I am learning about impedance. So sorry any stupid question here...

Consider this circuit:

enter image description here

I have an audio input, from 20 to 20 kHz.

I want to know how the AC voltage over AB will behave as the frequency increases from 20 to 20 kHz.

If my math is right, the impedance of the branch AB will be:

Zab = 1000 - j/(2πfC)

Plotting the imaginary part for the frequency range I see that the "length of that vector" (I don't know how to call it... I mean sqrt(a^2 + b^2)) of Zab will fall from 8kΩ to 1kΩ.

Because the RC group is a voltage divider with R1, I imagine that at 20 Hz we will see a volt divider between R1 (100) and 8kΩ and at 20 kHz, we will see a voltage divider between R1 (100) and 1kΩ.

So, will Vab decrease from 8/9 of the input level to 1/2 of the input level as the frequency increases?

Is this what will happen?

  • \$\begingroup\$ At high-frequency Xc = 0 so the voltage divider gain is R/(R1+R) = 0.9V/V \$\endgroup\$
    – G36
    Commented Jun 15, 2018 at 14:36

2 Answers 2


It's just easier to find the transfer function Vout/Vin =

$$\dfrac{R_2 + \frac{1}{j\omega C}}{R_1+R_2 + \frac{1}{j\omega C}}$$

$$=\dfrac{1+j\omega R_2C}{1+ j\omega C(R_1+R_2)}$$

At DC the T.F. reduces to 1 and at high frequencies it reduces to: -


If you want in-between values you need to get rid of the "j" term underneath by multiplying numerator and denominator by the denominator's complex conjugate....

.... Or do what every one esle does and use a simulator: -

enter image description here

R1 = 100 ohms, R2 = 1000 ohms and C1 = 1 uF


For a simple analysis that can be done by inspection, with a little help from the calculator on your phone perhaps, you can simply calculate the turnover point given by (R1 + R2) = 1.1k and 1uF => 1/(6.3*R*C) = 144Hz.

As pointed out above, the circuit has no attenuation at DC and attenuation of 1/1.1 = -.8dB (-1dB is good enough for almost all purposes - you remember that 10% is about 1dB you can do this bit just by looking). 144Hz is the midpoint between those two - about -0.4dB

EDIT : as the OP commented below I should have used R2 (1k) and 1uF (without R1, because it is R2 and C that are the lower part of the divider). That gives 159Hz, not 144Hz.

  • \$\begingroup\$ What do you mean by the turnover point? why do you include R1 on that calculation if, in theory, we are looking at the output at AB? \$\endgroup\$
    – Duck
    Commented Jun 15, 2018 at 20:19
  • \$\begingroup\$ the turnover frequency is the frequency where R = Xc in a CR circuit. If it is a series circuit, that is the frequency where the level at the junction of R and C is -3dB w.r.t. "in band" (i.e. 0dB, very low of very high frequency, depending which way up things are). You are right - I should have taken R2 and C, as that is the lower half of the divider. \$\endgroup\$
    – danmcb
    Commented Jun 16, 2018 at 18:13
  • \$\begingroup\$ you're welcome! I find it helps to try to see things in the the simplest possible way ;-) \$\endgroup\$
    – danmcb
    Commented Jun 16, 2018 at 20:17

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