Unable to create Bode plot of H(s)=(s+10)/(s(s+5)^2)
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% Case 1. T(s)=(s + 10)/[s(s + 5)^2]
% Case 1. expanded: (s+10)/(s^3 + 10s^2 + 25s)
num = [1 10];
den = [1 10 25 0];
sys = tf(num,den)
bode(sys)
grid on
Using the above code results in 2 error messages:
Error using DynamicSystem/bode (line 104) Line 104: throw(E)
A and B must be vectors.
Error in Bode_Pole_squared (line 10) -> This refers to: bode(sys)
bode(sys)
The code works without the 's' in the denominator (using (s+5)^2 only.)
2 Kommentare
Walter Roberson
am 21 Dez. 2024
I suspect that the problem is something along the lines of you having your own height.m function that is interfering with the MATLAB function. However, I did some tracing in the code, and I cannot see at the moment how the graphics are actually built, and I do not see any calls to height() or similar functions.
You should use
dbstop if caught error
and run again, and see where the problem is showing up.
Akzeptierte Antwort
Hassaan
am 21 Dez. 2024
Bearbeitet: Hassaan
am 21 Dez. 2024
% Define numerator and denominator
num = [1 10];
den = [1 10 25 0];
% Create the transfer function
sys = tf(num, den);
% Display the transfer function
disp(sys);
% Plot the Bode plot
bode(sys);
grid on;
The error you're encountering is due to the fact that the denominator provided in the code is incomplete for representing the transfer function properly in MATLAB. Specifically, the denominator vector [1 10 25 0] corresponds to s^3 + 10s^2 + 25s, but this form may lead to numerical instabilities or issues when attempting to compute the Bode plot because the transfer function has a pole at s=0 (a DC pole).
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1 Kommentar
Paul
am 21 Dez. 2024
"The error you're encountering is due to the fact that the denominator provided in the code is incomplete for representing the transfer function properly in MATLAB."
As your own answer showed, there is no error.
How would one properly represent this transfer function in MATLAB?
"..., but this form may lead to numerical instabilities or issues when attempting to compute the Bode plot because the transfer function has a pole at s=0 (a DC pole)."
What is the basis of this statement?
Weitere Antworten (1)
Paul
am 21 Dez. 2024
As shown after the edit of @Walter Roberson, the code runs fine here. Do you see the same output after executing the following commands?
num = [1 10];
den = [1 10 25 0];
sys = tf(num,den)
which tf(num,den)
which bode(sys)
2 Kommentare
Paul
am 22 Dez. 2024
The problem probaby lies with a function on your path that is shadowing a same-named function used by the Control System Toolbox.
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