课件：信号与系统Chapter-2.ppt

1、 2 Linear Time-Invariant Systems2.1 Discrete-time LTI system:The convolution sum2.1.1 The Representation of Discrete-time Signals in Terms of Impulses2.Linear Time-Invariant Systemskknkxnxnxnxnxnxnx 2 2 1 1 0 1 1 2 2If xn=un,then 0kknnuLinear combinations of delayed impulse(Sifting property)2 Linear

2、 Time-Invariant Systems2.1.2 The Discrete-time Unit Impulse Response and the Convolution Sum Representation of LTI Systems(1)Unit Impulse(Sample)Response LTIxn=nyn=hn Unit Impulse Response:hn 2 Linear Time-Invariant SystemsmknkxnxSifting property represents xn as a superposition of scaled versions o

3、f shifted unit impulse n-k(2)Convolution Sum of LTI System LTIxnyn=?Solution:Question:n hnn-k hn-kxkn-k xkhn-kkkknhkxnyknkxnx 2 Linear Time-Invariant SystemsTime-invariantAdditivityScalingmknkxnx 2 Linear Time-Invariant SystemsHere,hkn denotes the response of the linear system to the shifted unit im

4、pulse n-k.LTIhkn=hn-k 2 Linear Time-Invariant Systems(Convolution Sum)Sokknhkxnyor yn=xn*hn(3)Calculation of Convolution SumTime Reversal:hk h-kTime Shift:h-k hn-kMultiplication:xkhn-kSumming:kknhkxnyExample 2.1 2.2 2.3 2.4 2.5 2 Linear Time-Invariant SystemsNote:A discrete-time LTI system is comple

5、tely characterized by its unit impulse response hnTo visualize the calculationConsider the two sequences xn=un，hn=anun，0a1，please calculate the convolution of the two signals yn=xn*hn10kh k n或 h n 0n1h -n h -k k1.Viewed as functions of k2.Time reversaln 0,No overlap between xk and hnkyn=00n1h -n 0n1

6、h -n x khk,0h nkn x knkConsider the two sequences xn=un，hn=anun，0a1，please calculate the convolution of the two signals yn=xn*hn3.Time shift4.Multiplication&SummingConsider the two sequences xn=un，hn=anun，0a1，please calculate the convolution of the two signals yn=xn*hnn 0,xk overlaps hnk0 nnkky na0n

7、1h -n 0n1h -n x khkkn,0h nknkOver all values of kn 0,xk overlaps hnk0 nnkky nan 0,No overlap between xk and hnkyn=00k1y k y nnConsider the two sequences xn=un，hn=anun，0a1，please calculate the convolution of the two signals yn=xn*hnalculate the convolution yn=RNn*RNn where1 01 0otherwiseNnNRnknN-101R

8、Nk 或 RNkn-(N-1)01RN-nkNRkyn=0nN-101RNnk-(N-1)RNk-n,k k 0 n 0,No overlap between RN k and RN nknN-101RNnk-(N-1)RNk-n,k10Nk 0 n N 1,RN k overlaps RN nk in 0，n 0 11nky nn NRk NRk,0NRnkn(1)nNnk,01NRnknNkn(1)nNalculate the convolution yn=RNn*RNn where1 01 0otherwiseNnNRn N1 2N2，No overlap again yn =0nN-1

9、01RNnk-(N-1)RNk-n,k221NkN NRk,122NRnkNnN kn(1)nNalculate the convolution yn=RNn*RNn where1 01 0otherwiseNnNRn n 0,No overlap between RN k and RN nkyn=0 0 n N 1，Overlaps in 0，n 0 11nky nn N1 2N2，No overlap againyn =01 01 0otherwiseNnNRnN-101kRNk*RNk2N-2N234123*NNRnRnnalculate the convolution yn=RNn*R

10、Nn where alculate the convolution yn=xn*hn where nx nu n nh nu n*nnu nu n kn kku ku nk0000nkn kknn11(1)nnnu nna u n)(e)(etututt)(e)()ee(1tuttuattt2.2 Continuous-time LTI system:The convolution integral2.2.1 The Representation of Continuous-time Signals in Terms of Impulsesotherwisett,00,1)(Define We

