ELEN 4810 Midterm Exam

Transcription

1 ELEN 4810 Midterm Exam Wednesday, October 26, 2016, 10:10-11:25 AM. One sheet of handwritten notes is allowed. No electronics of any kind are allowed. Please record your answers in the exam booklet. Raise your hand if you need additional scratch paper. The last page of the exam contains a table of common Discrete Time Fourier Transform pairs. There are a total of 4 questions. Good luck! Name: Uni:

2 1. Systems in Time and Frequency Domain. Consider a causal LTI system defined by the following block diagram: Notation. Above, the symbol denotes multiplication by a scalar α: if the input is w[n], the output is αw[n]. The symbol denotes a delay by one sample: if the input is w[n], the output is w[n 1]. Please answer the following questions about this system. (a) What is the impulse response h[n]? (b) For what choices of (α, β) is the system stable? (c) For any (α, β) such that the system is stable, what is the frequency response H(e jω ) of the system? (d) What is the output y[n] when the input is the oscillatory signal x[n] = ( 1) n? (1)

3 Answer to Problem 1: (a) The system can be represented by the difference equation y[n] = x[n] + αβy[n 2]. (2) In particular, if the input is x[n] = δ[n], then the output h[n] is zero for n < 0. h[0] = 1. For n > 0, x[n] = 0, and so h[n] = αβh[n 2]. Continuing, we have (αβ) n/2 n 0, n even, h[n] = (3) 0 else. (b) The system is stable if and only if αβ < 1, since h[n] = n=0 n 0, even αβ n/2 = αβ l = l=0 1 1 αβ αβ < 1 + else. (4) (c) The frequency response is H(e jω ) = = n 0, even (αβ) n/2 e jωn = ( αβe 2jω ) l l=0 1. (6) 1 αβe 2jω (5) (d) Here, the input is the complex exponential x[n] = e jπn. The output is y[n] = H(e jπ )e jπn = 1 1 αβ ( 1)n. (7)

4 2. Modulation and Sampling. Consider a system which performs the following operations to sample and then reconstruct a continuous-time signal w(t): (i) Obtain samples w d [n] = w(nt ), with sampling period T. (ii) Form an impulse train y s (t) = n= w d[n]δ(t nt ), (iii) Pass y s (t) through a filter with frequency response H(jΩ), to produce a continuous-time output y(t). We denote this entire operation by y = SRw}. Consider a continuous time signal x(t) with Fourier transform X(jΩ) shown below: (a) Suppose we choose the reconstruction filter H(jΩ) as T Ω < π/t H(jΩ) = 0 Ω π/t (8) and set y = SRx}. What is the largest sampling period T for which y(t) = x(t)? (b) Can we increase the sampling period by choosing a different filter H(jΩ)? What is the largest sampling period we can achieve? (c) Now suppose we modulate the input and output with complex exponentials, by setting x(t) = x(t)e jω1t, ỹ = SR x}, and y(t) = ỹ(t)e jω2t. Suppose we again fix T Ω < π/t H(jΩ) = (9) 0 Ω π/t We are free to choose Ω 1 and Ω 2, in order to achieve the largest possible sampling period T, while ensuring that y(t) = x(t). How should we choose Ω 1 and Ω 2?

5 Answer to Problem 2: (a) By the Nyquist criterion, the minimum sampling rate is Ω s = 2 3Ω 0 = 6Ω 0. The largest sampling period is T = 2π/Ω s = π/3ω 0. (b) Yes. If we choose a bandpass filter satisfying T 2Ω 0 < Ω < 3Ω 0 H(jΩ) = 0 else (10) we can increase the sampling period. In particular, if we set Ω s = Ω 0, the Fourier transform of x s (t) = x(t) k δ(t kt ) is as below: There is no aliasing, and the bandpass filter H(jΩ) correctly picks out the component corresponding to X(jΩ). Thus, we can increase the sampling period to T = 2π/Ω s = 2π/Ω 0. (c) Yes. If x(t) = x(t)e jω1t, using the fact that multiplication in time corresponds to convolution in frequency, we obtain X(jΩ) = X(j(Ω Ω 1 )). In particular, if we choose Ω 1 = 2.5Ω 0, then x is bandlimited, with bandlimit Ω 0 /2. We can exactly reconstruct x from samples taken with sampling rate at least 2 Ω 0 /2 = Ω 0. This corresponds to sampling period 2π/Ω 0. Setting Ω 2 = 2.5Ω 0 moves the frequency content of the reconstructed signal back to the interval [2Ω 0, 3Ω 0 ].

