MATLAB Answers

Why wont this run? "Array indices must be positive integers or logical values."

1 view (last 30 days)
Steven Butts
Steven Butts on 5 Feb 2020
Answered: Guillaume on 5 Feb 2020
clear all
close all
clc
L =.2;% H
C = 2.0*10^-6; %F
R1 = 1500; %Ohms
R2 = 1000; %Ohms
f = (1/(2*(pi)))*(L*C(((R1^2)*C-L)/((R2^2)*C-L)))^(1/2)

  1 Comment

Adam
Adam on 5 Feb 2020
C(((R1^2)*C-L)/((R2^2)*C-L))
is not sensible code. All that stuff in the parentheses is being interpreted as an index into C, which is scalar anyway, but even if it wasn't I doubt that evaluates to an integer index.
Did you just miss out a * operator?

Sign in to comment.

Answers (2)

Fangjun Jiang
Fangjun Jiang on 5 Feb 2020
Most likely typo
f = (1/(2*(pi)))*(L*C*(((R1^2)*C-L)/((R2^2)*C-L)))^(1/2)

  0 Comments

Sign in to comment.


Guillaume
Guillaume on 5 Feb 2020
"Why wont this run? "Array indices must be positive integers or logical values."
The clue is in the error message. Array indices... so you must have some indexing somewhere. Look at your equation:
.. C(((R1^2)*C-L)/((R2^2)*C-L))) ..
yep, you've got C(something). Looking at the something, it's very unlikely to be an integer index. Considering that C is scalar, it's very unlikely you meant to index C. Maybe you're missing an operator?
To make your expression more readable, I would recommend spacing out some terms. I would also recommend against using brackets when they're not needed:
f = 1/(2*pi) * L*C ?? sqrt((R1^2*C - L) / (R2^2*C - L));
%No idea what goes ^ there

  0 Comments

Sign in to comment.

Sign in to answer this question.