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### 20115ee113_1_Solution_HW3_revised

Course: EE 113, Fall 2011
School: UCLA
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2011 October Fall 18, 2011 EE113: Digital Signal Processing Prof. Abeer Alwan TA: Jianshu Chen, and Sen Li Solution to HW 3 1. (Problem 5.1(part a,b)) Solution: 1 (a) The system y (n) = 3 y (n 1) + x(n 1) is relaxed, which implies that for input x(n) = (n), the output y (n) = 0, n &lt; 0. 1 y (0) = y (1) + (1) = 0 3 1 y (1) = y (0) + (0) = 1 3 1 1 y (2) = y (1) + (1) = 3 3 . . . 1 y (n) = y (n 1) +...

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2011 October Fall 18, 2011 EE113: Digital Signal Processing Prof. Abeer Alwan TA: Jianshu Chen, and Sen Li Solution to HW 3 1. (Problem 5.1(part a,b)) Solution: 1 (a) The system y (n) = 3 y (n 1) + x(n 1) is relaxed, which implies that for input x(n) = (n), the output y (n) = 0, n < 0. 1 y (0) = y (1) + (1) = 0 3 1 y (1) = y (0) + (0) = 1 3 1 1 y (2) = y (1) + (1) = 3 3 . . . 1 y (n) = y (n 1) + (n 1) = 3 1 3 n1 we can see that systems impulse response is h(n) = 1 n1 , 3 n>0 n0 0, 1 (b) The system y (n) = 3 y (n 1) + x(n + 1), y (1) = 1 1 y (1) = 1 = 3 y (2) + (0) = y (2) = 0, then we can see that y (n) = 0, n < 1 1 y (0) = y (1) + (1) = 3 1 y (1) = y (0) + (2) = 3 1 3 1 3 1 y (2) = y (1) + (3) = 3 . . . 1 3 2 3 1 y (n) = y (n 1) + (n + 1) = 3 1 3 n+1 we can see that systems output is y (n) = ( 1 )n+1 , n 0. Thus, the impulse response of 3 the system is n+1 1 , n0 3 h(n) = 1, n = 1 0, n < 1 EE113 - Fall 2011 1 of 13 2. (Problem 5.8) Solution: MA(Moving Average): FIR LTI system is also called MA system. AR(Auto-Regressive): systems are described by dierence equations of the form M y (n) = ak y (n k ) + b0 x(n) k=1 ARMA: systems are described by dierence equations of the form M y (n) = N ak y (n k ) + k=1 bk x(n k ) k=1 (a) ARMA (b) ARMA (c) MA (d) ARMA 1 Note: for (d), we can write h(n) in a recursive form y (n) = 2 y (n 1) + 1 x(n 2) 2 3. (Problem 5.9 (part a,b,c)) Solution: (a) FIR (b) IIR (c) IIR 4. (Problem 5.14) Solution: We have a LTI system with impulse response: 1 15 1 h(n) = (n) + (n 1) (n 2) 2 8 2 (a) parallel cascade: Let 1 h1 (n) = (n) 2 15 1 h2 (n) = (n 1) (n 2) 8 2 to make sure h(n) = h1 (n) + h2 (n). 1 (b) series cascade: we know that h(n) have 3 nonzero value [ 2 , 15 , 1 ], and h(n) = h1 (n) 8 2 h2 (n), thus we assume that h1 (n) and h2 (n) have 2 nonzero value at position 0 and 1, EE113 - Fall 2011 2 of 13 ie, the nonzero values are h1 (0), h2 (0), h1 (1), h2 (1) h(n) = h1 (n) h2 (n) 1 = h1 (k )h2 (n k ) k=0 h(0) = h1 (0)h2 (0) = 1 2 (1) h(1) = h1 (0)h2 (1) + h1 (1)h2 (0) = h(2) = h1 (1)h2 (1) = 15 8 (2) 1 2 (3) 1 we can choose h1 (0) = 2, h1 (1) = 2 , h2 (0) = 1 , h2 (1) = 1 to satisfy the equations 4 above. 