数字信号处理a(双语)chapter6-ztransformb.ppt

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1、Chapter 6,z-Transform,6.4 The Inverse z-Transform6.5 z-Transform Theorems,Part B:Inverse ZT and ZT Theorems,6.4 The Inverse z-Transform,6.4.1 General Expression6.4.2 Inverse z-Transform by Table Look-Up Method6.4.3 Inverse z-Transform by Partial-Fraction Expansion6.4.4 Partial-Fraction Using MATLAB6

2、.4.5 Inverse z-Transform via Long Division6.4.6 Inverse z-Transform Using MATLAB,6.4.1 General Expression,6.4.1 General Expression,6.4.1 General Expression,逆z变换是一个对X(Z)Zn-1进行的围线积分，积分路径C是一条在X(Z)收敛环域（Rx-，Rx+）内以逆时针方向绕原点一周的单围线。,围线积分路径,直接计算围线积分比较麻烦，一般不采用此法求逆z变换，求解逆z变换的常用方法有：留数定律法查表法部分分式法长除法,6.4.1 General

3、 Expression,6.4.1 General Expression,利用留数定理计算围线积分,一阶极点的留数,N 阶极点的留数,1.1 General Expression,6.4.1 General Expression,6.4.1 General Expression,6.4.1 General Expression,6.4.1 General Expression,6.4.1 General Expression,6.4.1 General Expression,Example 6.12,6.4.2 Table Look-up Method,Look up Table 6.1 on

4、 Page 253,6.4.3 Inverse z-Transform by Partial-Fraction Expansion,A rational z-transform G(z)with a causal inverse transform g(n)has an ROC that is exterior to a circleHere it is more convenient to express G(z)in a partial-fraction expansion form and then determine g(n)by summing the inverse transfo

5、rm of the individual simpler terms in the expansion,6.4.3 Inverse z-Transform by Partial-Fraction Expansion,6.4.3 Inverse z-Transform by Partial-Fraction Expansion,Solutions:Step 1-Converting G(z)into the form of proper fractions by long divisionStep 2-Summing the inverse transform of the individual

6、 simpler terms in the expansion,6.4.3 Inverse z-Transform by Partial-Fraction Expansion,6.4.3 Inverse z-Transform by Partial-Fraction Expansion,6.4.4 Partial-Fraction Using MATLAB,r,p,c=residuez(num,den)develops the partial-fraction expansion of a rational z-transform with numerator and denominator

7、coefficients given by vectors num and denVector r contains the residuesVector p contains the polesVector c contains the constantsl,num,den=residuez(r,p,c)converts a z-transform expressed in a partial-fraction expansion form to its rational form,6.4.4 Partial-Fraction Using MATLAB,6.4.5 Inverse z-Tra

8、nsform via Long Division,6.4.5 Inverse z-Transform via Long Division,6.4.6 Inverse z-Transform Using MATLAB,The function impz can be used to find the inverse of a rational z-transform G(z)The function computes the coefficients of the power series expansion of G(z)The number of coefficients can eithe

9、r be user specified or determined automatically,6.4 The Inverse z-Transform6.5 z-Transform Theorems,Part B:Inverse ZT and ZT Theorems,6.5 z-Transform Theorems,6.5 z-Transform Theorems,6.5 z-Transform Theorems,6.5 z-Transform Theorems,6.5 z-Transform Theorems,6.6 Computation of the Convolution Sum of

10、 Finite-Length Sequences6.6.1 Linear Convolution Using Polynomial Multiplication6.6.2 Circular Convolution Using Polynomial Multiplication,Part C:Convolution,6.6.1 Linear Convolution Using z-Transform,Let denote xn,0 n L,a finite-length sequence of length L+1Let denote hn 0 n M,a finite-length seque

11、nce of length M+1We shall evaluate yn=xn*hn using z-transformNote:yn is a sequence of length L+M+1,6.6.1 Linear Convolution Using z-Transform,Let X(z)denote the z-transform of xn which is a polynomial of degree L in z-1,i.e.,X(z)=x0+x1 z-1+x2 z-2+xL z-L Let H(z)denote the z-transform of hn which is

12、a polynomial of degree M in z-1 H(z)=h0+h1 z-1+h2 z-2+hM z-M,6.6.1 Linear Convolution Using z-Transform,From the convolution property of the z-transform,it follows that the z-transform of yn is simply given by Y(z)=X(z)H(z)which is a polynomial of degree L+M in z-1,i.e.,Y(z)=y0+y1 z-1+y2 z-2+yL+M z-

13、(L+M),6.6.1 Linear Convolution Using z-Transform,WhereIn the above we have assumed xn=0 for nL hn=0 for nM,6.6.1 Linear Convolution Using z-Transform,(pp.278)Example 6.30 Linear Convolution of One-Sided Sequences Using the Polynomial Multiplication Method,6.6.2 Circular Convolution Using z-Transform

