版《数字信号处理(英)》课件Chat-6-z-tranform.ppt

1、Chat 6 z-tranform Definition z-Transforms Region of Convergence z-Transforms The inverse z-Transforms z-Transforms Properties The Transfer Function6.1 Definition and Properties v The DTFT provides a frequency-domain representation of discrete-time signals and LTI discrete-time systems.v Because of t

2、he convergence condition,in many case,the DTFT of a sequence may not exist.v As a result,it is not possible to make use of such frequency-domain characterization in these case.jjj(e)(e)(e)(3.88)YHXjj(e)e (3.85)nnHh n6.1 Definition and Properties(p227)v z-Transform may exist for many sequence for whi

3、ch the DTFT does not exist.v Moreover,use of z-Transform techniques permits simple algebraic manipulation.v Consequently,z-Transform has become an important tool in the analysis and design of digital filters.1.Definition ()(6.1)nng nG zg n zZRe()Im()where zzjz is complex variable6.1 Definition and P

4、roperties(p227)()(6.2)Zg nG z:jIfzre()(6.3)jnj nnG reg n r e1When z ()jj nnG eg n eDTFTRe zj Im zz=r e j r 11jjUnit circle0v For a given sequence,the set R of values of z for which its z-transform converges is called the region of convergence(ROC).6.1 Definition and Properties(p227)v The interpretat

5、ion of the z-transform G(z)as the DTFT of sequence gnr-n.(6.4)nng n r v We can choose the value of r properly even though gn is not absolutely summable.(6.5)ggRzRIn general,ROC can be represented as6.1 Definition and Properties(p227)Note:The z-transform of the two sequence are identical even though

6、the two parent sequence are different.Only way a unique sequence can be associated with a z-transform is by specifying its ROC.The DTFT G(ej)of a sequence gn converges uniformly if and only if the ROC of the z-transform G(z)of gn includes the unit circle.6.1 Definition and Properties(p227)Table 6.1

7、u n1zz n na u n1|1z nu n2(1)zz|1z zz a|za|0z 1na un zz a|za6.2 Rational z-Transforms(p231)M-the degree of the numerator polynomial P(z)N-the degree of the denominator polynomial D(z)1(1)0111(1)011(6.13)MMMMNNNNP zpp zpzp zH zD zdd zdzd z 1()0111011(6.14)MMN MMMNNNNp zp zpzpH zzd zd zdzd 1()010110011

8、(1)()(6.15)(1)()MMllN MllNNllllzzppH zzddzz6.2 Rational z-Transforms(p231)1()010110011(1)()(6.15)(1)()MMllN MllNNllllzzppH zzddzzllthe zerosthe poles of H(z),of H(z)vIn Eq.(6.15),there are M finite zeros and N finite poles vIf NM,there are additional N-M zeros at z=0.vIf NM,there are additional M-N

9、poles at z=0.6.3 ROC of Rational z-TransformsThe ROC of a rational z-transform is bounded by the location of its poles.The ROC of a rational z-Transform cannot contain any poles A sequence can be one of the following type:finite-length,right-sided,left-sided and two-sided.If the rational z-transform

10、 has N poles with R distinct magnitudes,then it has R+1 ROCs,R+1 distinct sequence having the same rational z-transform.a)The ROC of the z-transform of a finite-length sequence defined for Mn N is the entire z-plane except possibly z=0 and/or z=+6.3 ROC of Rational z-Transforms We have the following

11、 observation with regard to the ROC of a Rational z-Transform6.3 ROC of Rational z-Transformsb)The ROC of the z-transform of a right-sided sequence defined for Mn is the exterior to a circle in the z-plane passing through the pole furthest from the origin z=0.6.3 ROC of Rational z-Transformsc)The RO

12、C of the z-transform of a left-sided sequence defined for-n N is the interior to a circle in the z-plane passing through the pole nearest from the origin z=0.6.3 ROC of Rational z-Transformsd)The ROC of the z-transform of a two-sided sequence of infinite length is a ring bounded by two circle in the

13、 z-plane passing through two poles with no poles inside the ring.6.4 The Inverse z-Transform(p238)6.4.1 General Expression()(6.1)nnG zg n z-Cauchys integral theorem11()(6.27)nlnCClG z zdzg n zzdz蜒1111()(6.28)22nn lCClG z zdzg nzdzjj 蜒11()(6.29)2n lCzdznlj 6.4.1 General Expression1()ng nresidues of G

14、 z zat the poles inside C If the pole at z=0 of G(z)zn-1 is of multiplicity m.10()()()nmzG z zz1110111()()()1(1)!1()(1)!nmmnzmmzmResidues G z zat zdzG z zmdzdzmdz000 6.4.3 Partial-Fraction Expansion MethodA rational z-transform G(z)with a causal inverse transform gn has an ROC that is exterior 1(1)0

15、111(1)011MMMMNNNNP zpp zpzp zG zD zdd zdzd z-M N,P(z)/D(z)is an improper fraction 10(6.38)MNlllP zG zzD z-M N,P1(z)/D(z)is a proper fraction6.4.3 Partial-Fraction Expansion MethodSimple Poles 11(6.39)1NlllG zz 1(1)|(6.40)lllzzG z1()(6.41)Nnlllg nu n 6.4.3 Partial-Fraction Expansion MethodMultiple Po

