Difference Equations and Z transform

Difference Equation and Z Transform

Let’s recall a bit some basic foundations of differential equations and Laplace transform, the simplest linear ODE:

$$\begin{array}{r}\dot{x}=ax+bu\end{array}$$

Laplace transform is invented to solve such a DE, take the Laplace transform, we have

$$\begin{array}{r}sX(s)=aX(s)+bU(s)\end{array}$$

Rearrange we have

$$\begin{array}{r}\frac{X(s)}{U(s)}=\frac{b}{s+a}\end{array}$$

we can see that the Laplace transform helps us to ease the computation of convolution into product and then do inverse Laplace transform.

Now let’s move on to difference equations

$$\begin{array}{r}x[n]=ax[n-1]+bu[n]\end{array}$$

Take the z transform,

$$\begin{array}{r}X(z)=a{z}^{-1}X(z)+bU(z)\end{array}$$

Rearrange we have,

$$\begin{array}{r}\frac{X(z)}{U(z)}=\frac{b}{1-a{z}^{-1}}\end{array}$$

let $b=1-\gamma$ and $a=\gamma$, the difference equation becomes

$$\begin{array}{r}x[n]=\gamma x[n-1]+(1-\gamma )u[n]\end{array}$$

Integral by parts

Integral by parts not only works for continuous functions, there’s a analog for discrete functions. Let’s recall integral by parts for indefinite integral first,

$$\begin{array}{r}\int udv=uv-\int vdu\end{array}$$

for definite integral,,

$$\begin{array}{r}{\int }_a^{b}udv=uv{|}_a^b-{\int }_a^{b}vdu\end{array}$$

The analog for discrete function would be

$$\begin{array}{r}\sum _0^{N}f[n](g[n+1]-g[n])=f[N+1]g[N+1]-f[0]g[0]-\sum _0^{N}g[n+1](f[n+1]-f[n])\end{array}$$

Life is short, let’s skip the proof.