EE Midterm Test 1 - Solutions

Transcription

1 EE35 - Midterm Test - Solutios Total Poits: 5+ 6 Bous Poits Time: hour. ( poits) Cosider the parallel itercoectio of the two causal systems, System ad System 2, show below. System x[] + y[] System 2 The impulse respose of System is give to be h [] = 3 u[] ad the trasfer fuctio of System 2 is give to be H 2 (z) = 2z. (a) Is the overall system LTI? Is the overall system FIR or IIR? Briefly explai your reasoig. (b) Fid the trasfer fuctio of the overall system. Fid the poles of the overall system. (c) Fid the impulse respose of the overall system. Solutio: (a) Both System ad System 2 are LTI. Hece, their parallel itercoectio is also LTI. The trasfer fuctio of System 2 is give to be H 2 (z) =. It is also 2z give that both System ad System 2 are causal. Hece, the impulse respose of System 2 is h 2 [] = 2 u[]. The impulse respose of a parallel itercoectio of two systems is the sum of the impulse resposes. Hece, the impulse respose of the overall system is h[] = h [] + h 2 [] = 3 u[] + 2 u[]. () This impulse respose is obviously of ifiite legth. Hece, the overall system is IIR. (b) It is give that the impulse respose of System is h [] = 3 u[]. Recall that the z-trasform of the sigal α u[] is. Therefore, the trasfer fuctio of αz System (i.e., the z-trasform of h []) is H (z) = 3 z. (2)

2 The trasfer fuctio of a parallel itercoectio of two systems is the sum of the trasfer fuctios of the two systems. Hece, the trasfer fuctio of the overall system is H(z) = H (z) + H 2 (z) = 3 z + 2z. (3) The poles of the system are 3 = ad 2. 3 (c) As computed i part (a) above, the impulse respose of the overall system is give by (). 2. (2 poits) You are give four systems with impulse resposes h [], h 2 [], h 3 [], ad h [], respectively: h [] = δ[] h 2 [] = u[ ] h 3 [] = 2 u[] h [] = δ[ + ] + δ[ + 2]. (a) Sketch the sigals h [], h 2 [], h 3 [], ad h []. (b) Based o your sketches i part (a), decide which oes of the four systems are causal? (c) Based o your sketches i part (a), decide which oes of the four systems are BIBO stable? Briefly explai the reasoig for your aswers to parts (b) ad (c). Solutio: (a) The four sigals h [], h 2 [], h 3 [], ad h [] are show i Figure. (b) By defiitio, a system is said to be causal if its impulse respose is zero for all <. Hece, the systems with impulse resposes h [] ad h 3 [] are causal. The systems with impulse resposes h 2 [] ad h [] are ot causal. (c) By defiitio, a system is said to be BIBO stable if the sum of the magitudes of the impulse respose values is fiite, i.e., if = h[] <. Hece, the systems with impulse resposes h [] ad h [] are BIBO stable. The systems with impulse resposes h 2 [] ad h 3 [] are ot BIBO stable. 3. (2 poits) You are give a system with the impulse respose (.5) u[]. (a) Fid the differece equatio of the system (i.e., the equatio relatig the output sigal y ad the iput sigal x). (b) Fid the frequecy respose H(e jω ) of the system. (c) If the iput sigal is x[] = 3 cos( π )u[], fid the steady-state output sigal. Solutio: 2

3 h [] h 2 [] h 3 [] h [] Figure : Problem 2(a): h [], h 2 [], h 3 [], ad h []. (a) The impulse respose of the system is give to be h[] = (.5) u[]. Hece, the trasfer fuctio of the system is Therefore, Hece, H(z) =.5z = Y (z) X(z). () Y (z).5z Y (z) = X(z). (5) y[].5y[ ] = x[]. (6) (b) Substitutig z = e jω, the frequecy respose of the system is H(e jω ) =.5e jω. (7) (c) The iput sigal is give to be x[] = 3 cos( π )u[]. The frequecy of this iput sigal is ω = π. Evaluatig the frequecy respose at ω = π, we get H(e j π ) =. (8).5e j π Sice e jθ = cos(θ) + j si(θ), we get e j π = 2 j 2 =.77.77j. Hece, H(e j π ) =.5(.77.77j) = j =.98.65j =.3572e.5j (9) 3

