1
\$\begingroup\$

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!

\$\endgroup\$
2
  • \$\begingroup\$ At frequency at which reactance of a capacitor (Xc) is equal to the value of the resistor Xc = R, the output voltage is equal Vout=0.707*Vin (half of a input power). And this happens at frequencies equals F = 1/(2 pi RC). The voltage ratio Vout/Vin of a voltage divider that contains two resistors doesn't depend on frequency because the resistance of resistors does not change with frequency. But is this case the divider's voltage ratio change with frequency. Because R is unchanged but capacitor reactance Xc is changing with frequency Vou/Vin = R/(Xc + R). And I do not understand your question. \$\endgroup\$
    – G36
    Apr 30 '18 at 18:25
  • \$\begingroup\$ The description provided by @G36 seems to me like something you should carefully think about. The capacitor has a different reactance for different frequencies, so the voltage divider result will be frequency-dependent. That's one way of looking at it, if you are willing to accept the model concept of "reactance." That concept is useful because it makes understanding certain behaviors easier. But like all models there will be circumstances where that model is less useful or doesn't directly apply. There are superior models where reactance is a subset, but require more math (and data) to use. \$\endgroup\$
    – jonk
    Apr 30 '18 at 19:43
3
\$\begingroup\$

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.

\$\endgroup\$
2
\$\begingroup\$

You mustn't forget that the amplitude plot is but half of the info a bode plot gives you: the other half is the phase, and you're neglecting that this implies a different delay for the different spectral parts of your input signal.

\$\endgroup\$
2
\$\begingroup\$

If you know the Fourier components of a step function include all frequencies and you know the low pass filter attenuates at 6 dB per octave with a phase shift going from two decades below the breakpoint near zero , And 45° out the breakpoint and then approaching 90° two decades above the breakpoint of course the impedance of the resistance is equal to the impedance of the reactants and the apparent amplitude of that right angle triangle is .707 with a 45 degrees phase shift.

The Fourier spectrum and phase shift of the output of the low pass filter matches the transfer function of the filter itself

\$\endgroup\$
1
\$\begingroup\$

I understand the relation between the time constant RC to 1/(2*pi*f_c). I understand it means that there is a trade off between the size of the time constant and the capacitor cutoff frequency point. I understand that this point determines the "Phase shift" size etc.

You are talking about the transfer function here and not the response to a step. The transfer function is equivalent to the response of a "swept" sine wave at the input and not a step change. Step changes are handled differently in the frequency domain.

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?

Where you are going wrong is that you are trying to apply a step function to a transfer function. That has to be done differently. When you are in the time domain you cannot readily (without care) mix and match frequency domain and time domain stimuli and transfer functions.

When you are in the frequency domain you can quite readily mix the equivalent of a step to a transfer function. But, in the frequency domain, a "step" does not mathematically look like what it is in the time domain.

The two domains are closely related by "Laplace" but they can't intermingle so easily without transformation from one to the other.

  • Consider your RC low pass filter - it has a TF of \$\dfrac{1}{1+sCR}\$
  • A step in the frequency domain is \$\dfrac{1}{s}\$
  • You can multiply them in the f domain to get \$\dfrac{1}{1+sCR}\cdot\dfrac{1}{s}\$
  • And you can convert that back to the t domain using Laplace tables as per this similar example: -

enter image description here

Which is the normalized response of an RC low pass filter to a step change with a = 1/RC.

So, How does this make sense? Shoudnt the capacitor have one behavior for each point of time?

In the time domain we rely on I = C dv/dt for a capacitor and, in the frequency domain we rely on the impedance of a capacitor being \$\dfrac{1}{sC}\$.

Where s = jw hence Xc = \$\dfrac{1}{2\pi F C} \angle -90\$

\$\endgroup\$
1
\$\begingroup\$

A very wise person (ex Bell Labs and ex Tek scope developer) whose recent designs likely define your telephone data rates, said very simply

"Signals happen in time, not in frequency"

Building on that, I've concluded "Frequency is just a way to indicate the periodicity of the peak correlation."

\$\endgroup\$
2
  • \$\begingroup\$ Wise indeed.. What is his name? So I'll mention him if I'll quote him. Thanks \$\endgroup\$
    – user135172
    May 1 '18 at 7:29
  • \$\begingroup\$ I suspect he was merely quoting one of his Bell or Tek mentors. I did offer to him, as a return gift, my thoughts of "Harmonics do not exist, the appearance being the results of our selected math methods." \$\endgroup\$ May 3 '18 at 4:57
0
\$\begingroup\$

Thank you all for your enlightening answers. Helped so much! I would like to add to your answers the below two great videos. Together with your answers it gave me the full understanding so maybe it will help others:

Arthur Mattuck MIT 1

Arthur Mattuck MIT 2

Thanks!

\$\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.