The file attached is part of a lab I am writing a formal report on. Part of the report notes that -3dB frequency occurs at 1/(R2*C). Can someone explain how this can be derived?
Thanks
Thanks
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....and what is the well-known and established name of the whole circuit?Also, what is the output as a fraction of the input as a function of frequency?
LOLshteii01...thanks for the thought provoking and engaging insight.
Are you sure that you have picked up the right answer from your textbook?LOL
20*log10(|H(jw)|)=-3 dB
where w=2*pi*cutoff frequency of your filter
I really like my textbooks. They are full of answers.
They said -3 dB.Are you sure that you have picked up the right answer from your textbook?
Don`t forget the DC gain Ao of the first-order lowpass.
Thus, the cut-off frequency will be at a point where |H(jw)| is 3 db below Ao (rather than |H(jw)|=-3 dB).
It depends, of course, on the DC gain Ao.They said -3 dB.
Maybe you should discuss the difference with the OP. They might benefit.
See.Blah....to many questions. LvW...this is obviously a lowpass filter. shteii01...thanks for the thought provoking and engaging insight. WBahn, this problem is pretty straight forward. We know all voltages but Vi and Vo -->
1) (o-Vi)/Ra + (0 - Vo)/(R2||C) = 0.
2) R2 = 2*R1
3) R2||C = 1/(1/R1 + jwc)
Damnit I found my error. Factoring is the problem here!
The cutoff frequency of the filter and the point at which |H(jw)| = -3 dB are not always one and the same. The cutoff frequency of the filter is almost always interpretted as being where the filter response to -3dB below the nominal passband response. That is what is being sought in this case.LOL
20*log10(|H(jw)|)=-3 dB
where w=2*pi*cutoff frequency of your filter
I really like my textbooks. They are full of answers.
Daniel, WBahn gave you some general recommendations how to proceed.I think you need to slow down and be sure that you aren't making stupid mistakes (which we will ALL always make from time to time, but you want to go at a pace and a level of detail that results in the rate at which you make them being very, very low).
by Don Wilcher
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by Duane Benson
by Duane Benson