ECE310_fa2021_e02_sol

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University of Illinois Fall 2021 ECE 310 Profs. Zhao, Katselis, and Kamalabadi Midterm Exam 2 - Solution 8:30-10:00 P.M. Tuesday, Nov. 9, 2021. Please do not start the exam until the starting time. • No collaboration allowed: You are not allowed to share or collaborate on this exam and that all work should be your own. • You can use your handwritten notes in paper, printouts of your tablet notes, and printouts of the instructor’s slides or notes. • You can also use your notes in your tablet but you can only scroll through them, you cannot search through them by typing or writing. • Calculators and other electronic ways to do calculations, like Wolfram alpha, are not allowed. Neither is searching online. • Please follow the rules for online examinations detailed on the course website. GOOD LUCK!

1. (15 Pts.) Answer True or False to each of the following statements: (a) x [ n ] = cos( ω 0 n ) is an eigenfunction of a stable LTI system. True/False (b) Two different sequences of same length can have the same DFT. True/False (c) The ideal reconstruction filter is causal. True/False (d) Assume x [ n ] is a finite-duration sequence of length 20, and y [ n ] is obtained by zero-padding x [ n ] to length 32. That is, y [ n ] = x [ n ], for n = 0 , 1 , . . . , 19, and y [ n ] = 0, n = 20 , 21 , . . . , 31. Let { X [ m ] } 19 m =0 and { Y [ m ] } 31 m =0 be the DFT of { x [ n ] } 19 n =0 and { y [ n ] } 31 n =0 , respectively, then X [10] = Y [16]. True/False (e) Increasing the sampling period shrinks the corresponding DTFT. True/False Solution: (a) False : cosine can not be an eigenfunction because if a filter only covers one of the impulses in the frequency domain, the output would not be the scaled version of the cosine in the time domain. (b) False : DFT is a one-to-one thing; in other words, every different signal should correspond to a unique DFT. (c) False : Ideal reconstruction filter is non-causal. (d) True. (e) False: The corresponding DTFT would be expanded. 2. (15 Pts.) A causal LTI system is described by the difference equation: y [ n ] = y [ n - 2]+ x [ n ] - x [ n - 1]. (a) Determine the system’s transfer function H ( z ). (b) Determine the system’s unit impulse response h [ n ]. (c) Determine the system’s frequency response H d ( ω ); is H d ( ω ) = H ( z ) | z = e jω ? If not, explain why. Solution: (a) Z{ y [ n ] = y [ n - 2] + x [ n ] - x [ n - 1] } (1) = > Y ( z ) = z - 2 Y ( z ) + X ( z ) - z - 1 X ( z ) (2) = > H ( z ) = Y ( z ) X ( z ) = 1 - z - 1 1 - z - 2 = 1 1 + z - 1 (3) (b) h [ n ] = Z - 1 { 1 1 + z - 1 } = ( - 1) n u [ n ] , where | z | > | - 1 | (4)

(c) Because the ROC of H ( z ) does not include the unit circle, H d ( w ) 6 = H ( z ) | z = e jw . H d ( w ) = DTFT { h [ n ] } = DTFT { e jπn u [ n ] } H d ( w ) = 1 1 - e - j ( w - π ) + π ∞ X k = -∞ δ ( w - π - 2 kπ ) 3. (10 Pts.) Let X [ k ] be the 8 point DFT of x [ n ] = { 1 ↑ , 3 , 2 , 0 , 1 , 3 , 2 , 0 } . (a) Compute X [ k ] for k = 0 , 2 , 4 , 6. Solution: X [ k ] = 7 X n =0 x [ n ] e - j 2 nkπ 8 = (1 + e - jkπ )(1 + 3 e - j kπ 4 + 2 e - j kπ 2 ) = (1 + ( - 1) k )(1 + 3 e - j kπ 4 + 2 e - j kπ 2 ) X [0] = 12 X [2] = - 2 - 6 j X [4] = 0 X [6] = - 2 + 6 j (b) Use properties of DFT to determine Y [ k ] for k = 0 , 2 , 4 , 6, where the corresponding sequence y [ n ] is y [ n ] = x [ h 3 - n i 8 ] · ( j ) n . Solution: X [ k ] = { 12 , 0 , - 2 - 6 j, 0 , 0 , 0 , - 2 + 6 j, 0 } z [ n ] = x [ h 3 - n i 8 ] ↔ Z [ k ] = W 3 k 8 X [ - k ] y [ n ] = z [ n ] W - 2 n 8 ↔ Y [ k ] = Z [ h k - 2 i 8 ] Y [ k ] = {- 6 + 2 j, 0 , 12 , 0 , - 6 - 2 j, 0 , 0 , 0 }

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4. (15 Pts.) Consider the following digital system ( p = π , w = ω , and W = Ω in the figure) x(n) T T (ideal) ya(t) D/A y(n) Hd(w) xa(t) with p p/2 2p10^3 0 1 1 0 w W Xa(W) Hd(w) Sketch X d ( ω ), Y d ( ω ), and Y a (Ω) and clearly label the axes, for: 1) T = 1 8 × 10 3 2) T = 1 4 × 10 3 Solution: Recall the relation between angular frequency Ω and normalized frequency ω : ω = Ω T, where T is the sampling period. Then, for each case, we have the following frequency values: 1) Ω 0 = 4 π · 10 3 rad s ⇒ ω 0 = π 2 rad sample , Ω 1 = 8 π · 10 3 rad s ⇒ ω 1 = π rad sample 2) Ω 0 = 4 π · 10 3 rad s ⇒ ω 0 = π rad sample , Ω 1 = 8 π · 10 3 rad s ⇒ ω 1 = 2 π rad sample Figure 1 shows the frequency spectrum for each sampling period. (a) 1a (b) 1b (c) 1c (d) 2a (e) 2b (f) 2c Figure 1: Question 4. Frequency spectrum for each sampling period T .

