Abide said:
Homework Statement
Find the Impulse response of this system
3y[n]+4y[n-1]+y[n-2] = x[n] + x[n-1]
Homework Equations
Eigenvalues = -1/3 and -1
hc[n] = k1[-1/3]n+k2[-1]n
3h[n] +4h[n-1] +h[n-2] = δ[n] + δ[n+1]
The Attempt at a Solution
I know that normally we would plug in hc[n] for two values of n into our given equation giving us a system of two equations for k1 and k2
However, since the answer for this inolves δ[n] + δ[n+1] I feel like there is a different approach.
I'm not really sure where to go after finding hc[n]
The solution you are coming to comes from solving the homogeneous equation:
3y[n]+4y[n-1]+y[n-2] = 0
The assumption is made that y[n]=c^n is a solution and all possible c's are found by substitution (the eigenvalues). The complete homogeneous solution is then a linear combination of these c^n factors.
The impulse is a unit input for a brief instant and then no further input for all time. So after the unit input is supplied, the system must satisfy the homogeneous solution. We can regard the impulse as happening an infinitesimally short time prior to n=0. Then the impulse response can be regarded as equivalent to the homogeneous solution, with the coefficients found by finding the first few y[n] from:
3y[n]+4y[n-1]+y[n-2] = delta[0]
All this I know you know but is said to point out the assumption made here: the system must have been at rest before the impulse was applied.Now we have this:
3y[n]+4y[n-1]+y[n-2] = x[n] + x[n-1]
Had it been this:
3y[n]+4y[n-1]+y[n-2] = x[n]
we could have proceeded as above with x[n] = delta[0] to find the impulse response but instead we have the single impulse causing input at time 0 and time 1.
Recall the assumption we made in finding the impulse response at the top:- the system had to be at rest. Now we apply a single impulse at time 0, x[0] = 1, all is good with zero system state. But then x[1] = 1 too but now system state is not zero. So we cannot say the impulse response will be the same as the homogeneous solution.Instead, you need to notice that this is an LTI system (linear and time invariant) which means if you apply an input (x1+x2) to the system the response is the sum of the responses of the system to x1 and x2 separately.
If we apply an impulse to:
(a) 3y[n]+4y[n-1]+y[n-2] = x[n] + x[n-1]
we are actually applying two impulses to:
(b) 3y[n]+4y[n-1]+y[n-2] = x[n]
one happening at time 0 and one happening at time 1. So we can find the impulse response to the system (b) and add a time delayed version of the same response to get the total response to system (a).You found the impulse response to (b) in your message:
ha[n] = k1[-1/3]^n+k2[-1]^n
Now we add a time delayed version for system (a):
hb[n] = ha[n] + ha[n-1]
= k1(1/3)^n + k2(-1)^n + u[n-1]*(k1(1/3)^(n-1) + k2(-1)^(n-1))
= k1(1/3)^n + k2(-1)^n + u[n-1]*(k3(1/3)^n + k4(-1)^n))
= k1(1/3)^n + k2(-1)^n + k3(1/3)^n + k4(-1)^n) + k5*delta(0)
= k6(1/3)^n + k7(-1)^n + k5*delta(0)
Generate the first three impulse response values for (a) and solve for k6, k7, k5. You'll find k6=1/3, k7=0, k5=0. I've verified this against a Z-transform solution (I am very rusty!) so I know it is correct.
A note on u[n-1]. This is a unit step function equal to 1 beginning at n=1 and above and zero below. It is used as a gate to make sure the second impulse response does not start until time n=1. In the step where u[n-1] disappears, note that n=0 generates some value in the gated part, so I add k5 to zero out that value at time n=0 only.