Final Exam January 31, 014 Signals & Systems (151-0575-01) Prof. R. D Andrea & P. Reist Solutions Exam Duration: Number of Problems: Total Points: Permitted aids: Important: 150 minutes 7 problems 50 points One double-sided A4 sheet. Questions are not equally weighted. The weighting of each question is given on the cover page for that question. Answers must be justified, except where otherwise stated. Use only the provided sheets for your answers. Additional sheets are available from the supervisors if required.
Page Final Exam Signals & Systems Problem 1 5 points Consider a causal, linear, time-invariant system described by the difference equation y[n] = ay[n 1] + bx[n] where x[n] is the input, y[n] is the output, and a and b are real numbers. 1. For which values of a 0 and b 0 is the system bounded-input bounded-output stable? ( pt). Now, let a = 3/3, b = 1, and the input to the system be given by ( π x[n] = cos n + π ) 3 for all n. What is the output y[n] for all n? (3 pt) Hint: The following trigonometric table might be useful. θ 0 π/6 π/4 π/3 π/ sin θ 0 1 3 1 cos θ 1 3 1 0 tan θ 0 3 3 1 3 - Solution 1 1) The transfer function of the system is given by b H(z) = 1 az 1, which has a pole at z p = a. A causal LTI system is stable if and only if all the poles of H(z) lie inside the unit circle. It follows that a < 1. ) The output of an LTI system to a sinusoidal input can be obtained by analyzing the frequency response H(Ω) of the system. If an input x[n] = A cos(ω 0 n + Θ)
Final Exam Signals & Systems Page 3 is applied to the system for all times n, the output y[n] is given by y[n] = H(Ω 0 ) A cos ( Ω 0 n + Θ + H(Ω 0 ) ) where H(Ω 0 ) and H(Ω 0 ) are respectively the magnitude and phase response at frequency Ω 0. The given input has the frequency Ω 0 = π/. Therefore, ( π ) b H = 1 ae = b jπ/ 1 + j a. Evaluating for a = 3/3 and b = 1 yields the following magnitude and phase response, respectively: ( π ) = b 3 = 1 + a H H( π ( a ) ) = (b) (1 + ja) = 0 arctan = π 1 6. It follows that the output y[n] for the given input x[n] is, for all times n, given by y[n] = ( π 3 cos n + π ).
Page 4 Final Exam Signals & Systems Problem 8 points Part A The continuous-time sinusoid x(t) = 0.5 + cos ( πt + π ) ( 8π ) + 3 cos 7 3 t is uniformly sampled with sampling time T s discrete-time sequence x[n]. seconds, resulting in the 1. For what range of T s can x(t) be sampled without aliasing effects? (1 pt). Let T s = 1/4 seconds. What is the fundamental period N 0 of the resulting periodic sequence x[n]? ( pt) Part B Let x[n] be a different periodic sequence with fundamental period N 0 = 6. The Discrete Fourier Series (DFS) coefficients c k of x[n] are c 0 = 1, c 1 = 0.5, c = 1, c 3 = 0, c 4 = 1, c 5 = 0.5. 1. Determine the constants A, B, C, and D, such that the sequence x[n] is given by (3 pt) x[n] = A + B cos (Cn) + D cos (Cn).. You take the Discrete Fourier Transform of x[n] over two of its fundamental periods N 0 : X[k] = N 0 1 n=0 x[n]w kn N 0, with W N0 = e jπ/ N 0. State the sequence X[k] as a function of N 0 and the previously-defined DFS coefficients c 0,..., c 5. ( pt)
