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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]}] URL: ,

 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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