信号英文版课件chapter

1、Content10.1 The Z transform10.2 The ROC for ZT10.3 The inverse ZT10.4 Omit10.5 Omit10.5 Properties of the ZT10.6 Some common ZT pairs10.7 Analysis and characterization of LTI systems using ZT10.8 System function algebra and block diagram representation10.9 The Unilateral (Bilateral ) ZTThe Z transfo

2、rmContinuousDiscreteDiscrete ak and X(ej ) with period of 2 .tjkaeFSk0 tj)j(XeFT st)S(XeLT njkaeFSk0 nj)e(XeFTj n)Z(XzZT where jrezThe Z transform10.1 The Z transform In section 3.2, for a discrete LTI system with hn, if input is zn, then yn=H(z) zn ( The dominance condition also must be satisfied )

3、.At that time, H( z) was referred to as the system function. nnz )n(h)z(HWhere The Z transform If z=ej (i.e. |z|= r =1), then H(z)=H(ej ) was discrete FT of h(n), or frequency response of the discrete system.1. Definition of ZTor denote it as nn znxzXnxZ)3 .10()()()z(XnxzThe Z transform i.e. ZT of s

4、equence xn is a power series of complex variable z-1, the coefficients is corresponding value of sequence xn.2. The relationship between ZT and discrete FT z=rej Therefore X(z) is the FT of xnr-n, that is X(z)=F ( xnr n ) (10.7) nnjjrenxreXzX then)()()( nnjne )rnx(The Z transformIf r =1 or |z|=1 = X

5、(ej ) = F(xn)3. Z plane (complex Z plane) jez)z(X j s-planeRezImzz-plane runitcircler=1original ( =0, =0)r=1, =0 axisj axisPositive Rez axisunit circleleft half planeright half planein unit circle (r 1)out unit circle (r 1)z=rej The Z transformAs the LT exists the ROC, the ZT also. But theROC of ZT

6、is a ring or out circle or in circle. If the ROC includes the unit circle, then FT of xn is4. Calculating examplesIf |z| |a|, the series converges, then jezj)z(X)e(Xaz azzanuaZzXn 1111)()(1,azz 0nnnza)nua(Z )1(.zaza122 The Z transformIf |z|2, xn is right sequence, If |z| 1/3 (z 0), xn is left sequen

7、ce, If 1/3 |z| 2 , xn is two sided sequence,and this case the ROC includes unit circle, so the FT also converges. ( see Figure 10.12 )Assignments (P798): 10.6, 10.7. The inverse ZT For calculating the inverse ZT, have followingmethod: Inversion integral; Partialfraction expansion; Power series expan

8、sion.1) Inversion integral (contour integral） Both sides multiply by zn-1 and integrate along any closed contour in the ROC . nnznxzX)( The inverse ZT nnnndzzznxdzzzX11)()( nnndzznxdzznx11 jdzz 21 According to complex variable integral )41.10( )()( 2111zXZdzzzXjnx n Example: given1 ,)5 . 0)(1(12)(2

9、zzzzzzXdetermine xn.The inverse ZTSolution: X(z)zn-1 there are three poles: (0, 0.5, 1) dzzzXjnxn1)(21 31 ,5 . 0,012 )5 . 0)(1(12Renuzzzzzszn112)5 . 0)(1(125 . 01202nuzzzznzzzzznz )5 . 0( 9825 . 012 112nunnuzzzznzn The inverse ZT2) Partial-fraction expansion. First using PFE, second using ZT pair (

10、be careful of the ROC ) z all ,1n 1 ,1111 zzzznu1 ,11111 zzzznuazazzaznuan ,111azazaazaznuan ,111The inverse ZTExampleSolution: min2 ,21:,2523)( 2nxDeterzzzzzXGiven )5 . 0z)(2z(2z3)z(X )2 , 5 . 0(: ,5 . 05 . 022zzz 1)5 . 0(5 . 0)2( 1 nununxnnThe inverse ZT3) Power-series expansion i.e. X(z) is a pow

11、er series of z . The coefficients in this series are xn. )5 . 0)(2( 23)( zzzzXor 2) (0.5,|:z| 5.02 zzzz)5 . 0(1)2(5 . 0 1nununxnn nnznxzX)3 .10( )( The inverse ZTExample 10.12 consider X(z)= 4z2+ 2+ 3z-1, |z|: ( 0, ) From eq. (10.3) obtain xn= 4n+2 + 2n + 3n-1ExampleIt can be expanded by long divisi

