# RC network (Filter) output calculation

I have an RC circuit with a transfer function $$h(s)=\frac{1}{1+0.0033s}$$
Is it possible to calculate theoretical output of this network for a set of input voltages? For this how do I solve this equation? Any clue for what transformation to apply and how?

• Are you aware it is a low pass filter with CR = 0.0033? Jun 9, 2015 at 14:50
• What kind of "set of input voltages"? Different amplitudes and/or different frequencies? Do you know the meaning of "s"? Which equation do you want to "solve"? Why do you expect to apply a transformation?
– LvW
Jun 9, 2015 at 15:12
• DO NOT allow this to make you lazy or to stop using your brain - WOW and Farout!- and Aha wander around the site. Jun 9, 2015 at 15:22
• If you do not know the meaning of "s" then I very much wonder why you ask this question in the first place. Maybe you should first educate yourself in electrical network theory before asking us to solve your homework ? Jun 9, 2015 at 15:23
• Sidk, you write down a formula (you call it transfer function) without knowing the meaning of the symbols used. More than that, you mention "different amplitudes". Question: Do you know that a transfer function is a LINEAR (small-signal) function ? And do you know what this means?
– LvW
Jun 9, 2015 at 18:29

The system represented by this transfer function can react to any number of input functions (like a ramp, a sine wave or any other function you could think of), so the output depends on the transfer function of the system (which we have) and the transfer function of the input signal (which we do not have).

Assuming the "set of input voltages" you referred to are just DC voltages of different values, the input function which translates that is the step function. Physically it can be a DC source being turned on at a given moment. Your system (mathematically, at least) is equivalent to this one:

simulate this circuit – Schematic created using CircuitLab

In the time domain the step function works as follows:

$d(t)=1\qquad,\,t>0$
$d(t)=0\qquad,\,t<0$

In the frequency domain (which is what we're working with here, since the variable is 's') we have (by the Laplace transform):

$D(s)=\dfrac{1}{s}$

The output can be obtained straight in the time domain (by using the convolution operator) or in the frequency domain and then converted to the time domain by applying the inverse Laplace transform:

$v_o(t)=h(t)*v_i(t)$
(convolution)

$V_o(s)=H(s).V_i(s)$

Having established that the input is a voltage step:

$V_i(s)=V.D(s)=V.\dfrac{1}{s}$
$H(s)=\dfrac{1}{1+0.0033s}$
$V_o(s)=V.\dfrac{1}{s}.\dfrac{1}{1+0.0033s}$

Taking the inverse Laplace transform we obtain:

$v_o(t) = V.(1-e^{-t/0.0033}) =V.(1-e^{-t/RC})$

The simulation of the circuit presented before (that represents the function you asked about), for a DC voltage of 10V is below:

By changing the voltage (to answer your question) the curve remains the same but the final voltage is the voltage of your DC source.

You can see that the formula for $v_o(t)$ is exactly the one presented in the graph.

In the time domain, there are limitations to how far you can go by calculation. The reason is simple and it's called "convolution".

Basically it's a mathematical function that is typically used as below (all in time domain): -

The output signal of a filter = the input CONVOLVED with the impulse response of the filter.


Here is a picture of a pulse input being convolved with the impulse response of a 1st order low pass filter: -

The top 3 images are relevant to what I'm talking about i.e. y(t) = x(t) $\star$ h(t) where $\star$ means convolution.

It's "kind of like" sliding the impulse responce (left or right) across the input to get the output shape. It works with any inputs of course. Here is an example in real time: -

The bottom three pictures show a good way of making this easier (once you've learnt how to convolve in your mind's eye) when the input is a square shape and relies on being able to differentiate the input to make the act of convolution easier to follow. You then integrate the output to get back to "the right answer".

But it's a lot of hard work and much easier to think of stuff in the frequency domain using jw or s.