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24th November 2017, 10:21 AM  #21  
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Join Date: May 2007

Quote:
40dB/decade (and 12dB/octave) is the ultimate slope. It will be less than this near the corner frequency. We still have no context, and only an assurance that the filter is Butterworth from someone who might not know what Butterworth means. 

24th November 2017, 02:44 PM  #22  
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Quote:
However as someone remarked, the OP may not actually know the possible range of Q values for his filter. 

25th November 2017, 10:02 AM  #23 
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He told us that it is Butterworth i.e. no peaking. I don't know how reliable this information is.

25th November 2017, 03:31 PM  #24 
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Location: Buenos Aires  Argentina

The response approximates/approaches the ideal 12dB/oct well above the cutoff frequency, and it becomes very close , say, 3 or 4 octaves above ... or below .... , BUT at the transition frequency (sorry, I am freely translating from my native Spanish, might not use the "popular" word, please try to focus on the concept I´m trying to refer to) slope is not "perfect" by any means, we don´t have a sharp angle breakpoint but transition is gradual.
Exactly one octave above we are still in the gradual slope area so result won´t be *exactly* 12 dB down but a different (although admittedly close) value. I guess the problem choice was not random, but to force synctronX do all the Math and justify it, instead of just answering the "trivial solution" such as offered by Abraxalito. "Trivial solution" in this case being the Math definition of it, not the everyday language meaning of "trivial", of course.
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25th November 2017, 07:29 PM  #25 
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@JMFahey,
I thought I found the solution in the first post, or, got close to what I was given. The information given was limited, so I had to do it with what I know. @Mark, I do not know the filter Q. @DF96, that is all the information that I was given. @PRR, Yes, I think this is also correct. The formula in the OP was all that I could come up with originally, so the .8V could be rounded up to 1V. Thanks for your contributions. I'll let you know what I find out. Cheers,
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3rd December 2017, 05:10 AM  #26 
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Post Script:
Here is the actual question: "A typical 2pole low pass Butterworth filter has an output voltage of 4V at 1200 Hz input. Assuming this is beyond the cutoff frequency, what is the output voltage at in input frequency of 2400 Hz." Answer Group: a) 8V b) 2V c) 1V <<<< d) 4V But still. I did get that correct even using the wrong formula and after thinking and Abraxilito, PRR, DF96. JMFahey, It was the trival 40dB/Decade = 12dB/Octave. Mark, Rayma, MarcelvdG Thanks for your clarifications also. Cheers,
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4th December 2017, 10:57 AM  #27 
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So it was a simple textbook or exam question, with multiple choice answers  and all except one being obviously wrong for anyone who understand the question.

4th December 2017, 12:21 PM  #28 
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Join Date: Mar 2009
Location: Buenos Aires  Argentina

Funny, but you *should* have disclosed the full information from the beginning instead of sending us on a wild Math chase all over the place.
FWIW even "1V" is NOT the exact answer, just "the less wrong" one.. We were thinking fractions of 1% differences while the actual question was "choose between approximate answer and gross 200% , 400% and 800% errors" Oh well.
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4th December 2017, 01:19 PM  #29 
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Join Date: May 2007

Yes, don't try to learn electronics from whoever set this question as he clearly gets confused by details.

4th December 2017, 10:15 PM  #30 
diyAudio Member

And even with the question rephrased...it still brings up uncertainties. I'm in the process of trying to recreate this Comparator question. It's one we haven't seen before, but it was on the final that I just finished.
Here is a schematic that I drew up from memory and Aol=120,000. The standard method of find the Vref doesn't work. Vref = V*R2/R1 + R2. Then we have the output voltage: And another if Vin is 10Vp, which input voltages will have positive max outputs? I'm pretty sure voltage gain in the comparator = Aol. I can't recall what all the answer groups were. I tried simulation, but got very different answer using AC input and Function Generator and still can't figure it out. It's a real pisser. Cheers
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