore
H(ejω) =
Y (ejω) X(ejω)
= (y[0] + y[1]e−jω)(1 − (1/2)e−jω)
= y[0] + (y[1] −
y[0])e−jω^ −
y[1]e−^2 jω
It follows that h[0] = y[0], h[1] = y[1] − (1/2)y[0], h[2] = −(1/2)y[1], and h[n] = 0 for all other n values. Since we were given h[1] = 0, we have y[1] = (1/2)y[0]. We also know that H(ejπ/^2 ) = 1, so 1 = y[0] + (1/2)y[1]. Finally, y[0] = 4/5 and y[1] = 2/5, leading to h[0] = 4/5 and h[2] = − 1 /5, completing the problem.
Problem 4 (cont)
(4B)
Find the Fourier Series representation for y[n] when x[n] = cos
(π 2 n
Solution: First, since x[n] = (1/2)(ejπn/^2 + e−jπn/^2 ), it follows that the Fourier series coefficients for x[n] are a− 1 = a 1 = 1/2 and a− 2 = a 0 = 0. It also follows that
y[n] =
k=
bkejkπn/^2
where bk = akH(ejkπ/^2 ) so b− 1 =
H
jπ 2
1 − (1/2)j
b 1 =
H
jπ 2
1 + (1/2)j
We can also solve for the simplified form of y[n], i.e.,
y[n] =
2 e−jπn/^2 2 − j
2 ejπn/^2 2 + j
(2 + j)e−jπn/^2 + (2 − j)ejπn/^2
4 cos(
nπ 2 ) + 2 sin(
nπ 2
