# UNIVERSITY OF CALIFORNIA, SAN DIEGO Electrical & Computer Engineering Department ECE Fall 2009 Linear Systems Fundamentals

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1 UNIVERSITY OF CALIFORNIA, SAN DIEGO Electrical & Computer Engineering Department ECE Fall 2009 Linear Systems Fundamentals MIDTERM EXAM You are allowed one 2-sided sheet of notes. No books, no other notes, no calculators. PRINT YOUR NAME Marc-Antoine Parseval des Chênes Signature 1 T T x(t) 2 dt = a k 2 Student ID Number...,a 2, a 1, a 0, a 1, a 2,... Problem Weight Score 1 30 pts pts pts pts 10 Total 100 pts 100 Please do not begin until told. Write your name on all pages. Show your work. Use back of previous page and attached scratch sheets as needed. Tables 3.1 and 3.2 from the textbook are attached to the back of the exam. Good luck! 1

2 Problem 1 (30 points) For each part, check the appropriate boxes. Justify your answers. Each answer is worth 6 points. (a) Let x[n] = δ[n] + 2δ[n 1] + 3δ[n 2]. Let x e [n] be the even part of x[n] and let x o [n] be the odd part of x[n]. Check the box next to the correct statement: x e [0] = 1 2 and x o[0] = 1 2 x e [1] = 3 2 and x o[1] = 1 2 X x e [ 2] = 3 2 and x o[ 2] = 3 2 x e [2] = 2 and x o [2] = 1 Since x e [n] = x[n]+x[ n] 2, we see that x e [0] = 1, x e [1] = 1, x e [ 2] = 3 2, x e[2] = 3 2. So, the only possible answer is the third one. Recalling that x o [n] = x[n] x[ n] 2, we confirm that x o [ 2] = 3 2. This confirms the choice of the third answer. 2

3 Problem 1 (cont.) (b) Consider the signal x[n] = e j 2π 9 (n 1) + δ[n 6k + 2]. The smallest positive integer N such that x[n + N] = x[n] for all integer values of n is: 9 12 X No such N Note that the first term in x[n], namely x 1 [n] = e j 2π 9 (n 1), has fundamental period N 1 = 9 because it is simply a time-shifted version of the periodic discrete-time exponential signal e j 2π 9 n with fundamental period 9. The second part of x[n], namely x 2 [n] = δ[n 6k + 2], has fundamental period N 2 = 6 because is simply a time shift of the periodic discrete-time impulse train δ[n 6k] with fundamental period 6. Therefore, the fundamental period N of x[n] = x 1 [n] + x 2 [n], is the least common multiple of N 1 and N 2 ; that is, N = lcm(n 1, N 2 ) = lcm(9, 6) = 18. 3

4 Problem 1 (cont.) (c) Let x(t) = 2e 3t be the input to a continuous-time LTI system with impulse response h(t) and transfer function H(s). Suppose the corresponding output of the system, y(t), satisfies y(ln 2) = 8. Check the box next to the correct statement: y(t) = 4e t X y(1) = e 3 H(3) = 1 8 H(3) = 2 None of the above Since the system, which we denote by S, is LTI, we have Therefore, x(t) = 2e 3t S y(t) = 2H(3)e 3t. y(ln 2) = 2H(3)e 3ln 2 = 2H(3)2 3 = 16H(3). Since the problem states that y(ln 2) = 8, we conclude that H(3) = 1 2, implying that y(t) = e 3t. Therefore, y(1) = e 3. 4

5 Problem 1 (cont.) (d) A discrete-time LTI system is described by y[n] = n+1 k=n 1 x[k]. Check the appropriate box indicating whether or not the system satisfies the specified property: True False X Stable X Causal The input-output relationship can be written as y[n] = x[n 1]+x[n]+x[n+1]. If the input is bounded, say x[n] B <, this means that y[n] = x[n 1] + x[n] + x[n + 1] x[n 1] + x[n] + x[n + 1] 3B <, where the first inequality follows from the generlized triangle inequality. Thus, a bounded input produces a bounded output, so the system is stable. On the other hand, since y[n], the output at time n, depends on x[n+1], the input at time n + 1, the system is not causal. 5

6 Problem 2 (36 points) Consider the discrete-time LTI system S 1 defined by the input-output relation: y[n] = x[n] + 2x[n 1] x[n 2] 2x[n 3]. (a) Determine and sketch precisely the impulse response h 1 [n]. (6 points) The impulse response is immediately seen to be h 1 [n] = δ[n] + 2δ[n 1] δ[n 2] 2δ[n 3]. The sketch of h 1 [n] is now easy to draw and is left to the reader. 6

7 Problem 2 (cont.) (b) Determine and sketch precisely the step response s 1 [n]. (6 points) The step response s 1 [n] is the running sum of the impulse response, meaning that n s 1 [n] = h 1 [k]. By inspection, this translates to s 1 [n] = 0, n < 0, s 1 [0] = 1, s 1 [1] = 3, s 1 [2] = 2, s 1 [n] = 0, n > 2. The sketch of s 1 [n] is now easy to draw and is left to the reader. 7

