7.17. Determine the z-transform and ROC for the following time signals: Sketch the ROC, poles, and zeros in the z-plane. X(z) = x[n]z n.
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1 Solutions to Additional Problems 7.7. Determine the -transform and ROC for the following time signals: Sketch the ROC, poles, and eros in the -plane. (a) x[n] δ[n k], k > 0 X() x[n] n n k, 0 Im k multiple Re Figure P7.7. (a) ROC (b) x[n] δ[n + k], k > 0 X() k, all Im k multiple Re Figure P7.7. (b) ROC
2 (c) x[n] u[n] X() n0 n, > Im Re Figure P7.7. (c) ROC (d) x[n] ( 4) n (u[n] u[n 5]) X() 4 n0 ( ) n 4 ( 4 ) 5 4 [ ) 5 ] 5 ( 4 4 ( 4 ), all 4 poles at 0, pole at 0 5 eros at 4 ejk π 5 k 0,,,, 4 Note ero for k 0 cancels pole at 4
3 Im 0.5 Re Figure P7.7. (d) ROC (e) x[n] ( 4) n u[ n] X() 0 n (4) n n0 ( ) n 4 4, < 4 Im 0.5 Re Figure P7.7. (e) ROC (f) x[n] n u[ n ] X() n n ( ) n ( ) n
4 Im, < Pole at Zero at 0 Re Figure P7.7. (f) ROC (g) x[n] ( ) n X() n ( ) n + n0 + ( ) n 5 6 ( )( ), < < Im Re Figure P7.7. (g) ROC (h) x[n] ( ) n u[n]+ ( 4) n u[ n ] 4
5 X() n0 ( ) n + n ( ) n 4 +, > and < 4 No region of convergence exists Given the following -transforms, determine whether the DTFT of the corresponding time signals exists without determining the time signal, and identify the DTFT in those cases where it exists: (a) X() 5 +, > ROC includes, DTFT exists. X(e jω ) 5 + e jω (b) X() 5 +, < ROC does not include,, DTFT does not exist. (c) X() ( )(+ ), < ROC does not include,, DTFT does not exist. (d) X() ( )(+ ), < < ROC includes, DTFT exists. X(e jω ) e jω ( e jω )(+e jω ) 7.9. The pole and ero locations of X() are depicted in the -plane on the following figures. In each case, identify all valid ROCs for X() and specify the characteristics of the time signal corresponding to each ROC. (a) Fig. P7.9 (a) X() C( ) ( + 4 )( ) There are 4 possible ROCs () > 4 x[n] is right-sided. 5
6 () < < 4 x[n] is two-sided. () < x[n] is left-sided. (b) Fig. P7.9 (b) X() C( 4 ) ( e j π 4 )( e j π 4 ) There are possible ROCs () > x[n] is right-sided. () < x[n] is two-sided. (c) Fig. P7.9 (c) X() ( )( + )( + 9 )C, < 6 x[n] is stable and left-sided Use the tables of -transforms and -transform properties given in Appendix E to determine the -transforms of the following signals: (a) x[n] ( ) n u[n] n u[ n ] a[n] ( ) n u[n] b[n] n u[ n ] x[n] a[n] b[n] A() B() X(), >, < X() A()B() ( )( ), < < (b) x[n] n (( n ( ) u[n] ) n ) 4 u[n ] a[n] ( ) n u[n] A(), > 6
7 b[n] ( ) n u[n] 4 c[n] b[n ] x[n] n[a[n] b[n]] B(), > 4 4 C() 4 X() d d A()B() X() ), > (c) x[n] u[ n] X() ) (, < (d) x[n] n sin( π n)u[ n] x[n] n sin( π n)u[ n] X() d ( ) d + [ ] + ( ) ( + ) + + ( + ) (e) x[n] n u[n] cos( π 6 n + π/)u[n] a[n] 9 n u[n] 9 A() b[n] cos( π 6 n + π )u[n] [ cos( π 6 n) cos(π) sin(π 6 n) sin(π) ] u[n] b[n] X() A()B() B() cos( π )( + cos( π 6 )) ( 9 cos( π sin( π )( sin( π 6 )) 6 )+ cos( π 6 ) )+ cos( π )( + cos( π 6 )) sin( π ) sin( π 6 ) cos( π 6 )+ 7.. Given the -transform pair x[n] to determine the -transform of the following signals: (a) y[n] x[n ] 6 with ROC < 4, use the -transform properties 7
8 y[n] x[n ] Y () X() 6 (b) y[n] (/) n x[n] y[n] ( )n x[n] Y () X() 4 (c) y[n] x[ n] x[n] y[n] x[ n] x[n] Y () X( )X() (d) y[n] nx[n] y[n] nx[n] Y () d d X() ( 6) (e) y[n] x[n +]+x[n ] y[n] x[n +]+x[n ] Y () ( + )X() + 6 (f) y[n] x[n] x[n ] y[n] x[n] x[n ] Y () X() X() ( 6) 7.. Given the -transform pair n n u[n] X(), use the -transform properties to determine the time-domain signals corresponding to the following transforms: (a) Y () X() Y () X() y[n] ( )n x[n] ( )n n n u[n] (b) Y () X( ) Y () X( ) y[n] x[ n] n n u[ n] 8
9 (c) Y () d d X() Y () d d X() ( d d X() ) y[n] (n )x[n ] (n ) n u[n ] (d) Y () X() (e) Y () [X()] Y () X() y[n] (x[n +] x[n ]) [ (n +) n+ u[n +] (n ) n u[n ] ] y[n] Y () X()X() y[n] x[n] x[n] n y[n] u[n] k k (n k) n k k0 n u[n] n [ k n nk + k 4] k0 7.. Prove the following -transform properties: (a) Time reversal x[n] X( ) y[n] x[ n] Y () x[ n] n let l n n l X( ) x[l]( ) l (b) Time shift x[n n o ] no X() y[n] x[n n o ] Y () x[n n o ] n let l n n o 9 n
