数字信号处理第3章

1、Understanding DSP, Second Edition1Chapter 3:z-TransformUnderstanding DSP, Second Edition2DSP2Contents3.1 The z-Transform3.1.1 Poles and Zeros on the z-Plane and Stability3.1.2 The ROC of z-Transform3.1.3 The Properties of z-Transform3.2 The Inverse z-Transform3.2.1 General Expression of Inverse z-Tr

2、ansform3.2.2 Inverse z-Transform by Partial-Fraction ExpansionUnderstanding DSP, Second Edition33Z TransformZT of the sequence x(n) is defined as:Z, complex variableInverse ZT:Understanding DSP, Second EditionZ TransformExample Determine the z-transform for . with constant . Solution:If , then the a

3、bove summation converges. Thus,)(nx)()(nunxn01)()()()(nnnnnnnzznuznxzX1|1z| | 11)()(101z,zzzzzXnnUnderstanding DSP, Second Edition55Z TransformWe have to ask if X(z) converge?For any x(n), the region of converge (ROC) of ZT is the set of values of the complex variable z for which the z-transform uni

4、formly converges.i.e.: z: X(z) existsDifferent x(n) with different ROC may have the same ZT.So, the ROC of each X(z) should be defined.Understanding DSP, Second Edition Consider thatIfthen is exist. In other words, if the series is absolutely convergent, then it uniformly converges to an analytical

5、function in its ROC. Such a condition is sufficient and necessary for the z-transform of a sequence to exist.)(nx01)()()()(nnnnnnznxznxznxzX0 1)()( )()( | )(nnnnnnnn|znx|znx|znx|znx|zX|)(zXUnderstanding DSP, Second Edition The convergence condition for the z-transform is as follows: (1)If the limit

6、of the nth root of the absolute value of the nth term in the following summationsatisfiesor equivalently,)(nx0 1)()()( | )(nnnnnn|znx|znx|znx|zX|0|)(|nnznx1| )(lim1|)(lim11n/nn/nnnx|z|znx|1/1| )(|lim |rnxznnROC for the z-transform of a General SequenceIf the z-transform of a sequence is exist, thenU

7、nderstanding DSP, Second Editionthen the seriesuniformly converges. (2) ConsideringIfor equivalently,then the series 0)(nnznx11)()(nnnn|znx|znx|1| )(lim|)(lim11n/nn/nnnx|z|znx|2/1| )(|lim/1|rnxznn1)(nn|znx| uniformly converges.Understanding DSP, Second Edition (3) Combining the above results, we see

8、 thatIts ROC is shown in the following Figure:)(nx2101 | r :ROC ,)()( )()(rzznxznxznxzXnnnnnnUnderstanding DSP, Second Edition1010Z Transform: Pole-zero PlotROC:ROC is determined by |z|=r, in terms of the theory of complex variable function, it can be a circular band: r1|z|r2In the ROC, X(z) is an a

9、nalytic function, and the pole of X(z) is out of ROC, with the pole as the edge.r1 can be zero, r2 can be .If r2r1，it means ROC is not exist, neither the ZT.Understanding DSP, Second Edition1111Z Transform: Pole-zero PlotSystem has ZT as:Understanding DSP, Second Edition12Z Transformcausal sequence

10、:When nr1, outside of radius r1.12Understanding DSP, Second Edition13Z TransformExample 1:Determine the ZT of x(n):13)()(1nuanxnUnderstanding DSP, Second Edition14Z TransformSolution:14The PoleUnderstanding DSP, Second Edition15Z Transformanticausal Sequence:When n 0, x(n)=0;X(z) only contains the p

11、ositive indexes of z.The ROC: |z|r2, inside of radius r2.15Understanding DSP, Second Edition16Z TransformExample 2:Determine the ZT of x1(n):16Understanding DSP, Second Edition17Z TransformSolution:17The PoleUnderstanding DSP, Second Edition18Z TransformAbout these two examples, if b=a:But:18=Unders

12、tanding DSP, Second Edition19Z TransformIf b=a:X1(z) has the same form with X(z), except for the ROC.That implies that the ROC insures only one ZT of x(n).Different ROC means different ZT.ROC plays an important role in system analysis.19Understanding DSP, Second Edition20Z TransformTwo-side Sequence

