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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.

Image alt text

$$ \frac{1}{\sqrt{LC}} = 10001 \rightarrow LC = 1/100020001 sec^{2}$$

$$ \frac{R}{L} = 2Hz$$$$ \frac{1}{RC} = 2Hz$$

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 comment threads

What's the purpose? (4 comments)
Unreadable image deleted. (2 comments)
Equations (1 comment)
"Bandwidth"? (1 comment)
"Bandwidth"?
Canina‭ wrote almost 2 years ago · edited almost 2 years ago

A bandpass filter doesn't have a specific bandwidth. It has a bandwidth at a given degree of attenuation. Since the edges aren't infinitely sharp, you have to specify the attenuation at the limits of the desired passband. 2 Hz (centered at 10 kHz) at -3 dB is very different from, say, 2 Hz (centered at 10 kHz) at -20 dB.