Signals and Systems. Problem Set: From Continuous-Time to Discrete-Time

Transcription

1 Sigals ad Systems Problem Set: From Cotiuous-Time to Discrete-Time Updated: October 5, 2017

2 Problem Set Problem 1 - Liearity ad Time-Ivariace Cosider the followig systems ad determie whether liearity ad time-ivariace are give. a) y[] = u[] 2 b) y[] = au[] + b with costat parameters a, b c) y[] = u[] d) y[] = y[ 1] + u[] e) y[] = u[2( 1)] f) y[] = u[k]s[k] Problem 2 - Classificatio of systems A system G is described by the differece equatio y[] = u[]. Determie whether the system is a) memoryless, b) causal, c) liear, d) time-ivariat, e) stable. Problem - Graphical methods a) The system represeted by G i Fig. 1 is assumed to be time-ivariat. Whe the iputs to the system are u 1, u 2, ad u, the outputs of the system are y 1, y 2, ad y as show. Show that the system is ot liear. b) Fig. 2 displays the outputs {y 1 []} ad {y 2 []} of a ukow system G with give iputs {u 1 []} ad {u 2 []}. Based o the plots, determie whether the system could be liear ad time-ivariat. Problem 4 - Impulse respose Cosider a LTI system G with kow impulse respose {h[]} = {..., 0, 0, 2, 2,, 1, 0,... }. For the give iput {u[]} = {..., 0, 0, 2, 1, 1, 0, 0,... }, choose the correct output sequece from a)-c): a) {y A []} = {..., 0, 0, 0, 2, 4, 7, 6, 4, 1, 0... } b) {y B []} = {..., 0, 0, 0, 4, 6, 10, 7, 4, 1, 0... } c) {y C []} = {..., 0, 2, 4, 9, 8, 7, 2, 0, 0, 0... }. 2

3 u 1 [] G y 1 [] u 2 [] G y 2 [] u [] G y [] Figure 1 u 1 [] G y 1 [] u 2 [] G a) y 2 [] Figure 2

4 Problem 5 - Stability of a LTI system Cosider a discrete-time LTI system with impulse respose h[] give by h[] = α s[], where s[] is the uit step sequece. Uder what coditio is this system stable? Problem 6 - Covolutio Evaluate y = h u, where h ad u are show i Fig. h[] u[] Figure Problem 7 - Properties of covolutio Show that a) x δ = x b) {x[]} {δ[ 0 ]} = {x[ 0 ]} c) x s = { x[k]} d) {x[]} {s[ 0 ]} = { 0 x[k]} 4

5 Sample Solutios Problem 1 - Solutio a) We will show that the system is ot liear with a couter-example. Cosider the iput sequece u = u 1 + u 2. The output y = Gu is give by: y[] = G{u 1 + u 2 }[] = (u 1 [] + u 2 []) 2 = u 1 [] 2 + 2u 1 []u 2 [] + u 2 [] 2 y 1 [] + y 2 [] = G{u 1 }[] + G{u 2 }[] = u 1 [] 2 + u 2 [] 2. Time-ivariace ca be verified by provig that a time-shift i the iput sequece with some costat delay k causes the same delay i the output sequece. Let u 2 [] = u 1 [ k] ad {y 2 []} = {u 2 [] 2 }, {y 1 []} = {u 1 [] 2 }. The {y 2 []} = {u 2 [] 2 } = {u 1 [ k] 2 } = {y 1 [ k]} The system is ot liear, but the system is time-ivariat. b) Let y 1 [] = au 1 [] + b, y 2 [] = au 2 [] + b, u[] = α 1 u 1 [] + α 2 u 2 []. The But y[] = au[] + b = a(α 1 u 1 [] + α 2 u 2 []) + b. α 1 y 1 [] + α 2 y 2 [] = α 1 au 1 [] + α 1 b + α 2 au 2 [] + α 2 b = a(α 1 u 1 [] + α 2 u 2 []) + (α 2 + α 1 )b, thus the superpositio property does ot hold, ad the system is ot liear. Let u 2 [] = u 1 [ k] ad {y 2 []} = {au 2 [] + b}, {y 1 []} = {au 1 [] + b}. Time-ivariace ca be verified with {y 2 []} = {au 2 [] + b} = {au 1 [ k] + b} = {y 1 [ k]}. The system is ot liear, but the system is time-ivariat. c) Let y 1 [] = u 1 [], y 2 [] = u 2 [], u[] = α 1 u 1 [] + α 2 u 2 []. Superpositio of u = u 1 + u 2 does ot hold: y[] = G{u}[] = G{u 1 + u 2 }[] = u 1 [] + u 2 [] u 1 [] + u 2 [] = G{u 1 }[] + G{u 2 }[]. Let u 2 [] = u 1 [ k] ad {y 2 []} = { u 2 []}, {y 1 []} = { u 1 []}. Time-ivariace ca be verified with {y 2 []} = { u 2 []} = { u 1 [ k]} = {y 1 [ k]}. The system is ot liear, but the system is time-ivariat. 5

