Chapter 6(Z transforms)

,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,Chapter 6 Z Transforms,The definition of Z transform and its region of convergence,Properties of Z transform,Make the connections between transfer functions, difference equations, and impulse responses,Introduce the methods of finding inverse Z transform,Explain how to identify the zeros and poles of a filter,Study first and second order system examples,6.1 Z Transform Basics,The definition of the Z transform for a digital,x (n),is:,With the complex variable z , it can be expressed as,Z Transform Basics,In practice , if the initial sampling signal can be determined for n = 0 (,the causal sequence,),，,the z-transform can be defined as :,z Transform Region of Convergence,Z transform lie in the z domain,，,which is a frequency domain containing complex numbers. The z transform may not be defined for all value of z in the z domain.,The values of z for which it is defined from its,region of convergence,.,The condition of z transform,uniform convergence,is:,Z transformation can be calculated directly by definition, but must also shown its convergence.,Compute z Transform,Example 6.1(P171),：,Find the,z transform X (,z,),of the signal,x,(,n,),(,n,),.,This z transform is defined for all values of z, so its region of convergence includes,all z.,Compute z Transform,Example 6.2(P171),：,Find the z transform X (,z,) of the signal,x,(,n,),(,n,1),.,z,transform is defined as long as z,0, so its region of convergence is,all z except z =0.,Compute z Transform,Example,：,Find the z transform X (,z,) of the signal,x,(,n,),R,N,(,n,),.,Its region of convergence is,:,Compute z Transform,Example 6.3,：,Find the z transform X (,z,) of the signal,x(n),u,(n).,Its region of convergence is,:,Compute z Transform,Example 6.5(P173),：,Find the z transform X (,z,) of the signal,Its region of convergence is :,Compute z Transform,Example,：,Find the z transform X (,z,) of the signal,x,(,n,),a,n,u,(,n,),.,Its region of convergence is :,Compute z Transform,Example,：,Find the z transform X (,z,) of the signal,x,(,n,),-,a,n,u,(-,n,-1),.,Its region of convergence is :,Compute Z transform,example,：,Find the z transform X (,z,) of the signal,x (n) = a,| n |, where,|,a,|,aRx,Z domain differential,nx(n,),invariance,convolution,x(n,)*,y(n,),X(z)Y(z,),Common region,Basic properties of Z transform (right unilateral,),Example,6.6(P175),：,Find the z transform X (,z,) of the signal,x (n) = 2 u (n-2),solution,：,Compute z Transform,example,：,For the z transformation of signal of,solution,：,Compute z Transform,Using the,properties,of the z transform :,Compute z Transform,example,：,Find the z transform X (,z,) of the signal,x,(,n,)=,u,(,n,)-,u,(,n,-3),.,Its region of convergence:,| z | 0,Compute z Transform,solution,：,example,：,If,x(n,)=,a,n,u(n,) ,h(n,)=,b,n,u(n,)- ab,n-1,u(n-1),，,y(n,) =,x(n,)*,h(n,),，,find the z transform of,y(n,).,The region of convergence: if |a | |b |,，,|z | |a |,if |b | |a |,，,|z | |b |,Compute z Transform,solution,：,exercise,：,Find the,z transform X (,z,) of following signal.,（,1,）,（,2,）,（,3,）,Compute z Transform,Exercise solution,（,1,）,（,2,）,（,3,）,Right sequence,Left sequence,6.2 Transfer functions,the definition of the transfer function,：,Transfer functions,Term-by-term transformation of a general difference equation,Transfer functions,The transfer function,：,Example 6.8(,P177,),：,Find the transfer function of the described by the difference equation,.,solution,：,The,z transformation,The transfer function,Transfer functions,Example 6.9(,P177,),：,Find the transfer function of the described by the difference equation,Transfer functions,solution,：,Example 6.10(,P177,),：,Find the transfer function of the described by the difference equation,.,Transfer functions,This non-recursive equation corresponding to the transfer function is:,Difference equation can be converted to transfer function, the same, transmission function can also be converted to difference equations,Transfer functions,Example 6.11(,P178,),：,Find the difference equation of transfer function,solution,：,Transfer functions,The inverse z transformation,：,Transfer functions,Example 6.12(,P178,),：,Find the difference equation of transfer function,solution,：,Transfer,functions,Then,：,The inverse z transforms,：,Moving to the right two samples, the difference equation is,：,Three kind of methods to descript the system,：,Difference