10 l ( l no X() x[l] (l+no) x[l] l ) no (c) Multiplication by exponential sequence α n x[n] X( α ) y[n] α n x[n] Y () α n x[n] n n n X( α ) x[n]( α ) n (d) Convolution Let c[n] x[n] y[n] x[n] y[n] C() X()Y () (x[n] y[n]) n n n p ( p ( x[p] p ( x[p]y[n p] n ) n y[n p] (n p) ) } {{ } Y () ) x[p] p Y () } {{ } X() X()Y () p (e) Differentiation in the -domain. nx[n] X() d d X() x[n] n n 0
11 d d X() Therefore nx[n] Differentiate with respect to and multiply by. nx[n] n n d d X() 7.4. Use the method of partial fractions to obtain the time-domain signals corresponding to the following -transforms: + (a) X() 7 6, > ( )(+ ) x[n] is right-sided X() A + B + A + B 7 6 A B X() + + x[n] [( )n ( ] )n u[n] (b) X() ( )(+ ), < same as (a), but x[n] is left-sided x[n] [ ( )n +( )n ] u[ n ] (c) X() ( )(+ ), < < same as (a), but x[n] is two-sided x[n] ( )n u[ n ] ( )n u[n] (d) X() x[n] is two-sided +, < < X() + A + B +
12 A + B A B X() + + x[n] ( )n u[n] () n u[ n ] (e) X() 4 6, > 4 x[n] is right-sided X() A +4 + B 4 A + B 4 4A +4B X() [ 49 x[n] ( 4)n + 47 ] 4n u[n] (f) X() , < x[n] is left-sided X() x[n] δ[n ] + [( ] )n ( ) n u[ n ] (g) X() 4, > x[n] is right-sided X() ( ) x[n] δ[n +]+[( ) n ] u[n +] 7.5. Determine the time-domain signals corresponding to the following -transforms: (a) X() , > 0
13 x[n] δ[n]+δ[n 6]+4δ[n 8] (b) X() 0 k5 k k, > 0 x[n] 0 k5 δ[n k] k (c) X() ( + ), > 0 X() (δ[n]+δ[n ]) (δ[n]+δ[n ]) (δ[n]+δ[n ]) x[n] δ[n]+δ[n ]+δ[n ] + δ[n ] (d) X() , > 0 x[n] δ[n +6]+δ[n +]+δ[n]+δ[n ] + δ[n 4] 7.6. Use the following clues to determine the signal x[n] and rational -transform X(). (a) X() has poles at / and, x[],x[ ], and the ROC includes the point /4. Since the ROC includes the point /4, the ROC is < <. A X() + B + ( ) n x[n] A u[n] B( ) n u[ n ] ( ) x[] A A x[ ] B( ) B ( ) n x[n] u[n] ( ) n u[ n ] (b) x[n] is right-sided, X() has a single pole, and x[0],x[]/. x[n] c(p) n u[n] where c and p are unknown constants. x[0] c (p) 0
14 x[] c (p) p x[n] ( ) n u[n] (c) x[n] is two-sided, X() has one pole at /4, x[ ],x[ ] /4, and X() /. A X() + B 4 c ( ) n x[n] A u[n] B(c) n u[ n ] 4 x[ ] Bc x[ ] 4 Bc c B X() A + 4 A 5 4 x[n] 5 4 ( ) n u[n] + () n u[ n ] Determine the impulse response corresponding to the following transfer functions if (i) the system is stable, or (ii) the system is causal: (a) H() ( )(+ ) H() + + (i) h[n] is stable, ROC < <, ROC includes. h[n] () n u[ n ]+( )n u[n] (ii) h[n] is causal, ROC > h[n] [ n +( )n ] u[n] 4
15 (b) H() 5 6 H() + + (i) h[n] is stable, ROC <, ROC includes. h[n] [ () n ( ) n ] u[ n ] (ii) h[n] is causal, ROC > h[n] [ n +( ) n ] u[n] (c) H() H() 4 ( 4 ) (i) h[n] is stable, ROC > 4, ROC includes. h[n] 6n( 4 )n u[n] (ii) h[n] is causal, ROC > 4 h[n] 6n( 4 )n u[n] 7.8. Use a power series expansion to determine the time-domain signal corresponding to the following -transforms: (a) X() 4, > 4 X() ( 4 ) k k0 5
16 x[n] ( 4 )k δ[n k] k0 { ( ) n 4 n even and n 0 0 n odd (b) X() 4, < 4 X() 4 () k x[n] k0 (k+) (k+) k0 (k+) δ[n +(k + )] k0 (c) X() cos( ), > 0 Note: cos(α) k0 ( ) k (k)! (α)k X() x[n] k0 k0 k0 ( ) k (k)! ( ) k ( ) k (k)! 6k ( ) k δ[n 6k] (k)! (d) X() ln( + ), > 0 Note: ln( + α) ( ) k (α) k k k0 X() x[n] ( ) k ( ) k k ( ) k δ[n k] k k k 6
17 7.9. A causal system has input x[n] and output y[n]. Use the transfer function to determine the impulse response of this system. (a) x[n] δ[n]+ 4 δ[n ] 8 δ[n ], y[n] δ[n] 4δ[n ] X() Y () 4 H() Y () X() h[n] [5( )n ( 4 )n ] u[n] (b) x[n] ( ) n u[n], y[n] 4() n u[n] ( ) n u[n] X() + Y () ( )( ) H() Y () X() h[n] [0() n 7( ] )n u[n] 7.0. A system has impulse response h[n] ( ) n u[n]. Determine the input to the system if the output is given by H() (a) y[n] δ[n 4] Y () 4 X() Y () H() 4 5 x[n] δ[n 4] δ[n 5] 7
