I've posted the result which I got from scilab:When you read my contributions carefully, you will understand that THIS function applies for Q<0.5 only.
Pleaase, read all the contributions in this thread again and try to UNDERSTAND!
Comment: Regarding comparison of the various terms I repeat here part of my response#15:
If you have the actual transfer function (derived from the circuit) in the same form you can compare the various terms and derive the expressions for ωp and Q
http://s10.postimg.org/d9hjzlpeh/Scilabbb.png
It wasn't a flat line, but it appears that the response time is very fast.
It's the only sense I make out of it. So should I assume that the settling time is 0.003 seconds?
The transfer function which I derived from the circuit shown in the link below;
http://www.electronics-tutorials.ws/filter/fil51.gif
which has values: R2=R1=10000Ω,C1=10nF,C2=1nF, is;
H(jw) = R2C1jw / [(R2C2jw+1)(R1C1jw+1)]
Low frequency cut off is = 1/2∏R1C1 = 1591 Hz = ωl
High frequency cut off is = 1/2∏R1C1 = 15915 Hz = ωh
Bandwidth B = 14324 Hz
Center Frequency ω0= sqrt(ωl*ωh) = 5032
Quality Factor = ω0/B = 0.5032
Or maybe it's ω0/B = 2∏ω0/B = 2∏*5032/10000=3.16
As you see, the quality factor is above 0.5
I have the actual transfer function (derived from the circuit) H(jw). So can't you compare the various terms and derive the expressions for ωp and Q for me?
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