Derivation of cutoff frequency for RC circuit?

[yosimba2000]

How is it derived?

The cutoff frequency is Fc = 1/2piRC. R = resistance, C = capacitance.

I read somewhere it has to do with Laplace Transforms, but I'm not sure where to go from here. It kind of irks me to just accept this equation without some proof.

Thanks!

 

[BvU]

Cutoff freq is where C impedance and R impedance are equal. I. e. where $R = {1\over \omega C}$ . No need for Laplace to solve this. See e.g. here

 

[Jeff Rosenbury]

Cutoff freq is where C impedance and R impedance are equal. I. e. where $\R = {1\over \omega C}$ . No need for Laplace to solve this. See e.g. here

The cut off frequency is measured at -3dB, which is the ½ power point. When the impedance of R and C are equal, ½ the power flows in each. Thus the above.

 

[LvW]

There are different formulations for the cut-off w_c:
* R=1/(w_c*C) is correct, applies for 1st order RC circuits only
* In general for the denominator D(jw): IM(D(jw_c))=R(D(jw_c)); this is the frequency where the phase shift is exactly -45 deg.
* This is identical to the requirement for the magnitude of the transfer function: H(jw_c)=H(0)/SQRT(2); this is identical to the well-known 3dB requirement
* In the s-domain: w_c is the value of the magnitude of the phasor in the s-plane between the origin and the pole location.

 

[meBigGuy]

How is it derived?
The cutoff frequency is Fc = 1/2piRC. R = resistance, C = capacitance.

Lets assume a series capacitor feeding a resistor.
First you have to define the corner frequency in a general sense. As above, it is at -3dB, which is when the power dissipated in the load is reduced to half.
The power dissipated in the load is reduced to half when the R and Zc are equal. You should derive that as an exercise.

From above, when R equals Zc (impedance of the capacitor) that is considered the corner frequency for a simple RC circuit.

for a capacitor, Z = 1/(2*pi*f*C)

So to derive the formula, just set R equal to 1/(2*pi*f*C) and solve for f