12、on: i.e. X(z)=2+0.5z-1+1.25z-2+0.875z-3+. or x0=2=1+(-0.5)0 x1=0.5=1+(-0.5)1 x2=1.25=1+(-0.5)2 . xn=1+(-0.5)n un), 1(: ,5 . 05 . 05 . 02)( 22 zzzzzzXconsiderThe inverse ZT If ROC is |z|010.5.3 Scaling in the Z domainRROCzXznnxnz )(00Properties of the ZTWhere z0 is any constant in z-plane. Compare: e

13、s0tx(t) X(s-s0) shifting in the S Sequence xn multiplication by exponential sequence be equivalent to scaling in z-domain. Poof: X(z/z0)=R|z| : ROC)(000zzXnxznR |:z/z| )(00nnzznx nnnznxz0=z z0nxzn|z|: |z0|RProperties of the ZTExample: we knowInfer1 ,111 zznuazaznuan ,111000 ),(zRROCzzXnxzn RROCzXnxn

14、 ),()1(0j0ez If 1 0 zRROC ),ze(Xnxe then00jnj Properties of the ZTRezImz ounitcircleRezImz 0unitcircleo Pole-zero patternof xnPole-zero patternof ej 0n xnProperties of the ZT10.5.4 Time Reversal x-nX(1/z), ROC=1/R Example： we know10.5.5 Time Expansion(1) Definition of time expansion), 1( : ,111 zznu

15、)1, 1 ( : ,111 11 zzzznuthenz nx)k(xn/k, If n is a multiple of k0,If n is not a multiple of kProperties of the ZT kkzkRROCzXnx1)( , )2( az ),z(Xnx Let 0123xn2134n4321012n3 64xn/2zeros insertExampleeven and ,0n ,2/nx )0( odd is n ,0 ny and Calculate Y(z)Properties of the ZT 002/)(nnnnznxznyzY 0nnznx)

16、z(X and)a( z3xz2xz1x0 x32 Solution:)( 32 1 0)( 32azxzxzxxzX )( 32 1 0)(Y 642bzxzxzxxz Compare (a) with (b)az ),z(X)2/nx(Z)z(Y2 In the same way, ZT of yn = x2n is 2 ,)()(21)(azzXzXzY Properties of the ZT10.5.7 Convolution property (1) Difference in time domain Consider an LTI system: yn=hn xnif hn= n

17、- n-1 (the first difference)then yn= xn-xn-1 Look at z-domain ROC: all z, except the originthen222111:)(:)(RROCzXnxRROCzXnx，；，212121: ),()(RRROCzXzXnxnx From convolution property Difference and partial accumulation.11)( zzHnhRROCzXzzY )()1 ()(1 ProofProperties of the ZT(2) Partial accumulation10.5.8

18、 Differentiation in the ZDomain nknxnunxnw With the possible deletion of z=0 and/or possible addition of z=1.1 ,11 1 zznuz nkzzRROCzXznxnw1 ),(11 1RROCdzzdXznnxz ,)(ExampleProperties of the ZTCompare: t x(t) - dX(s)/ds ROC: RExample:)()( dzzdXzdzdzdzdznxnm mmmmds)s(Xd)1()t (xt ?2nunZnnuZ，1 ,11 1 zzn

19、uz 1 ,111)( 2111 zzzzdzdznnuthen 1 ,1)1 ()( 3 112112 zzzzzdzdznunAnother ExampleProperties of the ZT10.5.9 The initial and final value theorem (1) If xn=0, n0, and the numerator order denominator order of X(z), then(2) If xnX(z), ROC: (1, ), and (z-1)X(z), ROC: 1, )10.5.10 Summary of properties (Tab

20、le 10.1)(0zXLimxz )()1()x( 1zXzLimthenz ProofExample ProofSome common ZT pairsz all ,1n 1 ,1111 zzzznu1 ,11111 zzzznuazazzaznuan ,111azazaazaznuan ,111Analysis LTI systems using ZTIf the input to a LTI system is xn=z0n, - n |pole|max of H(z), then yn=H(z0) z0n, - n 10.7.1 CausalityA causal LTI syste