18 (b) y[n] u[n]+ ( ) n u[n] Y () + + X() Y () H() x[n] δ[n]+ 6 u[n]+4 ( )n u[n] 7.. Determine (i) transfer function and (ii) impulse response representations for the systems described by the following difference equations: (a) y[n] y[n ] x[n ] Y () ( ) X() H() Y () X() h[n] ( )n u[n ] (b) y[n] x[n] x[n ] + x[n 4] x[n 6] Y () ( + 4 6) X() H() Y () X() h[n] δ[n] δ[n ] + δ[n 4] δ[n 6] (c) y[n] y[n ] 5y[n ] x[n]+x[n ] ( Y () ) 5 (+ )X() H() Y () X() h[n] ( 4 5 ) [( 45 )n + 4 ] n(45 )n u[n] 8
19 7.. Determine (i) transfer function and (ii) difference-equation representations for the systems with the following impulse responses: (a) h[n] ( 4) n u[n ] h[n] H() ( )( ) n u[n ] Y () X() Taking the inverse -transform yields: y[n] 4 y[n ] x[n ] 4 (b) h[n] ( ) n u[n]+ ( ) n u[n ] ( ) n ( ) n h[n] u[n]+ u[n ] H() Y () X() Taking the inverse -transform yields: y[n] 5 6 y[n ] + 6 y[n ] x[n]+ x[n ] x[n ] (c) h[n] ( n ( ) u[n ] + ) n 4 [cos( π 6 n) sin( π 6 n)]u[n] Taking the -transform yields: y[n] ( H() + 4 cos( π 6 ) sin( π 6 ) cos( π 6 )+ 6 +( 5 ) 8 + ( 6 ) 4 + ( + ) 4 + ( 6 + ) 6 4 ) y[n ] + ( x[n]+ ( 5 8 ) x[n ] + ( 6 4 ) y[n ] 4y[n ] ) x[n ] + x[n ] 9
20 (d) h[n] δ[n] δ[n 5] H() 5 Taking the -transform yields: y[n] x[n] x[n 5] 7.. (a) Take the -transform of the state-update equation Eq. (.6) using the time-shift property Eq. (7.) to obtain q() (I A) bx() where q() Q () Q (). Q N () is the -transform of q[n]. Use this result to show that the transfer function of a LTI system is expressed in terms of the state-variable description as H() c(i A) b + D q[n +] Aq[n]+bx[n] q() A q()+bx() q()(i A) bx() q() (I A) bx() y[n] cq[n]+dx[n] Y () c q()+dx() Y () c(i A) bx()+dx() H() c(i A) b + D (b) Determine transfer function and difference-equation representations for the systems described by the following state-variable [ ] descriptions. [ ] Plot the pole and ero locations in the -plane. (i) A 0 0 [ ], b, c, D [] 0 H() c(i A) b + D
21 Imaginary Part Real Part Figure P7.. [ (b)-(i) Pole-Zero ] [ Plot] (ii) A [, b, c 4 0 ], D [0] H() Imaginary Part Real Part Figure P7.. (b)-(ii) Pole-Zero Plot
22 (iii) A [ ] [, b ] [, c 0 ], D [0] H() Imaginary Part Real Part Figure P7.. (b)-(iii) Pole-Zero Plot 7.4. Determine whether each of the systems described below are (i) causal and stable and (ii) minimum phase. (a) H() ero at: poles at: 5 4, 4 (i) Not all poles are inside, the system is not causal and stable. (ii) Not all poles and eros are inside, the system is not minimum phase. (b) y[n] y[n ] 4y[n ] x[n] x[n ] eros at: 0, poles at: ±
23 (i) Not all poles are inside, the system is not causal and stable. (ii) Not all poles and eros are inside, the system is not minimum phase. (c) y[n] y[n ] x[n] x[n ] H() ( ) eros at: 0, poles at: ± (i) Not all poles are inside, the system is not causal and stable. (ii) Not all poles and eros are inside, the system is not minimum phase For each system described below, identify the transfer function of the inverse system, and determine whether it can be both causal and stable. (a) H() H() ( 4) ( ) H inv () ( ) ( 4) poles at: 4 (double) For the inverse system, not all poles are inside, so the system is not causal and stable. (b) H() 8 00 H inv () poles at: (double) For the inverse system, all poles are inside, so the system can be causal and stable. (c) h[n] 0 ( ) n ( u[n] 9 ) n 4 u[n] ( ) H() ( + )( + 4 ) H inv () ( + )( + 4 ) ( )
24 poles at: 0, For the inverse system, not all poles are inside, so the system cannot be both causal and stable. (d) h[n] 4 ( ) n u[n ] 0 ( ) n u[n ] ( + H() ) ( )( ) H inv () ( )( ) ( + ) pole at: For the inverse system, all poles are inside, so the system can be both causal and stable. (e) y[n] 4y[n ] 6x[n] 7x[n ]+x[n ] H inv () X() Y () ( )( + ) poles at: 7 ± j For the inverse system, all poles are inside, so the system can be both causal and stable. (f) y[n] y[n ] x[n] H inv () pole at: 0 For the inverse system, all poles are inside, so the system can be both causal and stable A system described by a rational transfer function H() has the following properties: ) the system is causal; ) h[n] is real; ) H() has a pole at j/ and exactly one ero; 4) the inverse system has two eros; 5) n0 h[n] n 0;6)h[0]. By ) and ) 4
