0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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:

$$\begin{align}y[0]&=b_0\\y[1]&=b_1-a_1b_0\\ y[2]&=-a_1(b_1-a_1b_0)\\ y[3]&=a_1^2(b_1-a_1b_0)\\\vdots\end{align}$$

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:

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

|improve this answer|||||
$\endgroup$

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.