6 3. Discrete Fourier Transform. Consider two length L signals x 1 [n] and x 2 [n]: Let N L be an integer, and x i [n] 0, n = 0, 1,..., L 1, x i [n] = 0, n < 0, n L. i = 1, 2, where 1 and 2 are nonnegative integers. ( V i [k] = exp j 2πk ) i, (11) N Notation. In what follows, D denotes a delay by samples: D x[n] = x[n ]. denotes linear convolution between two sequences. Please answer the following questions about this system. (a) Suppose we wish to compute D 1 x 1 D 2 x 2, i.e., we want y[n] = (D 1 x 1 D 2 x 2 ) [n] n = 0,..., M, (12) where M is the largest integer m for which (D 1 x 1 D 2 x 2 ) [m] 0. What is the smallest N for which this occurs? State your answer in terms of L, 1 and 2. (b) Now suppose we are allowed to reorder the samples of the output y[n], to create a new output ȳ[n]. I.e., we can choose some mapping n (n) and set ȳ[n] = y[n (n)]. What is the smallest N we can achieve?

7 Answer to Problem 3: Write Write = By the cyclic shift property of the DFT, w[n] = DFT 1 N X 1[k]X 2 [k]}. (13) y[n] = w[ n mod N ]. (14) (a) The smallest N is N = 2L Provided N 2L 1, w = x 1 x 2. As long as N 2L 1+, the cyclic shift (14) by does not cause the nonzero entries of w to wrap around, and so it is equivalent to a linear shift by. (b) The smallest N is 2L 1. As long as N is at least as large as the length 2L 1 of x 1 x 2, w[n] = x 1 x 2. y[n] consists of a cyclically shifted version of w[n]. By reordering the samples of y[n], we can obtain D 1 x 1 D 2 x 2 = D x 1 x 2 }.

8 4. A system with upsampling. Consider the following system, in which an input x[n] is upsampled by a factor of 4, that is x[n/4], if n/4 Z, x 4 [n] = (15) 0, otherwise. The resulting signal is filtered first by a linear time invariant system with frequency response A(e jω ), and then by a linear time invariant system with impulse response h[n] = δ[n 1], and then the resulting signal is downsampled by a factor of 4 to produce the output y[n]: Please answer the following questions about this system. (a) Under what conditions on A(e jω ) is the overall system S LTI? Please make your conditions as broad as possible for full credit. (b) Under the conditions you derived in part (a), what is the overall frequency response H S (e jω ) of the system S? (c) Suppose we make the specific choice A(e jω ) = What is the impulse response h S [n] of the overall system S? 4 ω < π 4 0 π 4 ω π. (16) (d) How would your answer to (c) change if we replace h[n] with δ[n 8]?

9 Answer to Problem 4: (a)-(b) Notice that X 4 (e jω ) = X(e j4ω ), and so, if we write v[n] for the output of the delay (i.e., the signal that is downsampled to produce y[n]), we have We have Y (e jω ) = 1 4 V (e jω ) = e jω A(e jω )X(e j4ω ). (17) 3 e j(ω 2πl)/4 A(e j(ω 2πl)/4 )X(e j(ω 2πl) ) (18) l=0 = X(e jω ) 1 3 e j(ω 2πl)/4 A(e j(ω 2πl)/4 ) 4 l=0 }} H S (e jω ) In the final line, we have used the fact that X(e jω ) is 2π-periodic. Because the system acts by multiplication in the frequency domain, it is always linear and time-invariant. [[A note on grading: a common answer is to choose A(e jω ) = 0 for π 4 < ω π to suppress aliasing. This answer is correct, in the sense that it is sufficient for the system to be LTI. However, it is not the broadest possible. This answer received almost all of the credit (1.5 pts / 2 pts for part (a)). With this answer, the overall frequency response becomes simpler: H S (e jω ) = 1 4 e jω/4 A(e jω/4 ). This answer received full credit (2 pts / 2 pts) for part (b). The most common mistake in (a)-(b) was not accounting for downsampling in H S (e jω ), e.g., writing H S = e jω. This is not correct.]] (c) The overall frequency response is H S (e jω ) = e jω/4. Taking the inverse DTFT, we obtain (19) h S [n] = 1 π e jω(n 1/4) dω (20) 2π π = = 1 2πj(n 1/4) ( e jπ(n 1/4) e jπ(n 1/4)) (21) sin(π(n 1/4). (22) π(n 1/4) (d) We have h S [n] = δ[n 2].

10 Scratch paper:

11 Scratch paper:

12 Discrete-Time Fourier Transform Pairs:.