21、m has hn=0, for n |a|, including infinity, then system is causal. Dominance condition is that z0 must belongs to the ROC of H(z).Analysis LTI systems using ZT2）Let H(z) is rational, if and only if (a) the ROC is |a| circle outside the outer- most pole. （b）the order of numerator the order of denomina

22、tor. Then system is causal.Example:81412)(223zzzzzzH2|,2115.011)(11zzzzHNot causal CausalAnalysis LTI systems using ZT10.7.2 Stability- equivalent to hn absolutely summable, i.e. FT of hn converges.(1) An LTI stable system, the ROC of H(z) must include unit circle |z|=1 .(2) An LTI stable causal sys

23、tem, its all of the poles of rational H(z) lie inside the unit circlei.e. |zj|1. RezImzr =1 RezImzr =1 Analysis LTI systems using ZT10.7.3 LTI system characterized by linear constant-coefficient difference equationsMkkkNkkkzXzbzYza00)()(From equationH(z) and hn or inverse.MkkNkkknxbknya00GenerallyNk

24、kkMkkkzazbzXzYzH00/)()()(ZT to both sidesAssignments (P800): 10.16ExampleSystem function algebra10.8.1 H(z) for interconnection systems Series and parallel form (omit)Feedback form:According to Mason EquationH1(z)H2(z)+xnyn+)()(1 )()()()(211zHzHzHzXzYzHSystem function algebra10.8.2 Block diagram rep

25、resentation for causal LTI systems Three basic operations:Example 10.28 (one order system) The causal LTI system Its difference equation is yn yn-1 = xn System block diagram:aZ-1+14111)( zzHSystem function algebraFor multi-order system, block diagram can represent as direct or cascade or parallel .E

26、xample: Given(a) Write the difference equation.(b) Draw three forms of block diagram.z-11/4+xnyn+8 . 0 ,)2 . 0)(8 . 0()5 . 0()( zzzzzzHSystem function algebra8 . 0 ,)2 . 0)(8 . 0()5 . 0()( zzzzzzH21116. 015 . 01)( )( zzzzHa 15 . 0 216. 0 1 nxnxnynynySolution:(b) From Mason e.q. direct form:21116. 01

27、5 . 01)( zzzzHSystem function algebra(2) H(z) rewrite as ( cascade form )2 . 0z)(8 . 0z()5 . 0z( z)z(H z1+xnyn0.5+z10.16 111z2 . 01z5 . 01z8 . 011xn0.8z1+z1+yn0.5+0.2System function algebra(3) For H(z), by performing a partial-fraction expanding: Parallel form 112 . 015 . 08 . 015 . 0)( zzzH+xnyn0.8

28、z1+0.50.2z1+0.5Assignments(P800): 10.18The Unilateral ZT1. Definitionso that, if xn=0, n0, UZ xn = Z xn if xn 0, na.Example 10.33 Let xn=an+1 un+1 0)(nnuzznxzXnxROC is always the exterior of a circle .aazz z ,1 xn Z X(z)then1azazazannn ,1 xn uZ (z) but X101The Unilateral ZTExample 10.34 Determine xn

29、=?Solution:Notes: (1) UZ need not label ROC, UZ1 X(z) is always causal signal.)311)(411(653)( 111 zzzzXnxLetuz11)3/1(12)4/1(11)( zzzX )3/1(2)4/1( nunxnn (2)( ,)()()( zXnxifandzqzpzXIfuz The Unilateral ZT then the degree of p(z) the degree of q(z), otherwise, UZ-1 X(z) does not exist.For example Z Z

30、-1 X(z) = an+1un+1 ( xn 0, n= -1 ) i.e. the UZ-1 of the X(z) does not exist.3. Time convolution of UZT azzzX 2)()(x 11zXnunLetuz)(x 22zXnunuz)()(2121zXzXnunxnunxuz thenThe Unilateral ZT4. Time shifting of UZT xn-1un = xn-1un-1 + x-1 n xn+1un = xn+1un+1 x0 n+1xn-1un z-1X(z)+x-1 = z-1( X(z)+x-1z ) xn-2un z-2( X(z)+x-1z+x-2z2 ) And xn+1un z X(z) -x0z = z ( X(z)-x0