CH2计算机控制技术（英文）

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1、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 2,,Discrete-time Systems,Analysis,Discrete-time systems,,Theory of the z-transform,,Signal sampling and reconstruction,,Pulse transfer function of sampled-data systems,,Stability , transient response and steady-state error,,,What is a discre

2、te time system?,They are systems in which the inputs and outputs are described by discrete samples in time domain.,Discrete Time,,System,u,k,y,k,k,y,k,k denotes the sampling instant at time t=kT,s,Inputs and outputs are not continuous in time but instead are sampled at t=kT,s,where T,s,is the sampli

3、ng interval.,,Sampling frequency = 1/T,s,Hz or 2,p,/T,s,rad/s.,Continuous time,Discrete samples,k,u,k,1,2,3,,Discrete-Time Systems,A Discrete-Time System,transforms discrete-time inputs to discrete-time outputs.,,The output at a particular time index depends on both the input at specific index value

4、s and output values at previous indices.,,In contrast to a continuous-time system whose operation is described (or modeled) by a set of differential equation, a discrete-time system can be described by a set of,difference equations,(,差分方程,),.,,How do you describe the input-output behavior of discret

5、e time systems?,Do so with,difference equations,instead of differential equations,Examples of difference equations (DE),1,st,order DE :,2,nd,order DE :,3,rd,order DE :,Compare with ordinary differential equations (ODE),1,st,order ODE :,2,nd,order ODE :,,Converting ODEs to difference equations,Approx

6、imate,by,Hence, 1,st,order ODE will lead to 1,st,order difference eqns,2,nd,order ODE will lead to 2,nd,order difference eqns,Thus, easy to see how continuous time systems can be converted into approximate discrete time models.,,Transform Methods,In linear time-invariant (LTI) continuous-time system

7、s, the,Laplace transform,,can be used in system analysis and design.,,,In linear time-invariant discrete-time systems, the,z-transform,,is utilized in the analysis of the system described by difference equations.,,What is,z,-transform ?,,Signal Sampling,,,x(t,),,载波器,脉冲调制器,x*,（,t,）,,x,(,t,),t,x,*,(,t

8、,),t,x,(,t,),x,*,(,t,),S,Sampling Switch,,The L-transform of,x,*(t):,=1 + e,-Ts,+ e,-2Ts,+ …,Example: Unit step signal,x,(t) =1(t).,Transcendental function !,,X(z),is called the,z-transform,of discrete signal,x,*(t). Since,x*(t),,is a sampled series from the signal,x(t),,we may also say,X(z),is t

9、he z-transform of,x(t).,Hence the following notation:,Let,e,Ts,= z,, then,In some cases,,x(kT),is written simply as,x(k).,,,The Unilateral Z transform,(,单侧,z,变换）,In control systems analysis, we use the unilateral z transform.,,Justified because in control systems, we only deal with signals that a

10、re causal.,2,….,k,1,0,Due to the infinite sum, convergence is an important issue. Ideally, the region of convergence (ROC) should be stated. ROC refers to the region on the complex plane on which the transform exists,,Bilateral z Transforms,(,双侧,z,变换）,,Given a sample sequence,,{… x(-2), x(-1), x(0),

11、 x(1), x(2), …},,,we define the bilateral Z-transform as,,Example : Unit Impulse,The discrete version of an unit impulse (with delay),,d,(t-t,0,), is defined to be,By definition of the z-transform :,If k,0,= 0, D(z)=1!,,Example,：,impulse series,,,Example : Unit Step,2,….,k,1,0,A step sequence,,Regio

12、n of convergence is |,z,| > 1.,Pole at z = 1,,2,….,k,1,0,a<1,2,….,k,1,0,a>1,Region of convergence is |z| > a,Pole at z = a,What does this tells us about the relationship between stability and poles?,Power series,,2,….,k,1,0,a ramp sequence,How to show this ?,,For unit-step signal,：,Multiply both sid

