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1 ECE-4: Sigals ad Systems Summer Solutios - Homework # PROBLEM A cotiuous time sigal is show i the figure. Carefully sketch each of the followig sigals: x(t) a) x(t-) b) x(-t) c) x(t+) d) x( - t/) e) x(t)*( (t+/) - (t-/)) f) (x(t) + x(-t))*u(-t) g) x(t/ - ) + x(t/ -) h) x(t-)*u(t-) - t - a) b) x(t-) x(t+) x(-t) t t - - t - c) x(t+) x(t+) t t - - d) x(t+) x(t/ + ) t t - - x(-t/ + ) t - Istructor: Daiel Llamocca

2 ECE-4: Sigals ad Systems Summer e) f) x(t) x(-t) x(t) - t - - t - - t - x(t)*( (t+/) - (t-/) x(t) + x(-t) u(-t) - t - - x() (t-/) - t - - t (x(t) + x(-t))*u(-t) - t h) x(t-) u(t-) x(t-)*u(t-) 4 5 t t 4 t - - Istructor: Daiel Llamocca

3 ECE-4: Sigals ad Systems Summer g) x(t-) x(t/-) - x(t-) t t x(t/-) t t 4 x(t/-) + x(t/ - ) t Istructor: Daiel Llamocca

4 ECE-4: Sigals ad Systems Summer PROBLEM The discrete-time sigals x [] ad x [] are show i the figure. Carefully sketch each of the followig sigals: x [] x [] a) x [] + x [-] b) x [] - u[-] c).5*x [] + (-) x [] d) x []*u[-] e) x [-] f) x [-] + u[+] g) x [-] [-] h) x [-] a) b) x [] x [] x [-] u[-] x [] + x [-] 4 x [] - u[-] Istructor: Daiel Llamocca

5 ECE-4: Sigals ad Systems Summer.5*x [] c) d) u[-] (-) x [] x [] x []*u[-] -.5*x [] + (-) x [] e) x [-] x [-] Istructor: Daiel Llamocca

6 ECE-4: Sigals ad Systems Summer f) x [-] x [-] u[+] u[+] x [-] + u[+] g) h) x [-] x [-] x [-] [-] / Istructor: Daiel Llamocca

7 ECE-4: Sigals ad Systems Summer PROBLEM Determie whether the followig sigals are periodic, ad for those which are, fid the fudametal period (T for cotiuous time sigals ad N for discrete-time sigals) ad the fudametal agular frequecies ( for cotiuous time sigals ad for discrete-time sigals). You must specify the uits of these quatities. a) x[] = cos((8/5)* ) b) x[] = si((7/5)* ) c) x(t) = si(t) + cos(t) d) x[] = si((/5)* )*si((/)* ) e) x(t) = si(t)u(t-) f) x(t) = si(t)u(t) + si(-t)*u(-t) a) N = m/(8/5 ) N = (5/4)m The, we choose m = 4 N = 5 samples, = /5 rads/sample. b) N = m/(7/5 ) N = (/7)m The, we choose m = 7 N = samples, = /5 rads/sample. c) For periodicity: x(t) = x(t + T) si(t) + cos(t) = si(t + T) + cos(t + T) We eed: T = k, ad T = r, where k,r are itegers The, we see that: T = k = ( /)r k = r The smallest umbers k,r that satisfy this coditio are k =, r =. The, the sigal x(t) is periodic with period T = secs, = rad per secod d) For periodicity: x[] = x[ + N] si( /5)si( /) = si( /5 + N/5)si( / + N/) We eed: N/5 = k, ad N/ = r, where k,r are itegers The, we see that: N = k = 6r 5k = r The smallest umbers k,r that satisfy this coditio are k =, r = 5. The, the sigal x[] is periodic with period N = samples, = /5 rads/sample. * Optioal: If we otice that: si( /5)si( /) = si( /5 + k)si( / + r) The, we eed: N/5 = k, ad N/ = r, where k,r are itegers The, k=, r=5, ad N = 5k = r = 5 samples, = /5 rads/sample. e) The sigal is zero oly for t. Therefore, x(t) is o-periodic. f) The sigal is o-periodic: x(t) x(t +T) x(t) x(t) T t T x(t+t) Istructor: Daiel Llamocca