25 poles at ± j By 4) H inv H H has two poles. e±j π H() By 5) h[n] n n0 A( C ) cos( π )+ 4 A( C ) + 4 h[n] n H() n0 Since H() A( C ) +, 4 H() 0 implies C h[n] [ A h[0] A H() h[n] ( ) n cos( π n) A + 4 [( ) n cos( π n) ( ) n sin( π ] n) u[n] ( ) n sin( π ] n) u[n] (a) Is this system stable? The poles are inside, so the system is stable. (b) Is the inverse system both stable and causal? No, the inverse system has a pole at, which is not inside. (c) Find h[n]. h[n] [( ) n cos( π ( ) n n) sin( π ] n) u[n] (d) Find the transfer function of the inverse system. H inv () H() + 4 5
26 7.7. Use the graphical method to sketch the magnitude response of the systems having the following transfer functions: (a) H() H() H(e jω ) ( + j 7 8 )( j 7 8 ) (e jω + j 7 8 )(ejω j 7 8 ) Im 7 8 Re Figure P7.7. (a) Graphical method. 4.5 P7.7 (a) Magnitude Response 4.5 H(e jω ) Ω Figure P7.7. (a) Magnitude Response (b) H() + + 6
27 H() + + H(e jω ) ejω + e jω + e jω poles at: 0, (double) eros at: e ±j π Im Re Figure P7.7. (b) Graphical method. P7.7 (b) Magnitude Response H(e jω ) Ω Figure P7.7. (b) Magnitude Response (c) H() + +(8/0) cos( π 4 ) +(8/00) 7
28 H() + ( 9 0 ej 4 π )( 9 0 e j 4 π ) H(e jω ) e jω + e jω e jω + (8/0) cos( π 4 )ejω + (8/00) eros: poles: 9 0 e±j π 4 e jπ Im 4 Re Figure P7.7. (c) Graphical method. P7.7 (c) Magnitude Response 5 4 H(e jω ) Ω Figure P7.7. (c) Magnitude Response 7.8. Draw block-diagram implementations of the following systems as a cascade of second-order sec- 8
29 tions with real-valued coefficients: (a) H() ( 4 ej π 4 )( 4 e j π 4 )(+ 4 ej π 8 )(+ 4 e j π 8 ) ( ej π )( e j π )( 4 ej 7π 8 )( 4 e j 7π 8 ) H() H ()H () H () cos( π 4 ) + 6 cos( π ) + 4 H () + cos( π 8 ) + 6 cos( 7π 8 ) X() Y() cos( ) cos( ) 4 cos( 7 8 ) cos( ) Figure P7.8. (a) Block diagram. (b) H() (+ ) ( π ej )( π e j ) ( 8 )( π 8 ej )( π 8 ej )(+ 4 ) H() H ()H () H () H () cos( π ) X() Y() cos( ) Figure P7.8. (b) Block diagram. 4 9
30 7.9. Draw block diagram implementations of the following systems as a parallel combination of secondorder sections with real-valued coefficients: (a) h[n] ( n ( ) u[n]+ j ) n ( u[n]+ j ) n ( u[n]+ ) n u[n] H() + j + + j H() H ()+H () X() Y() 0.5 Figure P7.9. (a) Block diagram. (b) h[n] ( ) ej π n ( 4 u[n]+ ) 4 ej π n ( u[n]+ ) 4 e j π n ( u[n]+ ) e j π n 4 u[n] H() ej π ej π + 4 e j π + e j π H() H ()+H ()
31 X() Y() /6 Figure P7.9. (b) Block diagram Determine the transfer function of the system depicted in Fig. P7.40. X() H () H () Y() H () Figure P7.40. System diagram H () 4 H () H () H() H ()H ()+H ()H ()
32 7.4. Let x[n] u[n + 4]. (a) Determine the unilateral -transform of x[n]. x[n] u[n +4] X() u X() n0 n x[n] n n0 (b) Use the unilateral -transform time-shift property and the result of (a) to determine the unilateral -transform of w[n] x[n ]. w[n] x[n ] u W () x[ ] + x[ ] + X() W () Use the unilateral -transform to determine the forced response, the natural response, and the complete response of the systems described by the following difference equations with the given inputs and initial conditions. (a) y[n] y[n ] x[n], y[ ], x[n] ( )n u[n] X() + Y () ( Y ()+ ) X() Y () ( ) +X() Y () + }{{ X() }}{{} Y (n) () Y (f) () Natural Response Y (n) () y (n) [n] ( ) n u[n] Forced Response 6 4 Y (f) 5 () [ ( 6 y (f) [n] ) n + 4 ( ) n ] u[n] 5 5
33 Complete Response y[n] [ ( 6 ) n ( ) n ] u[n] (b) y[n] 9y[n ] x[n ], y[ ],y[ ] 0, x[n] u[n] X() Y () ( Y ()+ ) X() 9 Y () ( 9 ) 9 + X() Y () 9 + X() 9 }{{ 9 }}{{ } Y (n) () Y (f) () Natural Response y (n) [n] 6 Forced Response Y (n) () [( Y (f) () y (f) [n] Complete Response y[n] ) n ( [ ( + ) n ] u[n] + ) n ( ) n ] u[n] [ 9 4 ( ) n 4 ( ) n ] u[n]+ 6 [( ) n ( ) n ] u[n] (c) y[n] 4 y[n ] 8y[n ] x[n]+x[n ], y[ ],y[ ], x[n] n u[n] X() Y () ( Y ()+ ) ( Y ()+ ) X()+ X() 4 8 Y () ( 4 8 ) (+ )X() Y () + ( + )X() 8 ( )( ) ( }{{} )( + 4 ) }{{} Y (n) () Y (f) () Natural Response Y (n) ()
