2020_M_sol (2)

MCMASTER UNIVERSITY Department of Electrical and Computer Engineering October 29, 2020 COMP ENG 4TL4 and ECE 6TL4 (Digital Signal Processing) Instructor: Dr. Ratnasingham Tharmarasa MIDTERM TEST Duration: 2 hours This is an opened book exam. This examination paper consists of 12 pages and 5 questions. The number in brackets at the start of each question is the number of points the question is worth. Answer all questions. • Answer each question on a separate letter-size paper. • All answers must be hand-written. • Use the following naming convention for your files: StudentNumber QuestionNumber (e.g., 400654321 Q2). – If you have answered on more than one page for a question, then the name of the second page should be StudentNumber QuestionNumber PageNumber (e.g., 400654321 Q2 p2). • Write your name and student number on each page. • Upload a scanned copy or clear photos of answer sheets to avenue by 8:30 pm on Oct 29, 2020. Academic integrity statement: By submitting this work, I certify that the work represents solely my own independent efforts. I confirm that I am expected to exhibit honesty and use ethical behaviour in all aspects of the learning process. I confirm that it is my responsibility to understand what constitutes academic dishonesty under the Academic Integrity Policy. Page 1 of 12

page 2 Numbers from Student Number Fill in the following table to use the values from this table whenever you see a number start with “#” in any of the questions. Example: If the student number is 400123456, D 1 = 6, D 2 = 5, D 3 = 4, D 4 = 3, D 5 = 2, D 6 = 1. Digits D9 D8 D7 D6 D5 D4 D3 D2 D1 Your student number

page 4 Figure 1 . Signal histogram Student Levels Student 1 -0.875 -0.625 -0.375 -0.125 0.125 0.375 0.625 0.875 Student 2 -1.000 -0.750 -0.500 -0.250 0 0.250 0.500 0.750 Student 3 -1.000 -0.750 -0.500 -0.250 0 0.250 0.500 0.750 Table 1 . Quantized levels 2. (a) [6] Find the output at n = 10 + # D 2 + # D 3 for input x 1 [ n ] = a n u [ n ] to a Linear Time-Invariant (LTI) system with impulse response h 1 [ n ] = 10 δ [ n ] + 5 δ [ n - 2] + 2 δ [ n - # D 1]. Show your work . h 1 [ n ] x 1 [ n ] y 1 [ n = 10 + # D 2 + # D 3]=? Answer: #D1 = 1, #D2 = 1, #D3 = 2, hence n = 13 y [ n ] = x [ n ] * h [ n ] (5) = ∞ summationdisplay k = -∞ x [ k ] h [ n - k ] (6) = 10 x [ n ] + 5 x [ n - 2] + 2 x [ n - 1] (7) y [13] = 10 a 13 + 5 a 11 + 2 a 12 (8) (b) [4] Find an input x 2 [ n ] that produces the same output as the one you obtained in part (a) for a different LTI system with impulse response h 2 [ n ] = 2 a n u [ n ]. Show your work .

page 5 Figure 2 . Error histogram of student 1 Figure 3 . Error histogram of student 2 Figure 4 . Error histogram of student 3

page 7 3. (a) [10] Given signal x [ n ] = e j π 8 n + 2 e - j π 8 n + 2 e j π 4 n + e - j π 4 n and the system transfer function (DTFT of the impulse response) H ( ω ) = braceleftbigg (2 + # D 4) e j 2 ω , | ω |≤ π/ 6 0 , π/ 6 < | ω | < π , (21) find the response y [ n ]. Show your work . Answer: #D4 = 5 The cut-off frequency of the low pass filter is π/ 6. Hence, the last two terms of the x [ n ] will be filtered out. y [ n ] of input e jωn is H ( ω ) e jωn . y [ n ] = 7 e j 2( π/ 8) e j π 8 n + 14 e - j 2( π/ 8) e - j π 8 n (22) (b) [5] Find the DTFT at ω = 0 (i.e., X (0)) of the signal x [ n ] = parenleftBig sin( ω 1 n ) πn parenrightBigparenleftBig sin( ω 2 n ) πn parenrightBig (23) with ω 1 < ω 2 . Show your work . Note that sin( ω c n ) πn DT F T ←--→ braceleftbigg 1 , | ω | < ω c 0 , ω c < | ω | < π (24) Answer: Consider the signal x [ n ] = x 1 [ n ] x 2 [ n ] with DTFT X 1 ( ω ) , X 2 ( ω ) respectively, where x 1 [ n ] = sin( ω c 1 n ) πn x 2 [ n ] sin( ω c 2 n ) πn (25) Using the convolution property of DTFT x [ n ] ←→ 1 2 π integraldisplay π - π X 1 ( τ ) X 2 ( ω - τ ) dτ (26) Setting ω = 0 and noting that X 1 ( ω ) and X 2 ( ω ) are even symmetric, X (0) = 1 2 π integraldisplay π - π X 1 ( τ ) X 2 ( - τ ) dτ (27) = 1 2 π integraldisplay π - π X 1 ( τ ) X 2 ( τ ) dτ (28) = 1 2 π integraldisplay ω c 1 - ω c 1 1 dτ = ω c 1 π (29)

