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EECS 50 Discrete Time Signals and Systems Final Exam

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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 Score1. (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]2. (10 pts) Consider the LTI system T1 with input x1[n] and output y1[n] = 1 2(x1[n] − x1[n − 1]). (a) (1 pt) Find the impulse response h1[n]. (b) (2 pts) Find the frequency response H1(Ω). (c) (1 pt) Identify the poles and zeros of the filter. (d) (1 pts) Sketch the magnitude response jH1(Ω)j. (e) (1 pt) What type of filter is this? (f) (2 pts) Sketch the phase response \H1(Ω). (g) (2 pts) Write the difference equation for an LTI system T2 whose frequency response is H2(Ω) = H1(Ω − π) Solution: (a) Plug in x1[n] = δ[n], we have h1[n] = 12(δ[n] − δ[n − 1]). (b) H1(Ω) = 1 2(1 − e−jΩ). (c) H1(z) = 1 2(1 − z−1) = z−1 2z . The zero is at z = 1 and the pole is at z = 0. (d ) As H1(Ω) = 12e−jΩ=2(ejΩ=2 − e−jΩ=2) = je−jΩ=2 sin(Ω=2), we have jH1(Ω)j = j sin(Ω=2)j and it is plotted as follows. −4 −3 −2 −1 0 1 2 3 4 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

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