Homework_9_solution

Homework_9_solution - 1 EE113 Homework 9 Problem 14.5...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 EE113 Homework 9 Problem 14.5 Consider the following sequences: sin( π n) 3 (a) x(n) = . n π −j π sin( 4 n) . (b) y (n) = e 4 · n π sin( π n) 6 (c) w(n) = e−j 4 n · . n (d) z (n) = x(n)y (n) from parts (a) and (b). Find the DTFTs of the following variations: (a) x(2n). (b) y (2n). (c) w(3n). (d) z (2n). (e) cos( π n)x(2n). 3 (f) (−1)n y (2n). (g) sin( π n) 4 y (n). n Solution: This problem set is about the down-sampling of the signals. (a) We know that the DTFT of x(n) is ω X (ejω ) = π · rect π 3 Thus, the DTFT of x1 (n) = x(2n) is 1 X1 (e ) = 2 1 jω X (ej ω−2πk 2 )= k=0 ω 2 π 3 π · rect 2 ω 2 + rect −π π 3 (b) We know that the DTFT of y (n) is ω π Y (ejω ) = e−j 4 · π · rect π 4 Thus, the DTFT of y1 (n) = y (2n) is Y1 (ejω ) = 1 2 1 Y (ej ω−2πk 2 π ) = e −j 4 · k=0 ω 2 π 4 π · rect 2 + rect ω 2 −π π 4 (c) We know that the DTFT of w(n) is W (ejω ) = π · rect ω+ π 4 π 6 Thus, the DTFT of w1 (n) = w(3n) is W1 (ejω ) = November 21, 2011 1 3 2 W (ej k=0 ω−2πk 3 )= π · rect 3 ω 3 + π 6 π 4 + rect ω 3 − 2π 3 π 6 + π 4 + rect ω 3 − 4π 3 π 6 + π 4 DRAFT 2 (d) The DTFT of z (n) = x(n)y (n) is Z (ejω ) = X (ejω ) ◦ Y (ejω ) ω = π · rect π = e −j 4 · π 3 π ◦ e−j 4 · π · rect π 7π · tri 2 12 ω 7π 12 − ω π 4 π tri 12 ω π 12 Thus, the DTFT of z1 (n) = z (2n) is Z1 (ejω ) = 1 2 1 Z (ej ω−2πk 2 ) k=0 π = e −j 4 · ω 2 7π 12 π 7π · tri 4 12 − π tri 12 ω 2 π 12 + 7π tri 12 ω 2 −π 7π 12 − π tri 12 ω 2 −π π 12 (e) The DTFT of x2 (n) = cos( π n)x(2n) = cos( π n)x1 (n) is 3 3 X2 (ejω ) = π π 1 X1 ej (ω− 3 ) + X1 ej (ω+ 3 ) 2 (f) The DTFT of y2 (n) = (−1)n y (2n) = ejnπ y1 (n) is Y2 (ejω ) = Y1 (ej (ω−π) ) (g) The DTFT of y3 (n) = sin( π n) 4 y (n) n π = ej 4 · y 2 (n) is π π Y3 (ejω ) = ej 4 · Y (ejω ) ◦ Y (ejω ) = e−j 4 · November 21, 2011 π2 · tri 4 ω π 2 DRAFT 3 Problem 14.10 Let x(n) = cos sin π (n − 1) π 8 n· 4 n−1 Find the DTFTs of the following sequences: (a) x(4n). (b) (−1)n−2 · x(3n). (c) n2 x(2n). (d) x(−n + 1). (e) x(−3n). (f) (−1)n−1 · x(n + 2) · sin π 2 (n − 1) . Solution: Note that X (ejω ) = π −j ( ω − π ) 4 e · rect 2 ω− π 4 + e−j (ω+ 4 ) · rect π π 8 ω+ π 4 π 8 (a) The DTFT of x1 (n) = x(4n) is X1 (ejω ) = 1 4 3 X (ej ω−2πk 4 ) k=0 (b) The DTFT of x2 (n) = (−1)n · x(3n) = ejnπ · x(3n) is X2 (ejω ) = 1 3 2 X (ej ω−π −2πk 3 ) k=0 (c) The DTFT of x3 (n) = n2 x(2n) is X3 (ejω ) = j d d2 DTFT(nx(2n)) = − 2 dω dω 1 2 1 X (ej ω−2πk 2 ) k=0 (d) The DTFT of x4 (n) = x(−n + 1) is X4 (ejω ) = e−jω X (e−jω ) (e) The DTFT of x5 (n) = x(−3n) is X5 (ejω ) = 1 3 (f) The DTFT of x6 (n) = (−1)n−1 · x(n + 3 − 1) · sin X6 (ejω ) = e−jω · November 21, 2011 2 X (ej −ω−2πk 3 ) k=0 π 2 (n − 1) is π π 1 j 3ω e X ej (ω−π− 2 ) − ej 3ω X ej (ω−π+ 2 ) 2j DRAFT 4 Problem 14.13 Let x(n) = sinc( π n). Plot the DTFT of 3 π π n− 6 3 y (n) = (−1)n x(−n + 2) cos Solution: The DTFT of x(n) is X (ejω ) = 3 · rect ω π 3 Notice that y (n) = (−1)n−2 x(−(n − 2)) cos π (n − 2) 6 Thus the DTFT of y (n) is Y (ejω ) = e−j 2ω · November 21, 2011 π π 1 X e−j (ω−π− 6 ) + X e−j (ω−π+ 6 ) 2 DRAFT 5 Problem 14.25 Consider the sequence x(n) = 0.5n u(n). Evaluate the following quantities without finding X (ejω ): (a) X (ej 0 ). (b) X (ejπ ). (c) 1 2π (d) 1 2π π −π π −π X (ejω )dω . |X (ejω )|2 dω . Solution: Note that ∞ jω x(n)e−jnω , X (e ) = x(n) = n=−∞ π 1 2π X (ejω )ejnω dω −π (a) We get ∞ ∞ j0 X (e ) = 0 .5 n = 2 x(n) = n=−∞ n=0 (b) We get ∞ X (ejπ ) = ∞ x(n) · (−1)n = n=−∞ (−0.5)n = n=0 2 3 (c) We get x(0) = 1 2π π X (ejω )dω = 1 −π (d) By the Parseval’s theorem, we get 1 2π November 21, 2011 ∞ π |X (ejω )|2 dω = −π ∞ |x(n)|2 = n=−∞ 0.25n = n=0 4 3 DRAFT 6 Problem 15.17 Given the four pole-zero distributions shown in Fig. 15.17, which ones correspond to low-pass, high-pass, band-pass, or band-stop filters? Solution: Refer to example 15.7 Geometric interpretation. (a) is a high-pass filter. (b) is a low-pass filter. (c) is a band-pass filter. And (d) is a band-stop filter. November 21, 2011 DRAFT 7 Problem 15.25 Find the frequencies that are present in the sequences x(3n), x( n ), and x2 (n) when x(n) = 2 cos(ωo n). Solution: The DTFT of x(n) is X (ejω ) = π [δ (ω − ωo ) + δ (ω + ωo )] (a) The DTFT of x1 (n) = x(3n) is π X1 (e ) = 3 2 jω = π 3 δ k=0 ω − 2πk − ωo 3 +δ ω − 2πk + ωo 3 2 [δ (ω − 2πk − 3ωo ) + δ (ω − 2πk + 3ωo )] k=0 Hence, the frequency of x1 (n) is 3ωo . (b) The frequency of x2 (n) = x(n/2) is ωo /2. (c) The frequency of x2 (n) = cos2 (ωo n) = [cos(2ωo n) + 1]/2 is 0 (the direct current) and 2ωo . November 21, 2011 DRAFT 8 Problem 15.28 Find a constant-coefficient difference equation to describe an LTI system whose impulse response sequence is given by 1 2 h(n) = n−1 sin π n u(n − 2) 4 What is the frequency response of the system? Solution: Notice that 1 · 2 1 2 n−2 h(n) = 1 · 2 1 2 n−2 = sin π π (n − 2) + u(n − 2) 4 2 cos π (n − 2) u(n − 2) 4 The z -transform of the system is H (z ) = 1 1 − 2 z −1 cos π z −2 4 · 2 1 − z −1 cos π + 1 z −2 4 4 One constant-coefficient difference equation can be y (n) = cos π 1 1 1 π y (n − 1) − y (n − 2) + x(n − 2) − cos x(n − 3) 4 4 2 2 4 The frequency response of the system is H (ejω ) = November 21, 2011 1 1 − 2 e−jω cos π e−j 2ω 4 · 2 1 − e−jω cos π + 1 e−j 2ω 4 4 DRAFT 9 Problem 15.29 A stable LTI ARMA system is described by the difference equation M N y (n) = ak y (n − k ) + k=1 bk x(n − k ) k=0 Let h(n) denote the impulse response sequence of the system. Find the difference equation that corresponds to the system with impulse response sequence h′ (n) = (−1)n h(n). Solution: The frequency response of the system is H (ejω ) = 1− N −jkω k=0 bk e M −jkω k=1 ak e The frequency response of the new system h (n) = (−1)n h(n) = ejnπ h(n) is ′ H (ejω ) = = 1− N −jk(ω −π ) k=0 bk e M −jk(ω −π ) k=1 ak e 1− N jkπ −jkω e k=0 bk e M jkπ e−jkω k=1 ak e One constant-coefficient difference equation can be M k=1 November 21, 2011 N ak ejkπ y (n − k ) + y (n) = bk ejkπ x(n − k ) k=0 DRAFT ...
View Full Document

This note was uploaded on 01/24/2012 for the course EE 113 taught by Professor Walker during the Fall '08 term at UCLA.

Ask a homework question - tutors are online