8 Problem 2 (cont.) (c) Suppose that S 1 is concatenated with the LTI system S 2 that has the impulse response h 2 [n] = u[n] u[n 5]. Call the resulting system S. Determine and sketch precisely the impulse response h[n] of the system S. (12 points) The overall impulse response h[n] satisfies h[n] = h 1 [n] h 2 [n]. Using the graphical approach discussed in class, we see that h[n] = 0, for n < 0 and n > 7. The values in the range 0 n 7 can be found by inspection to be: h[0] = 1 h[4] = 0 h[1] = 3 h[5] = 1 h[2] = 2 h[6] = 3 h[3] = 0 h[7] = 2 The sketch of h[n] is now easy to draw and is left to the reader. 8

9 Problem 2 (cont.) (d) The system S 1 is not invertible. Find two distinct input signals x 1 [n] and x 2 [n] that produce the same output y[n]. Specify x 1 [n], x 2 [n], and y[n] precisely. (12 points) Recall that for system S 1, we have y[n] = x[n]+2x[n 1] x[n 2] 2x[n 3]. Let x 1 [n] = 0, for all n. The corresponding output is easily seen to be y 1 [n] = 0, for all n. Let x 2 [n] = 1, for all n. The corresponding output is easily seen to be y 2 [n] = 0, for all n. Hence, the two distinct inputs x 1 [n] and x 2 [n] produce the same output, namely y[n] = 0, for all n. This proves that the system S 1 is not invertible. 9

10 Problem 3 (24 points) Let x[n] be a real discrete-time signal with fundamental period N = 6. Denote the Fourier series coefficients of x[n] by a k. Suppose x[n] satisfies the following properties: (i) 6 n=1 x[n] = 6 (ii) a 1 = 2 (iii) a 2 = j (iv) a 3 = 1 (a) Determine the values of the Fourier series coefficients a k, k = 0, 1,...,5 and list them below. (12 points) From the analysis equation, we see that a 0 = 1 6 n=<6> x[n] = 6 x[n] = 1. Since x[n] is stated to be real, the Fourier Series coefficients satisfy the conjugate symmetry property. Using facts (ii), (iii), and (iv), along with the conjugate-symmetry property and the periodicity of the a k, we deduce that n=1 a 1 = a 1 = a 5 = 2 a 2 = a 2 = a 4 = j a 3 = a 3 = a 3 = 1 One period of the coefficients a k is therefore given by: a 0 = 1, a 1 = 2, a 2 = j, a 3 = 1, a 4 = j, a 5 = 2. 10

11 Problem 3 (cont.) (b) Express x[n] in the form (8 points) x[n] = A 0 + A k sin(ω k n + φ k ). k=1 Using the results of part (a), we have x[n] = a 0 + a 1 e j 2π 6 n + a 2 e j 4π 6 n +a 3 e j 6π 6 n + a 4 e j 8π 6 n + a 5 e j 10π 6 n = 1 + 2e j π 3 n + je j 2π 3 n +e jπn je j 2π 3 n + 2e j π 3 n. Grouping terms, applying Euler s Formula, and using the fact that cos(θ) = sin(θ + π 2 ), we find x[n] = 1 + 2(e j π 3 n + e j π 3 n ) + j(e j 2π 3 n e j 2π 3 n ) + e jπn = cos( π 3 n) 2 sin(2π 3 n) + cos(πn) = sin( π 3 n + π 2 ) 2 sin(2π 3 n) + sin(πn + π 2 ) Note that e jπn = cos(πn) + j sin(πn) = cos(πn), since sin(πn) = 0, n. 11

12 Problem 3 (cont.) (c) Determine (4 points) 2 n= 3 x[n] 2 From Parseval s Relation, we know 1 6 n=<6> x[n] 2 = k=<6> a k 2. The sums can be evaluated over any length-6 interval of integers, such as n = 3, 2, 1, 0, 1, 2. Therefore, 2 n= 3 x[n] 2 = 6 5 a k 2 k=0 = 6( j j ) = 6( ) = 6 12 =

13 Problem 4 (10 points) Consider a continuous-time system S with frequency response { 1 for ω 101 H(jω) = 0 for ω > 101. When the input to this system is a signal x(t) with fundamental period T = π/5 and Fourier series coefficients a k, the output signal y(t) satisfies y(t) = x(t). Determine the values of k for which a k must be equal to zero. Since the fundamental period of x(t) is T = π 5, the fundamental frequency is ω 0 = 2π T = 10. Therefore, x(t) has Fourier Series representation x(t) = a k e jkω0t = a k e jk10t. The corresponding output y(t) satisfies y(t) = b k e jkω 0t = a k H(jkω 0 )e jkω 0t = For the specified frequency response H(jω), we see that { ak for 10k 101 b k = 0 for 10k > 101. a k H(jk10)e jk10t. That is, y(t) = 10 k= 10 a k e jk10t. Since y(t) = x(t), if follows that a k = 0 for k >

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