4 Hece, the steady-state output sigal is y[] = 3(.3572) cos( π.5)u[] =.76 cos(π.5)u[]. (). (8 poits) Cosider the sigal x[] = δ[] + δ[ + ]. We wat to verify Parseval s theorem for this sigal. Recall that Parseval s theorem says that the eergy i the timedomai sigal is equal to the eergy i the frequecy-domai DTFT. I mathematical terms, Parseval s theorem says that if the DTFT of x[] is X(e jω ), the = x[]x [] = 2π X(e jω )X (e jω )dω. 2π Verify Parseval s theorem for the give iput sigal x[] usig the followig procedure: (a) Fid the DTFT X(e jω ) of the give sigal x[]. (b) Fid = x[]x []. (Hit: The easy way to do this is by first sketchig the sigal x[]x [].) (c) Fid 2π 2π X(e jω )X (e jω )dω usig the DTFT X(e jω ) that you foud above i part (a). (d) Check that your aswers to parts (b) ad (c) are equal. Solutio: (a) We are give the sigal x[] = δ[] + δ[ + ]. Takig the z-trasform, we have X(z) = + z. Hece, the DTFT of x[] is X(e jω ) = + e jω. () (b) The sigal x[] is real. Hece, x [] = x[]. Therefore, x[]x [] = x[]x[]. Sice x[] is at = ad = ad zero elsewhere, the sigal x[]x[] is also at = ad = ad zero elsewhere. See Figure 2. Therefore, = x[]x [] = 2. (2) (c) I part (a) above, we foud the DTFT to be X(e jω ) = + e jω. Hece, Therefore, X(e jω )X (e jω ) = ( + e jω )( + e jω ) = + e jω + e jω + = 2 + e jω + e jω. (3) 2π X(e jω )X (e jω )dω = 2π [2 + e jω + e jω ]dω. () 2π 2π It ca be easily see that 2π e jω dω = 2π e jω dω =. Hece, () simplifies to 2π X(e jω )X (e jω )dω = 2π 2π 2π 2dω = 2 2π 2π = 2. (5)

5 x[] x * [] x[]x * [] Figure 2: Problem (b): Fidig x[]x []. (d) I both parts (b) ad (c), the aswer was 2. Hece, the aswers to parts (b) ad (c) were ideed equal. Thus, Parseval s theorem has bee verified for the give sigal, i.e., the eergy i the time-domai sigal is equal to the eergy i the frequecy-domai DTFT. 5. (8 poits + 6 Bous Poits) Cosider a causal system with the differece equatio y[] =.2y[ ].36y[ 2] + x[]. (a) Fid the trasfer fuctio H(z) of the system. Fid the regio of covergece (ROC) of H(z). (b) By the method of guessig expoetials, it ca be show that the impulse respose of the system is of the form Fid the costats α, α 2, ad k. h[] = c α u[] + c 2 k α 2u[]. (c) (For extra credit - 6 Bous Poits) Fid the costats c ad c 2 i the impulse respose. Solutio: (a) Takig the z-trasform of both sides of the give differece equatio, we get Y (z) =.2z Y (z).36z 2 Y (z) + X(z). (6) Hece, the trasfer fuctio of the system is H(z) = Y (z) X(z) = +.2z +.36z 2. (7) 5

6 The poles of the system are the roots of the quadratic equatio +.2z +.36z 2 =. This quadratic equatio has a repeated root at.6 with multiplicity 2. The system is causal. Hece, the ROC of H(z) is the exterior of the circle with ceter at the origi ad radius.6. (b) Sice H(z) has a twice repeated root at -.6, the impulse respose is of the form Hece, α = α 2 =.6 ad k =. h[] = c (.6) u[] + c 2 (.6) u[]. (8) (c) To fid the costats c ad c 2, we eed to obtai two equatios by umerically iteratig the differece equatio for = ad =. We get From (8), we also have h[] =.2h[ ].36h[ 2] + x[] = h[] =.2h[].36h[ ] + x[] =.2 (9) h[] = c (.6) + c 2 (.6) = c h[] = c (.6) + c 2 (.6) =.6c.6c 2. (2) This gives us the two equatios Solvig these two equatios, we get c = c 2 =. c =.6c.6c 2 =.2. (2) 6