Comments on Band pass filter given cutoff frequency and bandwidth
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Band pass filter given cutoff frequency and bandwidth
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I have decided to try design a band-pass filter with a cut-off frequency of 10kHz and bandwidth of 2 Hz.
$$ \frac{1}{\sqrt{LC}} = 10001 \rightarrow LC = 1/100020001 sec^{2}$$
$$ \frac{R}{L} = 2Hz$$$$ \frac{1}{RC} = 2Hz$$Here is what I have done but if I plug the values to WolphramAlpha
it says it doesnt have any solutions.What am I doing wrong?
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−4
My mistake was taking the third relationship $$ \frac{1}{RC} $$ which says how to find the bandwidth of a parallel RLC filter but we have a series RLC filter.
Now solving the system : $$ LC = 100020001 , \frac{R}{L} = 2 $$
gives us the correct ratio between C1 and R1 and L1 with more detail:
C1:L1 = 100020001:1
C1:R1 = 100020001:2
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