6 d) Let y[] = Gu[], y 1 [] = Gu 1 [], y 2 [] = Gu 2 [] ad u[] = α 1 u 1 [] + α 2 u 2 []. We wat to show that y[] = α 1 y 1 [] + α 2 y 2 [] for all by iductio. With the iitial coditio y[0] = y 1 [0] = y 2 [0] = 0 the above equatio is fullfilled for = 0. We ow show that if the above equatio is fullfilled for = k 1, the it also holds for = k. We have y[k] = ky[k 1] + u[k] = ky[k 1] + α 1 u 1 [k] + α 2 u 2 [k] = ky[k 1] + α 1 (y 1 [k] ky 1 [k 1]) + α 2 (y 2 [k] ky 2 [k 1]) = α 1 y 1 [k] + α 2 y 2 [k] + k(y[k 1] α 1 y 1 [k 1] α 2 y 2 [k 1]). }{{} =0 by the iductio hypothesis Sice both the basis ad the iductive step have bee performed, by mathematical iductio, we have show that y[] = α 1 y 1 [] + α 2 y 2 [] holds for all ad therefore the system is liear. Let u 2 [] = u 1 [ k] ad {y 2 []} = {y 2 [ 1] + u 2 []}, {y 1 []} = {y 1 [ 1] + u 1 []}. The {y 2 []} = {y 2 [ 1] + u 2 []} = {y 2 [ 1] + u 1 [ k]} = {y 2 [ 1] + y 1 [ k] ( k)y 1 [ k 1]} = {(y 2 [ 1] y 1 [ k 1]) + y 1 [ k] + ky 1 [ k 1]} {y 1 [ k]}. Therefore, the system is ot time-ivariat, but it is liear. e) Let y 1 [] = u 1 [2( 1)], y 2 [] = u 2 [2( 1)], u[] = α 1 u 1 [] + α 2 u 2 []. The y[] = u[2( 1)] = α 1 u 1 [2( 1)] + α 2 u 2 [2( 1)] = α 1 y 1 [] + α 2 y 2 []. Let u 2 [] = u 1 [ k] ad {y 2 []} = {u 2 [2( 1)]}, {y 1 []} = {u 1 [2( 1)]}. The y 2 [] = u 2 [2( 1)] = u 1 [2( 1) k] = u 1 [2( k 1) + k] u 1 [2( k 1)] = y 1 [ k]. The system is liear, but it is ot time-ivariat. f) Let y 1 [] = Gu 1 [], y 2 [] = Gu 2 [], u[] = α 1 u 1 [] + α 2 u 2 []. For < 0, y[] = 0 = α 1 y 1 [] + α 2 y 2 [] trivial. For 0, y[] = = u[k] = (α 1 u 1 [k] + α 2 u 2 [k]) α 1 u 1 [k] + α 2 u 2 [k] = α 1 y 1 [] + α 2 y 2 [] 6

7 Let u 2 [] = u 1 [ l] ad {y 2 []} = G{u 2 []}, {y 1 []} = G{u 1 []}. The y 2 [] = = = u 2 [k] = l m= l l m=0 u 1 [m] u 1 [m] + = y 1 [ l] + u 1 [k l] 1 m= l 1 m= l u 1 [m] u 1 [m] y 1 [ l], where we used k l = m. The system is liear, but it is ot time-ivariat. Problem 2 - Solutio a) Sice the output value at time depeds oly o the iput value at time, the system is memoryless. b) Sice the output does ot deped o future iput values, the system is causal. c) Let y 1 [] = u 1 [], y 2 [] = u 2 [], be the system outputs to arbitrary iputs u 1 [] ad u 2 []. The, with u [] := α 1 u 1 [] + α 2 u 2 [], we have: y [] = u [] = α 1 u 1 [] + α 2 u 2 [] = α 1 y 1 [] + α 2 y 2 []. Because the system satisfies the superpositio property, we coclude that the system is liear. d) Let u 2 [] := u 1 [ + 0 ] with 0 0 for all, the y 1 [ + 0 ] = ( + 0 )u 1 [ + 0 ] = ( + 0 )u 2 [] y 2 [] = u 2 []. Because a time-shift of 0 i the iput does ot correspod to a time-shift of 0 i the output, the system is ot time-ivariat. e) As we are dealig with a liear system, we ca use the defiitio of stability from the lecture otes. Let u[] = s[], which is clearly bouded. The y[] = s[]. Thus the output grows without boud despite a bouded iput. The system is therefore ot stable. 7