equation, impulse response and transfer function,.,transfer function,H,(,Z,),X,(,z,),Y,(,z,),H,(,z,),X,(,z,),Z,domain,Difference equation,x,(,n,),y,(,n,),impulse response,h,(,n,),x,(,n,),y,(,n,),h,(,n,)*,x,(,n,),Time domain,Transfer functions,Taking the z-transform of the convolution equation,Z-transform :,It is very important,：,H(Z) is z transform of,h(n,),；,in other words,，,The filter transfer function is the z-transform of its impulse response.,Transfer functions,Transfer function and impulse response,Example 6.13(,P180,),：,Find the filter transfer function,of the filter,Transfer functions,solution,：,Calculating the output of the filter,Transfer function,H,(,Z,),X,(,z,),Y,(,z,),H,(,z,),X,(,z,),Z,domain,Transfer functions,To compute the output of the system with transfer function is simpler than the convolution ,the disadvantage is calculating the z-transform and inverse z transform.,the cascade and parallel of transfer functions,Transfer functions,Example 6.14(,P181,),：,Find the differential equation of the cascade system shown as follows.,Transfer functions,solution,：,Transfer functions,Example 6.15(,P182,),：,Find the transfer function of the cascade system.,Transfer functions,solution,：,The,difference equations of the two filters respectively,Transfer functions,Therefore, the transfer function of the two filters, respectively,The transfer function of the system,Example 6.16(,P183,),：,Find the transfer function of the cascade system,.,Transfer functions,solution,：,Standard form,：,All exponents of z in the z transform be positive, and the coefficient of the highest power term in both the numerator and the denominator be one.,Suppose NM,，,Transfer function is,：,6.3 Inverse z Transforms,Standard Form,Example 6.17,(P184),：,Express,the following transfer function in standard form,solution,：,first step,：,make,the exponents of all delay items,positive.,Second step,：,ensure that the highest power denominator coefficient is 1.,The highest power in standard form is called its,degree,.,In a,proper rational function, the degree of the numerator is less than or equal to the degree of the denominator.,In a,strictly proper rational function, the degree of the numerator is less than the degree of the denominator.,The degree of the numerator of an,improper rational function,is greater than the degree of its denominator.,Standard Form,Some simple inverse z transformation can be,obtained,by checking,“,Basic z transforms,(P174),”,table.,Simple,Inverse z Transforms,Example 6.19(,P186,),：,Find the inverse z transform of the following functions.,solution,：,with “,Basic z transforms,”,:,Example 6.20(P186),：,Find the inverse z transform of the following functions.,Simple Inverse z Transforms,solution,：,check,Basic z transforms,：,Where,：,so,：,Simple Inverse z Transforms,Inverse z transform is,：,Example 6.21(,P186,),：,Find the difference equation and impulse response of system,Simple Inverse z Transforms,solution,：,the,difference equation is,：,the impulse response is obtained by calculated inverse z-transform of transfer function,：,H(Z) is delay,two units, compared with the inverse z transform of function,z,/(,z,+0.25).,By checking,“,Basic z transforms,”,table , obtained the inverse z transform of,z,/(,z,+0.25),is,：,Simple Inverse z Transforms,the impulse response,Example 6.22(,P187,),：,Digital filter input/output as follows,Find the difference equation and impulse response of the,filter .,Simple Inverse z Transforms,Solution,：,The system transfer function is :,Simple Inverse z Transforms,Checking table,：,Example 6.23(,P188,),：,Find the inverse z transform of the following functions.,Simple Inverse z Transforms,solution,：,The inverse z transform is :,Inverse z Transforms by Long Division,If z-transform can written in the form of power series,The sequence of values x (n) is the coefficient of power series in,z,-n,.,Normally,available long division can deduce expansion to power series of,X,(,z,).,Example,：,Using the,power series expansion method to find the inverse z transform,Inverse z Transforms by Long Division,Solution,：,for the convergence region of X(Z) is a round of external,，,so ,The corresponding sequence should be the,right sequence,then,：,That is, X(Z) is,：,Inverse z Transforms by Long Division,Therefore, the inverse z transform of X(Z) is :,Example 6.24(,P188,),：,Find the inverse z transforms of,Inverse z Transforms by Long Division,Inverse z