# transient response of simple digitalized RC low-pass filter

In http://en.wikipedia.org/wiki/Bilinear_transform#Example, digital version of simple RC low-pass filter is presented:

$$\frac{1 + z^{-1}}{(1 + 2RC / T) + (1 - 2RC / T) z^{-1}}$$

where $T$ is sampling interval. Let me reorganize by $A = 2RC/T$, and get the following:

$$y(n) = \frac{x(n) + x(n-1)+(A-1)y(n-1)}{A+1}$$

But I cannot see how I would get transient and steady response back from this difference equation. Can anyone help here?

If you assume a zero initial condition (i.e., if the system is switched on at $n=0$, then $y[-1]=0$), then the system is fully described by its impulse response, i.e. its response to the unit impulse $x[n]=\delta[n]$ (which is zero everywhere, except at $n=0$, where it is $1$).

Write the system equation as

$$y[n]=b_0x[n]+b_1x[n-1]-a_1y[n-1]\tag{1}$$

In your example $b_0=b_1$. If you apply $x[n]=\delta[n]$, you get the following response:

\begin{align}y&=b_0\\y&=b_1-a_1b_0\\ y&=-a_1(b_1-a_1b_0)\\ y&=a_1^2(b_1-a_1b_0)\\\vdots\end{align}

So we get for the impulse response

$$h[n]=\begin{cases}0,&n<0\\b_0,&n=0\\ (-a_1)^{n-1}(b_1-a_1b_0),&n>0\end{cases}\tag{2}$$

The general input-output relation is described by the convolution of the input signal and the impulse response:

$$y[n]=\sum_{k=0}^{\infty}h[k]x[n-k]\tag{3}$$

E.g., the step response can be easily computed from the impulse response:

$$a[n]=\sum_{k=0}^{n}h[k]\tag{4}$$