信号与系统 Chapter 2 Time Domain Analysis of Continuous Signals

《信号与系统 Chapter 2 Time Domain Analysis of Continuous Signals》由会员上传分享，免费在线阅读，更多相关内容在行业资料-天天文库。

1、Chapter2TimeDomainAnalysisoftheContinuousSystemsξ2.1TheDifferentialEquationofSystemsξ2.1.1LinearConstant-CoefficientDifferentialEquationy(n)(t)+an-1y(n-1)(t)+…+a1y(1)(t)+a0y(t)=bmf(m)(t)+…+b0f(t)RCdy(t)/dt+y(t)=f(t)Ex2.1.1:GetiL(t)Equation?Solution:iL(t)+iC(t

2、)=iS(t)iC(t)=CduC(t)/dtuC(t)+R1iC(t)=LdiL(t)/dt+R2iL(t)diL2(t)/dt2+(R1+R2)/LdiL(t)/dt+1/(LC)iL(t)=R1/LdiS(t)/dt+1/(LC)iS(t)ξ2.1.2TheClassicSolutionofDifferentialEquationy(n)(t)+an-1y(n-1)(t)+…+a1y(1)(t)+a0y(t)=bmf(m)(t)+…+b0f(t)t>0:y(t)=yp(t)+yh(t)yh(n)(t)+an

3、-1yh(n-1)(t)+…+a1yh(1)(t)+a0yh(t)=0pn+an-1pn-1+…+a1p+a0=0yp(t)aparticularsolutionyh(t)ahomogeneoussolutionRoot:p1,p2,…pnp1≠p2≠…≠pn:yh(t)=c1ep1t+c2ep2t+…+cnepntp1=p2:yh(t)=c1ep1t+c2tep1t+…+cnepnty(i)(0+)=yp(i)(0+)+yh(i)(0+)y(0+)=yp(0+)+yh(0+)y(t)=yp(t)+yh(t)C1

4、C2…Cn=?Ex2.1.2:y(2)(t)+5y(1)(t)+6y(t)=f(t)f(t)=e-tε(t)y(0-)=3.5,y’(0-)=-8.5Gety(t)=?y(t)=(e-2t+2e-3t+.5e-t)ε(t)Solution:y(0+)=3.5,y’(0+)=-8.5y(t)ContinuousNo:δ(t)ξ2.2Zero-InputResponseofContinuousSystemsξ2.2.1Zero-InputResponseoftheSimpleSystems1.First-OrderS

5、ystemsf(t)=0:(p-λ)=0=yx(0+)eλty(1)(t)-λy(t)=f(t)yx(t)=Ceλtpole:p=λ=yx(0-)eλtb.λ1=λ2=λ2.Second-OrderSystemsy(2)(t)+a1y(1)(t)+a0y(t)=f(t)a.λ1≠λ2Prof:(p-λ1)(p-λ2)=0yx(t)=C1eλ1t+C2eλ2t(p-λ)2=0yx(t)=(C1+C2t)eλty’x(t)-λyx(t)=Ceλtd(e-λtyx(t))/dt=Cyx(t)=(C1+C2t)eλtyx

6、(0-),yx’(0-)ξ2.2.2Zero-InputResponseoftheSystemsy(n)(t)+an-1y(n-1)(t)+…+a1y(1)(t)+a0y(t)=bmf(m)(t)+…+b0f(t)Third-rootλk:(Ck1+Ck2t+Ck3t2)eλktA(p)=0Second-rootλj:(Cj1+Cj2t)eλjtFirst-rootλi:Cieλityx(t)=∑Cieλit+∑(Cj1+Cj2t)eλjt+∑(Ck1+Ck2t+Ck3t2)eλkt+…Ex2.2.1:y(3)(

7、t)+5y(2)(t)+8y(1)(t)+4y(t)=f(1)(t)+3f(t)y(0-)=3y’(0-)=-6y’’(0-)=13Getyx(t)=?yx(t)=(e-t+(2-t)e-2t)ε(t)Solution:ξ2.3Zero-StateResponseofContinuousSystemsf(t)yf(t)y(n)(t)+an-1y(n-1)(t)+…+a1y(1)(t)+a0y(t)=bmf(m)(t)+…+b0f(t)ξ2.3.1ImpulseResponsef(t)=δ(t)a.h(1)(t)-

8、λh(t)=kδ(t)h(t)=Ceλtε(t)=keλtε(t)b.h(2)(t)-2λh(1)(t)+λ2h(t)=kδ(t)t>0:h(2)(t)-2λh(1)(t)+λ2h(t)=0h(t)=(C1+C2t)eλtε(t)=kteλtε(t)δ’(t)C1=0δ(t)C2=kt>0h(1)(t)-λh(t)=0(Ceλt)ε(1)(t))=kδ(t)3Ex2.3.