34 y (n) [n] 8 Forced Response Y (f) () y (f) [n] Complete Response y[n] [ ( ) n ( ) n ] u[n] [ ()n ( 5 [ ()n 0 + ) n + 4 ( 4 ) n ] u[n] ( ) n ( ) n ] u[n] 04 4 Solutions to Advanced Problems 7.4. Use the -transform of u[n] and the differentiation in the -domain property to derive the formula for evaluating the sum n a n assuming a <. n0 X() x[n] n n0 let x[n] a n u[n] d d X() nx[n] (n ) n d d X() d ( ) d a Evaluate at d ( ) d a n0 n n n0 n(n )x[n] (n ) n x[n] (n ) n n a n (n ) na n (n ) n a n na n n0 ( n a n d d n0 }{{} d d X() n0 a ) d d a ( a) + a ( a) 4 ( nx[n] (n ) a )
35 a a ( a) A continuous-time signal y(t) satisfies the first-order differential equation d y(t)+y(t) x(t) dt Use the approximation d dt y(t) [y(nt s) y((n )T s )] /T s to show that the sampled signal y[n] y(nt s ) satisfies the first-order difference equation y[n]+αy[n ] v[n] Express α and v[n] in terms of T s and x[n] x(nt s ). y[n] y[n ] +y[n] x[n] T ( ) s y[n] + y[n ] x[n] T s T s y[n] +T s y[n ] T s +T s x[n] α +T s T s v[n] x[n] +T s The autocorrelation signal for a real-valued causal signal x[n] is defined as r x [n] x[l]x[n + l] Assume the -transform of r x [n] converges for some values of. Find x[n] if R x () ( α )( α) where α <. l0 r x [n] x[n] x[ n] Let y[n] x[ n] r x [n] y[k]x[n k] let p k k k 5 x[ k]x[n k]
36 r x [n] x[p]x[n + p] p l x[l]x[n + l] x[l]x[n + l] l0 since x[l] 0, for l<0 R x () X( )X() ( ) α α Implies X() α x[n] α n u[n] The cross-correlation of two real-valued signals x[n] and y[n] is expressed as r xy (n) x[l]y[n + l] l (a) Express r xy [n] as a convolution of two sequences. r xy [n] x[n] y[ n], see previous problem (b) Find the -transform of r xy [n] as a function of the -transforms of x[n] and y[n]. R xy () X()Y ( ) A signal with rational -transform has even symmetry, that is, x[n] x[ n]. (a) What constraints must the poles of such a signal satisfy? x[n] x[ n] Implies X() X( ) X() X( ) 6
37 This means if there is a pole at o, there must also be a pole at o. Hence poles occur in reciprocal pairs. (b) Show that the -transform corresponds to a stable system if and only if n x[n] <. The reciprocal poles are ( ) o, o assume they take the following form: o r o e jθo o e jθo r o If o is inside, its -transform is right-sided and stable. For the pole at o, its corresponding -transform is either right-sided unstable, or left-sided stable. For convergence, the ROC must include the unit circle,, which means the -transforms are exponentially decaying as they approach, respectively. (c) Suppose Determine the ROC and find x[n]. X() (7/4) ( (/4) )( 4 ) For the system to be stable, the pole at 4 must be right-sided, and the pole at 4 must be left sided so their -transforms are exponentially decaying as they approach, respectively. This implies the ROC is 4 < < Consider a LTI system with transfer function H() a a, a < Here the pole and ero are a conjugate reciprocal pair. (a) Sketch a pole-ero plot for this system in the -plane. Im Let a a e j then a * a j e a a Re Figure P7.48. (a) Pole-Zero plot. 7
38 (b) Use the graphical method to show that the magnitude response of this system is unity for all frequency. A system with this characteristic is termed an all-pass system. H(e jω ) a e jω e jω a a e jω e jω a As shown below, H(e jω ) for all Ω. Im H (e j ) +a a a a Re + Im a H (e j ) +a a a Re + Figure P7.48. (b) Magnitude Response. 8
39 H (e j ) H (e j ) + Figure P7.48. (b) Magnitude Response. (c) Use the graphical method to sketch the phase response of this system for a. H(e jω ) ejω e jω arg { H(e jω ) } arg { } { ejω arg e jω } π + arg { e jω } { arg e jω } Im Im pole ero 0.5 Re Re Figure P7.48. (c) Pole/Zero graphical method. 9
