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i.sac
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Hi,
Can someone explain me how the cutoff formula for passive filters is derived.
Why is it 1/2*pi*R*C?
Thanks
Can someone explain me how the cutoff formula for passive filters is derived.
Why is it 1/2*pi*R*C?
Thanks
The cutoff frequency ωc = 1/RC is the frequency where the output power is halved, i.ei.sac said:Can someone explain me how the cutoff formula for passive filters is derived. Why is it 1/2*pi*R*C?
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.psparky said: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.
gnurf said: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.
A passive filter is an electronic circuit that uses passive components such as resistors, capacitors, and inductors to filter out unwanted frequencies in a signal. It does not require an external power source to function.
A passive filter works by using the inherent properties of its components to attenuate certain frequencies in a signal. For example, a low-pass filter will allow low frequencies to pass through while blocking high frequencies.
The cutoff frequency of a passive filter is the frequency at which the filter begins to attenuate the signal. It is also known as the corner frequency or -3dB frequency.
The cutoff frequency of a passive filter can be derived by using the following formula: fc = 1/(2πRC) for a low-pass filter, where fc is the cutoff frequency, R is the resistance value, and C is the capacitance value. For other types of filters, different formulas may be used.
Passive filters are commonly used in audio systems, power supplies, and communication systems to remove unwanted noise or interference from the signal. They can also be used in electronic circuits to shape the frequency response of a signal.