2nd Order difference eqn ZIR ZSR

[trojsi]

Homework Statement

I have the following difference equation;

$y[n] -1.7y[n-1] -0.72y[n-2]=x[n]$

with aux conditions; $y[-1]=1, y[-2]=-2$

input; $x[n] = (0.7)^{n}u[n]$

I used the recursive method to get 5 consecutive values of the impulse response of the system and also 5 consecutive values of the system response.

I need to determine the ZIR response analytically and therefore I obtained the general solution and after, the particular solution below;$Gen. sol = \frac{10}{9}p^{n} - \frac{5}{4}q^{n}$

$part. sol = \frac{-216}{35}(\frac{10}{9})^{n} - \frac{164}{35}(\frac{5}{4})^{n}$

I assume that if in the process I put the zero as input and used the initial conditions, the particular solution would be the ZIR.

Homework Equations

I also need to prove that ZSR + ZIR = system output response. The system response values can be obtained from the recursive method. Is the impulse response(recursive method) the same as the ZSR of the system?

I tried to obtain the ZSR analytically but I did not manage to find a good source. I would appreciate any help. thanks

 

[aralbrec]

I assume that if in the process I put the zero as input and used the initial conditions, the particular solution would be the ZIR.

The particular solution is the part of the solution due to the input *not* being zero.

When the input is zero, your system is:

$y[n] -1.7y[n-1] -0.72y[n-2]=0$
with aux conditions; $y[-1]=1, y[-2]=-2$

Which is actually your homogeneous solution with given initial conditions. That would be your ZIR.

I also need to prove that ZSR + ZIR = system output response. The system response values can be obtained from the recursive method. Is the impulse response(recursive method) the same as the ZSR of the system?

Yes it is, if the input is an impulse. If the input is more than an impulse, a convolution will have to be done.