Solutions of Chapter 5 Part 1/2

Transcription

1 Page 1 of 8 Solutios of Chapter 5 Part 1/2 Problem Usig the defiitio, compute the -trasform of x[] ( 1) (u[] u[ 8]). Sketch the poles ad eros of X[] i the plae. Solutio: Accordig to the defiitio, X[] x[] 7 0 ( 1) 7 0 ( 1 ) 1 ( 1 )8 1 ( 1 ) 1 8 ; ROC > Sice the time-domai sigal is fiite i duratio, the regio of covergece should iclude the etire -plae, except possibly 0. X[] appears to have eight fiite eros ad oe fiite pole. The eight eros are the eight roots of uity, or e j2πk/8 for k 0,1,,7. The apparet pole is at 1. However, there is also a ero 1(k 4) that cacels this pole. Thus, there are actually o fiite poles ad oly seve fiite eros, e j2πk/8 for k 0,1,2,3,5,6,7. The sketch of the eros are plotted i figure S5.1-1, where the uit circle is also plotted for referece. 1 Im() -1 0 Re() 1-1 Figure 1: Fig5.1-1: Pole-ero plot for x[] ( 1) (u[] u[ 8]) Problem Fid the iverse -trasform of the followig system. Solutio: (a) X[] 4 ( 2)( 3) X[] 2 2 3

2 Page 2 of 8 x[] [2(2) (3) ] u[] (b) X[] 4 ( 2)( 3) 2/ /3 3 X[] x[] 2 [ 3 δ[] + (2) 1 ] 3 (3) u[] (c) (d) (e) (f) X[] X[] X[] X[] Multiply both sides by ad let. This yields (g) X[] Multiply both sides by ad let. This yields e 2 2 ( e 2 )( 2) 1 e X[] e 2 2 x[] [ (e) 2 (2) ] u[] ( 1) x[] δ[ 1] 2δ[ 2] + δ[ 3] ( 1)( 2)( 3) 5/ /2 3 X[] [ 5 x[] 2 7(2) + 9 ] 2 (3) u[] ( + 1)( 2) k ( 2) k + 0 k 3 X[] ( 2) 2 x[] [3( 1) 3(2) + 2(2) ] u[] ( 0.2)( 0.8) k ( 0.8) k k 1 X[] ( 0.8) 2 x[] [(0.2) (0.8) + 52 ] (0.8) u[]

3 Page 3 of 8 (h) X[] ( 2) We use pair 12c with A 1,B 2,a 0.5, γ 1. Therefore (i) r 2,β cos 1 (0.5) π 3,θ ta 1 ( 1 3 ) π 3 x[] 2(1) cos( π 3 + π ) u[] 2cos(π π 3 ) u[] X[] Multiply both sides by ad let. This yields Settig 1 o both sides yields ( ) 1 + A + B A A B 1.85 B 0.9 X[] 1 + ( 0.9) For the secod fractio o right side, we use pair 12c with A 1,B 0.9,a 0.3, γ 0.5. Therefore (j) r 10,β cos 1 ( ) 2.214,θ ta 1 ( ) x[] δ[] + 10(0.5) cos( ) u[] X[] Multiply both sides by ad let. This yields Settig 0 o both sides yields 2(3 23) ( 1)( ) A + B A A B 25 B 4 X[] 2 (2 4) For the secod fractio o right side, we use pair 12c with A 2,B 4,a 3, γ 5. Therefore (k) 17 r 2,β cos 1 ( 3 5 ) 0.927,θ ta 1 ( 1 4 ) 0.25 [ ] 17 x[] (5) cos( ) u[] X[] ( + 2)( ) A + B

4 Page 4 of 8 Multiply both sides by ad let. This yields Settig 0 o both sides yields A A B 25 B 6.83 X[] ( ) For the secod fractio o right side, we use pair 12c with A 1,B 6.83,a 2.5, γ 5. Therefore (l) X[] r 2,β cos 1 (0.5) π 3,θ ta 1 ( ) 3π 4 x[] [(2) + 2(5) cos( π3 3π4 ] ) u[] Multiply both sides by ad let. This yields Settig 0 o both sides yields ( ) ( 1)( 2) k k 2 ( 2) ( 2) k 1 k k k 2 1 X[] ( 2) ( 2) 3 x[] [1 3(2) 2 (2) + 14 ] ( 1)(2) u[] Problem Fid x[] by expedig as a power series i 1. X[] r ( r) 2 Solutio: Performig log divisio o X[] yields X[] r 2 2r + r 2 r + 2(r )2 + 3( r )3 + Therefore, x[0] 0,x[1] r,x[2] 2r 2,x[3] 3r 3,, ad which is exactly as pair 8 i Table 5.1. x[] r u[]

