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

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1、Chapter3TimeDomainAnalysisoftheDiscrete-TimeSystemsξ3.1TheDifferenceEquationofSystemsξ3.1.1LinearConstant-CoefficientDifferenceEquationEx3.1.1:ifonepairrabbitcanbornonepairrabbitinamonth,morethantwomonthsoldrabbitcanbornnewone,ifwehaveonepairrabbitinfirstmo

2、nth,Howmanypairrabbitswegetinnmonths?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)=0011235813213455y(n)=2*y(n-2)+y(n-1)-y(n-2)y(0)=0;y(1)=1;ξ3.1.2TheClassicSolutionofDifferentialEquationy(k)+an-1y(k-1)+…+a1y(k-n+1)+a0y(k

3、-n)=bmf(k)+…+b0f(k-m)k>=0y(k)=yh(k)+yp(k)yh(k)+an-1yh(k-1)+…+a1yh(k-n+1)+a0yh(k-n)=0λ(n)+an-1λ(n-1)+…+a1λ+a0=0yp(k)aparticularsolutionyh(k)ahomogeneoussolutionRoot:λ1,λ2,…λnλ1≠λ2≠…≠λn:yh(k)=c1λ1k+c2λ2k+…+cnλnkyh(k)=c1λ1k+c2kλ1k+…+cnλnkλ1=λ2:Ex3.1.2:y(k)+3y(

4、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)10-14-920-4184Ex3.1.3: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)002-1034-98258n=[0:6];x=((-2).^n).*U(n);y=zeros(1,

5、length(n));y(1)=0;y(2)=0;fork=3:length(n)y(k)=x(k)+x(k-1)-3*y(k-1)-2*y(k-2);endystem(n,y);ylabel('y(n)');6ξ3.2Zero-InputResponseofDiscreteSystemsξ3.2.1Zero-InputResponseoftheSimpleSystemsf(k)=0:yx(k):yh(k)y(k)+an-1y(k-1)+…+a1y(k-n+1)+a0y(k-n)=bmf(k)+…+b0f(k

6、-m)y(k)-λy(k-1)=f(k)yx(k)=Cλk=y(0)λk1.First-OrderSystemsf(k)=0pole:p=λ=λ*λ*…*λy(0)y(k)=y(0)λk2.Second-OrderSystemsyx(k)=C1λ1k+C2λ2ky(k)+a1y(k-1)+a0y(k-2)=f(k)yx(k)=(C1+C2k)λky(k)=λy(k-1)=λ*λy(k-2)a.λ1≠λ2b.λ1=λ2=λξ3.2.2Zero-InputResponseoftheSystemsy(k)+an-1

7、y(k-1)+…+a1y(k-n+1)+a0y(k-n)=bmf(k)+…+b0f(k-m)f(k)=0λ(n)+an-1λ(n-1)+…+a1λ+a0=0Ex3.2.1:y(k)+0.25y(k-2)=f(k-1)-2f(k-2)yx(0)=2yx(1)=3Getyx(k)=?0.5k(2cos(kπ/2)+6sin(kπ/2))ε(k)Solution:2.00003.0000-0.5000-0.75000.12500.1875-0.0313ξ3.3Zero-StateResponseofDiscret

8、eSystemsy(k)+an-1y(k-1)+…+a1y(k-n+1)+a0y(k-n)=bmf(k)+…+b0f(k-m)h(k)f(k)yf(k)y(-1)=y(-2)=…=0ξ3.3.1ImpulseResponsef(k)=δ(k)a.h(k)-λh(k-1)=δ(k)k>0:h(k)-λh(k-1)=0h(k)=λkε(k-1)h(k)=Cλkε(k-1)h(0)=δ(0)+λh(-1)