As a start, we will look at the following two filters.

A previous lesson tought us that the
filter on the left has a cut-off frequency of 159kHz. The
X_{C} of the 1n capacitor is 1k at 159kHz. So the
cut-off freqency of the filter on the right is also 159kHz. However, at
10kHz, X_{C} will be 15.9k. The output voltage will be
1V∙(X_{L}/(Z_{L+C})). Again, we
must use some complex math to
prove that:

**Z _{L+C} =
|X_{L} - X_{C}|**.

At 10kHz this will be |62.8 - 15.9k| = 15837.2. The output voltage will be 1V∙(62.8/15837.2) = 0.00397V. This means that a CL filter suppresses unwanted freqencies much better than RL filters. Let's calculate how much better.

To do this we calculate the output voltage of both filters at one
tenth of the cut-off frequency, i.e. 15.9kHz. At this frequency,
X_{C}=10k and X_{L}=100Ω.

V_{out,RL}=1V∙(X_{L}/√(X_{L}^{2}+R^{2}))
= 1V∙(100/1005k) = 0.0995V. This is 10 times less than the input
voltage.

V_{out,CL}=1V∙(X_{L}/|X_{L}-X_{C}|)
= 1V∙(100/9.9k) = 0.0101V. This is 100 times less than the input
voltage.

Calculating the output voltages at one hundreth of the cut-off frequency (i.e. 1.59kHz) gives:

V_{out,RL}=1V∙(X_{L}/√(X_{L}^{2}+R^{2}))
= 1V∙(10/1000.05) = 0.0099995V. This is 100 times less than the input
voltage.

V_{out,CL}=1V∙(X_{L}/|X_{L}-X_{C}|)
= 1V∙(10/99.99k) = 0.0001V. This is 10000 times less than the input
voltage.

We see that the attenuation of an RL filter is 10 times (or 20dB) per decade, and the attenuation of a CL filter is 100 times (or 40dB) per decade!

The same is true for RC and LC filters.

Let's now calculate the cut-off frequency of an LC filter. We
already know that X_{C}=X_{L}. It
can now be easily proven that:

Filling in L=1mH and C=1nF gives f=159kHz (but we already knew that).