Could anyone tell me what it means in the code(xfer1(i) = xfer(5,1))?

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a11=-400;
a12=0.077;
a13=2200*(-1.3*10^-5)/(1.6*10^-5);
a14=2200*(-2*10^-4)/(2*1.6*10^-5);
a31=0.975*10^6/(12537.5*1055.1);
a33=-1/6.04;
a41=0.025*10^6/(31355.1*1055.1);
a43=1/15.1;
a44=-1/15.1-2/0.84;
a54=-1/15.1+2/0.84;
a55=-2/0.84;
b1=2200/(1.6*10^-5);
b4=2/0.84;
A=[a11,a12,a13,a14,a14;-a11,-a12,0,0,0;a31,0,a33,-a33,0;a41,0,a43,a44,0;a41,0,a43,a54,a55];
b=[0;0;0;b4;0];
w=logspace(-3,4,10000)
w = 1x10000
0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010 0.0010
for i=1:length(w)
AB = [j*w(i)-a11 -a12 -a13 -a14 -a14; a11 j*w(i)+a12 0 0 0; -a31 0 j*w(i)-a33 a33 0; -a41 0 -a43 j*w(i)-a44 0;-a41 0 -a43 -a54 j*w(i)-a55];
xfer = inv(AB)*b;
xfer1(i) = xfer(5,1);
end
loglog(w,abs(xfer1));
title('\fontsize{16}Coolant Temperature Transfer Function-\delta\theta_{2}(\omega)/\delta\theta_{in}(\omega)')
ylabel('\fontsize{13}Gain')
xlabel('\fontsize{13}Frequency (rad/sec)')
This code calculates inv[SI-A]*b, the transfer function's gain.
Could you please explain what the code (xfer1(i) = xfer(5,1)) means?
Does this mean that the code computes the transfer function for the fifth row?
  1 Commento
Mathieu NOE
Mathieu NOE il 27 Mar 2024
seems to me (according to size(A)) you have a 5 input , 5 output model
with b you select the 4th input
then you plot the 5th output (and that is xfer1)

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Risposta accettata

Aquatris
Aquatris il 27 Mar 2024
Modificato: Aquatris il 27 Mar 2024
So from the context, it seems like 'xfer' is the complex value of your transfer function at a particular 'w', it is a 5x5 matrix. Then, since you are only interested in the effects of 1st input on the 5th state, you only store the 'xfer(5,1)'. Keep in mind, rows corresponds to states/outputs, columns correspond to inputs.
So to answer your question, the code calculates the TF for all input and outputs, but only stores the one from 1st input to 5th output with '(xfer1(i) = xfer(5,1)) '.
% so your system transfer function TF is
TF = [xfer(1,1) xfer(1,2) xfer(1,3) xfer(1,4) xfer(1,5);
xfer(2,1) xfer(2,2) xfer(2,3) xfer(2,4) xfer(2,5);
xfer(3,1) xfer(3,2) xfer(3,3) xfer(3,4) xfer(3,5);
xfer(4,1) xfer(4,2) xfer(4,3) xfer(4,4) xfer(4,5);
xfer(5,1) xfer(5,2) xfer(5,3) xfer(5,4) xfer(5,5)];

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