1、零输入零状态举例,例：系统方程为 y(k) + 3y(k 1) + 2y(k 2) = f(k) 已知激励f(k)=2k , k0，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。,解：（1）yzi(k)满足方程yzi(k) + 3yzi(k 1)+ 2yzi(k 2)= 0yzi(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)= 1yzi(1)= 3yzi(0) 2yzi(1)=3 特征根
2、为1= 1 ，2= 2,解为 yzi(k)=Czi1( 1)k+Czi2(2)k 将初始值代入 并解得 Czi1=1 , Czi2= 2yzi(k)=( 1)k 2( 2)k , k0,（2）零状态响应yzs(k) 满足,yzs(k) + 3yzs(k 1) + 2yzs(k 2) = f(k) yzs(1)= yzs(2) = 0 递推求初始值 yzs(0), yzs(1)，yzs(k) = 3yzs(k 1) 2yzs(k 2) + 2k , k0yzs(0) = 3yzs(1) 2yzs(2) + 1 = 1yzs(1) = 3yzs(0) 2yzs(1) + 2 = 1,分别求出齐次解和特解，得yzs(k) = Czs1(1)k + Czs2(2)k + yp(k)= Czs1( 1)k + Czs2( 2)k + (1/3)2k 代入初始值求得Czs1= 1/3 , Czs2=1 yzs(k)= ( 1)k/3+ ( 2)k + (1/3)2k , k0,
