Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science. Fall Problem Set 11 Solutions.

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1 Massachusetts Istitute of Techology Departmet of Electrical Egieerig ad Computer Sciece Issued: Thursday, December 8, : Discrete-Time Sigal Processig Fall 005 Problem Set 11 Solutios Problem 11.1 (OSB 11.3) Note: The aswers i the back of the book may ot be correct i your versio of the textbook. We factor X(e jω ) ito: As a first attempt, we take X(e jω ) 5 4 cos ω (1 1 ) (1 e jω 1 ) ejω X(e jω )X (e jω ) X(e jω ) 1 1 e jω x[] δ[] 1 δ[ 1] which does ot satisfy the costraits x[0] 0 ad x[1] > 0. We ca modify the above choice by cascadig it with a all-pass system, which will ot affect the magitude squared of the Fourier trasform. Therefore we let X(e jω ) (1 1 ) e jω e jω which does satisfy all of the costraits. x[] δ[ 1] 1 δ[ ] Aother choice that works is to take the secod factor i X(e jω ) ad cascade it with ( e jω) : ( X(e jω ) 1 1 ) ( e ejω jω ) 1 e jω e jω x[] 1 δ[ 1] δ[ ] Note that this secod choice uses the zero at z, the cojugate reciprocal of the zero at z 1 i the first choice. Cojugate reciprocal zeroes yield the same Fourier trasform magitude (up to a scalig).

2 Problem 11. The iverse DTFT of jimy (e jω )} is the odd part of y[], deoted by y o []. y o [] DTFT 1 [j3 siω + j si 3ω] [ 1 DTFT 1 ( 3e jω 3e jω + e j3ω e j3ω)] 1 (3δ[ + 1] 3δ[ 1] + δ[ + 3] δ[ 3]) Sice y[] is real ad causal, y[] y o []u[] + y[0]δ[] y[0]δ[] 3δ[ 1] δ[ 3] To determie y[0], we use the fact that Y (e jω ) 3, i.e., ωπ Y (e jω ) y[]( 1) ωπ y[0] y[0] 1 Therefore, y[] δ[] 3δ[ 1] δ[ 3] Problem 11.3 (OSB 11.5) I the frequecy domai, the Hilbert trasform is a 90 phase shifter: H(e jω j, 0 < ω < π ) j, π < ω < 0 To fid the Hilbert trasform of each sequece, we will take the Fourier trasform, multiply by H(e jω ), ad take the iverse Fourier trasform. (a) x r [] cos ω 0 X r (e jω ) πδ(ω ω 0 ) + πδ(ω + ω 0 ), X i (e jω ) H(e jω )X r (e jω ) π < ω π jπδ(ω ω 0 ) + jπδ(ω + ω 0 ), π < ω π x i [] siω 0

3 3 (b) x r [] siω 0 X r (e jω ) π j δ(ω ω 0) π j δ(ω + ω 0), X i (e jω ) πδ(ω ω 0 ) πδ(ω + ω 0 ), x i [] cos ω 0 π < ω π π < ω π (c) x r [] is the impulse respose of a ideal low-pass filter with cut-off frequecy ω c : x r [] si(ω c) π X r (e jω 1, ω < ω c ) 0, ω c < ω π j, 0 < ω < ω c X i (e jω ) j, ω c < ω < 0 0, ω c < ω π x i [] 1 π 0 ω c je jω dω 1 π 1 cos(ω c) π ωc 0 je jω dω Problem 11.4 The DTFT of y 1 [] is give by Y 1 (e jω ) X(e jω )e jθ(ω), π < ω < π The DTFT of y [] is give by Y (e jω X(e jω )e j(θ(ω) π/), 0 < ω < π ) X(e jω )e j(θ(ω)+π/), π < ω < 0 jx(e jω )e jθ(ω), 0 < ω < π jx(e jω )e jθ(ω), π < ω < 0