40 ero Figure P7.48. (c) Zero phase response. pole Figure P7.48. (c) Pole phase response. H(e j ) Figure P7.48. (c) Phase response. (d) Use the result from (b) to prove that any system with a transfer function of the form H() P k a k a k a k < 40
41 corresponds to a stable and causal all-pass system. Since a k <, if the system is causal, then the system is stable. H(e jω ) H(e jω ) P k a k ejω e jω a k a e jω e jω a a e jω e jω a a pe jω e jω a p P H(e jω a k ) ejω e jω a k k H(e jω ) a e jω a e jω e jω a e jω a a pe jω e jω a p The system is all-pass. (e) Can a stable and causal all-pass system also be minimum phase? Explain. For a stable and causal all-pass system, a k < for all k. Using P: H() a a The ero is a Which implies a > This system cannot also be minimum phase Let H() F ()( a) and G() F ()( a) where 0 <a< is real. (a) Show that G(e jω ) H(e jω ). G() H() G(e jω ) H(e jω ) H(e jω ) 4 a a }{{} all-pass term a a
42 Since a a, see prob 7.48 (b) Show that g[n] h[n] v[n] where V () a a V () is thus the transfer function of an all-pass system (see Problem P7.48). G() H()V () V () a a a a g[n] h[n] v[n] (c) One definition of the average delay introduced by a causal system is the normalied first moment k0 d kv [k] k0 v [k] Calculate the average delay introduced by the all-pass system V (). V () a a a v[k] a k u[k ] aa k u[k] for k 0 v[0] a v [0] a for k v[k] a k a k+ v [k] a k + a k+ a (k )+(k+) a k (a + a ) kv [k] (a + a ) k(a ) k k0 k0 k0 +a4 a ( a ) v [k] a +(a + a ) (a ) k a + +a4 a a k a a, which also follows from Parseval s Theorem. k0 d kv [k] k0 v [k] 4
43 +a4 a ( a ) The transfer function of a LTI system is expressed as H() b M 0 k ( c k ) N k ( d k ) where d k <,k,,...,n, c k <,k,,...,m, and c M >. (a) Show that H() can be factored in the form H() H min ()H ap () where H min () is minimum phase and H ap () is all-pass (see Problem P7.48). H() ( c M ) b M 0 k ( c k ) N k ( d k ) b M 0 k ( c k )( c M ) N k ( d k ) }{{} H min() H min ()H ap () c M c M }{{ } H ap() (b) Find a minimum phase equalier with transfer function H eq () chosen so that H(e jω )H eq (e jω ) and determine the transfer function of the cascade H()H eq (). H eq () H min () so H()H eq () H ap () H()H eq () H ap () H ap () c M c M 7.5. A very useful structure for implementing nonrecursive systems is the so-called lattice stucture. The lattice is constructed as a cascade of two input, two output sections of the form depicted in Fig. P7.5 (a). An M th order lattice structure is depicted in Fig. P7.5 (b). (a) Find the transfer function of a second-order (M ) lattice having c and c 4. X() 0.5 A() Figure P7.5. (a) Lattice Diagram. B() 0.5 Y() 4
44 Y () 4 A() + B() A() X()( + ) B() X()( +) ( Y () 4 + ) 8 + X() H() Y () X() (b) We may determine the relationship between the transfer function and lattice structure by examining the effect of adding a section on the transfer function, as depicted in Fig. P7.5(c). Here we have defined H i () as the transfer function between the input and the output of the lower branch in the i th section and H i () as the transfer function between the input and the output of the upper branch in the i th section. Write the relationship between the transfer functions to the (i ) st and i th stages as [ Hi () H i () ] T() [ Hi () H i () where T() is a two-by-two matrix. Express T() in terms of c i and. ] H i () H i () + c i H i () H i () H i () c i + H i () [ ] [ ][ Hi () c i Hi () H i () c i H i () [ ] c i T() c i ] (c) Use induction to prove that H i () i H i ( ). i Hi () H i () H () + c H () c + H ( )+c H () H ( ) i k assume H k () k H k ( ) i k + Hk+ () Hk ()+c k+ H k () 44