1 Thus, we have h1 (n) = 2 (n) 2 (n 1) and h2 (n) = 1 (n) + (n 1) to make sure 4 h(n) = h1 (n) h2 (n) 5. (Problem 5.15) Solution: We can rewrite the system to be y (n) = 3n x(3n 1) cos 0, , n 5 n < 5 (a) linear = True Let y1 (n) = S{x1 (n)}, y2 (n) = S{x2 (n)} and x(n) = x1 (n) + x2 (n) y (n) = [x1 (3n 1) + x2 (3n 1)] cos = x1 (3n 1) cos n 3 n + x2(3n 1) cos n 3 3 = y1 (n) + y2 (n) (b) causal = False Let n = 1, we have y (1) = x(2) cos , which means the output of the system depends 3 on the future value of input. Thus, the system is not causal. (c) time-invariant = False Suppose input is x1 (n) = x(n n0 ), then y1 (n) = x(3(n n0 ) 1) cos (d) stable = True Assume x(n) Bx , and we know that | cos 3n 3n = y (n n0 ). | 1, then we have |y (n)| = |x(3n 1)|| cos n ||u(n)| 3 |x(3n 1)| Bx 6. (Problem 5.19) Solution: EE113 - Fall 2011 3 of 13 (a) We can express (n) = u(n) u(n 1), and the system is LTI, then we know that h(n) = S{ (n)} = S{u(n) u(n 1)} = S{u(n)} S{u(n 1)} = y (n) y (n 1) 1 2 = n n1 1 2 u(n) u(n 1) from above, we can see that h(n) = 0, n < 0, which indicates the system is CAUSAL. The length of h(n) is innite, which means the system is IIR. To determine the stability of the system, we do the following calculation: 1 2 h(n) = (n) n u(n 1), n 0 2 |h(n)| = 1 + n= n=1 1 4 =1+ n=1 n=0 n |2 n n 1 4 = = 1 2 | 4 < 3 Then we conclude that the system is BIBO STABLE. 1 (b) We can rewrite x(n) = 1 (n + 1) + (n) + 2 (n 1), then we can directly get the 3 output to be 1 1 y (n) = h(n + 1) + h(n) + h(n 1) 3 2 n+1 1 1n 1 = u(n + 1) u(n) + 3 2 2 n1 1 2 1 3 1 2 1 2 1 2 1 2 n 1 2 n1 n1 u(n) 1 2 u(n 1) n2 4 3 n+1 = 1 2 n n2 u(n 1) + 1 2 1 2 u(n + 1) + u(n 2) u(n) 1 2 u(n 1) u(n 2) 7. (Problem 6.1) Solution: To solve the problems here, we need to call upon the following properties of convolution. h(n) (n n0 ) = h(n n0 ) x(n) [h(n) w(n)] = [x(n) h(n)] w(n) (4) x(n) h(n) = h(n) x(n) EE113 - Fall 2011 4 of 13 Now, let us solve problem (a)(d). (a) x(n) h(n 1) = x(n) [h(n) (n 1)] = [x(n) h(n)] (n 1) = y (n) (n 1) = y (n 1) (b) x(n 1) h(n) = [x(n) (n 1)] h(n) = [ (n 1) x(n)] h(n) = (n 1)[ x(n) h(n)] = (n 1) y (n) = y (n) (n 1) = y (n 1) (c) x(n 1) h(n 2) = x(n 1) [h(n) (n 2)] = [x(n 1) h(n)] (n 2) = y (n 1) (n 2) = y (n 3) (d) x(n) h(n 3) = x(n) [h(n) (n 3)] = [x(n) h(n)] (n 3) = y (n) (n 3) = y (n 3) 8. (Problem 6.6) Solution: (a) Analytical Method Let x(n) and y (n) denote n+1 x(n) = 1 4 n1 y (n) = 1 2 u(n 2) u(n + 1) We are asked to evaluate the convolution of x(n) with y (n), which can be expressed as z (n) = x(n) y (n) + = x(k )y (n k ) k= + = k=2 EE113 - Fall 2011 1 4 k+1 1 2 nk1 u(n k + 1) 5 of 13 Figure 1: Graphical method to evaluate convolution in Problem 9 (Problem 6.6 in textbook). We notice that u(n k + 1) = 1 for k n + 1 and u(n k + 1) = 0 for k n + 2. Hence, we need to compute z (n) in two separate cases: i. If n + 1 < 2, i.e., n < 1 or equivalently, n 0, we have z (n) = 0. ii. If n + 1 2, i.e., n 1, then, we have n+1 k+1 1 2 nk1 1 2 n+1 n+1 1 2 k 1 2 n+3 n+1 1 2 k2 1 2 k z (n) = k=2 = = = = EE113 - Fall 2011 1 2 1 2 1 4 k=2 k=2 n+3 n1 k=0 n+2 1 1 2 n 6 of 13 In summary, can we write the convolution result as z (n) = 1 2 n+2 n 1 2 1 u(n 1) (b) Graphical Method We sketch out sequences x(k ), y (k ), and y (n k ) in Figure 1. We notice that there is overlap between x(k ) and y (n k ) only when n 1. Therefore, we have, for n 1, n+1 k+1 1 2 nk1 1 2 n+1 n+1 1 2 k 1 2 n+3 n+1 1 2 k2 1 2 k z (n) = k=2 = = = = 1 2 1 2 1 4 k=2 k=2 n+3 n1 k=0 n+2 1 1 2 n For n 0, the output z (n) = 0. (c) MATLAB Method Example of MATLAB code is shown below and the result is given in Figure 2. % Evaluate convolution by MATLAB clear all close all n = -2:10; x = (1/4).