Step response to a discrete differential equation

Question

How would you calculate the step response for the following equation if you assume initial rest:

$$y[n] - \frac{1}{4}y[n-1] = u[n]$$

I understand that a Z-transformation is needed but I do not know how to implement it or why it is needed.

Answer 1

Well, you don't necessarily need to solve it through Z-Transform, once there is a whole theory about solving constant coefficients linear difference equations, but surely is the easiest way.

A Z-transform is applied to describe discrete-time signals and/or systems in the frequency domain.

With help of a Z-transform table, easily found in the web, one can get:

$$ \begin{gather} \mathcal{Z}\left\{y[n] - \frac{1}{4}y[n-1]\right\}=\mathcal{Z}\left\{u[n]\right\}\\ Y(z)-\frac{1}{4}z^{-1}Y(z)=\frac{1}{1-z^{-1}}\\ Y(z)\left[1-\frac{1}{4}z^{-1}\right]=\frac{1}{1-z^{-1}}\\ Y(z)=\frac{z^{2}}{\left(z-\frac{1}{4}\right)\left(z-1\right)}\\ \end{gather} $$

Now, we start the procedure to apply the inverse Z-Transform and obtain the Step Response $y[n]$ by doing Partial Fractions Expansion:

$$ \begin{gather} \frac{Y(z)}{z}=\frac{z}{\left(z-\frac{1}{4}\right)\left(z-1\right)}=\frac{R_1}{\left(z-\frac{1}{4}\right)}+\frac{R_2}{\left(z-1\right)}\\ R_1=-\frac{1}{3}\\ R_2=\frac{4}{3}\\ Y(z)=-\frac{1}{3}\frac{z}{\left(z-\frac{1}{4}\right)}+\frac{4}{3}\frac{z}{\left(z-1\right)}\\ \end{gather} $$

And finally:

$$ \begin{gather} \mathcal{Z}\left\{Y(z)\right\}=\mathcal{Z}\left\{-\frac{1}{3}\frac{z}{\left(z-\frac{1}{4}\right)}+\frac{4}{3}\frac{z}{\left(z-1\right)}\right\}\\ y[n]=\frac{4}{3}u[n]-\frac{1}{3}\left(\frac{1}{4}\right)^{n} \end{gather} $$

So, your step response is:

$$y[n]=\frac{4}{3}u[n]-\frac{1}{3}\left(\frac{1}{4}\right)^{n}$$