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• • • High pass filter

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 XC 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, XC will be 15.9k. The output voltage will be 1V∙(XL/(ZL+C)). Again, we must use some complex math to prove that:

ZL+C = |XL - XC|.

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, XC=10k and XL=100Ω.

Vout,RL=1V∙(XL/√(XL2+R2)) = 1V∙(100/1005k) = 0.0995V. This is 10 times less than the input voltage.

Vout,CL=1V∙(XL/|XL-XC|) = 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:

Vout,RL=1V∙(XL/√(XL2+R2)) = 1V∙(10/1000.05) = 0.0099995V. This is 100 times less than the input voltage.

Vout,CL=1V∙(XL/|XL-XC|) = 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 XC=XL. It can now be easily proven that: Filling in L=1mH and C=1nF gives f=159kHz (but we already knew that).

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