8 ECE-4: Sigals ad Systems Summer PROBLEM 4 The systems that follow have iput x(t) or x[] ad output y(t) or y[] respectively. For each system, determie (ad justify) whether it is (i) memoryless, (ii) stable, (iii) causal, (iv) liear, ad (v) time ivariat. Recall that to disprove that a system has a certai property, all you eed is to come up with a couter-example. x(t) CT SYSTEM y(t) x[] DT SYSTEM y[] a) y(t) = cos(x(t)) b) y(t) = x(t/) c) y[] = *x[]u[] d) y[] = log ( x[] ) e) y[] = x[] + x[-] + x[+] f) y[] = x[] a) y(t) = cos(x(t)) It is memoryless: it oly depeds o the curret value of the sigal. Stability: If x(t) Mx <, t it should be that y(t) My <, t y(t) = cos(x(t) y(t) = cos(x(t) Therefore, the system is stable. Causality: It is causal, because it does ot deped o future values of the iput x(t). Liearity: If the iput to the system is ax A (t) + bx B (t), where a,b, are real umbers, the the output to the system should be ay A (t) + by B (t), where y A (t), y B (t) are the resposes of the system to x A (t) ad x B (t) respectively. If the iput to the system is ax A (t) + bx B (t), the: y(t) = cos(ax A (t) + bx B (t)) ay A (t) + by B (t) = acos(x A (t)) + bcos(x B (t)) We see that y(t) a(y A (t)) + b(y B (t)). Thus, the system is NOT liear. Time ivariace: We have a system y(t) = H(x(t)). The respose of the system to a shifted iput x(t-k) should be the same as if the output y(t) has bee shifted by k, i.e., y(t-k): x(t) S k x(t-k) H y (t) y(t) x(t) H S k y(t-k) these two should be the same Respose of system to shifted iput x(t-k): y (t) = cos(x(t-k)) Output y(t) shifted by k: y(t-k) = cos(x(t-k)) We see that y(t-k) = y (t). Thus, the system is time ivariat. Istructor: Daiel Llamocca

9 ECE-4: Sigals ad Systems Summer b) y(t) = x(t/) It is NOT memoryless: y() = x(), y() = x(.5). Stability: If x(t) Mx <, t it should be that y(t) My <, t y(t) = x(t/) x(t) Therefore, the system is stable. Causality: It is NOT causal, because it depeds o future values of the iput x(t): For example: y(-) = x(-.5), y(-4) = x(-). Liearity: If the iput to the system is ax A (t) + bx B (t), where a,b, are real umbers, the the output to the system should be ay A (t) + by B (t), where y A (t), y B (t) are the resposes of the system to x A (t) ad x B (t) respectively. If the iput to the system is ax A (t) + bx B (t), the: y(t) = ax A (t/) + bx B (t/)) ay A (t) + by B (t) = ax A (t/) + bx B (t/) We see that y(t) = ay A (t) + by B (t). Thus, the system is liear. Time ivariace: We have a system y(t) = H(x(t)) = x(t/). The respose of the system to a shifted iput x(t-k) should be the same as if the output y(t) has bee shifted by k, i.e., y(t-k). x(t) S k x(t-k) H y (t)= x(t/-k) y(t)=x(t/) x(t) H S k y(t-k)= x((t-k)/) Respose of system to shifted iput x(t-k): y (t) = x(t/-k) Output y(t) shifted by k: y(t-k) = x((t-k)/) We see that y(t-k) y (t). Thus, the system is NOT time ivariat. c) y[] = *x[]u[] It is memoryless: it oly depeds o the curret sample of the sigal. Stability: If x[] Mx <, it should be that y[] My <, y[] = x[]u[] x[] Therefore, the system is stable. Causality: It is causal, because it does ot deped o future samples of the iput x[]. Liearity: If the iput to the system is ax A [] + bx B [], where a,b, are real umbers, the the output to the system should be ay A [] + by B [], where y A [], y B [] are the resposes of the system to x A [] ad x B [] respectively. Istructor: Daiel Llamocca

