离散系统时域分析

1/4 特解:yp(k)=2k–2 , k≥0? 全解为: y(k)= yh+yp = (C1k +C0) (– 2)k + 2k–2, k≥0 ?代入初始条件解得: C1=1 , C0= – 1/4Р3.1 LTI离散系统的响应Р三、零输入响应和零状态响应Рy(k) = yzi(k) + yzs(k)Р当激励分解为外部激励和内部状态时? 完全响应= 零输入响应+零状态响应Рy(j) = yzi(j) + yzs(j) , j = 0, 1 , 2, …, n –1?设激励f (k)在k=0时接入系统,?通常以y(–1), y(–2) , …,y(–n)描述系统的初始状态。? yzs(–1) = yzs(–2) = …= yzs(–n) = 0?所以 y(–1)= yzi(–1),y(–2)= yzi(–2),…,y(–n)= yzi(–n)?利用迭代法分别求得零输入响应和零状态响应初始值 yzi(j)和yzs(j) ( j = 0, 1, 2 , …,n – 1)Р3.1 LTI离散系统的响应Р例3:某LTI离散系统的差分方程? y(k) + 3y(k –1) + 2y(k –2) = f(k)?已知激励f(k)=2k , k≥0,初始状态y(–1)=0, y(–2)=1/2, 求系统的零输入响应、零状态响应和全响应。Р解:(1)零输入响应yzi(k)? 满足方程 yzi(k) + 3yzi(k –1)+ 2yzi(k –2)= 0? 其初始状态 yzi(–1)= y(–1)= 0, yzi(–2) = y(–2) = 1/2? 首先递推求出初始值yzi(0), yzi(1) ? yzi(k)= – 3yzi(k –1) –2yzi(k –2)? yzi(0)= –3yzi(–1) –2yzi(–2)= –1 ? yzi(1)= –3yzi(0) –2yzi(–1)= 3