信号与系统 Chapter 5 The S-Domain Trans For Continuous Signals and Systems.ppt

1、Chapter 5 The S-Domain Transform For Continuous Signals and Systems,5.1 The Laplace Transform,5.1.1 Define,f(t) F(j),f(t) e-t F(+j),-+|f(t)| e-t dt,-+|f(t)| dt,5.1.2 Region of Convergence (ROC),-+|f(t)| e-t dt,a. f(t)= e-t(t) ROC: -,c. f(t)= et(t) -et(-t)ROC: ,If ROC include j axis F(j)=F(s)| s=j El

2、se F(j)not exist,b. f(t)= -e-t(-t) ROC: -,5.1.3 The Unilateral Laplace Transform,f(t)=f(t)(t) Causal,5.1.4The Unilateral Laplace Transform for Some Signal,f(t)F(s)ROC,(t) 1 -,(t) 1/s 0,e-at(t) 1/(s+a) -a,eat(t) 1/(s-a)a,Rule ?,cos0t s/(s2+02) 0,sin0t 0/(s2+02) 0,5.2 Properties of The Unilateral Lapl

3、ace Transform,f(t)=F(s) -1F(s)= f(t) ROC =R (0),5.2.1 Linearity,Result: F(s)=(e-2(s+2)-e-4(s+2)/(s+2) ROC: -2,Ex5.2.2:f(t)=e-2t(t-2)-(t-4) F(s)=?,5.2.2 Time Shifting,Ex5.2.1: Proof:,Ex5.2.3: Proof:,5.2.3 Shifting in the s-Domain,Ex5.2.4: f(t)=e-tcos(0t)(t) F(s)=?,Result: F(s)=(s+)/(s+ )2+02),5.2.4 T

4、ime Scaling,Result:,15,5.2.5 Conjugation,5.2.6 Convolution Property,Ex5.2.6: h(t)=e-t(t) f(t)=(t) yf(t)=?,5.2.7 Differentiation in the Time Domain,Ex5.2.7: f(t)=(t)-(t-2) F(s)=?,Result: yf(t)=(1-e-t) (t),Result:F(s)= (1-e-2s)/sROC: 0,5.2.8 Differentiation in the s-Domain,Ex5.2.8: f(t)=te-t (t) F(s)=

5、?,Result: F(s)=1/(s+)2,5.2.9 Integration in the Time Domain,Ex5.2.9: f(t)=t (t) F(s)=?,Result: F(s)=1/s2,5.2.10 The Initial- and Final- Value Theorems,if ROC include s=0,if lims-+sF(s) exist,Ex5.2.10: F(s)=1/(s+1) f(0+)=? f(+)=?,Result: f(0+)=1 f(+)=0,5.3 The Unilateral Laplace Inverse Transform,F(s

6、)f(t),1(t),s(1)(t),1/s(t),1/(s+a)e-at(t),1/(s+a) 2te-at(t),s/(s2+02) cos(0t)(t),0/(s2+02) sin(0t)(t),mn if si sj first order root,Ex5.3.1,Result:,f(t)=?,16,Ex5.3.2,f(t)=?,Result:,Ex5.3.3,f(t)=?,Result:,Ex5.3.4,f(t)=?,Result:,Ex5.3.5,f(t)=?,Result:,5.4 Analysis of The System Using the Unilateral Lapl

7、ace Transform,5.4.1 The Zero-State Response of est,yf (t)= est H(s),H(s),f(t),yf(t),f(t)= est,yf (t)= est *h(t),=-+es(t-) h()d,=est0-+e-s h()d,!=est(t),H(s),f(t),yf(t),5.4.2 The Zero-State Response of f(t),yf (t)= f(t)*h(t),Yf (s)= F(S) H(s),yf (t)= -1 Yf (s),Linear System:,Ex5.4.1,f1(t)=e-t(t) yf1(

