I think I'm somewhere half way through to understand it.
Considering an ordinary voltage divider like this:
the output voltage is independent of the frequency and is equal to the ratio of these two resistors.
Going through to passive filters there is somehow an analogy, which is a bit misleading.
I know that the capacitive reactance depends on the frequency according to the equation:
$$X_C = \frac{1}{2\pi fC} [\Omega]$$
So following the idea of the voltage divider, I would expect the output voltage to be at 50% of the input voltage, but it is not the case. For some reason you have to use the impedance like:
$$V_{out} = V_{in} \frac{X_C}{\sqrt{R^2 + X_C^2}} = V_{in}\frac{X_C}{Z}$$
and it leads to the output voltage of 70,7%, when the resistance and the capacitive reactance are equal.
What have I missed?
Best Answer
Here's what you assert (correctly): -
\$V_{OUT} = V_{IN} \dfrac{X_C}{\sqrt{R^2 + X_C^2}} = V_{IN}\dfrac{X_C}{Z}\$
Let's take the example of R and Xc being equal in magnitude. What does this make the denominator: -
\$\sqrt{R^2 + X_C^2} = \sqrt{2R^2}\$ (because Xc = R)
Therefore the denominator becomes \$\sqrt2\cdot R\$ and the R cancels with the R (or the Xc) in the numerator hence,
\$V_{OUT} = \dfrac{V_{IN}}{\sqrt2}\$
Pythagorous is the reason why the R term and Xc term are squared - an R and a C do not form a potential divider like an R and an R - the impedance of a capacitor is at right angles to the impedance of a resistor.