^(n+1).*(n>=2); y = (1/2).^(n-1).*(n >=-1); z = conv(x,y,same); %------------subplot(3,1,1); stem(n,x,k,Linewidth,2,Markersize,8); xlabel(\$n\$,Interpreter,Latex,Fontsize,16) ylabel(\$x(n)\$,Interpreter,Latex,Fontsize,16) axis([-2 15 0 max(x) ]) set(gca,Fontsize,12) %------------subplot(3,1,2); stem(n,y,k,Linewidth,2,Markersize,8); xlabel(\$n\$,Interpreter,Latex,Fontsize,16) ylabel(\$y(n)\$,Interpreter,Latex,Fontsize,16) axis([-2 15 0 max(y)]) set(gca,Fontsize,12) EE113 - Fall 2011 7 of 13 x(n ) 0.01 0 2 0 2 4 6 0 2 4 6 0 2 4 6 y (n ) 4.00 8 10 12 14 8 10 12 14 8 10 12 14 2.00 0 2 z (n ) n n 0.04 0.02 0 2 n Figure 2: MATLAB method to evaluate convolution in Problem 9 (Problem 6.6 in textbook). %------------subplot(3,1,3); stem(z,k,Linewidth,2,Markersize,8); xlabel(n) ylabel(z(n)) xlabel(\$n\$,Interpreter,Latex,Fontsize,16) ylabel(\$z(n)\$,Interpreter,Latex,Fontsize,16) set(gca,Fontsize,12) axis([-2 15 0 max(z)]) 9. (Problem 6.13) The result is shown in Figure 3 in the last page. 10. (Problem 6.15) Solution: Since x(n) is nonzero for 3 n 4, we can write the expression of z (n) as z (n) = x(n) y (n) = 4 x(k )y (n k ) = k= x(k )y (n k ) (5) k=3 Now we analyze (a)(d) one by one. (a) False. For n < 3, we have for any k that satises 3 k 4, n k < 3 k 3 (3) = 6 Since y (n) is nonzero only for 7 n 9, we thus have y (n k ) = 0, for n < 3 and 3 k 4 Then, for any n < 3, all the terms in the sum of (5) are zero. Hence, the claim that z (n) = 0 for n < 3 is true. However, the other part of the claim that z (n) = 0 for n > 12 is false. This is because for n > 12 and 3 k 4, we have n k > 12 k 12 4 = 8 EE113 - Fall 2011 8 of 13 In other words, n k 9, and y (n k ) can be nonzero for n k = 9. To see this, let us consider y (13), which can be expressed as 4 z (13) = x(k )y (13 k ) k=3 = x(3)y (16) + x(2)y (15) + x(1)y (14) + + x(3)y (10) + x(4)y (9) = x(4)y (9) where both x(4) and y (9) can be nonzero. Then, z (13) may not be zero. (b) False. z (0) must be zero. To see this, let us write z (0) as 4 4 z (0) = x(k )y (0 k ) = k=3 x(k )y (k ) (6) k=3 We already know that y (n) is nonzero only for 7 n 9, thus for 3 k 4, we have 4 k 3, which implies y (k ) = 0. Then, z (0) is zero. (c) False. Let us consider the example that x(n) = 1 for all 3 n 4, and y (n) = (1)n for 7 n 9. Then, z (5) can be expressed as 4 z (5) = x(k )y (5 k ) k=3 = x(3)y (8) + x(2)y (7) + + x(4)y (1) = x(3)y (8) + x(2)y (7) = 1 + (1) = 0 (d) True. Since (a)(c) are false, this claim is true. 11. (Problem 6.32) Solution: Let us rst evaluate the impulse response of the system. The system has to be relaxed to ensure it is LTI, i.e., y (n) should be zero before the input is added. If x(n) = (n), then y (1) = 0. The impulse