13、es with,,(-Tz),，,and obtain the z-transform of unit ramp function:,Take derivative with respect to z:,Proof:,,Example,：,Exponential function,(,指数函数,),,,x,(,t,)=,e,-,a,t,(a : constant parameter〕,,This is a geometric series with a common ratio of (,e,-,aT,,z,-1,),,，,When,|,e,-aT,,z,-1,,|,<1,，,this

14、series is convergent and can be written in closed form as follows:,,Example,：,Sinusoidal signal (,正弦信号,),,,,x,(,t,)=,sin t,,Properties of z-transform,Linearity:,,If,X,(,z,)=,Z,[,x,(,t,)],，,Then,,Real Translation (Time Shift,,实数位移定理,),,实数位移,定理,,若,,X,(,z,)=,Z,[,x,(,t,)],，,则,Proof,：,假定,k<0,时,x(kT)=0,,

15、y,k,k,0,1,2,3,4,5,6 . . .,y,k-2,k,0,1,2,3,4,5,6 . . .,shift,sequence {y,k,},sequence {y,k-2,},Appears like a,,delayed sequence,If,Z,{y,k,} = Y(z), then,Z,{y,k-2,} = z,-2,Y(z),,Example,：,x,( t )= t,2,, solve for,X,(z).,,Solution,：,x,( t )= t,2,,,x,(0)=0。,,,x,(t+T) = (t+T）,2,= t,2,+ 2Tt + T,2,,,x,

16、(t+T)-,x,(t) = T (2t + T),,Taking z-transform:,By time shift:,,,Complex Translation(,复平移定理,),,If,x,(,t,),,,X,(,z,),，,then,,,,,Example:,,,X(z)=?,,,Differentiation Law,(,z,域微分定理,,),Example:,x(t)=t,3,, find X(,z,).,Proof:,If,x,(,t,),,,X,(,z,),，,then,,,,Example:,,,X,(z)=?,Solution:,By time shift:,C

17、onsider {,x(k),} as a sampled series from signal,x(t)=t a,t-1,,with T=1 and Apply the differentiation law:,,If,x,,(,n,),,,X,(,z,),，,then,,,Z,[,a,n,,x,(,n,)]=,X,(,z,/,a,),,,a,: constant.,,Scaling Theorem,,(,z,域尺度定理,),Solution,：,,Example:,,Find the z-transform of,,,Initial Value Theorem,,(,初值定理,),P

18、roof,：,By the definition of z-transform:,,Final Value Theorem,(,终值定理,),Proof,：,By the definition of z-transform:,and,,By time shift:,,Final Value Theorem,Given the Laplace Transform, X(s) of signal,x,(t), the final value is given by,Given the definition of the z transform, X(z) of a sequence,,In

19、both cases, we assume that the signal is stable.,,The discrete convolution of two sampled sequences,x,(,nT,) and,y,(,nT,),,is defined as,,,，,,then,：,,,Z,[,x,(,nT,),*,y,(,nT,)]=,X,(,z,),Y,(,z,),,Convolution of Time Sequence,(,离散,卷积定理,,),Proof:,,Let,n-k=m,, then,m=-k,when,n=0,.,Substitute,n=m+k,into E

20、qn(1),,:,,Useful z-transform pairs,,By partial fraction expansion:,Taking inverse L-transform:,Taking z-transform:,Example,: the case when the signal is given by its L-transform.,,Inverse z transform,Given,X(z),,what is the sequence,x(k),?,Power series expansion,(,幂级数法,),,long division to obtain se

21、quence,,,Partial fraction expansion,(,部,分分式展开法，又称查表法,),,makes use of known z-transform pairs to find the closed form inverse solution,,,Inverse transform integral,(,反演积分法、留数法）,,makes use of Residue calculus,,,,Power series expansion,(,幂级数法,),,,X(z,),can be written as a rational function,：,,,,Using,l

22、ong division (,综合除法,),to obtain a sequence,,:,k,c,kT,x,=,),(,Then by definition:,,,)=?,(,),(,k,x,a,z,z,z,X,,,+,=,Solution:,Example:,,Example,: Given ,,Find the corresponding signal,x,*(t).,Solution:,,（2）,Partial fraction expansion,,,(,部,分分式展开法，又称查表法,),,,Decompose,X,(z)

