MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 0. 4ptOpt 6.003: Signals and s ystems-Fall 2003 Tuesday, October 14, 2003 Directions: The exam consists of 5 problems on pages 2 to 19 and work space on pages 20 and 21. Please make sure you have all the pages. Tables of Fourier series properties are supplied to you at the end of this booklet. Enter all your work and you answers directly in the spaces provided on the printed pages of this booklet. Please make sure your name is on all sheets. DO IT NOW!. All sketches must pe adequately labeled. Unless indicated otherwise, answers must be derived or explained, not just simply written down. This examination is closed book, but students may use one 8 1/2x 1l sheet of paper for reference. Calculators may not be used NAME: SOLUTIONS Check your section Section Time Rec. Instr 10-11 Prof. zue 2 l1-12 Prof. zue 1-2 Prof gr 411-12 Dr. Rohrs 12-1 Prof voldman Prof g Dr rohrs 1-12 Prof voldman Please leave the rest of this page blank for use by the graders [Problem No of points Score Grader 21 Total
1 2 3 4 5 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 0.4pt0pt 6.003: Signals and Systems—Fall 2003 Quiz 1 Tuesday, October 14, 2003 Directions: The exam consists of 5 problems on pages 2 to 19 and work space on pages 20 and 21. Please make sure you have all the pages. Tables of Fourier series properties are supplied to you at the end of this booklet. Enter all your work and your answers directly in the spaces provided on the printed pages of this booklet. Please make sure your name is on all sheets. DO IT NOW!. All sketches must be adequately labeled. Unless indicated otherwise, answers must be derived or explained, not just simply written down. This examination is closed book, but students may use one 8 1/2 × 11 sheet of paper for reference. Calculators may not be used. NAME: SOLUTIONS Check your section Section Time Rec. Instr. � 1 10-11 Prof. Zue � 2 11-12 Prof. Zue � 3 1- 2 Prof. Gray � 4 11-12 Dr. Rohrs � 5 12- 1 Prof. Voldman � 6 12- 1 Prof. Gray � 7 10-11 Dr. Rohrs � 8 11-12 Prof. Voldman Please leave the rest of this page blank for use by the graders: Grader 18 20 20 21 21 100 Problem No. of points Score Total
PROBLEM 1(18%) For the questions in this problem, no explanation is necessary. Consider the following three systems SYSTEM A: y(t)=r(t+2)sin(at +2), where w+0 SYSTEMB8=()(间+ SYSTEM C:=∑(2k+1-(l where and y are the input and output of each system. Circle YES or NO for each of the following questions for each of these three systems SYSTEM A SYSTEM B SYSTEM C Is the system linear YES NO YES NO YES Is the system time invariant YES YES NO YES Is the system causal? YES NO YES Is the system stable YES
� � PROBLEM 1 (18%) For the questions in this problem, no explanation is necessary. Consider the following three systems: SYSTEM A: y(t) = x(t + 2)sin(�t + 2), where � =→ 0 1 �n SYSTEM B: y[n] = − (x[n] + 1) 2 n SYSTEM C: y[n] = � x2 [k + 1] − x[k] � k=1 where x and y are the input and output of each system. Circle YES or NO for each of the following questions for each of these three systems. SYSTEM A SYSTEM B SYSTEM C Is the system linear ? Is the system time invariant ? Is the system causal ? Is the system stable ? YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO YES NO 2