response is the solution to the following dierence equation: 1 h(1) = y (1) = 0 h(n) = h(n 1) (n 2), 2 We can solve h(n) as 1 h(0) = h(1) (2) = 0 2 1 h(1) = h(0) (1) = 0 2 1 h(2) = h(1) (0) = 1 2 1 1 h(3) = h(2) (1) = (1) 2 2 1 12 h(4) = h(3) (2) = (1) 2 2 . . . h(n) = EE113 - Fall 2011 1 2 n2 u(n 2) 9 of 13 Next, we evaluate the response to the input in Figure 6.19 of the textbook. The input can be written in the following form: x(n) = (n + 2) + 2 (n + 1) 2 (n 1) + 2 (n 2) Then, the output becomes: y (n) = h(n) x(n) = h(n) [ (n + 2) + 2 (n + 1) 2 (n 1) + 2 (n 2)] = h(n) (n + 2) + 2h(n) (n + 1) 2h(n) (n 1) + 2h(n) (n 2) = h(n + 2) + 2h(n + 1) 2h(n 1) + 2h(n 2) = 1 2 n u(n) 2 1 2 n1 u(n 1) + 2 1 2 n3 u(n 3) 2 1 2 n4 u(n 4) 12. (Problem 6.34) Solution: We rst solve the implies response of the system. The system has to be relaxed to ensure it is LTI, i.e., y (1) = 0. The impulse response is the solution to the following dierence equation: 1 h(n) = h(n 1) + (n 1), 2 h(1) = y (1) = 0 Then, we can have 1 h(0) = h(1) + (1) = 0 2 1 h(1) = h(0) + (0) = 1 2 1 1 h(2) = h(1) + (1) = 2 2 12 1 h(3) = h(2) + (2) = 2 2 . . . Hence, we conclude that the impulse response is h(n) = 1 2 n1 u(n 1) Next, we evaluate the response to the two inputs respectively. EE113 - Fall 2011 10 of 13 (a) The response to the unit step function is y (n) = u(n) h(n) + = u(k )h(n k ) k= + = h(n k ) k=0 + = k=0 n1 = 1 2 k=0 0 1 =2 0 = EE113 - Fall 2011 1 2 nk1 u(n k 1) 1 2 nk1 if n 1 if n < 1 1 2 n+1 if n 1 if n < 1 1 2 n+1 u(n 1) 11 of 13 (b) The response to the one-side exponential sequence is y (n) = n 1 4 u(n) h(n) + = k= + = k=0 + = k=0 + = k=0 + = k=0 n1 = k=0 0 1 = 2 0 = EE113 - Fall 2011 1 2 k 1 4 u(k )h(n k ) 1 4 k 1 4 k 1 4 k 1 2 n+k1 h(n k ) 1 2 nk1 1 2 nk1 u(n k 1) u(n k 1) u(n k 1) n+k1 1 2 if n 1 if n < 1 n2 1 4 n1 if n 1 if n < 1 n2 1 4 n1 u(n 1) 12 of 13 Figure 3: Graphical method to evaluate convolution in Problem 10 (Problem 6.13 in textbook). EE113 - Fall 2011 13 of 13
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EE 131AProbabilityInstructor: Lara DolecekProblem Set 6Wednesday, November 9, 2011Due: Wednesday, November 16, 2011Reading: Chapter 5 of Probability, Statistics, and Random Processes by A. Leon-Garcia100 points total1. Coin tossing and two RVs. Pr
UCLA - EE - 132A
EE 131AProbabilityInstructor: Lara DolecekProblem Set 6Wednesday, November 9, 2011Due: Wednesday, November 16, 2011Reading: Chapter 5 of Probability, Statistics, and Random Processes by A. Leon-Garcia100 points total1. Coin tossing and two RVs. Pr
UCLA - EE - 132A
EE 131AProbabilityInstructor: Lara DolecekProblem Set 7Wednesday, November 16, 2011Due: Wednesday, November 23, 2011Reading: Chapter 5 of Probability, Statistics, and Random Processes by A. Leon-Garcia100 points total1. Expectation of a function o
UCLA - EE - 132A
EE 131AProbabilityInstructor: Lara DolecekProblem Set 7Wednesday, November 16, 2011Due: Wednesday, November 23, 2011Happy Thanksgiving!Reading: Chapters 5 and 6 of Probability, Statistics, and Random Processes by A. Leon-Garcia100 points total1.
