8-90 Signals and Sysems Profs. Byron Yu and Pulki Grover Fall 07 Miderm Soluions Name: Andrew ID: Problem Score Max 0 8 4 6 5 0 6 0 7 8 9 0 6 Toal 00
Miderm Soluions. (0 poins) Deermine wheher he following saemens or equaions are rue or false by circling he answer. No explanaion is needed. (a) The signal x[n] = sin(n) is periodic. True False (b) To obain he graph of y() = x( + ), we firs scale in ime by and hen shif i o he lef by. True False (c) The sysem wih inpu x[n] and oupu y[n] = x[n ] is memoryless. (d) cos() δ( π) = (e) sin(ω) = e jω e jω j Soluion True True True False False False (a) False. This is a DT signal and scaling i wih ineger makes i aperiodic. (b) True. (c) False. The oupu is dependen on pas inpu. (d) False. y() = cos( π) = cos() (e) True.
Miderm Soluions. (8 poins) Deermine wheher he following signals are periodic and jusify. If he signal is periodic, find he fundamenal period. (a) = e cos(π + ) + 4 (b) = e jω +e jw (c) x[n] = sin( π n) cos(n + ) (d) x[n] = sin ( πn) + sin( πn) 4 Soluion: (a) No periodic due o e. (b) Periodic because = cos(ω), and he fundamenal period is π ω. (c) No periodic since cos(n + ) is no periodic. (d) Periodic because sin ( πn) and sin( π n) are boh periodic. The fundamenal 4 periods of sin ( πn) and sin( π n) are and 8, so he fundamenal period of x[n] 4 is 8.
4 Miderm Soluions. ( poins) Deermine he oal energy and average power of he following signals., (a) = 5, 5 0, oherwise (b) = cos() sin() (c) x[n] = a n u[ n], a > Soluion: (a) Energy: E = d = 5 = d + d = = 5 = () d + (5 ) d = = = 5 4d + (5 + 0)d = = = (4) + (5 + = 5 ) 5 = = (4() 4( )) + (5(5 ) + (5 ) 5(5 )) = 4 + (50 + 98 80) = 4 + 8 = 80 Power: P = lim T T T = T d.
Miderm Soluions 5 Since T is large enough, we have, P = lim T = lim T = 0 T E 9 T (b) Noice ha, = cos() sin() = sin() Energy E = d = = = = = = 8 = = 4 sin ()d 4 ( cos(4))d ( cos(4))d
6 Miderm Soluions Power: (c) Energy Since a >, a P = lim T = lim T = lim T = lim T = lim T = lim T = 8. T T 6T T = T T = T T = T d ( cos(4))d 8 ( cos(4))d 6T ( 4 sin(4)) T = T 6T (T sin(4t )) 8 sin(4t ) T E = = = = n= n= 0 n= n=0 x[n] a n u[ n] a n a n. <, hence we can use infinie GP sum. E = a = a a Power Since he energy is finie, power P = 0.
Miderm Soluions 7 4. (6 poins) Plo he even and odd componens of he following signal: Label he axes appropriaely. 0 0 Soluion: Even componen x e () = ( + x( ))
8 Miderm Soluions, x( ) 0 x e () 0 Even componen x o () = ( x( )), x( ) 0 x o () 0
Miderm Soluions 9 5. (0 poins) Consider he following coninuous-ime signal: Skech he following signals. Label he axes appropriaely. (a) y () = x( + ) 0
0 Miderm Soluions (b) y () = x( 4 ) 0 Soluion: (a) Shif lef by, scale by 0.5 and flip by x-axis 0 7 6 5 4 4 5 6 7 (b) Shif righ by 4, flip by y-axis and scale y by
Miderm Soluions 4 0 7 6 5 4
Miderm Soluions 6. (0 poins) Consider a coninuous-ime LTI sysem H. Given inpu below, le y() = H{} be he oupu of he sysem. Assume = 0 and y() = 0 for values of no shown. y() For he inpu x () given below, plo he oupu y () = H{x ()} of he sysem. Label he axes appropriaely. x () 4 4
Miderm Soluions 0 Soluion: By looking a he graphs, x () = x( ) Since he sysem is linear and ime-invarian, we have y () = H{x ()} = H{ x( )} = y() y( ) See he plo below. y () 4 4 4
4 Miderm Soluions 7. ( poins) The following sysems have or x[n] as inpu and y() or y[n] as oupu. For each sysem sae wheher he following properies hold and jusify: ( poin) ( poin) Causal Sable ( poins) Linear ( poins) Time invarian (a) y() = { x[n] if x[n] 0 (b) y[n] = 0 if x[n] > 0 Soluion: (a) Causal, no sable, linear, no ime invarian Causaliy: Since y() only depends on he curren value of, i is causal. Sabiliy: Le M. Then, y() = = y() = M = M as. Hence he sysem is no sable. Lineariy: Le x () and x () be wo inpus, and le = ax () + bx () be a hird inpu, where a, b are arbirary consans. Then, Hence he sysem is linear. Time invariance: y () = x () y () = x () y() = = (ax () + bx ()) = ax () + bx () = ay () + by (). y() = = y( τ) = ( τ)x( τ).