Final Exam Signals & Systems Page 5 Solution A.1) The allowed range is 0 < T s < 3/8 seconds. A.) The fundamental period of the sum of three periodic sequences is given by the least common multiple (lcm) of the individual fundamental periods. The sampled signal is ( π x[n] = }{{} 0.5 + cos n + π ) 7 N 0 =1 }{{} N 0 =4 ( π + 3 cos ) 3 n. }{{} N 0 =3 The fundamental period of the signal is therefore N 0 = lcm(1, 4, 3) = 1. B.1) A periodic signal x[n] with fundamental period N 0 can be rewritten as: x[n] = N 0 1 k=0 c k e jkω 0n. (1) Evaluating (1) with the given DFS coefficients c 0,..., c 5 leads to x[n] = c 0 + c 1 e j π 3 n + c e j π 3 n + c 3 e jπn + c 4 e j 4π 3 n + c 5 e j 5π 3 n ( π ) ( ) π = c 0 + ( c 1 + c 5 ) cos 3 n + ( c + c 4 ) cos 3 n ( π ) ( ) π = 1 + cos 3 n cos 3 n. Therefore, A = 1, B = 1, C = π 3, and D =. Note that ej 5π 3 n = e j π 3 n and e j π 3 n (as well as e j π 3 n and e j 4π 3 n ) are complex conjugates. As you can see, the constants A, B, and D can be directly inferred from the DFS coefficients. B.) The Discrete Fourier Transform (DFT) is closely related to the DFS. If the DFT is applied to one fundamental period of the periodic sequence x[n] the following holds: X[k] = N 0 c k. In this case, the DFT is applied to two fundamental periods of x[n]. Therefore, the resulting sequence is X[k] = N 0 { c 0, 0, c 1, 0, c, 0, c 3, 0, c 4, 0, c 5, 0}.
Page 6 Final Exam Signals & Systems Problem 3 7 points Consider the continuous-time, linear, time-invariant system q(t) = Aq(t) + Bx(t) y(t) = Cq(t) + Dx(t) with input x(t) and output y(t). Assume that x(t) is piece-wise constant over intervals of T s : x(t) = x[k], kt s t < (k + 1)T s. The system is to be discretized such that where q[k + 1] = A D q[k] + B D x[k] () y[k] = C D q[k] + D D x[k] q[k] = q(kt s ), y[k] = y(kt s ). 1. Discretize the system by calculating the matrices A D, B D, C D, and D D for (4 pt) [ ] [ ] 0 1 A =, B =, 0 0 C = [ 1 0 ], D = 0. Now, consider the following MATLAB script, where the matrices Ad, Bd, Cd, and Dd represent A D, B D, C D, and D D respectively, and characterize a discrete-time, linear, time-invariant system as defined in (). Ad = [1 0 1; -1 0; -1 0 3]; Bd = [ 1 0] ; Cd = [1 0 0]; Dd = 0; M = obsv(ad,cd); % obsv(a,c) returns the observability matrix. r = rank(m); % rank(a) returns the rank of the matrix A.. What is the value of the variable r after executing the script? ( pt) 3. What conclusions can you draw from the value of r? (1 pt)
Final Exam Signals & Systems Page 7 Solution 3 1) We will calculate this with the matrix exponential. First, we assemble the matrix M [ ] 0 1 A B M = = 0 0 0 0 0 0 0 Notice that M is nilpotent, that is, M k = 0 for k 3. The discrete-time matrices A D and B D are calculated as [ ] AD B D = e MT s M k Ts k = k! k=0 = I + MT s + 1 M T s. Solving the above yields [ ] 1 Ts A D = 0 1 [ Ts + Ts B D = T s ]. The matrices C D and D D are C D = C = [ 1 0 ] D D = D = 0. ) The observability matrix is C 1 0 0 O = CA = 1 0 1 CA 0 0 4 and has rank. Therefore, r =. 3) For the system to be observable the observability matrix must be full rank (3). This is not the case, therefore the system is not observable.
