Today I got confused about Low pass filter transfer function. Let's assume we have a simple low pass RC filter . It is well known that at f=fc we expect -3dB (0.707) drop in Vin.

Everyone knows a LPF transfer function is:

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On the other hand we know that :

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Which results in :

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If we set f=fc we will have:

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But I expected 1/√2 =0.707 ! I am sure I am missing something but can not find that because this differs from the formula that I knew:

enter image description here.

  • \$\begingroup\$ Your result is correct if the input to the filter is \$e^{t/RC}\$, a real exponential. \$\endgroup\$ – Alfred Centauri Oct 26 '13 at 16:07

The problem has come from the substitution of s for 2*PI*f.

s is a substitution for j*w

thus the equation actually is: 1/(1+j)
If you resolve this back to magnitude: 1+j = sqrt(1^2 + 1^2) = sqrt(2)

resulting in: 1/sqrt(2) = 0.707 as you expect

So take a low-pass filter: RC


simulate this circuit – Schematic created using CircuitLab

$$ Vout = Vin*\frac{\frac{1}{j\omega C}}{R + \frac{1}{j\omega C}} $$ $$ Vout = Vin*\frac{1}{j\omega RC + 1} $$ $$ \frac{Vout}{Vin} = \frac{1}{j\omega RC + 1} $$ $$ \frac{Vout}{Vin} = \frac{1}{\sqrt{(\omega RC)^{2} + 1^{2}}} $$ From a magnitude point of view.

  • \$\begingroup\$ I thought s=jwf not j*w. Is that? \$\endgroup\$ – Aug Oct 26 '13 at 14:09
  • 1
    \$\begingroup\$ w (radians per second) is equal to 2*pi*f \$\endgroup\$ – JonRB Oct 26 '13 at 14:13

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