Miderm Soluions 5 Le ˆ = x( τ). ŷ() = ˆ Hence he sysem is no ime invarian. = x( τ) y( τ). (b) Causal, sable, no linear, ime invarian Causaliy: Since y[n] only depends on he curren value of x[n], i is causal. Sabiliy: Case : Le x[n] M 0. Then, y[n] = x[n] = y[n] = x[n] M Case : Le M > 0. Then, y[n] = = y() 0 { x[n] if x[n] 0 0 if x[n] > 0 Hence he sysem is sable. Lineariy: Counerexample: Le x [n] = 0 and x [n] = 0 be wo inpus and le x[n] = x [n] + x [n] be a hird inpu. This means ha x[n] = 0. y [n] = 0 y [n] = 0 y[n] = 0 y [n] + y [n] Hence he sysem is no linear.
6 Miderm Soluions 8. ( poins) Compue he convoluion y() = h(), where = u( + ) u( ) h() = u() u( ) Soluion: I is easier o solve his problem graphically. 0 0 Case : > no overlap beween wo figures. 0 Case: < < he overlap area is
Miderm Soluions 7 0 Case: 0 < < he overlap area is 0 Case4: < < 0 he overlap area is +
8 Miderm Soluions 0 Case5: < no overlap beween wo figures. 0 The complee signal, 0, < +, < 0 y() =, 0 <, < 0,
Miderm Soluions 9 9. ( poins) Evaluae he following convoluion: y[n] = e n u[n] (u[n] u[n 5]) Soluion: Le x[n] = u[n] u[n 5] and h[n] = e n u[n]. The convoluion sum is given by, y[n] = e k u[k](x[n k]) = k= e k x[n k] k=0 Observe ha x[n] = {, for all 0 n 4 0, oherwise Thus, x[n k] is also non-zero only in he range n 4 k n. And he summaion in convoluion is in he range 0 k <. So we look a he overlap of hese wo ranges of k o find he convoluion sum. Case : n < 0, i.e. no overlap beween n 4 k n and 0 k <. In his case, y[n] = e k x[n k] = 0 k=0 Case : n 0 bu n 4 < 0, i.e. 0 n < 4. In his case, here is some overlap beween n 4 k n and 0 k <. y[n] = = = e k x[n k] k=0 n e k x[n k] + k=0 n e k + k=0 = e (n+) e k=n+ e k 0 k=n+ e k x[n k]
0 Miderm Soluions Case : n 4 i.e. here is full overlap beween n 4 k n and 0 k <. Thus, he soluion is: y[n] = = e k x[n k] k=0 n k=n 4 e k = e n e5 e y[n] = 0, for all n < 0 e (n+), e for all 0 n < 4 e n e5 e, for all n 4
Miderm Soluions 0. (6 poins) Consider a discree-ime LTI sysem H which has he following inpuoupu relaionship: y[n] = 0.5x[n + ] + x[n] + 0.5x[n ] Draw he impulse response of he sysem, h[n]. Label he axes appropriaely. 0 Soluion: For impulse response, we subsiue x[n] = δ[n], which gives us, y[n] = 0.5 δ[n + ] + δ[n] + 0.5 δ[n ] h[n] n