How can I call an array of letters into a for loop?

1 vue (au cours des 30 derniers jours)
Patrick Scott
Patrick Scott le 11 Nov 2021
Commenté : Patrick Scott le 11 Nov 2021
I would like to setup this for loop to output each interation of with the coorasponding part of my homework. I can do it all individually but that doesn't seem efficient. Every way I have tried outputs to cooresponding integer and not the actual character (a,b,c,d)
I would like it to read:
The open loop poles for part c are:
ans =
0
-10
-10
Closed loop system is stable for part c.
Gain Margin for part c is: 20.000000.
Phase Margin part c is: 78.689008.
Time delay Margin for part c is: 1.386851.
k = [200 100 100 80];
a = [1 1 0 1];
z = [1 0.1 1 1];
w = 10;
s = tf('s');
C = [65 66 67 68];
letters = char(C)
for i=1:4
L = k(i)/ ((s+a(i))*(s^2 + 2*z(i)*w*s + w^2));
[top, bottom] = tfdata(L);
fprintf('The open loop poles for part %f are:\n', letters(i))
roots(bottom{1})
marg = allmargin(L);
if marg.Stable == 1
fprintf('Closed loop system is stable for part %f.', letters(i))
fprintf('\n\n')
else
fprintf('The closed loop system is not stable for part%f.', letters(i))
fprintf('\n')
end
fprintf('Gain Margin for part %f is: %f.\n', letters(i) , marg.GainMargin)
fprintf('Phase Margin for part %f is: %f.\n', letters(i) , marg.PhaseMargin)
fprintf('Time delay Margin for part %f is: %f.\n\n', letters(i) , marg.DelayMargin)
end

Réponse acceptée

Paul
Paul le 11 Nov 2021
If I understand the question ....
letters = 'ABCD'; % simpler
for ii = 1:4
fprintf('The open loop poles for part %s are:\n', letters(ii)) % use s not f as the format
end
The open loop poles for part A are: The open loop poles for part B are: The open loop poles for part C are: The open loop poles for part D are:
  1 commentaire
Patrick Scott
Patrick Scott le 11 Nov 2021
You are amazing. Thank you very much!

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Loops and Conditional Statements dans Help Center et File Exchange

Produits


Version

R2021b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by