8 Problem - Solutio a) From Fig. 1, we see that {u []} = {u 1 []} + {u 2 [ 2]}. Thus, if G is liear, the G{u []} = G{u 1 []} + G{u 2 [ 2]} = {y 1 []} + {y 2 [ 2]}, which is show i Fig. 4. We see however that {y []} {y 1 []} + {y 2 [ 2]}. Hece, the system is ot liear. b) The iput sequeces ca be represeted as summatio of shifted uit impulses δ u 1 [] = δ[ 2] u 2 [] = δ[] + δ[ 1]. Sice u 1 [] is a time-shifted uit impulse ad assumig time-ivariace of the system, we coclude that y 1 [] = h[ 2] With m = 2, we see that h[m] = y 1 [m + 2]. Assumig liearity ad time ivariace, the system output for u 2 [] is {y 2 []} = G{δ[]} + G{δ[ 1]} = {h[]} + {h[ 1]} = {y 1 [ + 2]} + {y 1 [ + 1]} = {..., 0, 0, 2, 1, 1, 0, 0,... } + {..., 0, 0, 0, 2, 1, 1, 0,... } = {..., 0, 0, 2,, 2, 1, 0,... }, which is the sigal i the plot. Sice we were able to fid a liear time ivariat system represetatio, we coclude that the system might be both liear ad time-ivariat. y 1 [] + y 2 [ 2] = y 1 [] + y 2 [ 2] Figure 4 8

9 Problem 4 - Solutio The iput sequece ca be rewritte as {u[]} = {..., 0, 0, 2, 1, 1, 0, 0,... } = {..., 0, 0, 1, 0, 0, 0, 0,... } 2 + {..., 0, 0, 0, 1, 0, 0, 0,... } + {..., 0, 0, 0, 0, 1, 0, 0,... } = 2{δ[]} + {δ[ 1]} + {δ[ 2]} Because the system is LTI, the output sequece is {y[]} = 2{h[]} + {h[ 1]} + {h[ 2]} = {..., 0, 0, 2, 2,, 1, 0, 0, 0,... } 2 + {..., 0, 0, 0, 2, 2,, 1, 0, 0,... } + {..., 0, 0, 0, 0, 2, 2,, 1, 0,... } = {..., 0, 0, 4, 6, 10, 7, 4, 1, 0,... }. Problem 5 - Solutio From the lecture we kow that a LTI system is stable iff h[k] <. Therefore the followig iequality must hold: h[k] = If α < 1, we ca write α k = 1 1 α < α k s[k] = α k <. Therefore, the system is stable if α < 1 ad ustable if α 1. Problem 6 - Solutio Per Equatio (2.5) of lecture 2, we ca write the output y of a LTI system G with impulse respose h ad iput u as y = h u = u[k]{h[ k]}. 9

10 I other words, the output of a DT LTI system, give by the covolutio of its iput ad impulse respose, is othig but the additio of scaled ad shifted impulse resposes. As u is oly o-zero o the rage [0, ], we rewrite the output as y = u[k]{h[ k]} = u[0]{h[ 0]} + u[1]{h[ 1]} + u[2]{h[ 2]} + u[]{h[ ]}. Give that u[] = 1 for 0 y = u[k]{h[ k]} = {h[ 0]} = {..., 0, 1, 1, 1, 0, 0, 0, 0,... } + {h[ 1]} +{..., 0, 0, 1, 1, 1, 0, 0, 0,... } + {h[ 2]} +{..., 0, 0, 0, 1, 1, 1, 0, 0,... } + {h[ ]} +{..., 0, 0, 0, 0, 1, 1, 1, 0,... } = {..., 0, 1, 2,,, 2, 1, 0,... }. Note that the same result is achieved if the sigals switch roles (h scalig shifted versios of u), i.e.: y = h[k]{u[ k]}. Problem 7 - Solutio a) Usig the defiitio of the covolutio ad the fact that δ[ k] = 0 for k we have x δ = x[k]{δ[ k]} = x b) Similarly, we have {x[]} {δ[ 0 ]} = x[k]{δ[ 0 k]} = {x[ 0 ]} (1) c) Agai usig the defiitio of the covolutio ad the fact that s[ k] = 0 for k > we have x s = x[k]{s[ k]} = { x[k]} d) Adjustig the previous result to iclude the additioal delay 0 we get 0 {x[]} {s[ 0 ]} = x[k]{s[ 0 k]} = { x[k]} 10