After trying much to get this data fit into wolfram; I got this f~= -3.30566-1.90852i and more solutions in this wolfram link:
https://www.wolframalpha.com/input/?i=|%28%2810*10^-9%29*%2822*10^-9%29*0.1*3300*%286.28*f*i%29^3%29%2F%281%2B%2822*10^-9%29*3300*%286.28*f*i%29^3%2B%2810*10^-9%29*0.1*%286.28*f*i%29^2%2B+%2822*10^-9%29*0.1*%286.28*f*i%29^2%2B%2810*10^-9%29*%2822*10^-9%29*0.1*3300*%286.28*f*i%29^3%29|%3D0.707
Have I made any errors and how do i proceed to get my frequency 3300 Hz??
I think that the problem is in Wolfram Alpha with complex numbers.
I tried simple transfer function 1/(1+1/iw) and it can't find solution for w at 0.707
If you assist by writing the denominator as sqrt(1+1/w^2) it finds the solution.
I know why it doesn't work anymore, and so does the OP. It's not the same circuit he was asking about; that's why it doesn't "work" anymore. The method I described was for his circuit, not for any circuit in general.
He already knew, as did I, that the response at very high frequencies was unity. He even included a plot in the first post. Therefore, it was clear that the 3 dB down point was where the voltage was .707*unity = .707
For your new example of a voltage divider on the output, then clearly we would need to multiply the gain at very high frequencies (which is not unity anymore, of course) by .707 to get the cutoff gain. But that's not what the OP had.
I don't deny that there are circuits where the method to find the cutoff would not be to solve the equation Vout/Vin = .707. In the general case of a simple high pass like the OP has, but where the gain at very high frequencies is not unity, but something else, call it A[sub]∞[/sub], then the equation to be solved would be Vout/Vin = .707*A[sub]∞[/sub].
You keep bringing up examples that are different from the circuit the OP wanted to analyze. For his circuit, the reason for and use of the factor sqrt(1/2) was understood by the OP, and I knew this and saw no need to digress from the answer to his specific question.
Well i wanted to bring out the point that 1/sqrt(2) does not always work.
... does not always work? What does this mean?
3. Solve T(s) for the required pass band frequency, call it w1. This could be 0 or infinity or anywhere in between.
4. Using w1, solve for the pass band gain G. This gives an exact value for G.
5. Set T(s) equal to G/sqrt(2).
6. Solve for the set of wc, the cutoff frequencies.
Also note that when we used Ampl(0) above that is because we plotted the amplitude function and noted that it was a low pass filter. However, there may be times when the filter is not so well behaved so we may actually have to pick the value of w for this calculation using a little guesswork. This wont happen for most functions, but it is a possibility. Of course there is also the possibility that we are using this for a limited frequency range and then we might again have to pick this number by hand.
And I'd like to clarify: As long as the response in the circuit is the same flat response with the same roll off, with a gain of say 2, my corner frequency can be calculated by equating it to Vout/Vin = 2*0.707 = 1.414. In that case to find my cut off frequency I should look at the frequency that corresponds to the point 20 log (1.414) = + 3 db, instead of the usual -3db which is in case of the filter which has unity gain at the desired pass band frequency. Is this right?
When i said guesswork, i meant guesswork. Simplifying down to a 2nd order filter we can eliminate the guesswork, but i dont know where you obtained the rule that we can only do 2nd order filters or less.
Did you detect such a "rule" anywhere?
To the benefit of the questioner, I have tried to give some practical hint`s that have been proofed to be very useful for designing filters in practice. That`s all.
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