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Final ExamSignals & SystemsJanuary 31, 2014(151-0575-01)Prof. R. D’Andrea & P. ReistSolutionsExam Duration:150 minutesNumber of Problems:7 problemsTotal Points:50 pointsPermitted aids:One double-sided A4 sheet.Important:Questions are not equally weighted. The weighting of eachquestion 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. Additionalsheets are available from the supervisors if required.

Page 2Final Exam – Signals & SystemsProblem 15 pointsConsider a causal, linear, time-invariant system described by the differenceequationy[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 6 0 and b 6 0 is the system bounded-inputbounded-output stable?(2 pt) 2. Now, let a 3/3, b 1, and the input to the system be given by ππ n x[n] 2 cos23for all n.What is the output y[n] for all n?(3 pt)Hint: The following trigonometric table might be useful.θ0 π/6 π/4 π/3 π/2sin θ 0cos θ 1tan θ 012 32 33 22 221 321120 3-Solution 11) The transfer function of the system is given byb,1 az 1which has a pole at zp a. A causal LTI system is stable if and only if allthe poles of H(z) lie inside the unit circle. It follows that a 1.H(z) 2) The output of an LTI system to a sinusoidal input can be obtained byanalyzing the frequency response H(Ω) of the system. If an inputx[n] A cos(Ω0 n Θ)

Final Exam – Signals & SystemsPage 3is applied to the system for all times n, the output y[n] is given by y[n] H(Ω0 ) A cos Ω0 n Θ H(Ω0 )where H(Ω0 ) and H(Ω0 ) are respectively the magnitude and phaseresponse at frequency Ω0 .The given input has the frequency Ω0 π/2. Therefore, π bbH . 21 ja1 ae jπ/2 Evaluating for a 3/3 and b 1 yields the following magnitude andphase response, respectively: π 3 b H 221 a2 a ππ H( ) (b) (1 ja) 0 arctan .216It follows that the output y[n] for the given input x[n] is, for all times n,given by π π y[n] 3 cosn .22

Page 4Final Exam – Signals & SystemsProblem 28 pointsPart AThe continuous-time sinusoid 8π π x(t) 0.5 cos 2πt 3 cost73 is uniformly sampled with sampling time Ts seconds, resulting in thediscrete-time sequence x[n].1. For what range of Ts can x(t) be sampled without aliasing effects?(1 pt)2. Let Ts 1/4 seconds. What is the fundamental period N0 of theresulting periodic sequence x[n]?(2 pt)Part BLet x̄[n] be a different periodic sequence with fundamental period N̄0 6.The Discrete Fourier Series (DFS) coefficients c̄k of x̄[n] arec̄0 1,c̄1 0.5,c̄2 1,c̄3 0,c̄4 1,c̄5 0.5.1. Determine the constants A, B, C, and D, such that the sequence x̄[n]is given by(3 pt)x̄[n] A B cos (Cn) D cos (2Cn).2. You take the Discrete Fourier Transform of x̄[n] over two of its fundamental periods N̄0 :X̄[k] 2XN̄0 1x̄[n]WN̄kn0 ,with WN̄0 e jπ/N̄0 .n 0State the sequence X̄[k] as a function of N̄0 and the previously-definedDFS coefficients c̄0 , . . . , c̄5 .(2 pt)

Final Exam – Signals & SystemsPage 5Solution 2A.1) The allowed range is 0 Ts 3/8 seconds.A.2) The fundamental period of the sum of three periodic sequences isgiven by the least common multiple (lcm) of the individual fundamentalperiods. The sampled signal is ππ 2π 3 cosx[n] {z}0.5 cosn n .273{z} N0 1{z}N0 4N0 3The fundamental period of the signal is therefore N0 lcm(1, 4, 3) 12.B.1) A periodic signal x̄[n] with fundamental period N̄0 can be rewrittenas:N̄0 1Xx̄[n] c̄k ejkΩ0 n .(1)k 0Evaluating (1) with the given DFS coefficients c̄0 , . . . , c̄5 leads toπ2π4π5πx̄[n] c̄0 c̄1 ej 3 n c̄2 ej 3 n c̄3 ejπn c̄4 ej 3 n c̄5 ej 3 n π 2π c̄0 (c̄1 c̄5 ) cosn (c̄2 c̄4 ) cosn33 π 2π 1 cosn 2 cosn .335ππTherefore, A 1, B 1, C π3 , and D 2. Note that ej 3 n e j 3 nπ2π4πand ej 3 n (as well as ej 3 n and ej 3 n ) are complex conjugates. As you cansee, the constants A, B, and D can be directly inferred from the DFScoefficients.B.2) The Discrete Fourier Transform (DFT) is closely related to the DFS.If the DFT is applied to one fundamental period of the periodic sequencex̄[n] the following holds:X̄[k] N̄0 c̄k .In this case, the DFT is applied to two fundamental periods of x̄[n]. Therefore, the resulting sequence isX̄[k] 2N̄0 {c̄0 , 0, c̄1 , 0, c̄2 , 0, c̄3 , 0, c̄4 , 0, c̄5 , 0}.

