• University of California, Irvine

• EECS 50

EECS 50 - Discrete Time Signals and Systems - Spring 2014

Final Exam

Student Name:

Student ID:

Instructions: The table below is for grading purposes only.

Problem No. Points

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Total Score

1. (10 pts) The frequency response of an LTI system is given by

H(Ω) = (1 + 4j sin(2Ω)) cos 3Ω 2  e-jΩ=2

(a) (4 pts) Find h[n].

(b) (3 pts) Find the output y1[n] if the input to the system is x1[n] = 1 + sin πn 2

(c) (3 pts) Find the output y2[n] if the input to the system is x2[n] which has DTFT

X2(Ω) = 1 - 2 cos(Ω)

Solution:

(a)

H(Ω) = (1 + 4j sin(2Ω)) cos 3Ω 2  e-jΩ=2

= (1 + 4j ej2Ω - e-j2Ω

2j

)(ej 3 2Ω + e-j 32Ω

2

)e-jΩ=2

= (1 + 2ej2Ω - 2e-j2Ω)1

2

(ejΩ + e-j2Ω)

=

1 2

ejΩ + 1

2

e-j2Ω + ej3Ω + 1 - e-jΩ - e-j4Ω

) h[n] = 1

2

δ[n + 1] + 1

2

δ[n - 2] + δ[n + 3] + δ[n] - δ[n - 1] - δ[n - 4]

(b) We will use the fact that complex exponentials are eigenfunctions of LTI systems.

x1[n] = 1 + sin(πn 2 ) = ej0n + 21j (ej πn 2 - e-j πn 2 ), so that

y1[n] = H(0)ej0n + 1

2j

[H(π

2

)ej πn 2 - H(-π

2

)e-j πn 2 ]

= 1 +

1 2j

[cos(3π

4

)e-jπ=4ej πn 2 - cos(3π

4

)ejπ=4e-j πn 2 ]

= 1 + cos(3π

4

) 1

2j

[ej(πn 2 - π4 ) - e-j(πn 2 - π4 )]

= 1 -

p2

2

sin(πn

2

-

π 4

) = 1 - 1

2

sin πn 2  + +1 2 cos πn 2 

We can also use convolution in time domain.

(c)

X2(Ω) = 1 - 2 cos(Ω) = 1 - (ejΩ + e-jΩ) ) x2[n] = δ[n] - δ[n + 1] - δ[n - 1]

y2[n] = x2[n] ∗ h[n]

= (δ[n] - δ[n + 1] - δ[n - 1]) ∗ (1

2

δ[n + 1] + 1

2

δ[n - 2] + δ[n + 3] + δ[n] - δ[n - 1] - δ[n - 4])

= (-δ[n + 1] + δ[n] - δ[n - 1]) ∗ (δ[n + 3] + 1

2

δ[n + 1] + δ[n] - δ[n - 1] + 1

2

δ[n - 2] - δ[n - 4])

= -δ[n + 4] + δ[n + 3] - 3

2

δ[n + 2] - 1

2

δ[n + 1] + 3

2

δ[n] - 5

2

δ[n - 1]

+

3 2

δ[n - 2] + 1

2

δ[n - 3] - δ[n - 4] + δ[n - 5]