Transforms by Long Division,Solution,：,That is , H(Z),：,Inverse z Transforms by Long Division,Inverse z-transform of each item ,we have the impulse response :,Example 6.25,(,P189,),：,With long division method to find the inverse z transform of,Inverse z Transforms by Long Division,Inverse z Transforms by Long Division,Solution,：,That is ,X(z,),：,Inverse z Transforms by Long Division,The impulse response :,Inverse z Transforms by Partial Fraction Expansion,If the z-transform expression is,rational fraction,，,(1) partial fraction expansion,(2) inverse z transform,If,H,(z,),can be expressed as,ratio of the two polynomial,When , and all of the poles are 1-order,，,then,H,(,z,),can be expressed with partial fraction expansion,：,residue,pole,Inverse z Transforms by Partial Fraction Expansion,Example,：,for the right sequence of the following z-transform.,Solution,：,Example 6.26(P192),：,Find the inverse z transform of,solution,：,therefore,：,Checking,“,basics z transform,”,table,：,Inverse z Transforms by Partial Fraction Expansion,Example,6.27(P193,),：,Find the inverse z transform of,solution,：,Inverse z Transforms by Partial Fraction Expansion,The standard form,：,Then,：,Inverse z Transforms by Partial Fraction Expansion,Example 6.28(,P193,),：,Find the inverse z transform of,Solution,：,Inverse z Transforms by Partial Fraction Expansion,6.4 Transfer Function and Stability,poles,：,are the value of z that make the denominator of a transfer function zero.,zeros,：,are the value of z that make the numerator of a transfer function zero.,Poles have the biggest effect on the behavior of a digital filter. Zeros tend to modulate , to a greater or lesser degree depending on their positions relative to the poles.,The poles and the zeros can be found if its transfer function is known.,Example 6.30(P198),：,Find the poles and zeros of the filter.,Poles and Zeros,Solution,：,Transfer into standard form,：,Find the roots of denominator polynomial,：,Poles and Zeros,so,：,z,0 is a zero; z,0.25and z,2 are two poles respectively.,Usually, mark poles with “,”, mark zeros with,“,”,.,Example 6.31(,P199,),：,Find and plot the poles and zeros of the,transfer function,Poles and Zeros,Solution,：,The standard form,：,Zeros : z,2,0,，,that is , it have two zeros,，,both located at z,0. Find the poles with,：,Poles and Zeros,Example 6.32(,P200,),：,The digital filter has zeros at -,0.2 and 0.4,poles at -0.7,j0.6, and a gain of 0.5.,(1)draw the poles and zeros of the filter.,(2) find the transfer function of the filter.,Poles and Zeros,Solution,：,Poles and Zeros,The transfer function is,：,It can be seen from the above example,，,specifying the gain of a filter along with its poles and zeros is enough to completely specify the system .,Of these system characteristics , the poles of the filter have the greatest influence on overall filter behavior, they determine not only whether or not a filter is stable but also the general type of output response the filter will produce.,Stability,As long as the input are bounded,，,the output from a,stable filter,will always settle to some regular behavior.,When the input stays at a constant level,the output will settle to a constant level,；,when the input is a sinusoid , a stable filter output will eventually become sinusoidal .,When a filter is unstable,，,outputs grow without bound.,The output from an unstable filter can change dramatically even when the input changes by only the smallest amount.,All useful filters are stable, and one important aspect of filter design is to guarantee stability.,Stability,The,unit circle,is a circle with radius one centered at the origin of the z plane.,A filter is stable if all its poles are inside the unit circle. A filter with poles on the unit circle is said to be marginally stable. A filter with poles outside the unit circle is unstable,.,Stability,If the magnitude of each pole is less than one , the filter is stable.,Stability,| z | 1,，,the impulse response grows without bound as n increases,，,as long as,a,0 and,a,0,Example,6.35,(,P205,),：,Digital filter transfer function as follows , find the pole-zero plot for the filter. Is the filter stable ? find the impulse response and the step response for the filter.,First Order Systems,Solution,：,Expressed as standard form:,There is a zero at z,0,，,pole at z,0.4. Since the pole is within the unit circle ,the filter is stable .,The impulse input has a z transform of