5. (15 Pts.) Consider the filtering system in Problem 4. Suppose that x a ( t ) is bandlimited to 4000 Hz. The system produces an output y a ( t ) such that Y a (Ω) = H a (Ω) X a (Ω), where H a (Ω) = 1 - | Ω | 4000 π , | Ω | ≤ 4000 π 0 , | Ω | > 4000 π (a) Determine the largest sampling period such that no aliasing occurs at the output of the sampler (ideal A/D converter). Solution: According to the Nyquist Theorem, F s ≥ 2 F max = 8000 Hz . Hence, T s ≤ 1 8000 = 0 . 125 msec. The largest sampling period is 0.125 msec. (b) Determine and plot H d ( ω ) for T = 10 - 4 sec. Solution: T = 0 . 1 msec, which is less than 0.125 msec. Hence, no aliasing happens at the A/D converter. Let’s also assume that we have an ideal D/A converter at the output end. The scaling effects from the A/D and D/A converters will cancel out with each other. Therefore, H d ( ω ) just needs to reproduce the scaling effects of H a (Ω). H a (0) = 1 and Ω = 0 in the CTFT domain corresponds to ω = 0 in the DTFT domain. As a result, H d (0) = 0. When | Ω | = 4000 π , H a (Ω) = 0. | Ω | = 4000 π in the CTFT domain corresponds to | ω | = 4000 π × 10 - 4 = 0 . 4 π . As a result, H d (0 . 4 π ) = H d ( - 0 . 4 π ) = 0. H a (Ω) decreases linearly as Ω goes from 0 to 4000 π or - 4000 π . Correspondingly, H d ( ω ) decreases linearly as ω goes from 0 to 0 . 4 π or - 0 . 4 π . Combining these infos together, we get: H d ( ω ) = 1 - | ω | 0 . 4 π , | ω | ≤ 0 . 4 π 0 , | ω | > 0 . 4 π

6. (15 Pts.) Consider the two finite-length sequences: x = {- 2 ↑ , 3 , 0 , 3 } and h = {- 2 ↑ , 5 , 1 , 5 , - 3 } (a) Let Y d ( ω ) = X d ( ω ) H d ( ω ), where X d ( ω ) , H d ( ω ) denote the DTFT of x [ n ] , h [ n ]. Compute y [ n ]. Solution: Y d ( w ) = X d ( w ) H d ( w ) ↔ y [ n ] = h [ n ] * x [ n ] y [ n ] = { 4 , - 16 , 13 , - 13 , 36 , - 6 , 15 , - 9 } (b) Let Y [ k ] = X [ k ] H [ k ], where X [ k ] , H [ k ] denote the 6-point DFT of x [ n ] , h [ n ]. Find y [ n ]. Solution: Y [ k ] 6 = X [ k ] 6 H [ k ] 6 ↔ y [ n ] = h [ < n > 6 ] ~ x [ < n > 6 ] y [ n ] = { 19 , - 25 , 13 , - 13 , 36 , - 6 } (c) True or False : Zero padding both sequences to length N = 7 is adequate to guarantee that linear and circular convolutions coincide. Solution: False Need to zero-pad to length 8. 7. (15 Pts.) A continuous-time signal x c ( t ) = cos (10 πt ) is sampled at a rate of 100 Hz for 5 seconds to produce a discrete-time signal x [ n ] with length L = 500. (a) Let X [ k ] be the L -point DFT of x [ n ]. At what value(s) of k will X [ k ] have the greatest magnitude?

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Solution: The ideal discrete-time sequence corresponds to: x [ n ] = x c ( nT ) = cos( πn 10 ) , ∀ n ∈ Z Since the sequence is sampled for 5 seconds only, it is truncated to n ∈ { 0 , . . . , 499 } . While the ideal cosine has a frequency response composed by pair of delta functions, the DTFT of its truncated version has the deltas replaced with sinc-like functions. Given the resemblance between the DTFT of the cosine and its truncated version, the greatest magnitudes for both DTFT correspond to the same normalized frequencies ω max : ω max = ± π 10 + 2 πk, k ∈ Z . Taking into account the frequency range ω ∈ [0 , 2 π ], the frequencies with the greatest magnitude correspond to - π 10 + 2 π = 19 π 10 and π 10 . Then, using the relation between DFT and DTFT: X [ k ] , X d 2 πk 500 , k ∈ { 0 , . . . , 499 } ω = 19 π 10 and ω = π 10 are mapped to k = 475 and k = 25, respectively. (b) Suppose that x [ n ] is zero-padded to a total length of N = 1024. At what value(s) of k does the N -point DFT have the greatest magnitude? Solution: By taking N = 1024 frequency samples and considering the uniform sampling of the DFT ( ω k = 2 πk N ), the frequency positions in which the maximum values appear ( 19 π 10 and π 10 ) do not correspond to particular samples k . In other words, there are no integer values k that correspond to 19 π 10 and π 10 . Thus, we approximate the frequencies with the highest frequency response values by rounding them to the nearest sampled frequencies ω k : 19 π 10 = 2 πk 1024 ⇒ k = 972 . 8 ≈ 973 π 10 = 2 πk 1024 ⇒ k = 51 . 2 ≈ 51