Fall 2003: Quiz 1 NAME: Work Page for Problem 1 Linearity System A: YES I(t)y1(t)=1(t+2)sin(at +2 2(t)g2(t)=2(t+2)sin(ut +2) r3(t)=ax1(t)+br2(t)→(t)=x3(t+2)sin(t+2) ac1(t+2)sin(at+2)+bc2(t+2)sin(at +2) ay1(t)+ by2(t) System B: NO if a[n]=0, then yn]≠0 System C: NO not linear due to the r k+1]term Time Invarience System A: NO the output has time varying gain System B: NO the output has time varying gain System C: NO the number of summed terms depends on n Causality System A: NO y(t) depends on (t+2) System B: YES yn depends only on the current value of n System C: NO yn] depends on the future value of an Stability System A: YES bounded az(t)→→ bounded y(t) System B: NO System C: NO ifl]=-1→yl=∑h=12,y→∞asn
� Fall 2003: Quiz 1 NAME: Work Page for Problem 1 Linearity System A: YES x1(t) ⇒ y1(t) = x1(t + 2)sin(wt + 2) x2(t) ⇒ y2(t) = x2(t + 2)sin(wt + 2) x3(t) = ax1(t) + bx2(t) ⇒ y3(t) = x3(t + 2)sin(wt + 2) = ax1(t + 2)sin(wt + 2) + bx2(t + 2)sin(wt + 2) = ay1(t) + by2(t) System B: NO if x[n] = 0, then y[n] = 0 → 2 System C: NO not linear due to the x [k + 1] term. Time Invarience System A: NO the output has time varying gain System B: NO the output has time varying gain System C: NO the number of summed terms depends on n Causality System A: NO y(t) depends on x(t + 2) System B: YES y[n] depends only on the current value of x[n] System C: NO y[n] depends on the future value of x[n] Stability System A: YES bounded x(t) −⇒ bounded y(t) 1 n System B: NO (−2 ) −⇒ √ as n −⇒ −√ System C: NO if x[n] = −1 −⇒ y[n] = n k=1 2, y −⇒ √ as n −⇒ √ 3
PROBLEM 2 (20%0) Consider a DT LTI system, H2 with a unit sample response h2/n= hn *hn+l], as shown below, where hn]=8n-8n-1. You may remember from one of the lectures that h(n can be viewed as the unit sample response of a DT lti system that acts as an edge detector The purpose of this problem is to develop an edge detector that is robust against additive noise h2[nl System H2 h2n dnI Part a. Assume that the input to the system, pIn] is as shown below, and there is no noise, i.e., d(n=0 and pIn]= n. Provide a labeled sketch of yIn, the output of the system 3-2-101 yn
−2 1 PROBLEM 2 (20%) Consider a DT LTI system, H2 with a unit sample response h2[n] = h[n]↔h[n+1], as shown below, where h[n] = �[n] − �[n − 1]. You may remember from one of the lectures that h[n] can be viewed as the unit sample response of a DT LTI system that acts as an edge detector. The purpose of this problem is to develop an edge detector that is robust against additive noise. h2[n] 1 1 System H2 x[n] + 0 p[n] h2[n] y[n] n −2 −1 1 2 d[n] −2 Part a. Assume that the input to the system, p[n] is as shown below, and there is no noise, i.e., d[n] = 0 and p[n] = x[n]. Provide a labeled sketch of y[n], the output of the system. 2 2 2 2 p[n] −4 −3 −2 −1 0 1 2 3 n −7 −6 −5 −4 −3 −1 2 3 4 5 6 7 y[n] n 2 −2 −2 2 4
Fall 2003: Quiz 1 NAME: Work Page for Problem 2 n]=a[n]* h2In USing DT convolution sum, yn is the sum of 3 copies of pIn shifted and scaled accordin to h2/nl pIn +1 2p pIn-11 The resulting gn is shown on page 4 5 Problem 2 continues on the following page
−2 −1 0 1 Fall 2003: Quiz 1 NAME: Work Page for Problem 2 x[n] = p[n], y[n] = x[n] ↔ h2[n] Using DT convolution sum, y[n] is the sum of 3 copies of p[n] shifted and scaled according to h2[n] −4 −3 −2 −1 0 1 2 3 p[n n 2 2 2 2 + 1] + −4 −3 2 3 −2p[n] n −4 −4 −4 −4 + −4 −3 −2 −1 0 1 2 3 p[n − 1] n 2 2 2 2 The resulting y[n] is shown on page 4. 5 Problem 2 continues on the following page
Part b. For the same input signal as Part a, now assume that the noise signal is d{n]=-6n+1 Provide a labeled sketch of the output y/n, i.e., the response to n]=pn +dn
