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

1、Chapter 2 Time Domain Analysis of the Continuous Systems,2.1The Differential Equation of Systems,2.1.1Linear Constant-Coefficient Differential Equation,y(n)(t)+an-1y(n-1)(t)+a1y(1)(t)+a0y(t)= bmf(m)(t)+b0f(t),RC dy(t)/dt+y(t)=f(t),Ex2.1.1: Get iL(t) Equation?,Solution:,iL(t)+iC(t)=iS(t),iC(t)=CduC(t

2、)/dt,uC(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),2.1.2 The Classic Solution of Differential Equation,y(n)(t)+a n-1 y(n-1)(t)+a 1 y(1) (t)+a 0 y(t)= b m f(m) (t)+b 0 f(t),t0: y(t)=y p (t)+y h (t),yh (n)(t)+an-1yh (n-1)(t)+a1yh(1)(t)+a0yh (t)=

3、0,pn+an-1pn-1+a1p+a0=0,yp (t) a particular solution,yh (t) a homogeneous solution,Root: p1,p2,pn,p1p2 pn:,yh(t)=c1ep1t+c2ep2t+cnepnt,p1=p2:,yh(t)=c1ep1t+c2 tep1t+cnepnt,y(i)(0+)=yp(i)(0+)+yh(i) (0+),y(0+)=yp(0+)+yh(0+),y(t)=yp(t)+yh(t),C1 C2 Cn=?,Ex2.1.2:y(2)(t)+5y(1)(t)+6y(t)=f(t) f(t)=e-t(t) y(0-)

4、=3.5,y(0-)=-8.5 Get y(t)=?,y(t)=(e-2t+2e-3t+.5e-t)(t),Solution:,y(0+)=3.5,y(0+)=-8.5,y(t) Continuous,No: (t),2.2 Zero-Input Response of Continuous Systems,2.2.1 Zero-Input Response of the Simple Systems,1. First-Order Systems,f(t)=0: (p-)=0,=yx(0+) et,y(1)(t)-y(t)=f(t),yx(t)=C et,pole: p=,=yx(0-) et

5、,b. 1= 2= ,2. Second-Order Systems,y(2)(t)+a1y(1)(t)+a0y(t)=f(t),a. 1 2,Prof:,(p-1) (p-2)=0,yx(t)=C1 e1t +C2 e2t,(p-) 2 =0,yx(t)=(C1 +C2 t)et,yx(t)- yx(t) =C et,d(e-tyx(t)/dt=C,yx(t)=(C1 +C2 t)et,yx(0-),yx(0-),2.2.2Zero-Input Response of the Systems,y(n)(t)+a n-1 y(n-1)(t)+a 1 y(1) (t)+a0 y(t)= bm f

6、(m) (t)+b0 f(t),Third-root k: (Ck1 +Ck2 t+ Ck3 t2)ekt,A(p)=0,Second-root j: (Cj1 +Cj2 t)ejt,First-root i: Cieit,yx(t)= Ci eit,+(Cj1 +Cj2 t)ejt,+(Ck1 +Ck2 t+ Ck3 t2)ekt,+,Ex2.2.1: y(3)(t)+5y(2)(t)+8y(1)(t)+4y(t)=f(1)(t)+3f(t) y(0-)=3 y(0-)= -6 y(0-)=13 Get yx(t)=?,yx(t)=(e-t+(2-t)e-2t)(t),Solution:,2

7、.3 Zero-State Response of Continuous Systems,f(t) y f (t),y(n)(t)+an-1y(n-1)(t)+a 1 y(1)(t)+a0 y(t)= bm f(m) (t)+b0 f(t),2.3.1 Impulse Response,f(t)= (t),a.,h(1)(t) -h(t)=k(t),h(t)= C e t(t),=k et(t),b.,h(2)(t) -2 h(1)(t)+ 2h(t)=k(t),t0: h(2)(t) -2 h(1)(t)+ 2h(t)=0,h(t)= (C1 +C2 t) et(t),=k t et(t),

8、(t) C1=0,(t) C2=k,t0 h(1)(t) -h(t)=0,(C et)(1)(t)= k(t),3,Ex2.3.1: Get h(t)=?,Solution1:,Solution2:,2.3.2 Zero-State Response of LTI System,(t),(t-),f()(t-),f(t),h(t),h(t-),f()h(t-),y f (t),=f(t)*h(t),2.3.3 Unit Step Response of LTI System,(t),h(t),(t),g(t),g(t) =-t h() d,= (t)*h(t),Ex2.3.2: y(3)(t)

