asymptotic magnitude bode plot for PI^alpha controller

how do i add the first three plot to get the forth plot
individual plot
a) for fractional zero
close all;hold off;alpha=0.8;Wcr=3.948;ki=3;w1=logspace(-2,log10(Wcr),1000);w2=logspace(log10(Wcr),2,1000);
Mag1= (20*log10(abs(ki)))*ones(1,numel(w1));Mag2=@(w2) 20*alpha*log10(w2);semilogx(w1,Mag1);hold on;
semilogx(w2,Mag2(w2));grid on;xlabel('Frequency');ylabel('Magnitude dB');
b)fractional pole at origin
clear all;hold off;alpha=0.8;Wcr=1;w1=logspace(-2,log10(Wcr),1000);w2=logspace(log10(Wcr),2,1000); Mag1= 0;Mag2=@(w2) -(20*alpha*log10(w2));
semilogx(w1,Mag1);hold on;semilogx(w2,Mag2(w2));grid on;xlabel('Frequency(rad/sec)');ylabel('Magnitude dB');

4 Kommentare

Do you have better code and better picture? Hard to read
What about this?
that is PI^alpha final equation,as per the above image they plot separately each plot 1)Kp, 2)s^alpha+ki and 3)1/s^alpha, from the PI^alpha equation and add the three plots to get the final plot
1) constant gain magnitude equation |T(jw)|= 20log|Kp|
2)fractional zero magnitude equation, a) w<Wcr,|T(jw)|= 20log|Ki|
b) w>Wcr,|T(jw)|= 20*alphalog*w
here Wcr=|Ki|^(1/alpha)
and 3) for fractional pole, 1/s^alpha,magnitude equation |T(jw)|=- 20*alphalog*w,here w=1
i plotted fractional zero and fractional pole separetly (as the code given above),as per the paper it says when you add the first three it result the final graph

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Ameer Hamza
Ameer Hamza am 27 Apr. 2020
Try this
Kp = 1;
Ki = 2;
alpha = 1.5;
w_cr = abs(Ki)^(1/alpha);
w = logspace(-1, 3, 1000);
mask = w < w_cr;
term1 = 20*log(Kp)*ones(size(w));
term2 = mask.*(20*log(Ki)) + (~mask).*(20*alpha*log(w));
term3 = -20*alpha*log(w);
combined = term1 + term2 + term3;
subplot(4,1,1);
semilogx(w, term1)
subplot(4,1,2);
semilogx(w, term2)
subplot(4,1,3);
semilogx(w, term3)
subplot(4,1,4);
semilogx(w, combined)

4 Kommentare

mask = w < Wcr only this condition is used
is w>Wcr condition in fractional zero not required???
I also used inverted mask (~mask) which is is equivalent to w>Wcr.
ok thanks
Glad to be of help.

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