−2 1 Part b. For the same input signal as Part a., now assume that the noise signal is d[n] = −�[n + 1]. Provide a labeled sketch of the output y[n], i.e., the response to x[n] = p[n] + d[n]. −7 −6 −5 −4 −3 −1 2 3 4 5 6 7 y[n] n 2 2 −1 −2 2 −3 6
Fall 2003: Quiz 1 NAME: Work Page for Problem 2 y[n]=an* ha[n (p{n+d{m])*h2l] (Pn*h2回])+(dn]*h2]) We have already found (p[n]* h2(n) in Part a. Now we need d(n]* h2n d(n]* hsIn Adding the above(dn]* h2[nd) to the answer in Part a, we get y[n for this part which is shown on page 6 7 Problem 2 continues on the following page
−1 −2 0 Fall 2003: Quiz 1 NAME: Work Page for Problem 2 y[n] = x[n] ↔ h2[n] = (p[n] + d[n]) ↔ h2[n] = (p[n] ↔ h2[n]) + (d[n] ↔ h2[n]) We have already found (p[n] ↔ h2[n]) in Part a. Now we need d[n] ↔ h2[n]. d[n] −2 0 1 n −1 −1 1 n d[n] ↔ hs[n] −1 2 −1 Adding the above (d[n] h2[n]) to the answer in Part a, we get y[n] for this part which is shown on page 6. ↔ 7 Problem 2 continues on the following page
Part c. In order to use system H2 as a part of an edge detector, we would like to add an LTI system Hs whose unit sample response, hsn is shown below. System Hs smoothes out effect of noise on rn. The overall system can be represented as below hs(nI System h hsIn h2In y Provide a labeled sketch of the overall output ys [n], when pIn] and d(n] are exactly the same in part b ys[n 2
−2 1 Part c. In order to use system H2 as a part of an edge detector, we would like to add an LTI system Hs whose unit sample response, hs[n] is shown below. System Hs smoothes out effect of noise on x[n]. The overall system can be represented as below: 2 hs[n] −2 −1 0 1 n 1 1 System Hs System H2 p[n] + x[n] hs[n] h2[n] ys[n] d[n] Provide a labeled sketch of the overall output ys[n], when p[n] and d[n] are exactly the same as in Part b. ys[n] −7 −6 −5 −4 −3 −1 2 3 4 5 6 7 n 2 1 −2 −2 2 2 −3 8
Fall 2003: Quiz 1 NAME: Work Page for Problem 2 sn= an]* hs[n* h2/n (x]*h2n])*h n=pn+d(n is as defined in Part b. Therefore, we already know(a[n]* h2n))from Part b n]* h2/n 2 Now we just need to convolve the above signal with hs n. We can do this convolution with nipping and sliding hs In hs(n-k 2 n-1 +1 The result is the desired ys [n] and is shown on page 8
−2 1 Fall 2003: Quiz 1 NAME: Work Page for Problem 2 ys[n] = x[n] ↔ ↔ hs[n] h2[n] = (x[n] ↔ h2[n]) ↔ hs[n] x[n] = p[n] + d[n] is as defined in Part b. Therefore, we already know (x[n] ↔ h2[n]) from Part b. −7 −6 −5 −4 −3 −1 2 3 4 5 6 7 x[n] ↔ h2[n] n 2 2 −1 −2 2 −3 Now we just need to convolve the above signal with hs[n]. We can do this convolution with flipping and sliding hs[n]. hs[n − k] 2 k 1 1 n − 1 n n + 1 The result is the desired ys[n] and is shown on page 8. 9
PROBLEM 3 (20%0) Consider the CT LTI system whose impulse response is given as h(t) h(t) y(t) The following two parts can be done independently. Part a. The input r(t), an impulse train starting at t= 2, is depicted below (1)(1)(1)(1 Provide a labeled sketch of the corresponding output y(t) y(t)
PROBLEM 3 (20%) Consider the CT LTI system whose impulse response is given as: h(t) x(t) h(t) y(t) t 1 −1 0 1 The following two parts can be done independently. Part a. The input x(t), an impulse train starting at t = 2, is depicted below: ����� x(t) (1) (1) (1) (1) t −1 0 1 2 3 4 5 Provide a labeled sketch of the corresponding output y(t). y(t) −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 t 2 1 10