13、:Contains right-side sequence and left-side sequence.So the ROC is defined as: r1|z|r2 or not exist if r2|a|, where a is the pole.(3) anticausal Sequence (n0): the ROC is |z|b|, where b is the pole.23Understanding DSP, Second EditionQuestions:1.Finite-length Sequences12 )()(21nn,znxzXnnnn2. Right-Si

14、ded Sequences1)()(nnnznxzX3. Left-Sided Sequences2)()(nnnznxzXROC forUnderstanding DSP, Second Edition The z-Transform of Basic Sequences (1) The Unit Sample Sequence (2) The Unit Step Sequence (3) The Anticausal Unit Step Sequence )(nx | 0 1)()(|z,znzXnn)()(nnx)()(nunx1 | 111)()()(101|z,zzzzznuzXnn

15、nn) 1()(nunx1 | 1)( )() 1()(11|z,zzzzznuzXnnnnnnUnderstanding DSP, Second EditionThe z-Transform of Basic Sequences (4) The Ramp Sequence Consideringand taking the derivative on two sides in this equation with respect to the variable , we getThis leads to )(nx)()(nnunx0)()(nnnnnzznnuzX1 | 1110|z,zzn

16、n1z1 | )(11)(210011|z,znzzznnnnn1 | ) 1()(1)(22110|z,zzzznzzXnnUnderstanding DSP, Second Edition27Common ZT PairsThe Table of common ZT pairs:27Understanding DSP, Second Edition|)1 () 1(|)1 ()(|)cos2(1)cos(1)(cos|)cos2(1)sin()(sin21121122101002210100bzbzbznunbazazaznunaazzazwazwanunwaazzazwazwanunwa

17、ROCTransformSequencennnnCommon ZT PairsUnderstanding DSP, Second Edition29The Properties of ZTLinear29ROC2 :ROC ),()(zYnyZROC1 :ROC ),()(zXnxZROC2ROC1 containing ROC with Understanding DSP, Second Edition. ) 1()()(nuanuanxnnExample: Find the z-transform for Solution: We letwhere)()()(21nxnxnx)()(1nu

18、anxn) 1()(2nuanxnandThe Properties of ZTUnderstanding DSP, Second EditionThenandBy the linearity property, we haveOn the other hand, we consider that)(nx| | :ROC1 )()(1az,azznuaZzXn| | :ROC2 )1()(2az,azanuaZzXn | 0 :ROC 1 )()()()(21z,azaazzzXzXnxZzX)() 1()()(nnuanuanxnnThe Properties of ZTUnderstand

19、ing DSP, Second Editionwhose z-transform is given by | 0 :ROC 1)()()(z,nZnxZzXThe Properties of ZTUnderstanding DSP, Second Edition2. Time shifting3. Frequency shifting (scaling in the z-domain)The Properties of ZTUnderstanding DSP, Second Edition4. Time Reversal IfthenHence, if the is then the ROC

20、of the is)(nxxzXnxZROC with ),()(x/zXnxZROC1 :ROC ),()(1xROCxxrzr | )(1zX. /1 | /1xxrzrThe Properties of ZTUnderstanding DSP, Second Edition35The Properties of ZT5. Differential6. Conjugation35Understanding DSP, Second Edition36The Properties of ZT7. Initial Value TheoremIf n0 and choose formula No.

21、1. Because X(z) have limited poles inside of C and zn-1 is analytic at z=0, but there are high order poles at z= for zn-1 when n is large.57Understanding DSP, Second Edition58Inverse ZT - General Expression of Inverse z-TransformNote:If the ROC is inside of a circle, usually we calculate x(n) at n0

22、and choose formula No.2. Because X(z) have limited poles outside of C and zn-1 is analytic at z=, but there are high order poles at z=0 for zn-1 when n is large. If the ROC is a ring, usually we use both of formula No.1 and No.2.58Understanding DSP, Second Edition59Inverse ZT - General Expression of

23、 Inverse z-TransformNote:Actually, n=0 is not the only edge for residue method. Now, we can reach one more general method:Extract X(z)=X0(z)zm, m is an integer. Therefore, X0(z) is analytic at both n=0 and n=. Then: X(z)zn-1 =X0(z)zmzn-1=X0(z)zn+m-1=X1(z)After determining the ROC of X(z) and C:We ca