X(z),1,，,the output Y(z) is,：,First Order Systems,The inverse z transform,：,The impulse response tends to zero for stable system.,The step input has a z transform of,X(z),z/(z,1),，,output Y(z),is,：,First Order Systems,So,，,step response is,：,First Order Systems,The difference equation of the filter is,：,Since the input step function x(n),1,，,the final value of the output in steady state may be found from:,The step response of stable system is tends to be a constant.,Example 6.36,(P208),：,The transfer function of a filter is as,a. determine the difference equation of the filter ,b. find the pole-zero plot and evaluate stability ,c. find and plot the impulse response.,First Order Systems,Solution,：,the difference equation is,：,First Order Systems,There is a zero at z,0,，,a pole at z,0.4. Since the pole is within the unit circle ,the filter is stable .,Pole-zero are found from,：,First Order Systems,The impulse response is,：,The impulse response tends to zero for stable system,.,Second Order Systems,As long as the two poles are both within the unit circle, the system,is stable, that is require:,Definition,：,Second Order Systems has two poles .,The transfer function for a simple second order system is,：,Second Order Systems,The second order system has the difference equation :,Example 6.37(,P210,),：,The second order system has the poles as below ,no zeros, and a gain of one. Is the system stable? What is the transfer function of the system?,Solution,：,The magnitude for these poles is,：,The poles fall outside the unit circle, and the system is unstable, the transfer function is,Second Order Systems,Example 6.38(P210),：,A second order,transfer function is as below,find the pole-zero plot and find the magnitude of each pole, plot the impulse response and step response for the system.,Solution,：,The transfer function is written in standard form as,：,It has no zeros,，,and two poles at,Second Order Systems,The magnitude of the first pole is :,Since the magnitude both smaller than one ,the system is stable.,The magnitude of the second pole is :,Second Order Systems,The difference equation for system is,：,The step response is,：,The impulse response is,：,The impulse response tends to zero for stable system.,The step response of stable system is tends to be a constant.,Second Order Systems,Example 6.39(P211),：,Find the poles and zeros for the following transfer function, and determine the magnitude of the poles.,Solution,：,The transfer function has no zeros. The,poles,Two poles at z,0.1 and z,0.5. find the two magnitudes,：,clearly,，,the system is stable.,Second Order Systems,Example (P212-216),：,Discuss the following transfer functions,determine the poles and step response,Second Order Systems,Solution,：,Second Order Systems,When the pole real part is negative, Impulse response sampled values alternating between positive and negative; When the pole real part is positive, the impulse response does not alternate.,The step response of stable system is tends to be a constant , The impulse response tends to zero for stable system.,conclusion,：,Magnitudes of poles have great effect on the time that system tends to the final value.,The poles is closer to the center of the unit circle, the system become stable is faster.,If the systems poles with different modulus values, The transient characteristics mainly depends on the big modulus of the poles.,Zero will change the impact of the pole. Zero is more close to the pole,，,the greater the impact on system , zero is more away from the pole, their impact is smaller, and even can be ignored.,First and Second Order Systems,Example 6.40(P217),：,Calculate the steady state output for,the system transfer function,Solution,：,The steady state output, respectively:,Example 6.42(P219),：,Present the impulse response and pole-zero plots for the system,These four system have identical poles, and zeros that come progressively closer to the positions. The impulse response decreases in amplitude as the zeros come closer to the poles.,Poles,：,0.8+0.5408j, 0.8-0.5408j,magnitude value,：,0.9656,Summary,The z transform of a signal x(n) is calculated from,：,Common z-transforms are listed in tables.,Each z transform has a region of convergence, it is a range of z values,，,The region of convergence can distinguish the same z-transform defined for different intervals (n = 0 or all n).,The factor,z,k,in the z domain indicates a dela