I've been having some problems with how to find the cutoff frequency and the phase shift for a simple RC low pass filter circuit. I've read this post, but I need a better understanding of what frequency response to fully be able to understand it.

So how is it that the cutoff frequency is \$f_c=\frac{1}{2\pi R C}\$? And how can i derive the phase shift being -\$\arctan(\omega RC)\$ ?

  • \$\begingroup\$ What are these "some problems" specifically? Look at my answer to the question (the 2nd one) and does what I say in the first line mean anything to you? Quote: "For a simple RC low pass filter, cut-off (3dB point) is defined as when the resistance is the same magnitude as the capacitive reactance" \$\endgroup\$ – Andy aka Jan 25 '18 at 14:58
  • \$\begingroup\$ Thanks for your answer. What I don't understand about your solution, is how did you get to R=1/(2πFC)? \$\endgroup\$ – Bassusour Jan 25 '18 at 15:04
  • \$\begingroup\$ Do you know what the reactive impedance of a capacitor is (common called capacitive reactance)? \$\endgroup\$ – Andy aka Jan 25 '18 at 15:11
  • \$\begingroup\$ Yea I do. And I just realized I'm stupid. Thanks for your help :) \$\endgroup\$ – Bassusour Jan 25 '18 at 15:12
  • \$\begingroup\$ Learning usually starts with stupidity but can sometimes involve it along the way at several points. Always best to be honest with yourself LOL. \$\endgroup\$ – Andy aka Jan 25 '18 at 15:14

Cut-off frequency or 3-dB frequency is defined as the frequency of the input signal at which, the magnitude of the output signal reduces to \$1/\sqrt2\$ of the input, or the power reduces to half ( i.e., by 3 dBs).

A simple RC circuit: enter image description here

$$V_{out} = \frac {V_{in} * -jX_c}{R-jX_c}$$

By our above definition, at cut-off frequency \$f_o\$ , \$ \frac {-jX_c}{R-jX_c} \$ should be equal to \$1/\sqrt2\$

i.e., $$\frac{-jX_c}{R -jX_c} = \frac{1}{\sqrt2} $$ taking magnitude of the complex expression: $$=> \frac {X_c}{\sqrt {R^2+X_c^2}} = \frac{1}{\sqrt2} $$ $$ => \frac {1}{\sqrt{(R^2/X_c^2)+1}} = \frac{1}{\sqrt2} $$ $$ => (R^2/X_c^2) = 1 $$ $$ => R = X_c $$ $$ => R = \frac{1}{Cw_o}$$ $$ => w_o = \frac {1}{RC}$$ $$ => f_o = 1/2 \pi RC$$

for complex numbers, phase of \$a+ jb = tan^{-1}(b/a)\$ , \$V_{out}\$ is a complex expression, Hence its phase \$ \phi \$ would be: $$ \angle \frac {tan^{-1}(-\infty)}{tan^{-1}(-X_c/R)} $$ $$ = \angle\frac {-tan^{-1}\infty}{-tan^{-1}(X_c/R)} $$ $$ = -\frac{\pi}{2} + tan^{-1}(X_c/R) $$ using the expression, \$ tan^{-1}x + tan^{-1}(1/x) = \pi/2 \$ $$ => \phi = -tan^{-1}(R/X_c) = -tan^{-1}(wRC)$$

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    \$\begingroup\$ I'm not rally sure where you got this from: ∠tan−1(−α)/tan−1(−Xc/R) And what excatly is α? Thanks \$\endgroup\$ – Bassusour Jan 26 '18 at 12:57
  • \$\begingroup\$ The real part = 0, imaginary part is -Xc. So phase is tan inverse of -infinity. I don't how to write " infinity ". So I put alpha symbol... :D \$\endgroup\$ – Mitu Raj Jan 26 '18 at 13:07
  • \$\begingroup\$ In your question, you are missing a -ve symbol in the expression for phase. It is there actually. Check wiki link :en.m.wikipedia.org/wiki/RC_circuit \$\endgroup\$ – Mitu Raj Jan 26 '18 at 13:09
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    \$\begingroup\$ This might be of help. \$\endgroup\$ – Shadow Jan 26 '18 at 14:34

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