11、 have the expression:kktkxtx)()()(Therefore:kktkxtx)()(lim)(0 2 Linear Time-Invariant Systemsmknkxnxdtxtx)()()(Sifting property)Pulse approximation(Sifting property)2 Linear Time-Invariant Systemskktkxtx)()(lim)(0Pulse approximation2.2.2 The Continuous-time Unit impulse Response and the convolution

12、Integral Representation of LTI Systems(1)Unit Impulse Response LTIx(t)=(t)y(t)=h(t)(2)The Convolution of LTI System LTIx(t)y(t)=?2 Linear Time-Invariant Systemsdtxtx)()()(Unit Impulse Response)A.LTI(t)h(t)x(t)y(t)=?dtxtx)()()(Because of dthxty)()()(So,we can get(Convolution Integral)or y(t)=x(t)*h(t

13、)2 Linear Time-Invariant SystemsTime-invariantAdditivityScalingmknhkxnyNote:A continuous-time LTI system is completely characterized by its unit impulse response h(t).(3)Computation of Convolution Integral Time Reversal:h()h(-)Time Shift:h(-)h(t-)Multiplication:x()h(t-)Integrating:dthxty)()()(Exampl

14、e 2.6 2.8 2 Linear Time-Invariant SystemsTo visualize the calculation)()(),(e)(),(*)(tuthtutxthtxt)(x)(hRegarded as a function of :x(t)x(),h(t)h()Reflection h()h()For t 0，)()(e)()(tuuthxttthtxe1de)(*)(0)()e1()(*)(tuthtxt)()(),(e)(),(*)(tuthtutxthtxtCalculate y(t)=p1(t)*p1(t)。)()(11tpp0.5t5.0t 5.01t1

15、t5.0t 5.0)()(11tpp01t1a)t 1b)1 t 0tttyt1d)(5.05.0)(1tp0.5-0.51t)(1py(t)=0)(1tp0.5-0.51t)(1pt 5.0t5.0)()(11tpp10t1c)0 1tttyt1d)(5.05.0y(t)=0 y(t)=p1(t)*p1(t)。)(1tp0.5-0.51t)(1pc)0 1tttyt1d)(5.05.0y(t)=0a)t 1b)1 t 0tttyt1d)(5.05.0y(t)=011-1)()(11tptpt y(t)=p1(t)*p1(t)。Exercise 1:u(t)u(t)Exercise 2:y(t

16、)=x(t)h(t)(tht201)(tyt20113tt3=r(t)trapezoid2.3 Properties of Linear Time Invariant SystemConvolution formula:dthxthtxty)()()(*)()(kknhkxnhnxny*h(t)x(t)y(t)=x(t)*h(t)hnxnyn=xn*hn 2 Linear Time-Invariant SystemsNote:The characteristics of an LTI system are completely determined by its impulse respons

17、e.(Holds only for LTI system)2.3.1 The Commutative PropertyDiscrete time:xn*hn=hn*xnh(t)x(t)y(t)=x(t)*h(t)x(t)h(t)y(t)=h(t)*x(t)2 Linear Time-Invariant Systems kky nx nh nx k h nkx nk h kh nx nContinuous time:x(t)*h(t)=h(t)*x(t)Proof:()()()()()()()()()y tx th txh tdx thdh tx tProof:Note:The output o

18、f an LTI system with input x(t)and unit impulse response h(t)is identical to the output of an LTI system with input h(t)and unit impulse response x(t).2.3.2 The Distributive PropertyDiscrete time:xn*h1n+h2n=xn*h1n+xn*h2nContinuous time:x(t)*h1(t)+h2(t)=x(t)*h1(t)+x(t)*h2(t)h1(t)+h2(t)x(t)y(t)=x(t)*h