UCLA - EE - 132A
EE 131AProbabilityInstructor: Lara DolecekProblem Set 8Wednesday, November 23, 2011Due: Monday, December 5, 2011Last Assignment!Please turn in this homework by 5 p.m. in the box in front of ProfessorsOce: Eng. IV, 56-147BReading: Chapters 5, 7 an
UCLA - EE - 132A
EE 131AProbabilityInstructor: Lara DolecekProblem Set 8Wednesday, November 23, 2011Due: Monday, December 5, 2011Last Assignment!Please turn in this homework by 5 p.m. in the box in front of ProfessorsOce: Eng. IV, 56-147BReading: Chapters 5, 7 an
Syracuse - PSY - 205
PSY205Chapter 2The Research Enterprise in PsychologyPsychology: The Scientific Approach- Theory: more established, general, established principle or law that has beentested- Hypothesis: Based off of theory. Educated guess.- Basic Assumtion: events
Syracuse - PSY - 205
PSY205Chapter 3Biological Bases of BehaviorThe Nervous System- Neurons: individual cells in the nervous system that receive, integrate, andtransmit information.- Glia: Glue- cells that provide structural support &amp; insulation for neurons.Neurons- S
Syracuse - PSY - 205
PSY205Ch 4Sensation &amp; PerceptionSensation and Perception: The Distinction- Sensation: Stimulation of sense organs- Perception: selection, organization, and interpretation of sensory inputThe Eye- Cornea: where light enters the eye- Lens: focuses t
Syracuse - PSY - 205
PSY205Ch. 5ConsciousnessThe Biological Clock- Biological rhythms- Circadian rhythms: 24-hour cycles- Light:o Retina Hypothalamus SCN Pineal glando Melatonin plays a key role in adjusting biological clockSCN- Suprachiasmatic Nucleus- Regulates c
Syracuse - PSY - 205
PSY205CH. 7Human MemoryMemory: Key Processes- Encoding- Storage- RetrievalEncoding: Getting Information into Memory- The role of attention- Selective attention = selection of inputo Cocktail party effectEncoding: Levels of Processing (Craik &amp; L
Syracuse - PSY - 205
PSY205Ch. 11DevelopmentAquatic Ape Theory- Elaine Moragno Orthodox evolution theory: Savannah Theory Problem: Humans have shed hair- Aquatic Ape Theoryo Evolution began when our earliest ancestors lived for a prolonged periodin a flooded, semi-aq
Syracuse - HST - 111
HST111 MID TERM REVIEW SHEETShort Answer questionsOf these 15 questions, 7 will appear on the actual midterm. You must answer 3 of them.Answers in a short paragraph only.Hundred Years WarJoan de ArcJacquerieAvignonHumanismJohannes GutenbergCosim
Syracuse - HST - 111
HST111: Assignment 2 DESCRIPTIONAnswer 1 of the following questions in an essay format:1. Why was the Thirty Years War fought? To what extent did politics determine theoutcome of the war?2. Henry of Navarre (Henry IV of France), Elizabeth I and Willia
Syracuse - HST - 111
HST111Assignment 1DescriptionAnswer ONE of the following questions:1. What were the economic, social and psychological effects of repeated attacks of plagueand disease?2. What were the intellectual and artistic hallmarks of the Renaissance?3. How a
Syracuse - PHI - 107
PHI1079/1/11ArgumentsSocratic Method in the EuthyphroThe value of PhilosophyArguments-Validity-Soundness-Deductive &amp; Inductive Arguments-PersuasivenessArguments-Argument in philosophy or in logic does not mean A disagreement or dispute (whichi
Syracuse - PHI - 363
PHI363JimandtheIndians(FromACritiqueofUtilitarianismbyBernardWilliams)JimfindshimselfinthecentralsquareofasmallSouthAmericantown.TiedupagainstthewallarearowoftwentyIndians,mostterrified,afewdefiant,infrontofthemseveralarmedmeninuniform.Aheavymaninaswe
Syracuse - PHI - 363
FINAL ESSAY QUESTIONPAPER DUE: MAY 7 2012 (Turn it in)Please write a paper of not more than 5 pages (or 1500 words) on ONE of the followingquestions:1. What do wealthy countries owe poorer countries? Answer with reference to foreign aid,immigration,
Syracuse - PHI - 363
On the Law of War and PeaceDe Jure Belli ac Pacisby Hugo GrotiusBook ICHAPTER 1: On War and RightOf War Definition of War Right, of Governors and of the governed, and of equals Rightas a Quality divided into Faculty and Fitness Faculty denoting Powe