8、t)=(e-t-e-2t) (t) f2(t)=t(t) yf2(t)=?,Result:,yf2(t)=(1/2t-1/4+1/4e-2t) (t),5.5 The Solution of Differential Equation Using the Unilateral Laplace Transform,y(2)(t)+a1y(1)(t)+a0y(t)=b2f(2)(t)+b1f(1)(t)+b0f(t),s2 Y(s)-sy(0-)- y(1)(0 -)+ a1(sY(s)- y(0-)+ a0Y(s) = (b2 s2 +b1s+b0)F(s),y(t)= -1 Y(s) = yf

9、(t)+ yx (t),Y(s)=( sy(0-)+y(1)(0 -)+a1y(0-)/(s2 +a1s+a0) Zero-Input +(b2 s2 +b1s+b0)F(s) /(s2 +a1s+a0) Zero-State,Ex5.5.1 y(t)+3y(t)+2y(t)=2f (t)+6f(t),y(0-)=2,y(0-)=1,f(t)=(t),y(t)=?,Result:,s2Y(s)-sy(0-)-y(0-)+3sY(s)-3y(0-)+2Y(s)=2sF(s)+6F(s),y(t)=(3+e-t-2e-2t) (t),Ex5.5.2,C= 1/ 2 F,R1=2,R2=2,L=2H

10、,u1(0-)=1V,u(1)1(0-)=2V,iS(t)=(t),u1(t)=?,Result:,16,5.6 The Analysis of The RLC System Using the Unilateral Laplace Transform,R:,u(t)=R*i(t),U(s)=R*I(s),L:,u(t)=L*di(t)/dt,U(s)=sL*I(s)-Li(0-),I(s)=1/(sL)U(s)+1/s i(0-),C:,i(t)=C*du(t)/dt,I(s)=sC*U(s)-Cu(0-),U(s)=1/(sC)I(s)+1/s u(0-),Ex5.6.1:,IL(s)=?

11、,sL,Ex5.6.2:,uC1(0-)=3V,uC2(0-)=0V,i1(t)=?,Result:,5.7 The Representation of the System,Differential Equation,y(2)(t)+a1y(1)(t)+a0y(t)=b1f(1)(t)+b0f(t),Block Diagram Representation,Ex5.7.1:,Draw Flow Diagram ?,Result:,5.8 The System Function and Characterization,5.8.1 Zeros and Poles,B(s)=0s jZero,A

12、(s)=0 p jPole,Poles and time domain response,17,System Frequency Character:,Ex5.8.1: |H(j )|, ()=?,Result:,5.8.2 Causality,2. Rational System function ROC:right-half plane toThe right of the rightmost pole,Ex5.8.2:,Res-1 Causal ?,Result:,Causal,Roc,x,x,h(t)=(k1e-t+k2e-2t) (t),Ex5.8.3:,Res-1 Causal ?

13、,Result:,h(t)=(k1e-(t+1)+k2e-2(t+1) (t+1),h(-1)=(k1+k2) 0,No Causal,1. h(t)=0 t0,5.8.3 Stability,H(s) LTI,1. if |f(t)|+ then |y(t)|+,BIBO (Bounded Input Bounded Output),2. Stable if only if ROC include the j-axis,3. Stable if the rational system is Causal and poles lie in the left-half of the s-plan

14、e,Stable,Ex5.8.4:,1) Res2,2)2 Res-1,Stable ?,Result:,x,x,x,x,1) Res2,No Stable,h(t)=(k1e-t+k2e2t) (t),2)2 Res-1,Stable,h(t)=k1e-t (t) +k2e2t (-t),Ex5.8.5:,Result:,1) The system is causal (LTI),2)H(S) rational ,Only 2 poles S1=-2, S2=4,3) If f(t)=1 then y(t)=0,4) h(0+)=4,H(s)=?,y(t)=estH(s),f(t)=1=e0t,0=y(t)=e0tH(0),