## Frequency filters Take a look at the diagram on the right. Assume that the voltage source supplies a 1V/10kHz signal (this means: the amplitude is 1V and the frequency is 10kHz = 10000Hz).

The impedance of inductor L will be XL = 2∙π∙104∙10-3 = 62.8Ω. The output voltage (voltage across inductor L) will be 1V∙(XL/(ZR+L)), where ZR+L is the total impedance of R and L. Because an inductor, just like a capacitor, causes a phase shift in in the current flow, we cannot just state that ZR+L = R + XL. Using some complex math we can prove that:

ZR+L = √(R2+XL2).

In our case ZR+L = √(1k2+62.82) = 1002Ω. Thus, the output voltage is 1V∙(62.8/1002) = 0.0627V.

(By the way, an inductor causes a +90 degrees phase shift, while a capacitor causes a -90 degrees phase shift.)

Now assume that the voltage source supplies a 1V/10MHz signal. The impedance of inductor L will then be XL = 2∙π∙107∙10-3 = 62.8kΩ. So ZR+L = √(1k2+62.8k2) = 62.8k. This is the same as XL, so the output voltage will be 1V. So we created a very simple frequency filter with just a resistor and a inductor.

In this case we created a so called high pass filter (HPF) since high frequency signals pass this filter easily while frequency signals are suppressed. If you swap R and L, you create a low pass filter (LPF).

Let's calculate the cut-off frequency of our filter. The cut-off frequency is the frequency at which R = XL => R = 2∙π∙f∙L => In our case f = 1k/(2∙π∙10-3) = 159kHz.