How do you find initial values for linear discrete time systems?

[Chandasouk]

Can someone the following steps of this solution for me?

http://imageshack.us/photo/my-images/828/80685231.jpg/

Mainly how they got their initial conditions? The 7.30 equation is just h[n-m] = bm which doesn't help me much.

For our problem, I know that n=3 and m =1 since the LHS highest degree is +3 and RHS highest degree is +1

then h[3-1] = h[2] = 6 because coefficient b1 which is in front of f[k+1] is 6. At least that is what I think. Correct me if I am wrong.

But then how do they obtain h[1] and h[0]? Why are we plugging in values for K?

 

[lippyka]

To obtain h[1] and h[0], we will use the equation in step 7.30, which is h[n-m]=bm, where n and m are the highest degrees of the LHS and RHS respectively. Since the highest degree on the LHS is 3 and the highest degree on the RHS is 1, n=3 and m=1. Plugging this into the equation, we get h[3-1]=h[2]=bm. Since b1 is 6, we can conclude that h[2]=6. Now, we can use the equations from steps 7.29 and 7.28 to obtain h[1] and h[0]. Step 7.29 gives us the equation h[k-1]+h[k+1]=2f[k], while step 7.28 gives us h[-1]=0. Plugging k=1 into both equations, we get h[0]+h[2]=2f[1] and h[-1]=0. Since we already know that h[2]=6, we can solve for h[0] by rearranging the equations. We get h[0]=2f[1]-6. Since f[1]=-3, we can conclude that h[0]=12. Finally, we can use step 7.29 again to obtain h[1]. Plugging k=2 into the equation, we get h[1]+h[3]=2f[2]. Since we know h[3]=0 (from step 7.28), we can solve for h[1]. We get h[1]=2f[2]. Since f[2]=1, we can conclude that h[1] = 2.