24、n get x(n) at n1-m by calculate the residue of X1(z) inside of C and then get -x(n) at n 1Both of the two poles are inside of C, use formula No.1.When n1-m=0, x(n)=0. When n1-m=0: 63Understanding DSP, Second Edition64Inverse ZT - General Expression of Inverse z-TransformSolution:(2) If ROC: z 1/3Bot

25、h of the two poles are outside of C, use formula No.2.When n1-m=0, x(n)=0. When n1-m=0: 64Understanding DSP, Second Edition65Inverse ZT - General Expression of Inverse z-TransformSolution:(3) If ROC :1/3 z 1Pole z=1/3 is inside of C, use formula No.1. When n0:Pole z=1 is outside of C, use formula No

26、.2. When n0:65Understanding DSP, Second Edition66Inverse ZT - General Expression of Inverse z-TransformSolution:(3) If ROC: 1/3 z r1):For ROC (|z|r2):69Understanding DSP, Second Edition70Inverse ZT - Part fractional methodFor ROC (r1|z|1, x(n) is causal sequence.(2) For ROC |z|1/3, x(n) is anticausa

27、l sequence.73Understanding DSP, Second Edition74Inverse ZT - Part fractional methodSolution:X(z) has two poles, z1=1 and z2=1/3.(3) For ROC 1/3|z|1, the x(n) is Two-side sequence.74Understanding DSP, Second Edition Starts withwhere and are positive integers and, and are real constants. Case 1: If ,

28、thenwhere for are simple poles and is a repeated pole with order . If all poles are distinct, then the xxNNMMR|z|R,zazazazbzbzbbzX 1)(221122110NMibiakzLN,k ,2, 1MN pzLxxLkpkLNkkkNNNNMNMNNR|z|RzzzBzzzAazazazzbzbzbzzX , )()( )(1k111112110Inverse ZT - Part fractional methodUnderstanding DSP, Second Edi

29、tionsecond summation disappears and the first summation contains terms. The expansion coefficients: Comparatively, the inverse z-transform for the expanded expression is easy to be obtained. If all poles are simple, then NLN,k,zzXzzAkzzkk , 2, 1 | )()(L,k,zzXzzdzdkLBpzzLpkLkLk , 2, 1 | )()()!(1Nkkzz

30、zZAzXZnx1k11)()(Inverse ZT - Part fractional methodUnderstanding DSP, Second EditionWhereExample Using the partial fraction expansion, find the inverse z-transform for Solution:whereand. 1 | ),505111/()(21zz.z.zX1 | 0.5)(z1)(z0.5)1)(z(z)(212z,zAzAzzX20.5)(z)(1)(z1 1 1zz|z|zzXA11)(z)(0.5)(z50 50 2.z.

31、z|z|zzXAxknkxknkkRznuzRznuzzzzZ | ),1()( | ),()(1Inverse ZT - Part fractional methodUnderstanding DSP, Second EditionTherefore,Thus,Example Determine the sequence from Solution: 1 | 0.5)(z1)(z2)(z,zzzX)()50(2)()50()(2 )()(1nu.nu.nuzXZnxnn)(nx. 3 | 2 ),212020/()(211zz.z.zzX3 | 2 ,23)2)(3(5 65)(212zzz

32、AzzAzzzzzzzXInverse ZT - Part fractional methodUnderstanding DSP, Second EditionwhereandTherefore,Thus,12)(z5)(3)(z3 3 1zz|zzXA13)(z5)(2)(z2 2 2zz|zzXA3 | 2 ,23)(zzzzzzX)(2) 1()3()()(1nunuzXZnxnnInverse ZT - Part fractional methodUnderstanding DSP, Second Edition Case 2: If thenwhere may be determin

33、ed by performing the long division. The proper rational part is expanded using the partial fraction expansion, which is identical to that in case 1.Example Determine the sequence from Solution: Decompose the given function into, MN 1)(2211)1(1100NNNMNMkkkzazazazczcczCzXxxLkpkLNkkkNMkkkR|z|RzzzBzzzAzC , )()(1k10kC1 | ,)21 ()23(121)(2121zz/z/zzzXInverse ZT - Part fractional methodUnderstanding DSP, Second EditionIf we denote the proper r