10 ECE-4: Sigals ad Systems Summer If the iput to the system is ax A [] + bx B [], the: y[] = (ax A [] + bx B [])u[] y[] = (ax A []u[]) + (bx B [])u[]) ay A [] + by B [] = a**x A []u[] + b**x B [] We see that y[] = a(y A []) + b(y B []). Thus, the system is liear. Time ivariace: We have a system y[] = H(x[]). The respose of the system to a shifted iput x[-k] should be the same as if the output y[] has bee shifted by k, i.e., y[-k]: x[] S k x[-k] H y []= x[-k]u[] y[]=x[]u[] x[] H S k y[-k]=x[-k]u[-k] Respose of system to a shifted iput x[-k]: y [] = x[-k]u[] Output y[] shifted by k: y[-k] = x[-k]u[-k] We see that y[-k] y []. Thus, the system is NOT time ivariat. d) y[] = log ( x[] ) It is memoryless: it oly depeds o the curret sample of the sigal. Stability: If x[] Mx <, it should be that y[] My <, y[] = log x[] Whe x[] approaches, log( x[] ) approaches towards mius ifiity. Thus, the system is NOT stable. Causality: It is causal, because it does ot deped o future samples of the iput x[]. Liearity: If the iput to the system is ax A [] + bx B [], where a,b, are real umbers, the the output to the system should be ay A [] + by B [], where y A [], y B [] are the resposes of the system to x A [] ad x B [] respectively. If the iput to the system is ax A [] + bx B [], the: y[] = log ( ax A [] + bx B [] ) ay A [] + by B [] = alog( x A [] ) + blog ( x B [] ) We see that y[] a(y A []) + b(y B []). Thus, the system is NOT liear. Time ivariace: We have a system y[] = H(x[]). The respose of the system to a shifted iput x[-k] should be the same as if the output y[] has bee shifted by k, i.e., y[-k]: Respose of system to a shifted iput x[-k]: y [] = log ( x[-k] ) Output y[] shifted by k: y[-k] = log ( x[-k] ) We see that y[-k] = y []. Thus, the system is time ivariat. Istructor: Daiel Llamocca

11 ECE-4: Sigals ad Systems Summer e) y[] = x[] + x[-] + x[+] It is NOT memoryless: it depeds o previous ad future samples of the sigal. Stability: If x[] Mx <, it should be that y[] My <, y[] = x[] + x[-] + x[+] x[] + x[-] + x[+] If x[] Mx <, x[-k] Mx <, The: y[] Mx + Mx + Mx y[] Mx Therefore, the system is stable. Causality: It is NOT causal, because it depeds o future samples of the iput x[]. Liearity: If the iput to the system is ax A [] + bx B [], where a,b, are real umbers, the the output to the system should be ay A [] + by B [], where y A [], y B [] are the resposes of the system to x A [] ad x B [] respectively. If the iput to the system is ax A [] + bx B [], the: y[] = (ax A [] + bx B []) + (ax A [-] + bx B [-]) + (ax A [+] + bx B [+]) y[] = a(x A [] + x A [-] + x A [+]) + b(x B [] + x B [-] + x B [+]) y[] = a(y A []) + b(y B []). Thus, the system is liear. Time ivariace: We have a system y[] = H(x[]). The respose of the system to a shifted iput x[-k] should be the same as if the output y[] has bee shifted by k, i.e., y[-k]: Respose of system to a shifted iput x[-k]: y [] = x[-k] + x[-k-] + x[-k+] Output y[] shifted by k: y[-k] = x[-k] + x[-k-] + x[-k+] We see that y[-k] = y []. Thus, the system is time ivariat. f) y[] = x[] It is memoryless: it oly depeds o the curret sample of the sigal. Stability: If x[] Mx <, it should be that y[] My <, y[] = x[] x[] As grows, teds to ifiity. Therefore, the system is NOT stable. Causality: It is causal, because it does ot deped o future samples of the iput x[]. Liearity: If the iput to the system is ax A [] + bx B [], where a,b, are real umbers, the the output to the system should be ay A [] + by B [], where y A [], y B [] are the resposes of the system to x A [] ad x B [] respectively. If the iput to the system is ax A [] + bx B [], the: Istructor: Daiel Llamocca

12 ECE-4: Sigals ad Systems Summer y[] = (ax A [] + bx B []) y[] = (ax A[ ]) + (bx B []) ay A [] + by B [] = a* x A []u[] + b* x B [] We see that y[] = a(y A []) + b(y B []). Thus, the system is liear. Time ivariace: We have a system y[] = H(x[]). The respose of the system to a shifted iput x[-k] should be the same as if the output y[] has bee shifted by k, i.e., y[-k]: x[] S k x[-k] H y []= x[-k] y[]= x[] H x[] S k y[-k]= -k x[-k] Respose of system to a shifted iput x[-k]: y [] = x[-k] Output y[] shifted by k: y[-k] = -k x[-k] We see that y[-k] y []. Thus, the system is NOT time ivariat. PROBLEM 5 Usig MATLAB, plot (with the commad 'stem') the followig sigals for = -4 to 4. Attach your MATLAB code to the plots. a) x[] =.6*(.95) b) x[] = cos(( /)* + /) + si(( /6)* + /5) c) x[] = A*cos( + ) for: i. A =.5, = /45, = /5 ii. A =.5, = /, = / iii. A =.5, = /, = /5 clear all; close all ; clc; % Geeratig vector of samples: = -4:4; % [ ] % a) x[] =.6*(.95)^ B =.6; r =.95; x = B*(r.^); figure; stem (,x,'.r'); title ('r =.95'); xlabel (''); % b) x[] = cos(pi*/ + pi/) + si(pi*/6 + pi/5) x = cos(pi*/ + pi/) + si(pi*/6 + pi/5); figure; stem (,x,'.k'); title ('sum of siusoids'); xlabel (''); % c) x[] = Acos(omega* + phi): A =.5; omega = *pi/45; phi = pi/5; xa = A*cos(omega* + phi); A =.5; omega = pi/; phi = pi/; xb = A*cos(omega* + phi); A =.5; omega = pi/; phi = pi/5; xc = A*cos(omega* + phi); figure; stem (,xa,'.b'); title ('Siusoid. A=.5, omega=\pi/45, phi= \pi/5'); xlabel (''); figure; stem (,xb,'.b'); title ('Siusoid. A=.5, omega=\pi/, phi= \pi/'); xlabel (''); figure; stem (,xc,'.b'); title ('Siusoid. A=.5, omega = \pi/, phi = \pi/5'); xlabel (''); Istructor: Daiel Llamocca

