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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)
Unreadable image deleted.
Olin Lathrop‭ wrote almost 2 years ago · edited almost 2 years ago

I've told you before, the last time only three days ago, to post images that we can actually read. Not only did you not fix that one, but now you're here doing it again. This time I deleted it since there is no point in having unreadable images here. I also downvoted due to the gross sloppiness. Clearly you didn't even bother to proofread what you posted. There is no period at the end of your "Here is..." sentence, the next sentence starts with a lower case letter, and you couldn't be bothered to put a space before the next sentence after that. This level of sloppiness is just plain rude to the volunteers you are asking for free help.

Olin Lathrop‭ wrote almost 2 years ago

You fixed the image, but you haven't fixed the other problems I mentioned above. And, you added new problems. You have various combinations of inductance and capacitance being equated to dimensionless quantities. It doesn't take special electronics knowledge to get units right. Sloppiness is not tolerated here. Fix it or I'll start closing your questions.