Question

[2pta The two graphs below show approximate phase Bode diagrams for transfer functions H(s). Each original function can be written in the form: For each given graph, determine the corresponding parameters of H s). Enter T" if a parameter is impossible to determine phase of H a. 180° phase of H 900 90° log ? log co -900 -900 -180° 3-2-1 0 1 2 3 3 -2-1 0 1 2 3 22 - jslv

Please answer all parts to 4 signifigant figures for a thumbs up, thanks!

Answer 1

Ans) As only phase plots given (no magnitude or gain plots given) ,we can't determine the constant 'K' value in both cases

a)

For figure 'a' the transfer function is $H(s)=K\frac{s}{(1+\frac{s}{w_{o}})}$ where 'w_o' is cut-off frequency where phase is 45 degrees i.e from figure $logw_{o}=1\Rightarrow w_{o}=10^{1}=10$

So $a=w_{o}=10$

hence

$H(s)=K\frac{s}{(1+\frac{s}{w_{o}})}=K\frac{s}{(1+\frac{s}{10})}=Ks^{1}(1+\frac{s}{10})^{-1}$

comparing with transfer function given in problem we have

$\mathbf{K='I',\: \: n=1,\: \: a=10,\: \: m=-1}$

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b)

ion is $H(s)=K\frac{1}{(1+\frac{s}{w_{o}})}$ where 'w_o' is cut-off frequency where phase is -45 degrees i.e from figure $logw_{o}=0\Rightarrow w_{o}=10^{0}=1$

So $a=w_{o}=1$

hence

$H(s)=K\frac{1}{(1+\frac{s}{w_{o}})}=K\frac{1}{(1+\frac{s}{1})}=Ks^{0}(1+\frac{s}{1})^{-1}$

comparing with transfer function given in problem we have

$\mathbf{K='I',\: \: n=0,\: \: a=1,\: \: m=-1}$