13 ECE-4: Sigals ad Systems Summer 5 4 r =.95 a) sum of siusoids b) Siusoid. A =.5, omega = /45, phi = /5 4 c)i Siusoid. A =.5, omega = /, phi = /.5 c)ii Siusoid. A =.5, omega = /, phi = /5 c)iii Istructor: Daiel Llamocca

14 ECE-4: Sigals ad Systems Summer PROBLEM 6 Let x(t) be the cotiuous-time complex expoetial sigal: x(t) = exp(j t) with fudametal frequecy =, ad fudametal period T = /. The discrete-time sigal x[] was geerated by uiformly samplig (takig equally spaced samples) the sigal x(t) with a samplig period T S (i secods) x[] = x(ts) = exp(j T S ) a. Show that x[] is periodic if ad oly if T S /T is a ratioal umber. b. If = /8, N = 4, what is the miimum umber of cycles of the origial complex expoetial (also called evelope cycles) that are required for x[] to be periodic? c. Oce you obtaied the miimum umber of evelope cycles, what is the samplig period (i secods)? a. exp(j T S ) = cos( T S ) + jsi( T S ) The complex expoetial has the same periodicity as the cosie ad sie fuctios: x[] = x[ + N] exp(j T S ) = exp(j (+N)T S )= exp(j T S )exp(j NT S ) NT S = m, = /T NT S /T = m For NT S /T = m to hold, i.e., for m to be a iteger, T S /T has to be a ratioal umber. b. N = 4, = /8 T = 6 secs. NT S /T = m 4*(T S /6) = m T S /6 = m/4 Ay iteger m will make T S /6 a ratioal umber, so we pick m =. c. Ts/6 = /4 Ts = 6/4 =.4 secs. PROBLEM 7 Usig MATLAB, plot (with the 'stem' commad) the followig expoetially damped siusoidal sigal for two differet values of r (oe positive ad oe egative). x[] = Br si( + ) Note that < r < (otherwise there is o expoetial decay). Rage: = -5 to 5, Fixed parameters: = /4, B =. Pick ad r judiciously so that a clear dampig o the siusoid ca be see i the plot. Attach your MATLAB code to the plots. clear all; close all; clc; = -5:5; B = ; phi = pi/4; omega = pi/6; r =.95; % < r < : decayig expoetial x = B*(r.^).*si(omega* + phi); figure; stem (,x,'.b'); title ('r =.95'); xlabel (''); z = B*((-r).^).*si(omega* + phi); figure; stem (,z,'.b'); title ('r =.95'); xlabel (''); Istructor: Daiel Llamocca

15 ECE-4: Sigals ad Systems Summer 5 r =.95 5 r = PROBLEM 8 The output of a discrete-time system is related to its iput x[] as follows: y[] = a x[] + a x[+] + a x[-] + a x[-] + a 4 x[-4] where a, a, a, a, a 4 are real values. Let the operator S k deote a system that shifts the iput x[] by k samples to produce x[-k]. a. Formulate the operator H for the system relatig y[] to x[]. The develop a block diagram represetatio for H, usig (i) cascade implemetatio, ad (ii) parallel implemetatio. b. Demostrate that the system is BIBO stable for all a, a, a, a, a 4 (real values) c. Uder what coditio (if ay) of the values a, a, a, a, a 4 is the system causal? d. Demostrate that the system is liear ad time-ivariat. a. A system y[] = ax[-k] is represeted by H = as k. The, the system y[] = a x[] + a x[+] + a x[-] + a x[-] + a 4 x[-4] is represeted by H = a S + a S - + a S + a S + a 4 S 4. H = a + a S - + a S + a S + a 4 S 4. CASCADE PARALLEL x[] S S S S a S - a S - a a a a 4 x[] S a S y[] a S S a y[] S 4 a 4 Istructor: Daiel Llamocca