19、1(t)+h2(t)h1(t)x(t)y(t)=x(t)*h1(t)+x(t)*h2(t)h2(t)Example 2.10 2 Linear Time-Invariant SystemsNote:A parallel combination of LTI systems can be replaced by a single LTI system whose unit impulse response is the sum of the individual unit impulse responses in the parallel combination.2.3.3 The Associ

20、ative PropertyDiscrete time:xn*h1n*h2n=xn*h1n*h2nContinuous time:x(t)*h1(t)*h2(t)=x(t)*h1(t)*h2(t)h1(t)*h2(t)x(t)y(t)=x(t)*h1(t)*h2(t)h1(t)x(t)y(t)=x(t)*h1(t)*h2(t)h2(t)2 Linear Time-Invariant SystemsNote:The unit impulse response of a cascade of two LTI systems does not depend on the order in which

21、 they are cascaded.However,the order in which nonlinear systems are cascaded can not be changed.If ，then 2 Linear Time-Invariant SystemsThe Time Shift PropertyThe Derivation PropertyIf ，then)()(*)(tythtx)()(*)()(*)(tythtxthtx)()(*)(tythtx)()(*)()(*)(000ttytthtxthttxNote:These properties can be use t

22、o simplify the calculation.()()()x tttT()x ttT0(1)(1)()()()()()()()()y tx th th tttTh th tTFrom the derivation property we know02T2Tt()h t 2 Linear Time-Invariant Systems0tTx(t)102T2Tt()h t=T2TT2T()y t3T2TT0t212T232TT3T2T0t()y t()()ty tyd 2 Linear Time-Invariant SystemsFrom the properties of linear

23、and time-invariant,we know 1）Differential or Difference property：If T x(t)=y(t)thenttyttxTd)(dd)(dIf Txk=yk then T xk-xk-1=yk-yk-1 2）Integral or Sum property：If Tx(t)=y(t)thend)(d)(yxTttIf Txk=ykthennynxTknkn 2 Linear Time-Invariant SystemsExample Consider an LTI system,we know that the input x1(t)l

24、eads to the output y1(t)，please determine the response of this system to the input x2(t)。The relation between x1(t)and x2(t)is as follows:d)()1()(11)1(12xtxtxtFrom the properties of linearity and time-invariance，we get the same relation between y2(t)and y1(t)d)()(11 2ytyt)1()e1(5.0)1(2tut2.3.4 LTI s

25、ystem with and without MemoryMemoryless system:Discrete time:yn=kxn,hn=kn Continuous time:y(t)=kx(t),h(t)=k(t)k(t)x(t)y(t)=kx(t)=x(t)*k(t)k n xnyn=kxn=xn*knImply that:x(t)*(t)=x(t)and xn*n=xn 2 Linear Time-Invariant Systemskknkxnxdtxtx)()()(2.3.5 Invertibility of LTI systemOriginal system:h(t)invers

26、e system:h1(t)(t)x(t)x(t)*(t)=x(t)So,for the invertible system:h(t)*h1(t)=(t)or hn*h1n=nh(t)x(t)x(t)h1(t)Example 2.11 2.12 2 Linear Time-Invariant Systems2.3.6 Causality for LTI systemDiscrete time system satisfy the condition:hn=0 for n0Continuous time system satisfy the condition:h(t)=0 for t0 2 L

27、inear Time-Invariant Systems*nhnxnykknhkxCausal Signal:xn=0 for n0 or x(t)=0 for t02.3.7 Stability for LTI system Definition of stability:Every bounded input produces a bounded output.Discrete time system:kkkhknxorknhkxnyIf|xn|B,the condition for|yn|A iskkh|AnythenkhifkhBkhknxnykkk|,|2 Linear Time-I

28、nvariant SystemsSufficient&necessaryContinuous time system:If|x(t)|B,the condition for|y(t)|A isdhtxordthxthtxty)()()()()(*)()(dh|)(|AtythendhifdhBdhtxtyttxtytyty0862 ss4221ss，ttKKty3221hee)(The characteristic roots areThen we get (1)Solve the homogeneous differential equation to get yh(t)y(t)+6y(t)

29、+8y(t)=0The characteristic equation ist0 (2)To determine the particular solution yp(t)to y(t)+6y(t)+8y(t)=x(t)yp(t)has the similar form of the input signal x(t)yp(t)=CetSubstitute yp(t)to the original system function,we gett0 The LTI system is given bywith initial conditions y(0)=1,y(0)=2.Please giv

30、e the response y(t)of this system to the input signal x(t)=e-t u(t).0),()(8)(6)(ttxtytytyC=1/3 (3)To get the complete solution A=5/2，B=11/6tttBAtytytye31ee)()()(42ph131)0(BAy23142)0(BAy0,e31e611e25)(42ttyttt The LTI system is given bywith initial conditions y(0)=1,y(0)=2.Please give the response y(t

31、)of this system to the input signal x(t)=e-t u(t).0),()(8)(6)(ttxtytytyUsing the initial conditions we knowThen we have1)With the same initial conditions,but different input signal x(t)=sin t u(t)，then y(t)=？2)Using the same input signal，but different initial conditions y(0)=0,y(0)=1,then y(t)=？Tota

32、l response=zero-input response+zero-state responseSolve the homogeneous differential equation to get yzi(t)Calculate the convolution x(t)*h(t)to get yzs(t)()()(zszitytyty)(*)()(zithtxty(2)The Convolution Method:Methods to solve the response of continuous-time LTI system 2 Linear Time-Invariant Syste

33、ms The Homogeneous Solution yh(t)(1)Different real characteristic roots s1,s2,sntsntstsnKKKtyeee)(2121h(2)Multiple real characteristic roots s1=s2=sn=stsnntststKtKKty 1 2 1heee)(3)Complex conjugate roots)sincos(e)sin cos(e)(11211h1tKtKtKtKtyinintti2/,jnisiii 2 Linear Time-Invariant SystemsZero-Input

34、 Response The zero-input response results only from the initial state of the system and not from any external drive.The LTI system is given by y(t)+5y (t)+6y(t)=4x(t),t0 with initial conditions y(0-)=1，y(0-)=3.Please determine the zero-input response yzi(t)of this system.0652 ss3221ss，ttKKty3221ziee

35、)(0,e5e6)(32zittytt y(0)=yzi(0)=K1+K2=1 y(0)=yzi(0)=2K13K2=3K1=6，K2=5The characteristic equation isThe characteristic roots are（different roots）0442 ss221 sstttKKty2221ziee)(0,e3e2)(22zitttytt（multiple roots）y(0)=yzi(0)=K1=1;y(0)=y zi(0)=2K1+K2=3 K1=2,K2=3 The LTI system is given by y(t)+4y (t)+4y(t

36、)=4x(t),t0 with initial conditions y(0-)=1，y(0-)=3.Please determine the zero-input response yzi(t)of this system.The characteristic equation isThe characteristic roots are0522 ssj21j2121ss，)2sin2cose)(21zitKtKtyt（y(0)=yzi(0)=K1=1y(0)=y zi(0)=K1+2K2=3 K1=1，K2=20),2sin22(cose)(zittttyt The LTI system

37、is given by y(t)+2y (t)+5y(t)=4x(t),t0 with initial conditions y(0-)=1，y(0-)=3.Please determine the zero-input response yzi(t)of this system.The characteristic equation isThe characteristic roots are(conjugate roots)Methods to solve the zero-state response yzs(t):1)Solve the differential equation wi

38、th initial state of zero.2)The：Calculate the convolution yzs(t)=x(t)*h(t)：The zero-state response is the behavior or response of a system with initial state of zero.It results only from the external inputs or driving functions of the system and not from the initial state.2 Linear Time-Invariant Syst

39、emsZero-State Response The LTI system is given by：y(t)+3y(t)=2x(t)with the impulse response h(t)=2e-3t u(t)and input signal x(t)=3u(t).Please determine the zero-state response yzs(t)of this system.d)()()()()(zsthxthtxtyd)(e2)(3=)(3tuut 0 00 d2e3=0)3(tttt 0 00 )e1(2=3ttt)()e12(=3tut The LTI system is

