Corner frequency of a pole?

[treez]

Could you confirm that in a RC divider as shown (attached), the corner frequency = 1/(2.pi.R.C) is that frequency at which the rms voltage at the divider point has gone down to (1/SQRT(2))*Vin(rms) ?

Also, at the corner frequency , the phase of the divider voltage is 45 degrees compared to the phase of vin

On page 197 of Marty Brown's "power supply cook book", he states that the phase is 45 degrees when the divider voltage is one half the input voltage...and this of course is incorrect, would you agree?

 

[Debdut]

For a low-pass filter configuration, the output of the above circuit should be measured across the capacitor C1.
Let's denote the input as V_i(s), the output as V_o(s), the resistance as R and capacitive impedance as 1/sC.
Then,
V_o(s)/V_i(s) = (1/sC) / (R + 1/sC)
or, V_o(s)/V_i(s) = 1/(1 + sRC)
or, V_o(w)/V_i(w) = 1/(1 + jwRC)

This is the transfer function of a low-pass filter whose mgnitude is maximum at w = 0, [V_o(0)/V_i(0) = 1/(1 + 0) = 1]
Thus the 3dB corner frequency will be the frequence at which
V_o(w₀)/V_i(w₀) = 1/sqrt(2)
or, AMP[1/(1 + jw₀RC)] = 1/sqrt(2)
or, 1/sqrt(1+w₀²R²C²) = 1/sqrt(2)
or, 1+w₀²R²C² = 2
or, w₀² = 1/(R²C²)
or, w₀ = 1/(RC)
or, f₀ = 1/(2.pi.RC)

The phase can be calculated as follows-
ARG[V_o(w)/V_i(w)] = ARG[1/(1 + jwRC)]
= ARG[(1 - jwRC) / (1 + w²R²C²)]
= arctan(-wRC/1) = -arctan(wRC)

At, w = w₀, phase = -arctan(w₀RC) = -arctan((1/RC) . RC) = -arctan(1) = -45 degrees

To answer your last question, the corner frequencies are called HALF-POWER POINTS, because the output power at that frequency is half the input power. But output voltage may not be half of input voltage!

 

[godfreyl]

Could you confirm that in a RC divider as shown (attached), the corner frequency = 1/(2.pi.R.C) is that frequency at which the rms voltage at the divider point has gone down to (1/SQRT(2))*Vin(rms) ?

Yes

Also, at the corner frequency , the phase of the divider voltage is 45 degrees compared to the phase of vin

Yes

On page 197 of Marty Brown's "power supply cook book", he states that the phase is 45 degrees when the divider voltage is one half the input voltage...and this of course is incorrect, would you agree?

Yes, dunno where he got that from.