Difference Equations

This page is very much a work in progress on showing how to find impulse responses of systems modeled with difference equations.

For second order difference equations with real-valued coefficients, there are three different possible forms of the homogeneous solution (and thus the impulse response) depending on the nature of the solutions of the characteristic polynomial:

$ \begin{array}{c|c}\mbox{Solutions of Characteristic} & \mbox{Form of Impulse Response}\\ \hline \hline\mbox{Two different real values} & K_1\gamma_1^n+K_2\gamma_2^n\\\mbox{Repeated real value} & K_1\gamma^n + K_2n\gamma^n\\\mbox{Complex Conjugate Pairs} & |\gamma|^nK\cos(n\angle \gamma + \phi)\mbox{ or}\\~ & |\gamma|^n\left(K_1\cos(n\angle \gamma)+K_2\sin(n\angle \gamma)\right)\end{array} $

where $ |\gamma| $ is the magnitude of the solution to the characteristic polynomial and $ \angle \gamma $ is the angle it makes with the positive real axis. There will also be a step function involved depending on the relative most-shifted versions of the input and output.

For the last case, it is often easier to get $ K_1 $ and $ K_2 $ using initial conditions and then relate them to the first form of the solution with:

$ \begin{align}K&=\sqrt{K_1^2+K_2^2}\\\phi&=-\mbox{atan}\left(\frac{K_2}{K_1}\right)\end{align} $

For instance, imagine something like $ y[n]-y[n-1]+1.69y[n-2]=x[n] $. The characteristic equation is $ \gamma^2-\gamma+1.69=0 $ which has roots of $ \gamma=0.5\pm 1.2j $. The magnitude of the root is $ |\gamma|=1.3 $ and the angle (picking the + version of the imaginary part) is $ \angle\gamma=\mbox{arctan(1.2/0.5)}=1.176 $. This means the form of the impulse response would be:

$ \begin{align}h[n]&= 1.3^nK\cos(1.176n + \phi)u[n]\mbox{ or}\\~&=1.3^n\left(K_1\cos(1.176n)+K_2\sin(1.176n)\right)u[n]\end{align} $