45 H k+ () c k+ Hk ()+H k () substitute H k () k H k ( ) H k+ () (k+) H k ( )+c k+ H k () H k+ () (k+) c k+ H k ( )+H k () H k+ ( ) (k+) c k+ H k ()+H k ( ) H ( ) k+ () (k+) (k+) c k+ H k ()+H k ( ) H k+ () (k+) H k+ ( ) Therefore H i () i H i ( ) (d) Show that the coefficient of i in H i () is given by c i. [ Hi H i ] [ c i c i ][ Hi H i ] H i c i Hi + H i H i () (i ) c i The highest order of ( )inh i () is(i), and the coefficients of i is (c i ), since H i () doesnot contribute to i, therefore the coefficient of i is H i () is given by (c i ). (e) By combining the results of (b) - (d) we may derive an algorithm for finding the c i required by the lattice structure to implement an arbitrary order M nonrecursive transfer function H(). Start with i M so that H M () H(). The result of (d) implies that c M is the coefficient of M in H(). By decreasing i, continue this algorithm to find the remaining c i. Hint: Use the result of (b) to find a two-by-two matrix A() such that [ Hi () H i () ] A() [ Hi () H i () ] [ ] [ ] Hi () Hi () T H i () H i () [ ] [ ] Hi () Hi () A H i () H i () where A T [ c i A ( c i ) c i let H() M b k k k0 () ] 45
46 H M () Where H() is given. From this we have c M b M. The following is an algorithm to obtain all c i s. () c M b M () for i M to, descending compute H i () and H i () from () get c i from H i () end Thus we will have c,c,...c M Causal filters always have a nonero phase response. One technique for attaining ero phase response from a causal filter involves filtering the signal twice, once in the forward direction and the second time in the reverse direction. We may describe this operation in terms of the input x[n] and filter impulse response h[n] as follows. Let y [n] x[n] h[n] represent filtering the signal in the forward direction. Now filter y [n] backwards to obtain y [n] y [ n] h[n]. The output is then given by reversing y [n] to obtain y[n] y [ n]. (a) Show that this set of operations is equivalently represented by a filter with impulse response h o [n] as y[n] x[n] h o [n] and express h o [n] in terms of h[n]. y[n] y [n] h[n] (x[n] h[n]) h[ n] x[n] (h[n] h[ n]) x[n] h o [n] h o [n] h[n] h[ n] (b) Show that h o [n] is an even signal and that the phase response of any system with an even impulse response is ero. h o [ n] h[ n] h[n] h[n] h[ n] h o [ n] h o [n] Which shows that h o [n] is an even signal. Since h o [n] is even, the phase response can be found from the following: H o (e jω ) H o (e jω ) h o [n]e jωn n n h o [n]e jωn 46
47 h o [ n]e jωn n n Ho (e jω ) H o (e jω ) arg { H o (e jω ) } 0 h o [n]e jωn (c) For every pole or ero at β in h[n], show that h o [n] has a pair of poles or eros at β and β. h o [n] h[n] h[ n] so H o () H()H( ) let H() c β H o () c c β β ( c)( c) ( β)( β )β Therefore, H o () has a pair of poles at β, β let H() β p H o () β β p p β ( β)( β ) p ( p)( p )β Therefore, H o () has a pair of eros at β, β 7.5. The present value of a loan with interest compounded monthly may be described in terms of the first-order difference equation ( where ρ + r/ 00 y[n] ρy[n ] x[n] ), r is the annual interest rate expressed as a percent, x[n] is the payment credited at the end of the n th month, and y[n] is the loan balance at the beginning of the n + st month. The beginning loan balance is the initial condition y[ ]. If uniform payments of $c are made for L consecutive months, then x[n] c{u[n] u[n L]}. 47
48 (a) Use the unilateral -transform to show that Y () y[ ]ρ c L n0 n ρ Hint: Use long division to show that L L n0 n Y () which implies y[ ]ρ X() ρ X() c L c ( L+) L c n0 n Y () y[ ]ρ c L n0 n ρ (b) Show that ρ must be a ero of Y () if the loan is to have ero balance after L payments. Y () y[ ]p + c L n0 n ρ The pole at p results in an infinite length y[n] in general. If the loan reaches ero after the L th payment, we have: y[n] 0,n L So, Y () L y[n] n n0 Thus we must have: L L y[ ]ρ c n ( ρ ) y[n] n to cancel the pole at p n0 n0 From polynomial theory, the first term is ero if f( ρ) 0, or ρ + r is a ero of Y () (c) Find the monthly payment $c as a function of the intitial loan value y[ ] and the interest rate r assuming the loan has ero balance after L payments. L y[ ]ρ c ρ n 0 n0 48
