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


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?



Your characteristic polynomial is indeed


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


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


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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.