Page 6Final Exam – Signals & SystemsProblem 37 pointsConsider the continuous-time, linear, time-invariant systemq̇(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 constantover intervals of Ts :x(t) x[k], kTs t (k 1)Ts .The system is to be discretized such thatq[k 1] AD q[k] BD x[k]y[k] CD q[k] DD x[k](2)whereq[k] q(kTs ),1. Discretize theDD for 0A 0y[k] y(kTs ).system by calculating the matrices AD , BD , CD , and(4 pt) 12,B ,C 1 0 ,D 0.02Now, consider the following MATLAB script, where the matrices Ad, Bd,Cd, and Dd represent AD , BD , CD , and DD respectively, and characterize adiscrete-time, linear, time-invariant system as defined in (2).Ad [1 0 1; -1 2 0; -1 0 3];Bd [2 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.2. What is the value of the variable r after executing the script? (2 pt)3. What conclusions can you draw from the value of r?(1 pt)

Final Exam – Signals & SystemsPage 7Solution 31) We will calculate this with the matrix exponential. First, we assemblethe matrix M 0 1 2A BM 0 0 2 0 00 0 0Notice that M is nilpotent, that is, M k 0 for k 3. The discrete-timematrices AD and BD are calculated as XM k TskAD BDM Ts e k!k 01 I M Ts M 2 Ts2 .2Solving the above yields 1 TsAD 0 1 2Ts Ts2BD .2TsThe matrices CD and DD are CD C 1 0DD D 0.2) The observability matrix is C1 0 0O CA 1 0 1 CA20 0 4and has rank 2. Therefore, r 2.3) For the system to be observable the observability matrix must be fullrank (3). This is not the case, therefore the system is not observable.

Page 8Final Exam – Signals & SystemsProblem 48 points1. The continuous-time sinusoid x(t) cos(2t) is uniformly sampledwith sampling time Ts 0.1 seconds to obtain the discrete-time sequence x[n]. Is x[n] a periodic sequence?(2 pt)2. Consider a stable, linear, time-invariant system, which is described bythe transfer functionH(z) 1.(z 0.25)(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.(2 pt)3. A signal w[n] is generated by drawing independent samples from aGaussian distribution with zero mean and variance 4. Calculate theexpected power of w[n] in the frequency band [0, π/2].(2 pt)4. The magnitude response H(ω) of a continuous-time filter is definedas follows:(1for 0 ω 2 rad/sec H(ω) 0.01 for ω 2 rad/sec.Using the Bilinear Transform and a sampling period of Ts 1, youdiscretize the continuous-time filter. Calculate the constants Mpass ,Mstop , and Ωc , such that the magnitude response H(Ω) of the resulting discrete-time filter is given by(2 pt)(Mpass for 0 Ω Ωc H(Ω) Mstop for Ωc Ω π.Hint: You may find the trigonometric table in Problem 1 useful.Solution 41) No, the resulting discrete-time sequence x[n] is not periodic.The resulting discrete-time sequence is x[n] x(nTs ) cos(2n/10). Thefundamental frequency Ω0 is therefore 1/5 radians/sample. A discrete-time

Final Exam – Signals & SystemsPage 9sinusoid is periodic if and only if Ω2π0 is a rational number, which is not thecase here.2) The system is stable and has poles at z 0.25, z 0.75 and z 1.5.Due to system stability, the region of convergence (ROC) must contain theunit circle. Furthermore, the ROC must not contain any poles. With theseconditions in mind, the ROC must be given byIm(z)1.50.75Re(z)3) The expected power in a the frequency band [Ω1 , Ω2 ] is given by1πΩ2ZSww (Ω)dΩ.Ω1In the case where w[n] is white noise, the power is equally spread acrossall frequencies, and its power spectral density function Sww (Ω) is equal tothe variance of the underlying probability density function for all Ω:Sww (Ω) XRww [k]e jΩk 4.k It follows that1πZπ/24 dΩ 2.04) The Bilinear Transform maps the imaginary axis in the s-plane to theunit 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 ωc is mapped to the discrete-