Page 8 Final Exam Signals & Systems Problem 4 8 points 1. The continuous-time sinusoid x(t) = cos(t) is uniformly sampled with sampling time T s = 0.1 seconds to obtain the discrete-time sequence x[n]. Is x[n] a periodic sequence? ( pt). Consider a stable, linear, time-invariant system, which is described by the transfer function H(z) = 1 (z + 0.5)(z 0.75)(z + 1.5). Draw and label the region of convergence of H(z). Hint: The region of convergence is a connected region. ( pt) 3. A signal w[n] is generated by drawing independent samples from a Gaussian distribution with zero mean and variance 4. Calculate the expected power of w[n] in the frequency band [0, π/]. ( pt) 4. The magnitude response H(ω) of a continuous-time filter is defined as follows: { 1 for 0 ω < rad/sec H(ω) = 0.01 for ω rad/sec. Using the Bilinear Transform and a sampling period of T s = 1, you discretize the continuous-time filter. Calculate the constants M pass, M stop, and Ω c, such that the magnitude response H(Ω) of the resulting discrete-time filter is given by ( pt) { M pass for 0 Ω < Ω c H(Ω) = M stop for Ω c Ω < π. Hint: You may find the trigonometric table in Problem 1 useful. Solution 4 1) No, the resulting discrete-time sequence x[n] is not periodic. The resulting discrete-time sequence is x[n] = x(nt s ) = cos(n/10). The fundamental frequency Ω 0 is therefore 1/5 radians/sample. A discrete-time
Final Exam Signals & Systems Page 9 sinusoid is periodic if and only if Ω 0 π is a rational number, which is not the case here. ) The system is stable and has poles at z = 0.5, z = 0.75 and z = 1.5. Due to system stability, the region of convergence (ROC) must contain the unit circle. Furthermore, the ROC must not contain any poles. With these conditions in mind, the ROC must be given by Im(z) 1.5 0.75 Re(z) 3) The expected power in a the frequency band [Ω 1, Ω ] is given by 1 Ω S ww (Ω)dΩ. π Ω 1 In the case where w[n] is white noise, the power is equally spread across all frequencies, and its power spectral density function S ww (Ω) is equal to the variance of the underlying probability density function for all Ω: It follows that S ww (Ω) = k= R ww [k]e jωk = 4. 1 π/ 4 dω =. π 0 4) The Bilinear Transform maps the imaginary axis in the s-plane to the unit circle in the z-plane by compressing the continuous-time frequencies < ω < to discrete-time frequencies π < Ω < π. In this question, the filter s continuous-time cutoff frequency ω c is mapped to the discrete-
Page 10 Final Exam Signals & Systems time cutoff frequency Ω c as follows: Ω c = arctan (ω c T s /) = arctan (1) = π 4 = π. The continuous- to discrete-time mapping does not alter the frequency response of the filter. Thus M pass = 1 and M stop = 0.01.
Final Exam Signals & Systems Page 11 Problem 5 5 points Consider a causal, discrete-time filter described by its frequency response H(Ω) = b 0 + b 1 e jω. 1. What is the impulse response h[n] of the filter? (1 pt). What are the coefficients b 0 and b 1 such that the following conditions are satisfied? (4 pt) H(π) = 1; A white noise input (zero mean, unit variance) results in a filter output y[n] with power spectral density S yy (Ω) = 5 + 4 cos(ω); b 0 > 0. Solution 5 1. The impulse response h[n] of the filter can be read directly from the (FIR) filter coefficients. The impulse response is h[n] = {b 0, b 1 }.. To solve for coefficients b 0 and b 1, we begin by expressing the given conditions in terms of b 0 and b 1 : With the condition that H(π) = 1, we have that b 0 b 1 = 1. From the condition imposed on the power spectral density, and the fact that the white noise input x[n] has a power spectral density S xx (Ω) = 1 for all Ω, it follows that S yy (Ω) = H(Ω) S xx (Ω) 5 + 4 cos(ω) = H(Ω) 1 By matching coefficients, we have that b 0 + b 1 = 5, and = b 0 + b 1 + b 0 b 1 cos(ω).
Page 1 Final Exam Signals & Systems b 0 b 1 = 4. Solving the above three conditions and choosing the only solution with b 0 > 0 yields b 0 =, b 1 = 1.
Final Exam Signals & Systems Page 13 Problem 6 10 points 1. The systems presented in this question are defined by difference equations, with input x[n] and output y[n]. Indicate whether these systems are linear and/or time-invariant by writing True (T) or False (F) in the appropriate boxes. You do not need to justify your answers. (3 pt) Grading: Six correct answers Five correct answers Four correct answers Three or fewer correct answers 3 points points 1 point 0 points. System 1 Linear Time-Invariant y[n] = nx[n] x[n 1] T F System Linear Time-Invariant y[n] = cos(x[n]) F T System 3 Linear Time-Invariant y[n] = x[n] + 1 F T
Page 14 Final Exam Signals & Systems. The systems presented in this question are discrete-time, causal, linear, time-invariant systems. Indicate whether these systems are bounded-input bounded-output (BIBO) stable by writing True (T) or False (F) in the appropriate boxes. You do not need to justify your answers. (3 pt) Grading: Three correct answers Two correct answers One or zero correct answers 3 points points 0 points. System 1 is given in state-space form: [ ] [ ] 0.1 5 0 q[n + 1] = q[n] + x[n] 0 0.5 10 [ ] y[n] = 1 0 q[n] System is described by its impulse response: h[n] = { 0 for n < 0 4e n + 1 otherwise System 3 is given by the pole-zero plot of its transfer function H(z): BIBO Stable T BIBO Stable F BIBO Stable Im(z) 1 Pole of H(z) Zero of H(z) F 0.5 1 0.5 0.5 Re(z) 0.5 1
Final Exam Signals & Systems Page 15 3. Consider the following MATLAB script: (4 pt) 1 Ntest = 10; Nin = 480; 3 Ntr = 10; 4 5 for i = 1: Ntest +1 6 Omega = pi *(i -1)/ Ntest ; 7 xn = cos ( Omega *(0: Nin -1)); 8 yn = Experiment ( xn ); 9 xn = xn(ntr +1: Nin ); 10 yn = yn(ntr +1: Nin ); 11 Xi = sum ( exp (-j * Omega * (0: length (xn ) -1) ).* xn ); 1 Yi = sum ( exp (-j * Omega * (0: length (yn ) -1) ).* yn ); 13 H(i) = Yi/Xi; 14 end Where the Experiment function is defined as: 1 > help Experiment Applies the discrete - time input xn to a system and 3 measures the output yn 4 - yn has the same length as xn 5 - The system begins at rest and is affected by noise Now briefly answer the following questions:
Page 16 Final Exam Signals & Systems 1. What is the purpose of the above script? To estimate frequency response H(Ω) from experimental data. Estimating H(Ω) is the first step in frequency-domain system identification.. What is the meaning of the value stored in H()? An estimate of the frequency response at frequency Ω = π/10. 3. What is the purpose of line 10 (yn = yn(ntr+1:nin))? To remove initial samples of yn and thus reduce the effects of transient behavior on H. 4. Qualitatively, what effect would substantially increasing Nin have on H? Nin is the length of the input sinusoid at frequency Ω i. By applying a longer input, the signal to noise ratio at frequency Ω i is increased, thus reducing the effect of noise on H(Ω i ).
