信号与系统英文课件：Chapter 2 Linear Time-Invariant Systems

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 L

2、inear 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 ve

3、rsions of 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 sh

4、ifted unit impulse 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

5、-time LTI system is completely 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 overla

6、p between xk and hnkyn=00n1h - n 0n1h - 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

7、the two signals yn = xn*hnn 0, xk overlaps hnk0 nnkky na0n1h - 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*hna

8、lculate the convolution yn = RNn* RNn where1 01 0otherwiseNnNRnknN-101RNk 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 con

9、volution yn = RNn* RNn where1 01 0otherwiseNnNRn N1 2N2，No overlap again yn = 0nN-101RNnk-(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 agai

10、nyn = 0N-101kRNk*RNk2N-2N234123 * NNRnRnnalculate the convolution yn = RNn* RNn where1 01 0otherwiseNnNRn 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 in

11、tegral2.2.1 The Representation of Continuous-time Signals in Terms of Impulsesotherwisett,00,1)(Define We have the expression: kktkxtx)()()( Therefore: kktkxtx)()(lim)(0 2 Linear Time-Invariant Systemsmknkxnxdtxtx)()()(Sifting property)Pulse approximation(Sifting property) 2 Linear Time-Invariant Sy

12、stemskktkxtx)()(lim)(0Pulse approximation2.2.2 The Continuous-time Unit impulse Response and the convolution 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 Re

13、sponse)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) 2 Linear Time-Invariant SystemsTime-invariantAdditivityScalingmknhkxnyNote: A continuous-time LTI system is completely characterized by its unit impulse response h(t).(3) Com

14、putation of Convolution Integral Time Reversal: h() h(-)Time Shift: h(-) h(t-)Multiplication: x()h(t-)Integrating: dthxty)()()(Example 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(

15、) For t 0，)()(e)()(tuuthxttthtxe1de)(*)(0)()e1 ()(*)(tuthtxt)()(),(e)(),(*)(tuthtutxthtxtCalculate y(t) = p1(t) * p1(t)。)()(11tpp0.5t5 . 0t 5 . 01t1t5 . 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

16、 . 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) = x(t) h(t)(tht201)(tyt20113tt3= r(t)trapezoid2.3 Properties of Linear Time Invariant SystemC

17、onvolution 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 response. (Holds only for LTI system)2.3.1 The Commutative PropertyDiscrete time: xn*hn=hn*xnh(t)

18、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 of an LTI system with input x(t) and unit impulse response h(t) is identical

19、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)*h1(t)+h2(t)h1(t)x(t)y(t)=x(t)*h1(t)+x(t)*h2(t)h2(t)Example 2.10 2 Linea

20、r 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 Associative PropertyDiscrete time: xn*h1n*h2n=xn*h1n*h2nContinuous time: x(

21、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 they are cascaded. However, the order in which nonlinear systems

22、 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 to simplify the calculation.( )( )()x tttT( )x ttT0(1)( 1)(

23、)( )( )( ) ( )()( )()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 and time-invariant, we know 1）Differentia

24、l 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) leads to the output y1(t) ，please

25、 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 system with and witho

26、ut 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)inverse syste

27、m: 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

28、 Linear 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| |,|

29、| | | | 2 Linear Time-Invariant 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 differenti

30、al equation to get yh(t) y(t)+6y(t)+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 bywi

31、th initial conditions y(0)=1, y (0)=2. Please give 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

32、 conditions y(0)=1, y (0)=2. Please give the response y(t) 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 sig

33、nal，but different initial conditions y(0) = 0, y (0) = 1, then y(t) = ？Total 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:Method

34、s to solve the response of continuous-time LTI system 2 Linear Time-Invariant Systems 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 )sinc

35、os(e)sin cos(e)(11211h1tKtKtKtKtyinintti2/,jnisiii 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. The LTI system is given by y (t)+5y (t) +6y (t) =4x(t), t0 with initial conditions y(0-) =

36、1，y (0-) = 3. Please determine the zero-input response yzi(t) of this system.0652 ss3221ss，ttKKty3221ziee)(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)(22zit

37、ttytt（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) =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

38、 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 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 char

39、acteristic equation isThe characteristic roots are(conjugate roots) Methods to solve the zero-state response yzs (t): 1) Solve the differential equation with 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 w

40、ith 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 SystemsZero-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)

41、 =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 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)(d

42、tthtth)(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 this system.0),( 3)(2)(6d)(dttxtxtytty)( 3)(2)(6d)(dttthtth)()(e)( 6tBtuAthtA= 16, B =3)( 3)(2)()( e6+ )()

43、( 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(1)(01)1(1)(tbtbtbtbthathathathmmmmnnn (nm)Methods to solve the response of discret

44、e-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 given by yk-0.5yk-1=uk,with the initial condition of y-1 =1 . Please use the i

45、terative 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.phkykyky2. The Classical Approach 2 Linear Time-Inv

46、ariant 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 system The LTI system is given by yk-5yk-1+6y

47、k-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 equation is0652 rr3, 221rrkkCCky3221hTh

48、e 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 similar form as the in

49、put 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 ukPlease

50、 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

51、response + zero-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) Dif

52、ferent real characteristic 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. knnkkhrCrCrCky2211knnk

53、khrkCkrCrCky1210j2, 1ejbarkCkCkykkh0201sincosThe characteristic roots areThe characteristic equation is C1=1，C2= -20232 rr2, 121rrkkCCky)2() 1(21zi21412021 12121CCyCCy0)2(2) 1(zikkykk The LTI system is given by yk+3yk-1+2yk-2=xk with initial conditions y-1=0， y-2= 1/2，. Please determine the zero-inp

54、ut response yzik of this system.（different roots ） C1 = 4, C2= 4kkCkCky)2()2(21zi022 121CCy142221CCy0,)2(4)2(4zikkkykkThe characteristic roots areThe characteristic equation is0442 rr The LTI system is given by yk+4yk-1+4yk-2=xk with initial conditions y-1=0， y-2= 1/2，. Please determine the zero-inp

55、ut response yzik of this system.（multiple roots ） 221 rr C1= 1，C2= 0 ，C5= 505 . 05 . 023rrrkrr2j3 , 21ej, 5 . 0kCkCCkyk2cos2sin)21(321zi22 121CCy14231CCy88321CCy0,2cos5)21(zikkkyk The LTI system is given by yk-0.5yk-1+yk-2 -0.5yk-3 =xk with initial conditions y-1 = 2， y-2= -1， y-3= 8. Please determi

56、ne the zero-input response yzik of this system.The characteristic equation isThe characteristic roots are(conjugate roots) Methods to solve the zero-state response yzs k: 1) Solve the differential equation with initial state of zero. 2) The ：Calculate the convolution yzs k= xk*hk ：The zero-state res

57、ponse 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 SystemsZero-State Response The LTI system is given by: 22 13kxkykykywith ,)21( 3kukxk)2(2) 1(k

58、ukhkkPlease determine the 解：zsnkhnxkynnnknknnkunu)2(2) 1()21( 3000,)41()2(6)21() 1( 300kkknnknknk)21(51)2(524) 1(2kukkk The causal LTI system is given by Please determine the impulse response hk. 22 13kxkykyky22 13kkhkhkhCausal LTI system h-1 = h-2 = 0，122 1300hhh3 1203 1 1 hhh 注意：选择初始条件的基本原则是必须将k的作

59、用体现在初始条件中二阶系统需要两个初始条件，可以选择h0和h1The characteristic equation isThen we have0232 rr2, 121rr0,)2() 1(21kCCkhkkUsing the initial values we know32 1 102121CChCCh C1=1，C2= 2)2(2) 1(kukhkk The causal LTI system is given by Please determine the impulse response hk. 22 13kxkykykyThe characteristic roots are2.

60、4.3 Block Diagram Representations of First-order Systems Described by Differential and Difference Equation(1) Discrete time system Basic elements: A. An adder B. Multiplication by a coefficient C. A unit delay 2 Linear Time-Invariant SystemsBasic elements: 2 Linear Time-Invariant SystemsAn adderMultiplicati