Page 10Final Exam – Signals & Systemstime cutoff frequency Ωc as follows:Ωc 2arctan (ωc Ts /2) 2arctan (1)π 24π .2The continuous- to discrete-time mapping does not alter the frequencyresponse of the filter. Thus Mpass 1 and Mstop 0.01.

Final Exam – Signals & SystemsPage 11Problem 55 pointsConsider a causal, discrete-time filter described by its frequency responseH(Ω) b0 b1 e jΩ .1. What is the impulse response h[n] of the filter?(1 pt)2. What are the coefficients b0 and b1 such that the following conditionsare satisfied?(4 pt) H(π) 1; A white noise input (zero mean, unit variance) results in a filteroutput y[n] with power spectral density Syy (Ω) 5 4 cos(Ω); b0 0.Solution 51. The impulse response h[n] of the filter can be read directly from the(FIR) filter coefficients. The impulse response ish[n] {b0 , b1 }.2. To solve for coefficients b0 and b1 , we begin by expressing the givenconditions in terms of b0 and b1 : With the condition that H(π) 1,we have that b0 b1 1.From the condition imposed on the power spectral density, and the factthat the white noise input x[n] has a power spectral density Sxx (Ω) 1for all Ω, it follows thatSyy (Ω) H(Ω) 2 Sxx (Ω)5 4 cos(Ω) H(Ω) 2 · 1 b20 b21 2b0 b1 cos(Ω).By matching coefficients, we have that b20 b21 5, and

Page 12Final Exam – Signals & Systems 2b0 b1 4.Solving the above three conditions and choosing the only solution withb0 0 yields b0 2, b1 1.

Final Exam – Signals & SystemsPage 13Problem 610 points1. The systems presented in this question are defined by difference equations, with input x[n] and output y[n]. Indicate whether these systemsare linear and/or time-invariant by writing True (T) or False (F) in theappropriate boxes. You do not need to justify your answers.(3 pt)Grading:Six correct answersFive correct answersFour correct answersThree or fewer correct answersSystem 1y[n] nx[n] x[n 1]System 2y[n] cos(x[n])System 3y[n] x[n] 13210pointspointspointpoints.Linear Time-InvariantTFLinear Time-InvariantFTLinear Time-InvariantFT

Page 14Final Exam – Signals & Systems2. The systems presented in this question are discrete-time, causal, linear,time-invariant systems. Indicate whether these systems are bounded-inputbounded-output (BIBO) stable by writing True (T) or False (F) in theappropriate boxes. You do not need to justify your answers.(3 pt)Grading:Three correct answers3 pointsTwo correct answers2 pointsOne or zero correct answers 0 points.System 1 is given in state-space form:"#" #0.1 50q[n 1] q[n] x[n]0 0.510hiy[n] 1 0 q[n]System 2 is described by its impulse response:System 3 is given by the pole-zero plot ofits transfer function H(z):Pole of H(z)Zero of H(z)10.5Re(z)0.5 0.5 1BIBO StableBIBO StableFIm(z) 0.5TF(0for n 0h[n] 4e n 1 otherwise 1BIBO Stable

Final Exam – Signals & Systems3. Consider the following MATLAB script:123Page 15(4 pt)Ntest 10;Nin 480;Ntr 120;4567891011121314for i 1: Ntest 1Omega pi *( i -1)/ Ntest ;xn cos ( Omega *(0: Nin -1));yn Experiment ( xn );xn xn ( Ntr 1: Nin );yn yn ( Ntr 1: Nin );Xi sum ( exp ( - j * Omega * (0: length ( xn ) -1) ) .* xn );Yi sum ( exp ( - j * Omega * (0: length ( yn ) -1) ) .* yn );H ( i ) Yi / Xi ;endWhere the Experiment function is defined as:12345 help ExperimentApplies the discrete - time input xn to a system andmeasures the output yn- yn has the same length as xn- The system begins at rest and is affected by noiseNow briefly answer the following questions: