建筑工業(yè)學院電信學院

f(f(k…，f(k+2)，f(k+1)，…，f(k-1)，f(k-2),1.df(t)limf(k)limf(tt)f(t)limf(t)f(t t tf(k)f(k1)f(k (k1)f(k)f(k)f(k k(k2.差分方y(tǒng)(k)+an-1y(k-1)+…+a0y(k-n)=bmf(k)+…+b0f(k-例差分方程的迭代y(k)+an-1y(k-1)+…+a0y(k-n)=bmf(k)+…+b0f(k-與微分方程經(jīng)典解類似，y(kyh(ky(k)+an-1y(k-1)+…+a0y(k-n)=1+an-1λ–1+…+a0λ–n=0定義定義差)–[af1(k)+bf2(k)]=af1(k)+b2f(k)=[f(k)]=[f(k)–f(k-1)]=f(k)–f(k-=f(k)–f(k-1)–[f(k-1)–f(k-2)]=f(k)–2f(k-1)+f(k-mf(k)=f(k)+b1f(k-1)+…+bmf(k-yy(k)+3y(k–1)+2y(k–2)= y(k)=–3y(k–1)–2y(k–2)+k=2y(2)=–3y(1)–2y(0)+f(2)=–k=3y(3)=–3y(2)–2y(1)+f(3)=k=4y(4)=–3y(3)–2y(2)+f(4)=–2.特解2.特解y(k)=yzi(k)+ 12(ki 12(2)求h(k)–h(k–1)–2h(k–2)=(λ+1)(λ–2)=h(k)=C1(–1)k+C2(2)k，h(0)=C1+C2=1,h(1)=–C1+2C2=解得C11/3,h(k)=(1/3)(–1)+ ,kkh(k)=[(1/3)(–1)k+(2/3)(2)k]激勵響應y(k)的特解kPmkmPm1km1P1kP0特征根均不為kr(PmkmPm1km1 Pak(a不等于特征根（PrkrPr1kr1P0)ak(a等于r重特征根cosksinkP1coskP2sink(特征根不等于ejh(k)–h(k–1)–2h(k–2)=h(–1)=h(–2)=h(k)=h(k–1)+2h(k–2)h(0)=h(–1)+2h(–2)+ =h(1)=h(0)+2h(–1)+ =例1y(ky(k-1)-2y(k-23.卷積和的定f3.卷積和的定f(k)=注意ii量，k為參變量。結(jié)果仍為k例f(k)f1(i)f2(k1.+…+f(i)δ(k–i)+f(i)(kf(k)f1(i)f2(k乘積f1(if2(k注意：k為參變量。例 22任意序列作用下的零狀態(tài)f根據(jù)h(k)的定義：f(i)δ(k-由疊加性：f(i)(kf(i)h(k-f(i)h(kfyy(kf(i)h(ki)卷積yzs(k)f(i)h(ki)f(k)h(ki▲■例：例：f(kakε(k),h(kbkε(k，求yzs(k)yzs(k)=ff(i)h(ki)ai(i)bki(k 當i0,ε(i0；當ik時，ε(kiy(k) a (k) kikikai kbb,a(k)i0bbk(k,af(kf(k)f1(i)f2(k=…+f1(-1)f2(k+1)+f1(0)f2(k)+f1(1)f2(k-+f1(2)f2(k-2)+…+f1(i)f2(k–i)+f(2)=…+f(-1)f(3)+f(0)f(2)+f(1)f(1)+f(2)f(0)+ f1(k0f1(1)f1(2)f2(k)={0，f2(0)，若：f(k)序 n1kn3kn1n3kn2n4例如：f(k)：0kh(k)：0ky(k)：0k例滿足乘法的三律：交換律,分配律,結(jié)合律f(k)﹡δ(k)=f(k)，f(k)﹡δ(k–k0)=f(k–f1(k–k1)﹡f2(k–k2)=f1(k–k1–k2)﹡[f(k)﹡f(k)]=f(k)﹡f(k)=f(k)﹡f 1 2例f(kf(k﹡f(k)，求f(2)12f2(–i元(–（4）f(i)乘f( 不進不進位乘f1(1) f1(2) f2(0) f1(1)f2(1)，f1(2)f2(1)，f1(3)f1(1)f2(0)，f1(2)f2(0)，f1(3)f(1)f(1)+f(2)f f1(3)f(1)f f(2)f(1)+f(3)f f(k)={0，f(1)f(0)，f(1)f(1)+f(2)f(0) f1(2)f2(1)+f1(3)f2(0)，f1(3)f2(1)，0f1(k0f1(1)f1(2)f2(k)={0，f2(0)，例f1(k0,215,0f2(k0,3,4,0,6, f(k 3，4，0，2，1，15，20，0，3，4，0，+f(k)={0,6,11,19,32,6,6，8，0， 6h1(k)=ε(k)，h2(k)ε(k–5)