23、 into,Partial fraction expansion,and then make use of known z-transform pairs to find the closed form inverse solution.,,Example:,Given,, x(kT)=?,Solution:,,Example:,,Given , find its inverse z transform.,,Solution,：,First decompose,X,(,z,)/,z into part fraction expa

24、nsion,x,(,nT,)=(-1+2,n,),,10,,,x,*,(,t,)=,x,(0),,(,t,),+,x,(,T,),,(,t,-,T,),+,x,(,2T,),,(,t,-,2T,),+…,,,,=,0+10,,(,t,-,T,)+,30,,(,t,-,2T,)+,70,,(,t,-3,T,)+…,,From the look-up table:,,=,x,(0) +,x,(T)z,-1,+,x,(2T)z,-2,+,…,,Multiply both sides with,z,k,-1,:,X,(,z,)z,k-1,=,x,(0)z,k-1,+,x,(T)z,k-2

25、,,+ … +,x,(kT)z,-1,+ …,,The above is a Laurent series, and,x,(kT) is the coefficient of z,-1,. Applying the relevant theorem in,theory of,complex variables:,,(3) Inverse transform integral,(,反演积分法、留数法）,,Where C is a closed contour in the ROC of,X(z).,,,By,Cauchy integral theorem,：,,where,z,i,,is th

26、e pole of,X(z)z,k-1,inside contour C.,,Residue calculus,If has a pole with multiplicity r, and another (m-r) poles at different locations, then,,Solution:,Example:,Given,,,Find the corresponding signal,x,*(t).,,Solution to Difference Equation,,—,Numerical solution:,,,Sequential Pr

27、ocedure,(,迭代法,),:,,compute the D.E. recursively from some initial values.,,used in computer solution of D.E.,,— Analytical solution:,,1. assuming some form for the solution with unknown constants and solve for the constants to match the initial conditions.,,2.,using z-transform,,The,z-transform Appr

28、oach,(z,变换法,),Determine the z-transform of the D.E. using the real translation property;,,Solve the algebraic equation for,Y(z);,,Obtain the solution of,y(k),by taking the inverse z-transform.,,Example,：,Find the solution to the following D.E.,,,,y(0) = -1,Solution,：,Taking the z-transform on both

29、sides of D.E.,,From the look-up table, we have,,,Discrete Transfer Function,Consider a first order difference equation is given by :,Taking Z-transforms on both sides :,a,,,b,are constants,Transfer function in z-domain!,Pole at z =,a,,,For a 2,nd,order difference equation :,Taking Z-transforms on bo

30、th sides :,Poles at z = -1, -2,,,,u,k,y,k,What is the unit step response of this discrete time system?,Z.T.,Note the need to set z aside,,Taking inverse z-transform :,The z-transform of the output :,for,k=0,1,2,3,4,,…,Output sequence :,Output samples getting larger for large k. Therefore system is u

31、nstable!,,Does not get,,larger with k,Does not get,,larger with k,,Grows,,larger with k,Recall that the poles are at z = -1 and -2.,Hence we conclude that poles with magnitudes |z| > 1 lead to unstable systems!,Poles with magnitudes |z| <1 are stable.,,Signal Sampling and Reconstruction,,Signal Samp

32、ling,,,x,(,t,),x,*,(,t,),S,Sampling Switch,(,a,),,t,x(t),(,b,),t,x*(t),,Obviously, is a periodic function,，,hence can be expanded into Fourier series:,,where is the sampling frequency.,,,,Hence,,Taking Laplace transform and using complex translation theorem:,Its sp

33、ectrum,（频谱）,can be given by,,-,,-,X(j ),0,0,-,-,(,a),,Spectrum of x(t).,,(,b),Spectrum of,,x*,（,t,）,（ ＞,2,）,,,Ideal low-pass filter,Spectrum preserved,,,Signal x(t) can be recovered,,Aliasing (,混叠,):,ω,s,,< 2,ω,max,Spectrum overlap,,,Signal distorted,,,Can not be recovered.,,Nyquist Sampling Th