9、+5 y(2)(t)+8 y(1)(t)+4y(t)=f(3)(t)+6f(2)(t)+10f(1)(t)+6f(t) Get h(t)=? g(t)=?,h(t)=(t)+ (e-t-2te-2t)(t),Solution:,g(t)=3/2-e-t+(1/2+t)e-2t)(t),6=c1+c2+5 10=4c1+3c2+c3+8 6=4c1+2c2+c3+4,?,Ex2.3.3: y(t)+2y(t)+2y(t)= 2f(t) Get h(t)=? g(t)=?,Solution:,See: Page59,Ex2.3.4: y(2)(t)+3 y(1)(t)+2y(t)=2f(1)(t)

10、+3f(t) y(0-)=y(0-)= 1 f(t)=e-t(t) Get yx(t)=? yf(t)=?,yx(t)=(3e-t-2e-2t)(t),Solution1:,yf(t)=(te-t +e-t-e-2t)(t),Solution2:,y(t)=yh(t)+yp(t),yh(t)=(C1e-t+C2e-2t)(t),y(0+)-y(0-)=2,yp(t)=kte-t(t),yp(t)=te-t(t),y(0+)-y(0-)=0,y(0+)=3,y(0+)=1,y(t)=(4e-t-3e-2t+ te-t )(t),(Natural+forced),(Transient+Steady

11、-state),(Zero-input+Zero-state),4,yx(t)=(3e-t-2e-2t)(t),yf(t)=y(t)-yx(t) =(te-t +e-t-e-2t)(t),2.4 Convolution Integral,2.4.1 Define,Ex2.4.1: f1 (t)=e-3t(t) ,f2 (t)=e-5t(t) f1 (t)*f2 (t)=?,Solution:,=1/2(e-3t-e-5t) (t),2.4.2 Graphical Interpretation,4. Integral from=t1 to=t2,Ex2.4.2:,y(t)=f(t)*h(t)=?

12、,1. Draw f1 () and f2 (-),2. Shift f2 (-) to f2 (t-),3. Multiply f1 () and f2 (t-),f(t)=(t) -(t-T) h(t)= t(t) -(t-2T),Solution:,T,f (t)*h (t)=,0 t0,0t (t-)d=1/2t2 0=tT,0T (t-)d=Tt-1/2T2 T=t2T,t-2TT (t-)d=Tt-1/2t2+3/2T2 2T=t3T,0 t=3T,t,Matlab Program : dt=0.01; T=1; t=-2*T:dt:6*T; ft=U(t)-U(t-T); ht=

13、t.*(U(t)-U(t-2*T); yt=conv(ft,ht); yt=yt*dt; subplot(311); plot(t,ft); ylabel(f(t); subplot(312); plot(t,ht); ylabel(h(t); subplot(313); y=zeros(1,length(t); for i=1:length(t) y(i)=yt(i+2*T/dt);%add -t0/dt end plot(t,y); ylabel(f(t)*h(t);,2.4.3 Properties of Convolution,1. f1 (t)*f2 (t)= f2 (t)*f1 (

14、t),2. f1 (t)*(f2 (t) *f3 (t)= (f1 (t)*f2 (t) *f3 (t),6.f1 (t-t1)*f2 (t-t2)= y(t-t1-t2),4. f(t)*(t)=f(t),5. f1 (t)*f2 (t)= f1 (-1)(t)*f2(1) (t),3. f1 (t)*(f2 (t) +f3 (t)= f1 (t)*f2 (t)+ f1 (t)*f3 (t),(f1 (t)*f2 (t)(1)= f1 (t)*f2(1) (t) (f1 (t)*f2 (t)(-1)= f1 (-1) (t)*f2 (t),Ex2.4.3: f(t)*(t- )=?,Ex2.

15、4.4: f(t) as figure; Get f(t)*f(t),Solution2:,Solution1:,Solution3:,Solution: f(t)*(t- ) =f(t- ),Define,Graph,Property,5,Ex2.4.5: y(2)(t)+3y(1)(t)+2y(t)=2f(1)(t)+3f(t) y(0-)=2,y(1)(0-)=0,f(t)=(t) y(t)=?,yx(t)=(-2e-2t+4e-t)(t),h(t)=(e-2t+e-t)(t),yf(t)=(3/2- 1/2e-2t - e-t)(t),Solution:,h(t)=2h(1)1(t)+3h1(t),2c1+c2=3 c1+c2=2,Delay(1),ha(t),hb(t),f(t),y(t),Ex2.4.6: ha(