4 4 Sice w[] y 1 [] + jy [], W(e jω ) Y 1 (e jω ) + jy (e jω ) X(e jω )e jθ(ω) (1 + 1), 0 < ω < π X(e jω )e jθ(ω) (1 1), π < ω < 0 X(e jω )e jθ(ω), 0 < ω < π 0, π < ω < 0 Therefore, ad sice e jθ(ω) 1, W(e jω ) 0, π < ω < 0 W(e jω ) X(e jω ), 0 < ω < π Problem 11.5 We fid the Fourier trasform of h[] ad the take its complex logarithm, h[] δ[] + αδ[ 0 ] H(e jω ) 1 + αe jω 0 Ĥ(e jω ) log ( 1 + αe jω ) 0 The power series expasio for log(1 + x) with x < 1 is give by: log(1 + x) k1 ( 1) k+1xk Lettig x αe jω 0 ad checkig that x αe jω 0 α < 1 as assumed, we obtai: Ĥ(e jω ) k ( 1) k+1αk k e jωk 0 k1 The complex cepstrum ĥ[] is foud by takig the iverse Fourier trasform of Ĥ(ejω ) ad idetifyig e jωk 0 δ[ k 0 ]: ĥ[] is plotted i Figure : ĥ[] k1 ( 1) k+1αk k δ[ k 0]

5 5 h[] α o o α α 3 3 α o 4 o α o Figure : Complex cepstrum ĥ[] for a echo system. Problem 11.6 The sequece x[] beig miimum-phase meas that x[] is also causal, so that x[] 0, < 0. As stated i the problem, miimum phase implies that the complex cepstrum ˆx[] is causal, i.e. ˆx[] 0, < 0. Thus the lower boud o the sum i equatio (1.34) becomes k 0 (because of ˆx[k]), while the upper boud becomes k (because of x[ k]). ( ) k x[] ˆx[k]x[ k], > 0 k0 Isolatig the k term from the sum ad solvig for ˆx[], 1 ( ) k x[] ˆx[]x[0] + ˆx[k]x[ k] k0 ˆx[] x[] 1 x[0] ( ) k k0 ˆx[k] x[ k], > 0 x[0] The equatio above is a recursio formula for ˆx[], > 0, while we kow that ˆx[] 0 for < 0: 0, < 0 1 ˆx[] x[] x[0] ( ) k x[ k] ˆx[k], > 0 x[0] k0 However, the recursio caot determie ˆx[0], so we must fid it through some other meas. Recall the iitial-value theorem for a causal sequece x[] such that x[] 0 for < 0: x[0] lim z X(z)

6 6 Sice ˆx[] is also zero for < 0, But ˆX(z) log X(z), so we have ˆx[0] lim z ˆX(z) ˆx[0] lim log X(z) z ( ) log lim X(z) z log (x[0]) Thus we ca determie ˆx[0] usig oly x[0], so the computatio is causal. Now suppose that ˆx[] is kow for Usig the recursio, we are able to calculate the ext value ˆx[ 0 ] from the kow past values of ˆx[] ad from the values of x[] for 0 0. To start the recursio, we determie ˆx[0], which is the used to determie ˆx[1], ad so o. ˆx[] ca therefore be recursively computed. Furthermore, the computatio of ˆx[ 0 ] for ay 0 0 oly ivolves values of x[] for 0 0, so the recursio ca be implemeted i a causal maer. Problem 11.7 (a) Similar to Problem 11., the iverse DTFT of ReX(e jω )} is the eve part x e [] of x[]. x e [] DTFT 1 [1 + 3 cos ω + cos 3ω] δ[] + 1 (3δ[ + 1] + 3δ[ 1] + δ[ + 3] + δ[ 3]) Sice x[] is real ad causal, it ca be uiquely determied from its eve part x e []: x[] x e []u[] x e [0]δ[] x[] δ[] + 3δ[ 1] + δ[ 3] (b) Let X(e jω ) ad ˆX(e jω ) deote the Fourier trasforms of the sequece x[] ad its complex cepstrum ˆx[]. X(e jω ) ad ˆX(e jω ) are related by: ˆX(e jω ) log X(e jω ) + j arg [ X(e jω ) ] where arg [ X(e jω ) ] deotes the cotiuous uwrapped phase. If x 1 [] x[ ], the X 1 (e jω ) X(e jω ), ad ˆX 1 (e jω ) log X 1 (e jω ) + j arg [ X 1 (e jω ) ] log X(e jω ) + j arg [ X(e jω ) ] ˆX(e jω )

7 7 Therefore ˆx 1 [] ˆx[ ] also. Statemet 1 is true. If x[] is real, the the Fourier trasform magitude X(e jω ) is a eve fuctio of ω, while the uwrapped phase is a odd fuctio of ω. The real part of ˆX(e jω ), which is the logarithm of X(e jω ), must be a eve fuctio, while the imagiary part of ˆX(e jω ) is equal to arg [ X(e jω ) ] ad must be a odd fuctio. Therefore ˆX(e jω ) is cojugate symmetric ad the complex cepstrum ˆx[] is real. Statemet is also true.