40、 given by Please determine the impulse response h(t)of this system.0),(2)(3d)(dttxtytty x(t)=(t)y(t)=h(t)，)(2)(3d)(dtthtth)(e)(3tuAtht)(2)(e3+)(edd33ttuAtuAtttA=2)(e2)(3tuthtThe characteristic root is s=3.And as nm,then we know The LTI system is given by Please determine the impulse response h(t)of

41、this system.0),(3)(2)(6d)(dttxtxtytty)(3)(2)(6d)(dttthtth)()(e)(6tBtuAthtA=16,B=3)(3)(2)()(e6+)()(edd66tttBtuAtBtuAttt)(e16)(3)(6tutthtThe characteristic root is s=6.And as n=m,then we knowx(t)=(t)()e()(1tuKthnitsii(nm)()()e()()(01tAtuKthjjnmjnitsii 将h(t)代入微分方程，使方程两边平衡，确定系数Ki，Aj)()()()()()()()(01)1(

42、1)(01)1(1)(tbtbtbtbthathathathmmmmnnn (nm)Methods to solve the response of discrete-time LTI system Using the initial values y1,y2,y2,yn and the input signal,the output can be iteratively given by01jkxbikyakyjmjini00jkxbikyajmjini1.Iterative Method 2 Linear Time-Invariant Systems The LTI system is g

43、iven by yk-0.5yk-1=uk,with the initial condition of y-1=1.Please use the iterative method to determine the output of this system.15.0kykukySubstitute the initial condition,we get:5.115.01 15.000yuy75.15.15.0105.0 1 1 yuy875.175.15.01 1 5.022yuyIterativelyShortcoming：Can not get the closed solution.p

44、hkykyky2.The Classical Approach 2 Linear Time-Invariant SystemsComplete solution=homogeneous solution+particular solutionyh k is determined by the characteristic roots of the homogeneous difference equation.yp k has the same form as the input signalMethods to solve the response of discrete-time LTI

45、system The LTI system is given by yk-5yk-1+6yk-2=x k with the initial conditions y0=0，y1=-1,and input xk=2k ukPlease determine the total response of this system.Then we get the homogeneous solution yhk as follows:(1)Solve the homogeneous difference equation yk-5yk-1+6yk-2=0The characteristic equatio

46、n is0652 rr3,221rrkkCCky3221hThe characteristic roots are0,2pkAkkyk The LTI system is given by yk-5yk-1+6yk-2=x k with the initial conditions y0=0，y1=-1,and input xk=2k ukPlease determine the total response of this system.(2)To determine the particular solution ypk to yk5yk1+6yk2=xkyp(t)has the simi

47、lar form as the input signal x(t)Substitute ypk to the original system function,we getA=2(3)To determine the complete solution yk C1=1，C2=10,232121phkkCCkykykykkk0021CCy1232 1 21CCy0,2321kkkykkk The LTI system is given by yk-5yk-1+6yk-2=x k with the initial conditions y0=0，y1=-1,and input xk=2k ukPl

48、ease determine the total response of this system.Using the initial conditions,we knowThen we get1)With the same initial conditions,but different input signal xk=sin0 k uk，then yk=？2)Using the same input signal，but different initial conditions y0=1,y1=1,then yk=？Total response=zero-input response+zer

49、o-state responseSolve the homogeneous difference equation to get yzi kCalculate the convolution xk*hk to get yzs k(2)The Convolution Method:Methods to solve the response of discrete-time LTI system 2 Linear Time-Invariant Systemszszikykyky*zikhkxky The Homogeneous Solution yh k(1)Different real char

50、acteristic roots r1,r2,rn(2)Multiple real characteristic roots r1=r2=rn(3)Complex conjugate roots 2 Linear Time-Invariant SystemsZero-Input Response The zero-input response results only from the initial state of the system and not from any external drive.knnkkhrCrCrCky2211knnkkhrkCkrCrCky1210j2,1ejb