16 ECE-4: Sigals ad Systems Summer b. Stability: If x[] Mx <, it should be that y[] My <, y[] = a x[] + a x[+] + a x[-] + a x[-] + a 4 x[-4] y[] a x[] + a x[+] + a x[-] + a x[-] + a 4 x[-4] y[] a x[] + a x[+] + a x[-] + a x[-] + a 4 x[-4] If x[] Mx <, x[-k] Mx <, The: y[] a Mx + a Mx + a Mx + a Mx + a 4 Mx y[] ( a + a + a + a + a 4 )Mx Therefore, y[] is stable. c. The term x[+] makes the system ocausal. The, for the system to be causal, we require that a =. d. Liearity: If the iput to the system is ax A [] + bx B [], where a,b, are real umbers, the the output to the system should be ay A [] + by B [], where y A [], y B [] are the resposes of the system to x A [] ad x B [] respectively. If the iput to the system is ax A [] + bx B [], the: y[] = a (ax A [] + bx B []) + a (ax A [+] + bx B [+]) + a (ax A [-] + bx B [-]) + a (ax A [-] + bx B [-]) + a 4 (ax A [-4] + bx B [-4]) y[] = a(a x A [] + a x A [+] + a x A [-] + a x A [-] + a 4 x A [-4]) + b(a x B [] + a x B [+] + a x B [-] + a x B [-] + a 4 x B [-4]) y[] = a(y A []) + b(y B []). Thus, the system is liear. Time ivariace: We have a system y[] = H(x[]). The respose of the system to a shifted iput x[-k] should be the same as if the output y[] has bee shifted, i.e., y[-k]: x[] S k x[-k] H y [] y[] x[] H S k y[-k] these two should be the same Respose of system to shifted iput x[-k]: y [] = a x[-k] + a x[-k+] + a x[-k-] + a x[-k-] + a 4 x[-k-4] Output y[] shifted by k: y[-k] = a x[-k] + a x[-k+] + a x[-k-] + a x[-k-] + a 4 x[-k-4] We see that y[-k] = y []. Thus, the system is time-ivariat. Istructor: Daiel Llamocca

17 ECE-4: Sigals ad Systems Summer PROBLEM 9 Cosider a series itercoectio of system as show below. The iput-output relatioship of each system is give by the followig equatios: x[] SYSTEM SYSTEM SYSTEM y[] System : y[] = x[-] System : y[] = ax[-] + b x[] + cx[+] System : y[] = x[-] Here a, b, c are real umbers. a. Fid the iput-output relatioship for the overall itercoected system. b. Uder what coditio (if ay) of the values a, b, c, is the overall system liear ad time-ivariat? c. Uder what coditio (if ay) of the values a, b, c, is the overall system causal? a. Give the system: x[] SYSTEM w[] SYSTEM y[] = z[-] z[] = aw[-] + b w[] + cw[+] w[] = x[-] w[-] = x[-+], w[+] = x[--] (the shift is o ). z[] = ax[-+] + b x[-] + cx[--] y[] = z[-] = ax[+] + (-) b x[] + cx[-] b. Liearity: If the iput to the system is kx A [] + rx B [], where k,r, are real umbers, the the output to the system should be ky A [] + ry B [], where y A [], y B [] are the resposes of the system to x A [] ad x B [] respectively. If the iput to the system is kx A [] + rx B [], the: y[] = a(kx A [+] + rx B [+]) + b (kx A [] + rx B []) + c(kx A [-] + rx B [-]) y[] = k(ax A [+]+ b x A []+ cx A [-])+ r(ax B [+] + (-) b x B [] + cx B [-]) y[] = k(y A []) + r(y B []). Thus, the system is liear for all real a, b, c. Time ivariace: We have a system y[] = H(x[]). The respose of the system to a shifted iput x[-k] should be the same as if the output y[] has bee shifted, i.e., y[-k]: x[] S k x[-k] H y [] y[] x[] H S k y[-k] Respose of system to shifted iput x[-k]: y [] = ax[-k+] + (-) b x[-k] + cx[-k-] Output y[] shifted by k: y[-k] = ax[-k+] + (-+k) b x[-k] + cx[-k-] For liearity, we eed: y[-k] = y[]. Thus, (-) b =(-+k) b. This oly happes whe b=. c. Causality requires the term x[+] to go. Thus, we eed a =. z[] these two should be the same SYSTEM y[] Istructor: Daiel Llamocca

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