14、,Let xn and hn be two length-N sequences defined for 0 n N-1 with X(z)and H(z)denoting their z-transformsLet ycn=xn N hn denote the N-point circular convolution of xn and hnLet yLn=xn*hn denote the linear convolution of xn and hn,6.6.2 Circular Convolution Using z-Transform,Let Yc(z)and YL(z)denote

15、the z-transforms of ycn and yLnIt can be shown that Yc(z)=(z-N-1)The modulo operation with respect to z-N-1 is taken by setting z-N=1(z-N-1)=z-1(z-N-1)=z-2,6.6.2 Circular Convolution Using z-Transform,(pp.279)Example 6.32 Circular Convolution of Causal Sequences Using the Polynomial Multiplication M

16、ethod,6.6.2 Circular Convolution Using z-Transform,6.6.2 Circular Convolution Using z-Transform,6.7 The Transfer Function,The transfer function is a generalization of the frequency response function.The convolution sum description of an LTI discrete-time system with an impulse response hn is given b

17、y,6.7 The Transfer Function,Taking the z-transforms of both sides we get,6.7 The Transfer Function,6.7.1 Definition,Hence,H(z)=Y(Z)/X(z)The function H(z),which is the z-transform of the impulse response hn of the LTI system,is called the transfer function or the system functionThe inverse z-transfor

18、m of the transfer function H(z)yields the impulse response hn,6.7.2 Transfer Function Expression,Consider an LTI discrete-time system characterized by a difference equationIts transfer function is obtained by taking the z-transform of both sides of the above equationThus,6.7.2 Transfer Function Expr

19、ession,Or,equivalently asAn alternate form of the transfer function is given by,6.7.2 Transfer Function Expression,Or,equivalently as1,2,M are the finite zeros,and 1,2,N are the finite poles of H(z)If N M,there are additional(N-M)zeros at z=0If N M,there are additional(M-N)poles at z=0,6.7.2 Transfe

20、r Function Expression,For a causal IIR digital filter,the impulse response is a causal sequenceThe ROC of the causal transfer function is thus exterior to a circle going through the pole furthest from the originThus the ROC is given by,6.7.2 Transfer Function Expression,Example-Consider the M-point

21、moving-average FIR filter with an impulse response,Its transfer function is then given by,6.7.2 Transfer Function Expression,The transfer function has M zeros on the unit circle at z=ej2k/M,0 k M-1There are M-1 poles at z=0 and a single pole at z=1The pole at z=1 exactly cancels the zero at z=1The R

22、OC is the entire z-plane except z=0,M=8,6.7.2 Transfer Function Expression,Example-A causal LTI IIR digital filter is described by a constant coefficient difference equation given by yn=xn-1-1.2xn-2+xn-3+1.3yn-1-1.04yn-2+0.222yn-3Its transfer function is therefore given by,6.7.2 Transfer Function Ex

23、pression,Alternate forms:,ROC:,Note:Poles farthest from z=0 have a magnitude,6.7.3 Frequency Response from Transfer Function,If the ROC of the transfer function H(z)includes the unit circle,then the frequency response H(ej)of the LTI digital filter can be obtained simply as follows:,For a real coeff

24、icient transfer function H(z)it can be shown that,6.7.3 Frequency Response from Transfer Function,For a stable rational transfer function in the form,the factored form of the frequency response is given by,6.7.3 Frequency Response from Transfer Function,It is convenient to visualize the contribution

25、s of the zero factor(z-k)and the pole factor(z-k)from the factored form of the frequency responseThe magnitude function is given by,6.7.3 Frequency Response from Transfer Function,The phase response for a rational transfer function is of the formThe magnitude-squared function of a real-coefficient t

26、ransfer function can be computed using,Geometric Interpretation of Frequency Response Computation,The factored form of the frequency response is convenient to develop a geometric interpretation of the frequency response computation from the pole-zero plot as varies from 0 to 2 on the unit circle,Geo

27、metric Interpretation of Frequency Response Computation,The geometric interpretation can be used to obtain a sketch of the response as a function of the frequencyA typical factor in the factored form of the frequency response is given by(ej-ej)where ej is a zero if it is zero factor or is a pole if

28、it is a pole factor,Geometric Interpretation of Frequency Response Computation,As shown below in the z-plane the factor(ej-ej)represents a vector starting at the point z=ej and ending on the unit circle at z=ej,As is varied from 0 to 2,the tip of the vector moves counterclockise from the point z=1 t

29、racing the unit circle and back to the point z=1,Geometric Interpretation of Frequency Response Computation,As indicated by the magnitude response|H(ej)|at a specific value of is given by the product of the magnitudes of all zero vectors divided by the product of the magnitudes of all pole vectors,G