page 8 (c) [5] DTFT of a continuous time signal sampled at the sampling rate f s , which is higher than the Nyquist rate, is X ( ω ) = braceleftbigg 1 + # D 1 , | ω | < 3 π/ 4 0 , 3 π/ 4 < | ω | < π (30) Sketch the DTFT obtained when the signal is sampled at half of the original rate (i.e., f s / 2). Show your work . Answer: #D1 = 1 When the sample rate is halved, we will get (before considering the aliasing) X ( ω ) = braceleftbigg 2 / 2 , | ω | < 2 × 3 π/ 4 0 , | ω | > 2 × 3 π/ 4 (31) However, with the aliasing we will have X ( ω ) = braceleftbigg 1 , | ω | < π/ 2 2 , π/ 2 < | ω | < π (32)

page 10 h [ n ] = 2( - 1) ( n - 1) u [ n - 1] + 3( - 1) ( n +6) u [ n + 6] (44) (d) [5] Find the output at n = 1 (i.e., y [1]) of a system with transfer function H ( z ) = 1 + 2 z - 2 (45) for input x [ n ] = ( - 1) n . Show your work . Hint: you may need to find the difference equation first. Answer: H ( z ) = Y ( z ) X ( z ) = 1 + 2 z - 2 (46) Y ( z ) = (1 + 2 z - 2 ) X ( z ) (47) y [ n ] = x [ n ] + 2 x [ n - 2] (48) y [1] = x [1] + 2 x [ - 1] (49) = ( - 1) 1 + 2( - 1) - 1 (50) = - 3 (51)

page 11 5. (a) [10] Consider the finite length sequence x [ n ] = (2 + # D 4) cos parenleftBig π 4 n parenrightBig 0 ≤ n ≤ 7 (52) i. Find the 8-point Discrete Fourier Transform (DFT) of x [ n ]. Show your work . Answer: #D4 = 3 x [ n ] = 5 2 ( e j 2 π 8 n + e - j 2 π 8 n ) 0 ≤ n ≤ 7 (53) = 1 8 (20 e j 2 π 8 n + 20 e - j 2 π 8 n ) 0 ≤ n ≤ 7 (54) = 1 8 (20 e j 2 π 8 n + 20 e j 2 π 8 7 n ) 0 ≤ n ≤ 7 (55) (56) So, by matching coefficent: Xk = { 0 , 20 , 0 , 0 , 0 , 0 , 0 , 20 } . ii. Find the value at k = 2 (i.e., X [2]) of 16-point DFT with zero-padding. Show your work . Answer: When we double the points by padding zeros, we double the points on the frequency domain. When the k is even, its value will be same as ¯ X ( k/ 2), hence X [2] = 20. (b) [5] Find the DTFT value at ω = π/ 4 of a finite length sequence that produces 4-point DFT X [ k ] = 2 δ [ k ] + 2 δ [ k - 3]. You do not need to simplify the answer. Show your work . Answer: x [ n ] = 1 4 3 summationdisplay k =0 X [ k ] e j 2 π 4 kn (57) = 1 4 (2 + 2 e j 2 π 4 3 n ) (58) = 1 2 (1 + 1 e j 2 π 4 3 n ) (59) X ( ω ) is X ( ω ) = 3 summationdisplay n =0 x [ n ] e - jωn (60)