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I'm working on some problems that instruct me to find the impulse response of a given system. For example

$$y(n)=4y(n-1)-4y(n-2)+x(n)-x(n-1)$$

The first step I take is to find the homogeneous solution of the equation like so:

$$ \lambda^n-4\lambda^{n-1}+4\lambda^{n-2}=0 \\\lambda^{n-2}(\lambda^2-4\lambda+4)=0 \\\lambda^{n-2}(\lambda-2)(\lambda-2)=0 \\\lambda = 2,2 \\y_{h}(n) = C_{1}(2)^n + C_{2}(2)^n $$

Is this correct homogeneous form for when you have two of the same values for lambda? I know it is when the values are different from each other.

I also have another question about this practice. When I've finished this step, I then plug n=0 and n=1 into the original equation to find the values of C1 and C2. How do I know how any times to plug a value into the original equation?

Thanks.

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Your characteristic polynomial is indeed

$$\lambda^2-4\lambda+4=(\lambda-2)^2$$

If you had two distinct roots $r_1$ and $r_2$ the general solution would have the form you suggested:

$$y(n)=C_1r_1^n+C_2r_2^n$$

However, in your case you have a double root $r=2$ which results in a general solution of the following form:

$$y(n)=C_12^n+C_2n2^n\tag{1}$$

The constants $C_1$ and $C_2$ are chosen to match the initial conditions and the input signal $x(n)$, which in order to compute the impulse response is a delta impulse $\delta(n)$. Assuming $y(-1)=y(-2)=0$ we get

$$y(0)=x(0)-x(-1)=1\\ y(1)=4y(0)+x(1)-x(0)=4+0-1=3$$

Comparing to (1) gives

$$y(0)=C_1=1\\ y(1)=2C_1+2C_2=3\Rightarrow C_2=1/2$$

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  • $\begingroup$ Even though the input goes to "n-1" I don't need to have a solution with an "n-1" term"? $\endgroup$ – codedude Sep 26 '14 at 19:40
  • $\begingroup$ @jollypianoman: Not sure I understand what you mean. You have a formula for $y(n)$, which is valid for all $n\ge 0$. $\endgroup$ – Matt L. Sep 27 '14 at 8:16
  • $\begingroup$ Nevermind, I was just going about it wrong. Thanks for your help :) $\endgroup$ – codedude Sep 27 '14 at 15:55

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