# Passive filters

1. Apr 24, 2013

### i.sac

Hi,

Can someone explain me how the cutoff formula for passive filters is derived.
Why is it 1/2*pi*R*C?

Thanks

2. Apr 24, 2013

### psparky

I may not have the exact answer, but I think I can get the conversation started. I'm referring to ideal filters like the ones on paper in college.

Lets talk in terms of ω=1/(RC)......same thing, just in radians.

If you take a simple RC passive filter and take the voltage across the capacitor....your transfer function will be
1/(1+JωRC)

This means that the output varies with omega (ω). Lets think in terms of bode plots as well.

When omega equals 1/(RC), you are simply left with 1/(1+J)
(1+J) equals 1.41 at an angle of 45 degrees. Take the inverse and you get .7 at angle -45.

20 * log .7 equals -3db which is why it drops 3 db at the break frequency.

Back to this 1/(1+JωRC) Don't forget that "J" in this equation simply equals 1<90 degrees.
When looking at bode plots, always look at your frequency at zero and infinity to get the basic shape of your bode.

So when omega equals zero....you get 1/1 or simply one. Voltage in equals voltage out.....gain of 1 or zero dB.

When omega equals infinity, you get a gain of nearly zero....so this sure looks like a low pass filter that breaks at the break frequency.

1/(1+JωRC) is simply a mathematical equation that dictates gain at any frequency. If you plug in any omega.....the exact gain and phase angle will be given thru the math from 0 to infinity.

The high pass filter is nearly the same concept, just the other way.
Do voltage division on a RC circuit across the resistor and you get JωRC/(1+JωRC)
Again, when omega equals RC, you get J/(1+J) which again equals .7 but at 45 degrees this time instead of minus 45 degrees (that's the phase angle btw).
So once again you get a minus 3 dB at the break.
Test your limits as zero for this JωRC/(1+JωRC) and you get zero gain.
Test limit at infinity and you get 1. Properties of a high pass filter.

Although the RC constant relates to a great number of things, I'm not sure this is actually derived, just simple voltage division that gives an output at a certain frequency. Or maybe I did actually derive it.

Active filters like the ones in op amps work almost exactly the same way.....except the unattenuated gain will be set by your RF/RA or (1+RF/RA).

Last edited: Apr 24, 2013
3. Apr 24, 2013

### gnurf

The cutoff frequency ωc = 1/RC is the frequency where the output power is halved, i.e

Pout/Pin = 0.5

Or in decibel,

10 log (Pout/Pin) = -3dB.

Replace power with P = V2/R to get

20 log (Vout/Vin) = -3dB → Vout/Vin = 1/√2

From inspection, you know that the RC filter's transfer function is

Vout/Vin = Z2/(Z1+Z2)

where Z1 = R and Z2 = 1/jωC are (in general) complex impedances. The -3dB we started with is related to the magnitude, or length, of the transfer function (the phase is irrelevant) which in this case reduces to finding the vector sum of the RHS denominator:

|Vout/Vin| = |1/(1 + jwRC)| = 1/√(1 + (ωRC)2)

Equate the above RHS with 1/√2 and solve for ω to confirm that it in fact equals 1/RC.

I agree—I think it would be a stretch to call any of this a derivation. My 'argument' is completely circular at least, but I'm too hungry to figure out if there's a way around that. Maybe one could take the route via the time constant τ = 1/ω and make a connection to the exponential nature of the (dis-)charging of the capacitor. Or something. I don't know.

4. Apr 24, 2013

### psparky

Wow, that's pretty cool stuff...you actually agree with what I said but in another way. Always nice to see the "10 log gain" (power) and "20 log gain" (voltage) as well.