HW5_sols_final

HW5_sols_final - UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN...

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Unformatted text preview: UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN Department of Electrical and Computer Engineering ECE 410 Digital Signal Processing Homework 5 Due Wednesday, October 1, 2008 Prof. Bresler / Prof. Jones 1. The diagram below represents a part of the computation in a 16-point decimation-in-time radix-2 FFT. Indicate the values of the three requested branch weights a, b and c and the indexes W, X, Y, and Z. Z = bit reversal of [0110] = 0110 = 6 Y = bit reversal of [1001] = 1001 = 9 X = 10 (outputs X[k] are presented in series from the top down) W=3 1 2. Determine if the following systems are (i) memoryless or with memory (ii) causal or non-causal (iii) time-invariant or time-varying (iv) linear or non-linear Note that no justification is required. (a) ∑ (i) with memory (ii) not causal (for any fixed integer a) if 0, then y[n] depends on x[n+a], where , so it’s non-causal if 0, then y[a-1] depends on x[-(a-1)+a] = x[1], so it’s non-causal (iii) time-varying (iv) linear ∑ ∞ sin (i) with memory (ii) causal (iii) time-invariant (iv) not linear 1 (i) with memory (ii) causal (iii) time-invariant (iv) linear 5| | (i) memoryless (ii) causal (iii) time-invariant (iv) non-linear (due to absolute value) ∑ (i) with memory (ii) causal (iii) time-invariant (iv) linear k, 0 (b) (c) (d) (e) 2 (f) 10 1 10 | | 1 10 1 (i) memoryless (ii) causal (iii) time-invariant (iv) non-linear Counterexample: let 8, 88 8 Since (g) 0 0 0 (i) memoryless (ii) causal (iii) time-invariant (iv) non-linear Counterexample: let 4 8 and let . 16 10 8 10 10 20 , it’s non-linear. 4, 41 4 Since 1 and let 3 0 1 04 , it’s non-linear. . 4 3 3. In this problem, let 0.9 . , where , where , where , where 0.7 1 3 and . 10 . 9. 1. (a) Find the DT convolution result: (b) Find the DT convolution result: (c) Find the DT convolution result: (d) Find the DT convolution result: Hint for parts c and d: There’s an easier way to compute this result instead of using the DT convolution formula directly! (a) Using convolution by hand and flipping h[n]: ∞ 0.7 ∞ 0.9 0.7 0.7 0.9 0.9 0.7 0.7 5 0.9 0.9 0.7 0.7 1 0.9/0.7 1 0.9/0.7 3.5 0.7 4.5 0.9 (b) Using convolution by hand and flipping x[n]: 0.9 0 0.9 0.9 0.9 0.9 0 0.9 1 0.9 0 9 0 9 0 9 0.9 0.9 1 0.9 0.9 0 0.9 1 0.9 1 0.9 1 0.9 0 9 0 9 10 1 10 0.9 0 9 4 (c) First, notice that: 1 5 1 1 10 9 1 5 1 1 10 9 Second, we can substitute and appropriately modify (shift & scale) the results from parts (a) and (b) to find : 1 5 1 10 9 1 0 0.9 0.7 0 0.9 1 0.9 1 1 9 9 0 0 1 0.9 Finally, putting the above two results together, we have: 1 1 0.7 0 0.9 0.9 0.9 1 0.9 0 1 0.9 1 0.9 0.9 0.7 0.9 2 9 9 0 1 9 9 0 1 0 0 0.9 (d) First, notice that by the properties of DT convolution: 3 5 0.9 0.7 1 2 3 5 0.9 1 0 0.7 2 2 2 5 4. Compute the linear convolution (a) , | | 1, , given 4, and | | 1, below: Method 1: flip ∞ ∞ 4 ∞ ∞ 1 1 / / 4 4 Method 2: flip ∞ ∞ 4 ∞ ∞ 1 4 4 To show that both of these answers are mathematically equivalent, we proceed to show that the first method’s result is exactly equal to the second method’s result: 4 1 1 4 4 4 1 1 4 4 4 Thus, we see that either way we start, no matter which function we initially flip, the result is exactly the same. 6 (b) Method 1: flip ∞ 1 , | | 1, 2 , | | 1, 1/ ∞ 1 ∞ ∞ 2 1 1 1 3 1 1 Method 2: flip ∞ ∞ 1 1 3 1 1 3 2 ∞ ∞ 1 1 1 3 1 1 1 1 3 3 3 1 1 3 1 1 1 We can easily see that both of the methods yield exactly the same result. 7 5. Determine if the systems characterized by the following relations are, with respect to the input, (i) linear or nonlinear, (ii) causal or noncausal, and (iii) shift-invariant or shift-varying. Justify your answers with proofs or counter-examples. a. y[n] = y[n − 12] + x[n − 2] + x[n − 1] Linear? Yes. This can be shown as follows: Let Let Now, let n n 12 a n n n 2 12 12 n n 2 2 a n n n 1 1 , for all n 1 , for all n b n (a) (b) 1 , for all n c , where b n 2 From (a), (b) and (c), we have: yn ay n by n y n 12 ay n 12 by n 12 , for all n d Now, we can define: To show linearity, we need to show that 0 for all n. Equation (d) implies that: 12 , for all n (e) Invoking the zero IC (initial conditions) assumption, we have: 0, for so that 0, for 13 0 f 13 0 Iterating (e) forward