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Student Support Forum: 'derivative of amplitude-frequency characteristic' topicStudent Support Forum > General > Archives > "derivative of amplitude-frequency characteristic"

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Author Comment/Response
devil
02/20/13 07:16am

I found a solution for my problem.

Thanks for the help Bill!


Here the Code that find the cut-off frequency of a given system:

fmin = 10^-1; fmax = 10^5;

R1 = 100*10^3;
R2 = 1*10^3;
C1 = 1*10^-6;

G[s_] := (s*R1*C1)/(1 + s*R1*C1)*1/(100 + s*R2*C1);

fac = Factor[G[s]];
Print["G = ", fac];

ampl[f_] =
Simplify[Sqrt[
Re[ComplexExpand[G[2*Pi*f*I]]]^2 +
Im[ComplexExpand[G[2*Pi*f*I]]]^2 ], f > 0,
Assumptions -> Element[f, Reals]];
LogLinearPlot[20*Log[10, ampl[f]], {f, fmin, fmax},
AxesLabel -> {"f [Hz]", "|G| [dB]"}, GridLines -> Automatic,
ImageSize -> 800, Frame -> False]
ampl[f_] = 20*Log[10, ampl[10^f]];

abl2[f_] = FullSimplify[D[ampl[f], {f, 2}]];
(*Plot[abl2[f],{f,Log[10,fmin],Log[10,fmax]},AxesLabel->{"10^f \
[Hz]","|G|'' [dB]"},GridLines->Automatic, Frame->False]*)
abl3[f_] = FullSimplify[D[abl2[f], {f, 1}]];
(*Plot[abl3[f],{f,Log[10,fmin],Log[10,fmax]},AxesLabel->{"10^f \
[Hz]","|G|''' [dB]"},GridLines->Automatic, Frame->False]*)

fg = Solve[
abl3[f] == 0 && f >= Log[10, fmin] && f <= Log[10, fmax] &&
Element[f, Reals], f];
Do[If[Abs[abl2[f /. fg[[i]]]] > 0.05,
Print["fg_", i, " = ", N[10^f /. fg[[i]], 5], " Hz"]], {i, 1,
Length[nst3]}]

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Subject (listing for 'derivative of amplitude-frequency characteristic')
Author Date Posted
derivative of amplitude-frequency characteristic devil 02/05/13 2:19pm
Re: derivative of amplitude-frequency character... Bill Simpson 02/05/13 6:43pm
Re: Re: derivative of amplitude-frequency chara... devil 02/06/13 2:22pm
Re: Re: Re: derivative of amplitude-frequency c... Bill Simpson 02/06/13 8:25pm
Re: Re: Re: Re: derivative of amplitude-frequen... devil 02/07/13 1:08pm
Re: derivative of amplitude-frequency character... devil 02/20/13 07:16am
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