30、eometric Interpretation of Frequency Response Computation,Likewise,fromwe observe that the phase response at a specific value of is obtained by adding the phase of the term p0/d0 and the linear-phase term(N-M)to the sum of the angles of the zero vectors minus the angles of the pole vectors,Geometric

31、 Interpretation of Frequency Response Computation,Thus,an approximate plot of the magnitude and phase responses of the transfer function of an LTI digital filter can be developed by examining the pole and zero locationsNow,a zero(pole)vector has the smallest magnitude when=,Geometric Interpretation

32、of Frequency Response Computation,幅度最小,幅度最大,To highly attenuate signal components in a specified frequency range,we need to place zeros very close to or on the unit circle in this range（零点谷值）Likewise,to highly emphasize signal components in a specified frequency range,we need to place poles very clo

33、se to or on the unit circle in this range（极点峰值）,Geometric Interpretation of Frequency Response Computation,Example:y(n)=x(n)-x(n-4),Question：,1、极点的位置对系统的稳定性会有影响吗？2、系统在Z域的稳定性条件是什么？,A causal LTI digital filter is BIBO stable if and only if its impulse response hn is absolutely summable,i.e.,We now dev

34、elop a stability condition in terms of the pole locations of the transfer function H(z),6.7.5 Stability Condition in Terms of the Pole Locations,6.7.5 Stability Condition in Terms of the Pole Locations,The ROC of the z-transform H(z)of the impulse response sequence hn is defined by values of|z|=r fo

35、r which hnr-n is absolutely summable,BIBO stable,Thus,if the ROC includes the unit circle|z|=1,then the digital filter is stable,and vice versaIn addition,for a stable and causal digital filter for which hn is a right-sided sequence,the ROC will include the unit circle and entire z-plane including t

36、he point z=An FIR digital filter with bounded impulse response is always stable,6.7.5 Stability Condition in Terms of the Pole Locations,On the other hand,an IIR filter may be unstable if not designed properlyIn addition,an originally stable IIR filter characterized by infinite precision coefficient

37、s may become unstable when coefficients get quantized due to implementation,6.7.5 Stability Condition in Terms of the Pole Locations,Example-Consider the causal IIR transfer functionThe plot of the impulse response coefficients is shown on the next slide,6.7.5 Stability Condition in Terms of the Pol

38、e Locations,6.7.5 Stability Condition in Terms of the Pole Locations,hn decays rapidly to zero value as n increasesThe absolute summability condition of hn is satisfied.Hence,H(z)is a stable transfer function.,Now,consider the case when the transfer function coefficients are rounded to values with 2

39、 digits after the decimal point:,6.7.5 Stability Condition in Terms of the Pole Locations,6.7.5 Stability Condition in Terms of the Pole Locations,increases rapidly to a constant value as n increasesHence,the absolute summability condition of is violated.Thus,is an unstable transfer function,The sta

40、bility testing of a IIR transfer function is therefore an important problemIn most cases it is difficult to compute the infinite sumFor a causal IIR transfer function,the sum S can be computed approximately as,6.7.5 Stability Condition in Terms of the Pole Locations,6.7.5 Stability Condition in Term

41、s of the Pole Locations,The partial sum is computed for increasing values of K until the difference between a series of consecutive values of SK is smaller than some arbitrarily chosen small number,which is typically 10-6For a transfer function of very high order this approach may not be satisfactor

42、yAn alternate,easy-to-test,stability condition is developed next,Consider the causal IIR digital filter with a rational transfer function H(z)given byIts impulse response hn is a right-sided sequenceThe ROC of H(z)is exterior to a circle going through the pole furthest from z=0,6.7.5 Stability Condi

43、tion in Terms of the Pole Locations,But stability requires that hn be absolutely summableThis in turn implies that the DTFT H(ej)of hn existsNow,if the ROC of the z-transform H(z)includes the unit circle,then,6.7.5 Stability Condition in Terms of the Pole Locations,Conclusion:All poles of a causal s

44、table transfer function H(z)must be strictly inside the unit circle The stability region(shown shaded)in the z-plane is shown below,6.7.5 Stability Condition in Terms of the Pole Locations,Re z,Im z,Unit Circle,stability region,Example-The factored form of is which has a real pole at z=0.902 and a r

45、eal pole at z=0.943Since both poles are inside the unit circle,H(z)is BIBO stable,6.7.5 Stability Condition in Terms of the Pole Locations,Example-The factored form of is which has a real pole on the unit circle at z=1 and the other pole inside the unit circleSince one pole is not inside the unit circle,H(z)is unstable,6.7.5 Stability Condition in Terms of the Pole Locations,第六章重点,Z变换、零极点求解、收敛域判定逆Z变换（部分分式法）传递函数几何作图法因果稳定性的Z域判决,Homework,Problems:6.2(a,b),6.5(a,b),6.7,6.8(a)(i,iv),6.13(a),6.16,6.44,6.81Matlab Exercises:M6.1(a),M6.5,