Final Exam Signals & Systems Page 17 Problem 7 7 points You wish to use time-domain system identification to identify a causal, stable, linear, time-invariant system with input x[n] and output y[n]. Your identification data set consists of the following input/output data, which were obtained during two separate experiments. In both experiments, the system was initially at rest. Experiment 1 Experiment n < 0 0 1 3 x 1 [n] 0 1 1 0 y 1 [n] 0 0 3 n < 0 0 1 3 x [n] 0 1 0 y [n] 0 Your colleague suggests that the system can be represented by the model y[n] 0.5y[n 1] = x[n] 1.5x[n 1]. 1. Let the input to the suggested model be x[n] := x 1 [n]. Calculate the model s output y[n] for n = [0, 1,, 3], assuming initial rest. (1 pt). Using only the data from Experiment 1 and the suggested model, estimate the ARX error ê 1 [n] for n = [0, 1,, 3]. ( pt) Hint: Recall that, in an ARX model, the effect of error is modeled as y[n] = B(q) A(q) x[n] + 1 A(q) e[n]. You think you can do better than your colleague by increasing the system order. You believe the system has the structure y[n] + a 1 y[n 1] = b 0 x[n] + b 1 x[n 1] + b x[n ]. 3. Write a MATLAB script that uses the ARX identification method to estimate the parameter vector Θ = [ ] a 1 b 0 b 1 b using the data from both Experiment 1 and Experiment. Do not use the inbuilt arx function. (4 pt)
Page 18 Final Exam Signals & Systems Solution 7 1. To calculate the suggested model s output, apply the given input x[n] to the CCDE. This results in y[0] = x[0] 1.5x[ 1] + 0.5y[ 1] = 0 + 0 = y[1] = x[1] 1.5x[0] + 0.5y[0] = 1 3 + 1 = 1 y[] = x[] 1.5x[1] + 0.5y[1] = 1 1.5 0.5 = 1 y[3] = x[3] 1.5x[] + 0.5y[] = 0 1.5 0.5 = y[n] = {, 1, 1, }.. Rearranging the given ARX model yields: ê[n] = A(q)y[n] B(q)x[n]. With respect to the suggested system, we have: Â(q) = 1 0.5q 1 Â(q)y[n] = y[n] 0.5y[n 1] ˆB(q) = 1 1.5q 1 ˆB(q)x[n] = x[n] 1.5x[n 1]. Finally, solving for the ARX error results in ê[n] = Â(q)y[n] ˆB(q)x[n] = (y[n] 0.5y[n 1]) (x[n] 1.5x[n 1]). Sequences x 1 [n] and y 1 [n] can then be applied to the above CCDE, as shown in the previous solution, to solve ê[n] for n = [0, 1,, 3], giving ê[n] = {0, 1, 1.5, 0.5}.
Final Exam Signals & Systems Page 19 Alternatively, the same result is achieved by using: E = Y DΘ Where: Y = y 1 = [ 0 3 ] 0 x 1 [0] 0 D = y 1 [0] x 1 [1] x 1 [0] y 1 [1] x 1 [] x 1 [1] y 1 [] x 1 [3] x 1 [] 0 0 = 1 0 1 1 0 1 Θ = [ ] a 1 b 1 b = [ 0.5 1 1.5 ] Note that both these solutions are independent of the results from question 1.
Page 0 Final Exam Signals & Systems 3. As shown in the lecture, an ARX system can be formulated as Y = DΘ, where Θ is a vector of coefficients corresponding to the coefficients in the system s constant coefficient difference equation. Matrices D and Y are given as 0 0... 0 x[0] 0... 0 y[0] 0... 0 x[1] x[0]... 0........ D = y[n 1] y[n ]... y[n N] x[n] x[n 1]... x[n M]........ y[k 1] y[k ]... y[k N] x[k] x[k 1]... x[k M] y[0] y[1]. Y =. y[n]. y[k] An ARX system can be estimated by using linear least-squares to solve Y = DΘ for the parameter vector Θ. The following MATLAB script is one possible implementation. 1 Y1 = [ ; 0 ; - ; -3]; Y = [- ; ; ; -]; 3 % -y[n -1] x[ n ] x[n -1] x[n -] 4 D1 = [ 0 0 0 ; 5-1 0 ; 6 0 1 1 ; 7 0 1 1 ]; 8 9 D = [ 0-0 0 ; 10 1-0 ; 11-0 1 - ; 1 - - 0 1 ]; 13 14 Y = [Y1 ; Y ]; 15 D = [D1 ; D ]; 16 17 Theta = inv (D *D)*D *Y;