I have read the linked explanation about the RC low pass filter circuit and understood most of it. I understand the relation between the time constant RC to 1/(2*pi*f_c). I understand it means there is a trade off between the size of the time constant and the capacitor cutoff frequency point. I understand this point determines the "Phase shift" size etc. But to be honest, nothing of it is making sense. I'll try to state a question here that should help me clarify "what did I miss here…"
So, my question is:
The Bode diagram shows the capacitor properties for each frequency. i.e the relation Vout/Vin (log of it times 20 but this doesn't matter to me here..).
But, if I apply a step function say at time t_0. Then, at this exact time, all frequencies are given. So, at this exact time, the capacitor should have all the properties of all frequencies. So Vout/Vin should get an infinite number of values at the exact time t_0. So, How does this make sense? Shouldn't the capacitor have one behavior for each point of time?
I really hope my question is making any sense for you.
Thanks!
Best Answer
Your confusion seems to come from thinking that the math suggests that there would be multiple possible outputs, but your intuition says that there should be only one. The answer is that there is only one output, and here is why.
An RC low pass filter is a linear system.
For linear systems, the following is true.
F(X1 + X2 + X3 ...) = Y1 + Y2 + Y3...
Which simply means that...
If input X1 produces output Y1.
And
If input X2 produces output Y2.
Then
input (X1 + X2) produces output (Y1 + Y2)
A step response, which does contain all frequencies, simply produces an output which is the sum of the response of the filter at each frequency. Note that even though there is an infinite number of elements being added together, their sum converges to a finite number at each point in time. That sum is an exponential decay which eventually reaches the value of the step input.