5 Page 5 of 8 Problem Fid the -trasform of the sigal illustrated i Fig. P Solve the problem i two ways as i Example 5.2d ad 5.4. Verify that the two aswers are equivalet. 4 x[] Figure 2: Fig5.2-2 Solutio: (a) Direct method: The sigal ca be writte as Therefore, x[] δ[ 1] + 2δ[ 2] + 3δ[ 3] + 4δ[ 4] + 3δ[ 5] + 2δ[ 6] + δ[ 7] X[] x[] 7 1 x[] (b) Usig the shift property: Write the sigal as Because u[] Also, because u[] x[] {u[] u[ 5]} + (8 ){u[ 5] u[ 9]} u[] 2u[ 5] + 8u[ 5] + u[ 9] 8u[ 9] u[] 2( 5)u[ 5] + ( 9)u[ 9] 2u[ 5] + u[ 9] 1, usig right-shift property, we have u[ m] 1 m 1 m ( 1) ( 1) 2, usig right-shift property, we have ( m)u[ m] 1 m ( 1) 2 m ( 1) 2

6 Page 6 of 8 Therefore, Z[] ( 1) ( 1) ( 1) 2 9 ( 1) 2 [ ( 1) + ( 1)] 1 7 ( 1) 2 [ ] 2 5 ( 1) + 9 ( 1) It is easy to verify that the two aswers are idetical. Problem (a) If x[] X[], show that x[k] X[] k0 1 (b) Use the result to derive pair 2 from pair 1 i Table 5.1. Solutio: (a) Sice k0 x[k] k0 Applyig time-covolutio property to the result yields Therefore x[k]u[ k] x[] u[] Z{x[] u[]} Z{x[]}Z{u[]} X[] 1 x[k] X[] k0 1 (b) Derive u[] 1 from δ[ k] k. Let x[] δ[] which yields X[] 0 1. Sice Applyig the result (a) yields u[] k0 u[] k0 δ[k] δ[k] X[] 1 1 Problem (a) Fid the output y[] of a LTID system specified by the equatio 2y[ + 2] 3y[ + 1] + y[] 4x[ + 2] 3x[ + 1] if the iitial coditios are y[ 1] 0,y[ 2] 1, ad the iput x[] (4) u[]. (b) Fid the ero-iput ad the ero-state compoets of the respose. (c) Fid the trasiet ad steady-state compoets of the respose.

7 Page 7 of 8 Solutio: (a) Write the system i delay form as 2y[] 3y[ 1] + y[ 2] 4x[] 3x[ 1] Note that y[] Y [] y[ 1] 1 Y [] y[ 2] 1 2Y [] + 1 y[ 1] + y[ 2] 1 2Y [] + 1 x[] X[] 0.25 x[ 1] 1 X[] Thus, the -trasform of the system is or Thus, 2Y [] 3 Y [] + 1 2Y [] ( )Y [] Y [] (3 2.75) ( )( 0.25) (3 2.75) 2( 0.5)( 1)( 0.25) 5/ /3 1 4/ Y [] From the iverse trasformatio, we have [ 5 y[] 2 (0.5) ] [ 1 3 (0.25) u[] (2) 4 ] 3 (4) u[] (b) From part (a), we have Y [] ( )Y [] 1 2 iitial coditio term Y [] iput term 2( 0.5)( 1) + (4 3) 2( 0.5)( 1)( 0.25) 2 2 (4 3) Y [] + 2( 0.5)( 1) 2( 0.5)( 1)( 0.25) ero iput respose ero state respose We exped both terms o the right-had side ito partial fractios as Y [] [ ] +[ ] ero iput respose ero state respose

8 Page 8 of 8 Thus, we have which agrees with (a). (c) y[] [ 1 2 (0.5) 1] u[] +[2(0.5) (0.25) ] u[] ero iput ero state y[] [ 5 2 (0.5) (0.25) ] u[] + 3 u[] trasiet compoets steady state compoet Problem Solve Prob if istead of iitial coditios y[ 1], y[ 2], you are give the auxiliary coditios y[0] 3/2 ad y[1] 35/4. Solutio: For iitial coditios y[0],y[1], we require the equatio i advaced form 2y[ + 2] 3y[ + 1] + y[] 4x[ + 2] 3x[ + 1] Note that y[] Y [] Thus, the -trasform of the system is y[ + 1] Y [] 3 2 y[ + 2] 2 Y [] x[] X[] 0.25 x[ + 1] X[] x[ + 2] 2 X[] ( 0.25) or 2[ 2 Y [] ] 3[Y [] 3 2 ] +Y [] 2( 0.25) ( )Y [] ( ) 0.25 Y [] ( 0.25)( 1)( 0.5) Y [] From the iverse trasformatio, we have [ 46 y[] (0.25) 25 ] 2 (0.5) u[]