信号与系统 Chapter 3 Time Domain Analysis of the Discrete-Time Systems.ppt

Chapter 3 Time Domain Analysis of the Discrete-Time Systems ξ3.1The Difference Equation of Systems ξ3.1.1 Linear Constant-Coefficient Difference Equation Ex3.1.1: if one pair rabbit can born one pair rabbit in a month, more than two months old rabbit can born new one, if we have one pair rabbit in first month, How many pair rabbits we get in n months? y(k)+an-1y(k-1)+… +a0y(k-n)= bmf(k)+bm-1f(k-1)+…+b0f(k-m) Solution: y(n)-y(n-1)-y(n-2)=0 0 1 1 2 3 5 8 13 21 34 55 y(n)=2*y(n-2)+y(n-1)-y(n-2) y(0)=0;y(1)=1; ξ3.1.2 The Classic Solution of Differential Equation y(k)+a n-1 y(k-1)+…+a 1 y (k-n+1)+a 0 y(k-n)= b m f (k)+…+b 0 f(k-m) k=0 y(k)=yh (k)+y p (k) yh(k)+an-1yh(k-1)+…+a1yh(k-n+1)+a0yh(k-n)=0 λ(n)+an-1λ(n-1)+…+a1λ+a0=0 yp (k) a particular solution yh (k) a homogeneous solution Root: λ1, λ2,…λn λ1≠λ2 ≠ … ≠ λn: yh(k)=c1λ1k+c2λ2k+…+cnλnk yh(k)=c1λ1k+c2kλ1k+…+cnλnk λ1=λ2: Ex3.1.2: y(k)+3y(k-1)+2y(k-2)=f(k) f(k)=ε(k),y(0)=1;y(1)=0; y(k)=? Solution: {3/2(-1)k-2/3(-2)k+1/6} ε(k) 1 0 -1 4 -9 20 -41 84 Ex3.1.3: y(k)+3y(k-1)+2y(k-2)=f(k) +f(k-1) f(k)=(-2)kε(k),y(0)=0;y(1)=0; y(k)=? Solution: {2(-1)k-2(-2)k+K(-2)k} ε(k) 0 0 2 -10 34 -98 258 n=[0:6]; x=((-2).^n).*U(n); y=zeros(1,length(n)); y(1)=0; y(2)=0; for k=3:length(n) y(k)=x(k)+x(k-1)-3*y(k-1)-2*y(k-2); end y stem(n,y); ylabel(y(n)); 6 ξ3.2 Zero-Input Response of Discrete Systems ξ 3.2.1 Zero-Input Response of the Simple Systems f(k)=0: yx(k) : yh(k) y(k) +a n-1 y(k-1)+…+a 1 y (k-n+1)+a 0 y(k-n)= b m f (k)+…+b 0 f(k-m) y(k)-λy(k-1)=f(k) yx(k)=C λk =y(0)λk 1. First-Order Systems f(k)=0 pole: p=λ =λ * λ*… *λy(0) y(k)=y(0)λk 2. Second-Order Systems yx(k)=C1 λ1k +C2 λ2k y(k)+a1y(k-1)+a0y(k-2)=f(k) yx(k)=(C1 +C2k) λk y(k)= λy(k-1)