49 c ρ L ρ y[ ]ρ c ρ y[ ] ρ L Solutions to Computer Experiments Use the MATLAB command plane to obtain a pole-ero plot for the following systems: (a) H() P7.54(a) Imaginary Part Real Part Figure P7.54. (a)pole-zero plot of H() (b) H()
50 P7.54(b) Imaginary Part Real Part Figure P7.54. (b) Pole-Zero plot of H() Use the MATLAB command residue to obtain the partial fraction expansions required to solve Problem 7.4 (d) - (g). P7.55 : Part (d) : r - p k 0 Part (e) : 50
51 r p 4-4 k 0 Part (f) : r - p k 0 Part (g) : r - p - 5
52 k Use the MATLAB command tfss to find state-variable descriptions for the systems in Problem 7.7. P7.56 : Part (a) : A B 0 C D Part (b) : A 6 0 B 0 5
53 C 5 0 D 5 Part (c) : A B 0 C 4 0 D Use the MATLAB command sstf to find the transfer functions in Problem 7.. P7.57 : Part (b)-(i) : Num Den
54 Part (b)-(ii) : Num 0 0 Den Part (b)-(iii) : Num Den Use the MATLAB command plane to solve Problem 7.5 (a) and (b). 54
55 P7.58(a) Poles Zeros of the inverse system.5 Imaginary Part Real Part Figure P7.58. (a) Pole-Zero plots of the inverse system. P7.58(b) Poles Zeros of the inverse system Imaginary Part Real Part Figure P7.58. (b) Pole-Zero plots of the inverse system. 55
56 7.59. Use the MATLAB command freq to evaluate and plot the magnitude and phase response of the system given in Example 7.. P Magnitude Omega Phase (rad) 0 0 Omega Figure P7.59. Magnitude Response for Example Use the MATLAB command freq to evaluate and plot the magnitude and phase response of the systems given in Problem
57 P7.60(a) Magnitude Omega Phase(rad) 0 0 Omega Figure P7.60. (a) Magnitude and phase response P7.60(b) 0.8 Magnitude Omega Phase(rad) 0 0 Omega Figure P7.60. (b) Magnitude and phase response 57
58 P7.60(c) Magnitude Omega.5 Phase(rad) Omega Figure P7.60. (c) Magnitude and phase response 7.6. Use the MATLAB commands filter and filtic to plot the loan balance at the start of each month n 0,,...L+ for Problem 7.5. Assume that y[ ] $0, 000, L 60, r 0. and the monthly payment is chosen to bring the loan balance to ero after 60 payments. From Problem 7.5: [ c ρ y[ ] ρ L y[ ] 0, 000 L 60 ρ + r.008 c 0.05 b [y[ ]ρ c, c(ones(, 59))] a, ρ] 58
59 0000 P Balance Month Figure P7.6. Monthly loan balance Use the MATLAB command psos to determine a cascade connection of second-order sections for implementing the systems in Problem 7.8. Part (a) : sos Part (b) : sos
60 7.6. A causal discrete-time LTI system has the transfer function H() ( ) ( +) ( j0.5889)( j0.5889)( j0.8)( j0.8) (a) Use the pole and ero locations to sketch the magnitude response. Im a tan a tan double a a double Re Figure P7.6. (a) Pole Zero plot symmetric a a Figure P7.6. (a) Sketch of the Magnitude Response. (b) Use the MATLAB commands ptf and freq to evaluate and plot the magnitude and phase response. 60
61 P7.6(b) 0.8 H(Omega) Omega < H(Omega) : rad 0 0 Omega Figure P7.6. (b) Plot of the Magnitude Response. (c) Use the MATLAB command psos to obtain a representation for this filter as a cascade of two second-order sections with real-valued coefficients. H() ( 0.4( + + ) )( ( + ) ) (d) Use the MATLAB command freq to evaluate and plot the magnitude response of each section in (c). 6
62 P7.6(d) H (Omega) Omega.5 H (Omega) Omega Figure P7.6. (d) Plot of the Magnitude Response for each section. (e) Use the MATLAB command filter to determine the impulse response of this system by obtaining the output for an input x[n] δ[n]. 0. P7.6(e) Impulse Resp Figure P7.6. (e) Plot of the Impulse Response. 6
63 (f) Use the MATLAB command filter to determine the system output for the input x[n] ( + cos( π4 ) n) + cos(π n) + cos(π4 n) + cos(πn) u[n] Plot the first 50 points of the input and output. 6 P7.6(f) 5 4 Input Output Figure P7.6. (f) System output for given input. The system eliminates and cos(πn) terms. It also greatly attenuates the cos( π 4 n) term. The cos( π n) term is also attenuated, so the output is dominated by cos( π 4 n). 6
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