Simple RC Highpass and lowpass filters are great! These have applications in all sorts of audio preamps / mixers etc, so there is REAL world application in these!

Often we would like to determine the frequency response of the RC circuit with both the magnitude and phase.

We can use Bode plots to help us! Take the following circuit for example. When the Input frequency is DC, the capacitor acts like a big resistor, and when the input frequency is very high, it acts as a short circuit.

So this is a low pass filter!

Here’s a vector diagram of the signals. Remember, the voltage of a capacitor always LAGS the input voltage.

I’m not sure of a good way to insert equations into wordpress… so here are the equations.

How do we determine the frequency response?

Firstly we write out the ‘resistor divider’ equation in the good old fashion way. Where Zc is the impedance of the capacitor.

The impedance of a capcitor is 1 / wC, but we include the j to represent that it introduces a phase shift of 90 degrees as shown in the vector above.

Expanded, we’re left with the equation H(jw) = 1 / (1 + jwCR)

You can download the following spreadsheet which is configured as a lowpass filter to see the equations in actual ACTION! It can also be configured as a high pass filter too. Just need to swap around the equations a bit…

LowPassExcel

Bode Plots, both magnitude and phase needn’t be hard, but if you’re a Uni or college student doing electrical engineering you’ll be facing these quite a lot! It’s worth understanding what’s actually going on rather than just blindly writing and solving equations!

Questions?

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Remember, the Output phase LAGS the input which is why it’s -arctan….. take a look at the vector diagram!

You can also see using the H(jw) = 1 / ( 1 + jwRC)

as w goes to infinity, we get

H() = 1 / j = -90 degrees

remember? your i’s and j’s?

j = 90 degs

-j = -90 degs

j*j = -180 degs

1/j = -90 degrees

j*-j = j * (1/j) = 0 degrees