16、les If the pole at z=v is of multiplicity L and the remaining N-L poles are simple.1111(6.45)1(1)N LLliililG zzvz101()1(1)()()!()()1L imiL iL iz vdzvzG zLivd ziL 6.5 z-Transform Properties(p246)()Zgg nG zROCR ()Zhh nH zROCR Conjugation Property()Zgg nGzROCR Time-Reversal Property(1/)1/ZggnGzROCR Lin

17、earity Property ()()Zghg nh nG zH zROC include RR 6.5 z-Transform Properties(p246)00(),excluding possibly the point 0 or nZgg n nzG zROCRzz Multiplication by an Exponential Sequence(/)|Zngg nG zROCR Differentiation Property()excluding possibily the point z=0 or z=ZgG zng nzdzROCR Time-Shifting Prope

18、rty6.5 z-Transform Properties(p246)Modulation theorem11 ()(/)2 include Zcghg n hnG v HzvvdvjROCRR Parsevals RelationConvolution Property ()()includeZghg nhnG z H zROCRR 11 ()(1/)2cng n hnG v Hvvdvj 6.7 The Transfer Function(p258)hnxnyn6.7.1 Definition (6.74)ky nh k x nk()()()(6.75)Y zH z X z()(6.76)

19、nnwhere H zh n z6.7.1 Definition()()(6.77)()Y zH zX z-system function or transfer function6.7.2 Transfer Function ExpressionFIR Digital Filter21()(6.78)Nnn NH zh n zFor a causal FIR filter,0N1N2,the ROC of H(z)is the entire z-plane,excluding the point z=0.Finite-Dimensional LTI IIR Discrete-Time Sys

20、tem00(2.76)NMkkkkd y nkp x nk6.7.2 Transfer Function Expression00()()(6.81)NMkkkkkkd z Y zp zX z()0101()()(6.15)()()MlN MlNllzpY zH zzX zdz 1(1)0111(1)011(6.13)MMMMNNNNY zpp zpzp zH zX zdd zdzd z6.7.2 Transfer Function Expression()0101()()(6.15)()()MlN MlNllzpY zH zzX zdz For a causal IIR filter,hn

21、is a causal,the ROC of H(z)is exterior to the circle going through the pole furthest from the origin.max ()kkzROC of H z6.7.3 Frequency Response from Transfer Function()()()reimH zHzjHzarg()()()jH zH zH z e1()arg()tan()imreHzH zHzIf the ROC of H(z)includes the circle()()(6.90)jjz eH eH zln/()()(6.91

22、)jz jH zH e6.7.3 Frequency Response from Transfer Function()()()jjjreimH eHejHeMagnitude function()0101()(6.92)()MjkjjN MkNjkkepH eede()10011001(6.93)MjkjjN MkNjkkMjkkNjkkepH eedeepde6.7.3 Frequency Response from Transfer FunctionPhase response0110argarg()()arg()arg()MNjjjkkkkpH eNMeedMagnitude-squa

23、red function for a real-coefficient rational transfer function21()()()()()()()jjjjjjz eH eH eH eH eH eH z H z220101()()()()kkMjjkjjjkNjjkkeepH eH eH edeev A causal LTI digital filter is BIBO stable if and only if its impulse response hn is absolutely summable.(6.102)nh n v We now develop a stability

24、 condition in terms of the pole locations of the transfer function H(z)v If the ROC includes the unit circle|z|=1,then the digital filter is stable,and vice versa.6.7.5 Stability Condition in terms of pole locationv A FIR digital filter with bounded impulse response is always stable.v On the other h

25、and,an IIR filter may be unstable if not designed properly.v An originally stable IIR filter characterized by in finite precision coefficients may become unstable when coefficients get quantized due to implementation6.7.5 Stability Condition in terms of pole locationExample 6.38:consider a causal II

26、R transfer function.121()1 1.8450.850586H zzz6.7.5 Stability Condition in terms of pole locationThe absolute summability condition of hn is satisfied.Hence,H(z)is a stable transfer function.6.7.5 Stability Condition in terms of pole location Now,consider the case when the transfer function coefficie

27、nts are rounded to values with 2 digits after the decimal point:121()1 1.850.85H zzz6.7.5 Stability Condition in terms of pole locationv In this case,the impulse response coefficient increases rapidly to a constant value as n increases.h nv Hence,the absolute summability condition of is violatedv Th

28、us,is an unstable transfer function()H z6.7.5 Stability Condition in terms of pole location6.7.5 Stability Condition in terms of pole location1)All poles of a causal stable transfer function H(z)must be strictly inside the unit circle.3)The ROC of transfer function of an LTI digital filter includes

29、the unit circle,then the filter is BIBO stable.2)All poles of a anticausal stable transfer function H(z)must be strictly outside the unit circle.Conclusions:Example:Determine the stable or causal of the following transfer function 5.05.0)()1(zzzzH22)()2(zzzzH22)()3(zzzzH(4)()0.52(0.5)(2)zH zzzzAnswe

30、r:(1)0.5 nh nu n5.01)Re(z)Im(zjstable and causal6.7.5 Stability Condition in terms of pole location6.7.5 Stability Condition in terms of pole location(2)2 nh nu n21)Re(z)Im(zjunstable and causal(3)21nh nun 21)Re(z)Im(zjstable and anticausal6.7.5 Stability Condition in terms of pole locationstable and no causal(4)()0.52(0.5)(2)zH zzzz2 0.5 213nnh nu nun 0.51)Re(z)Im(zj2Exercises6.1 6.2 6.13 6.20(a)6.25(a)(b)6.37 6.40(a)6.41 6.43(a)(b)6.48