34、eorem (,采样定理,),Nyquist Sampling Theorem,:,,One can recover a signal from its samples if the sampling frequency (,ω,s,,= 2π/T) is at least,,twice,the highest frequency (,ω,max,,) in the signal,,i.e.,,Put in another way:,,For a given sampling frequency,ω,s,, only when the highest frequency (,ω,max,,)

35、of the signal is no larger than,half,of sampling frequency (,ω,s,) can we recover the signal without any distortion,,i.e.,,Nyquist frequency,,Ideal low-pass filter,-,Ideal Reconstruction of Signal,After filtering:,,Impulse response:,Noncausal,！,,Can not be implemented physically.,,t/T,1,2,3,-1,-2,-3

36、,,Signal,,Reconstruction: a,polynomial extrapolation approach.,Using a,Taylor’s series expansion,about,t = nT,,,,We define,as the reconstructed version of,x(t).,Such a mechanism is called,data hold,,,,and,x,h,(t),is the output of the data hold.,,,If only the first term of the Taylor’s series is use

37、d, the data hold is called a,,zero-order hold,(,零阶保持器,),,i.e.,,If the first two terms of the Taylor’s series are used, it is the,first-order hold,(,一阶保持器,),,,i.e.,,We approximate the derivatives by,backward difference,.,,Zero-Order Hold,(,ZOH,,零阶保持器,),ZOH is the most commonly used data hold,，,it mai

38、ntains the sampled value for the whole sampling period,，,and output a staircase signal.,,x,h,(,t,),x,*,(,t,),x,*,(t),t,,ZOH,x,h,(,t,),t,,Taking Laplace transform:,,Hence the transfer function of ZOH is given by,Then,,Sampling and Hold,,x,h,(t),G,h,（s）,x*(t),x(t),Sampler,,,Data Hold,For ZOH:,,,Frequ

39、ency response of ZOH,Amplitude:,Phase:,,Frequency response of ZOH,T,High frequency components are attenuated,,,but can not be totally erased;,,Phase delay related to T.,,First-Order Hold (FOH,,,一阶保持器,),0,T 2T 3T …..,,Its frequency response:,The transfer function of FOH is given by,where,,Frequency r

40、esponse of FOH,Conclusion: FOH is not better than ZOH.,FOH,ZOH,,Pulse transfer function,(,脉冲传递函数,),,1.,,Open loop,Pulse transfer function,,G(s),r*( t ),,r ( t ),y*( t ),,,y ( t ),Pulse transfer function,(z,transfer function,, discrete,transfer function,) is defined as the ratio of the z-transform of

41、 output y*(t),,,or Y(z), to that of input r*(t), or R(z),，,i.e,., H(z)=Y(z)/R(z).,,,Any continuous-time signal r(t) sampled by an ideal sampler with period T will produce a train of pulse signal as:,If,,is input into G(s),,If the input is ,,Assuming that the continuous output c(t) is also

42、 sampled by an ideal sampler as that of input, then the output sample at t=nT,,is,,By the theorem of discrete convolution,：,G(z)=,Z,[,G(s),],,Generally G(z),,can be written as :,Caution,：,G(z) is determined by the structure and parameters of the discrete system, and is independent of the reference

43、input.,,Example,: find the z transfer function for the system with the following s transfer function:,Solution:,,Example,:,determine the pulse transfer function for the following open-loop sampled-data system :,r*( t ),,r ( t ),y*( t ),,y ( t ),Solution:,,Pulse transfer function of cascaded systems

44、,,Case 1: No sampler between two cascaded subsystems,G,1,(s),G,2,(s),,r*(t),,r ( t ),,y*(t),y(t),,The block diagram can be reduced to,G,1,(s)G,2,(s),,r*(t),,r (t),y*(t),y(t),Then,,Let,,Case 2: There is a sampler between two cascaded subsystems, and samplers are synchronized.,,y*(t),y(t),G,1,(s),G,2,

