# Finding the impulse response of a system without using Z-Transform

## Homework Statement

Find the impulse response of the following: Assume the system is initially at rest.

y[n] - (1/2) y[n-1] = x[n] + 2 x[n-1] + x[n-2]

## The Attempt at a Solution

To find the impulse response y[n]: we know that y[n] = homogenous solution + particular solution

so...

lamda - (1/2) = 0 => lamda= (1/2)

homogenous solution = C1*(1/2)^n

For the particular solution, we assume that a x[n] = delta[n]...am i correct?

so ... y[n] - (1/2) y[n-1] = delta[n] + 2 delta[n-1] + delta[n-2]

so, for n >2, delta[n] = 0

=> y[n] - (1/2) y[n-1] = 0, which means the particular solution is the homogenous solution?

Can someone help me with this?

Thanks

vela
Staff Emeritus
Homework Helper

## Homework Statement

Find the impulse response of the following: Assume the system is initially at rest.

y[n] - (1/2) y[n-1] = x[n] + 2 x[n-1] + x[n-2]

## The Attempt at a Solution

To find the impulse response y[n]: we know that y[n] = homogenous solution + particular solution

so...

lamda - (1/2) = 0 => lamda= (1/2)

homogenous solution = C1*(1/2)^n

For the particular solution, we assume that a x[n] = delta[n]...am i correct?

so ... y[n] - (1/2) y[n-1] = delta[n] + 2 delta[n-1] + delta[n-2]
Yes, that's correct.
so, for n >2, delta[n] = 0

=> y[n] - (1/2) y[n-1] = 0, which means the particular solution is the homogenous solution?
No, just because for n>2, the RHS is 0, it doesn't follow the particular solution is the homogeneous solution. If the difference equation were simply

y[n] - (1/2) y[n-1] = x[n]

then the impulse response would be the homogeneous solution. Just try calculating y, y, y, and y, and you'll see that's indeed the case. Now, obviously, the two extra terms in the given difference equation will have an effect on how the system responds, so the impulse response for the original equation can't be just the homogeneous solution.
Can someone help me with this?

Thanks
You might try calculating y, y, y, y, ... for the original equation when you set x[n]=δ[n] and see how the system responds. It helps not to simplify much so you can see how the delayed terms on the right affect the output.

Alternatively, you can use the principle of superposition to calculate what the output should be in general.

Thanks for the help...I figured it out....