with the IC’s (f) shows that w[n] = 0 for all n and thus, yn Hence the system is linear. Causal? Yes. Recall that the system is causal if y[n] depends only on x[m], for any m ≤ n. Hence, in this problem, this is true with respect to y[n] but not true with respect to y[n − 12]. In other words, causality of the given system depends on the “direction” in which the equation is iterated. However, as stated in the lecture notes, we always assume forward direction, i.e. we always consider the system with respect to y[n], unless when otherwise stated. So, with forward iteration y[n] depends on past values of the input, x[n-1] and x[n-2]. But, y[n] also depends on past values of the output. However, y[n-12] in turn satisfies 8 ay n by n , for all n. 12 24 13 14 Iterating this relationship establishes that y[n] only depends on past values of the input, i.e., on x[m], for m < n. Hence, the system is causal. Shift Invariant? Yes. Let 12 so that, substituting yields: 12 On the other hand, replacing by 12 We need to show that have: 1 in the given D.E. 1 2 2 . Then, 1 2 for all n. Subtracting equation (c) from (b), we 12 12 Now, define . Then, we have 12 , for all We need to show that 0 for all n. To do so, we invoke an argument about zero initial conditions – our assumption throughout. Assume that y[n] in the original D.E. was zero for n<N, and that the input x[n] was applied at time N. (N is arbitrary, could be negative) Hence, y[n]=0 for n < N and x[n]=0 for n < N. It follows that y[n-n0]=0 for n < N+n0 and that 0 for n < N+n0 . Because the system is assumed to be at rest until the input is applied, 0 for n < N+n0. Hence, 0 for n < N + n0. Iterating equation (d) from these zero initial conditions, yields 0 for all n. This establishes that , and the system is shiftinvariant. b. y[ n − 1] = 5 y[ n] + 3x[ n] + y[ n 2 ] Linear? Yes Let Let Now, let n 1 5y n yn 3 1 1 5 5 n n 3 3 n n n , for all n n , for all n (a) (b) 1 , where , for all n c Subtracting (a) and (b) from (c), and defining: 9 , for all n d we have: 1 5 e Next, we invoke the zero initial conditions assumption: 0 for a set of values on n such that the corresponding outputs are zero for all n, if the input is zero. It follows from (d) that 0 for the same set of values of n. Iterating (e) forward shows that w[n] = 0 for all n and thus, yn Hence the system is linear. NOTE: Finding exactly what set of values of n specify the initial conditions (or “state”) for this system is not required for the above proof. Causal? No As written, suggesting backward iteration, the system is clearly non-causal: e.g., y[4] depends on x[5]. However, even if iterated forward (which is our convention in this course unless otherwise stated), this system is non-causal. For example, y[2] depends on y[22], i.e., on y[4]. But y[4] depends on x[4], hence y[2] depends on x[4], so the system is non-causal. Shift invariant? No To test shift-invariance, we can directly see that the term y[n2] is not going to satisfy our requirement because: ay n by n , for all n. y ((n − n0 ) 2 ) ≠ y ( n − n0 ) c. y[n − 2] + 2 y[n] = nx[n − 1] + 23x[n] Linear? Yes Let Let n n 1 1 , where 2 n ax n 2 1 bx n 1 0 1 23 ax n 2 bx n , for all n c n n n n 2 , for all n 2 , for all n (a) (b) Now, let 1 yn n 2 From (a), (b) and (c), we have: yn ay n by n 1 yn 2 2 a yn 2 2 b yn 2 2 , for all n d Invoking the zero IC (initial conditions) assumption, we have: 0, for and then (d) implies that y 0 y 2 ay 0 by 0 3 y 0 2 ay n y 2 0, and iterating (d) forward shows that y n by n , for all n. Hence, the system is linear. Causal? Yes, because here depends on the input samples are not future samples with respect to the time instant n. 1 and , which Shift invariant? No the n coefficient would change the system nx[n − 1] Define x[n] = x[n − n0 ]. x [n] → y[n − n0 − 2] + 2 y[n − n0 ] = nx[n − 1 − n0 ] + 23 x[n − n0 ] y[n − n0 − 2] + 2 y[n − n0 ] = (n − n0 ) x[n − 1 − n0 ] + 23 x[n − n0 ] These are not the same, so the system is not shift-invariant. 6. Assume that the zero-state response of an LSI system to input x[n] =2-nu[n] is y[n]=(1/3)nu[n]. Use the system properties (linearity and shift-invariance) to find the system’s response h[n] to a unit pulse input. We simply need to create a unit pulse as the input and the output will be h[n] 1 1 ...
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