45、(s),,r*(t),,r ( t ),y,1,*(t),,Case 3: Open loop system preceded by a ZOH.,G,p,(s),,r*(t),,r ( t ),y*(t),y(t),,3.,Pulse transfer function of closed-loop discrete systems,,y*(t),G,1,(s),G,2,(s),H(s),r(t),,e(t),e*(t),,d(t),,b(t),,y(t),-,+,+,Figure: Linear,discrete system with disturbance,By assuming d(

46、t)=0, the diagram can be reduced to:,,,Figure: Linear,discrete system,By the definition of pulse transfer function:,G,1,(s),G,2,(s),H(s),r(t),,e*(t),,y*(t),,y(t),b(t),,Define the error pulse transfer function G,e,(z),,as:,,Hence the closed-loop pulse transfer function G,B,(z) is given by,,Now assume

47、 r(t)=0, and obtain the following diagram with disturbance as an equivalent input:,G,2,(s),G,1,(s),H(s),r(t)=0,,e*(t),,y*(t),-,y(t),,d(t),+,+,Figure: Linear,discrete system with disturbance,as input.,,,,Example: Consider the following sampled-data system:,G(s),H(s),,r(t),b*(t),y*(t),y(t),－,,,Analys

48、is of Discrete Systems,Transient response,,Stability,,Steady-state error,,1.,Transient response,,,,Closed-loop transfer function of a typical discrete system:,N(z),and,D(z),are monic polynomial of,z,.,,The unit step response is given by,By partial fraction expansion:,where,,(1) p,k,is real:,Case a

49、: p,k,=1,,y,k,(n) is a,constant sequence.,,The output series:,,Case b: 01, expanding geometric sequence.,,Case e: -1

50、geometric sequence with alternating signs.,,,Summary: transient response with,,a single real pole p,k,Im,Re,[,Z],,,f,,f,,d,,d,a,a,,c,,c,,b,,b,e,e,,(2),,p,k,is conjugate complex (in pairs),Then, c,k,and c,k+1,form a conjugate pair:,,,The magnitude of pole, |p,k,|, will determine whether the response

51、 is convergent or divergent.,The transient response:,,Case a: |p,k,|<1, damped sinusoidal sequence,Case b: |p,k,|=1, sinusoidal sequence,,Case c: |p,k,|>1,,,exponentially expanding sinusoidal sequence,,A larger means faster oscillation in the transient response.,Let,,θ,k,=ω,d,T,,then,,is th

52、e oscillating frequency of the response, and the period of oscillation,,is given by,The impact of the argument,,(3). Deadbeat system (,有限时间响应系统,),,When all the closed-loop poles are at the origin, the transient response will settle down within limited periods. Such a system is called deadbeat syst

53、em.,,The unit impulse response:,The transient process will die out after n periods. This property is never found in a continuous-time system.,,A very important qualitative property of a dynamic system is,stability,.,,Internal stability,is concerned with the responses at all the internal variables.,,

54、External stability,is concerned with the input-output relation.,,The most common definition of,appropriate response,is that for every,Bounded Input,, we should have a,Bounded Output,.,i.e.,,we call the system,BIBO stable,.,2. Stability Analysis,,,Linear Discrete System:,G（s）,,r(t),,y,*,(t),y (t),,_,

55、,If all closed-loop poles of a system is,inside,the unit circle, the system is,stable,.,,If at least one pole is,on or outside,the unit circle, the corresponding system is,not BIBO stable,.,1+,G,(z)=0,,,T,he stability boundary,of,discrete-time,systems (,in the,,z-plane,),is different from that,of co

56、ntinuous systems (,in the s-plane,).,,How does this happen?,Consider the following mapping (,from s to z):,,,z,,= e,Ts,,For any point in the s-plane: s =,σ,+jω,,,then after mapping, the point in z-plane is:,,,case 1,:,σ,=0，the imaginary axis in s-plane is mapped into the unit circle in z-plane – st

