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Finding the point of inflection on a curve

조회 수: 12 (최근 30일)
Wern-Juin Choy
Wern-Juin Choy 2012년 1월 2일
답변: Osita Onyejekwe 2016년 10월 31일
Hi everyone, I have a question.
Is there any code that allows one to find the point of inflection on a step response curve of a transfer function? An example is given below.
Ts = 0.0005;
%Continuous system
Ps = tf(2,[1,12,20.02]);
Pz = c2d(Ps,Ts,'zoh');
%response
figure(1)
hold on
step(Pz)
hold off

채택된 답변

Teja Muppirala
Teja Muppirala 2012년 1월 4일
You need to find where the 2nd derivative is zero. There are many different ways to approach this. This is one possible way:
Ts = 0.0005;
%Continuous system
Ps = tf(2,[1,12,20.02]);
Pz = c2d(Ps,Ts,'zoh');
%response
[y,t] = step(Pz);
plot(t,y);
hold on;
% Estimate the 2nd deriv. by finite differences
ypp = diff(y,2);
% Find the root using FZERO
t_infl = fzero(@(T) interp1(t(2:end-1),ypp,T,'linear','extrap'),0)
y_infl = interp1(t,y,t_infl,'linear')
plot(t_infl,y_infl,'ro');
Another way would be to find the roots of the step response of
tf([2 0 0],[1,12,20.02]);

추가 답변 (1개)

Osita Onyejekwe
Osita Onyejekwe 2016년 10월 31일
how do i do it for this?
x = 1:500; X = x; J = 1; Fs = 499; N = J*Fs; t = 0: 1/Fs : J; Fn = 3; % this control the number of cycles/periods %deltax = 0.0020; deltax = 1;
y_sine_25HZ = sin(Fn*2*pi*t); y = y_sine_25HZ ;

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