57、ability boundary.,,,case 2,:,σ,<0，the LHP of s-plane is mapped into the interior of the unit circle in z-plane – stability region.,,case 3,:,σ,>0，the RHP of s-plane is mapped into the exterior of the unit circle in z-plane – instability region.,s =,σ,+jω,,Re,Re,Im,Im,Mapping the s-plane into z-plane

58、,s-plane,z-plane,s =,σ,+jω,,Ways to check stability,,,Direct calculation: for simple cases;,,,Bilinear transform + Routh test;,,,Jury’s test,: similar to Hurwitz test in continuous-time case.,,,Other ways,:,root locus, Nyquist stability criterion, Lyapunov theorem, etc.,,Solution,：,,,The open loop

59、pulse transer function is,Example,: Check the stability of the following sampled-data system with T=1s.,,,r (t),y,*,(t),,y (t),,1+,G(z)=0,,z,2,+4.952z+0.368=0,,z,1,=-0.076 z,2,=-4.876,There is one pole outside the unit circle, hence the,,system is unstable.,The closed-loop C.E. is given by,,,Defi

60、ne,(,1),The two complex variables,z,and,w,can be written as,z=x+jy,,w=u+jv,(,2),(,3),Substitute (2) and (3) into (1):,Bilinear transform + Routh test,then,Bilinear transform,,,w-transform,,Case 1: x,2,+y,2,=1 , the unit circle in z-plane,,,,u=0 the imaginary axis in w-plane.,,,Case 2: x,2,+

61、y,2,<1, the interior of unit circle in z-plane,,,,u<0 the left half of w-plane.,,,Case 3: x,2,+y,2,>1, the exterior of unit circle in z-plane,,,,u>0 the right half of w-plane.,z=x+jy,,w=u+jv,,For Discrete-time systems: poles are inside unit circle (,z,plane)?,Stability ?,,

62、For Continuous-time systems: poles are on the left half plane (,w,domain) ?,Bilinear transform,,Routh test,,,Given a sampled-data system with,T,=1s .,,Check its stability for the case when,,K,=10, and find the critical gain K.,Example:,Solution,： ⑴,The closed-loop CE is:,,z,2,+2.31,z,+3=0,,By manua

63、l calculation:,,,1,=-1.156,＋,j,1.29,,2,=-1.156-,j,1.29,,Both poles are outside the unit circle,，,hence the system is unstable.,,C,(,s,),R,(,s,),,—,Open-loop pulse TF:,when,,K,=10, the closed-loop TF:,,CE: 1+G(z)=0,,,,z,2,-(1.368-0.368,K,),z,+(0.368+0.264,K,) =0,⑵,Open loop pulse transfer funct

64、ion:,The critical value of,gain,,K is :,,K,c,=2.4,Routh array:,,,w,2,0.632,K,2.736-0.104,K,,,w,1,1.264-0.528,K,0,,,w,0,2.736-0.104,K,,For stability, we need,After,w,-transform:,,,0.632,Kw,2,+(1.264-0.528K),w,+(2.736-0.104K) =0,0<,K,<2.4,,Stability Conditions for 2nd order monic C.E.,CE:,f(z)=z,2,+a

65、z+b=0,,,,Sufficient & Necessary Conditions for Stability:,f(1)>0,,f(-1)>0,,|f(0)|<,1,,3,.,Steady-state error in discrete-time systems,Consider a discrete-time system with unit feedback:,,G（s）,,r(t),,y (t),e,*,(t),-,The error transfer function:,,By final value theorem:,,Based on the number of open-lo

66、op poles at,,z=1, the open-loop system G(z) can be,categorize,d as Type 0, Type 1, Type 2,,……,The error is given by,,1),Unit step input,Define as position error constant.,Type 0:,Type 1+:,,2),Unit ramp input,Define the velocity error constant K,v,as,Then,,System Type 0,：,,K,v,=0,Type 1,：,,Where G,1,(z) has no pole at,,z=1.,Type 2,+ :,,,3),Acceleration input,,Define the acceleration error constant K,a,,as,,Type 0,,,1,:,K,a,= 0,Type 2,：,Type 3,+:,,Steady-state error versus Di