Error: Subscript indices must either be real positive integers or logicals.

In code below, I get error "Subscript indices must either be real positive integers or logicals." when I run MAIN CODE from command line of customrlc(1,2,3). However, when I just run from command line this code:

%R=1;L=2e-3;C=3e-6;
G = tf([1/(L*C)],[1 R/L 1/(L*C)])
bode(G), grid

It works. I'm not sure what is causing the error, any thoughts? Thanks.

MAIN CODE

function customrlc(R,L,C)  
close all
%clear all
%clc
A = [0 1/C; -1/L -R/L]; 
B = [0; 1/L]; 
Cc = [1 -1]; 
D = [0]; 
%Create a Matlab state-space model.
sys = ss(A, B, Cc, D); 
close 'all';  
%Compute and plot the step response. 
%figure; 
[y, t] = step(sys); 
subplot(2,2,1)
plot(t, y); 
grid 
xlabel('Time (sec)'); 
ylabel('Amplitude ');
title('Step Response'); 
%Compute and plot the impulse response. 
%figure; 
[y, t] = impulse(sys); 
subplot(2,2,2) 
plot(t, y);
grid 
xlabel('Time (sec)'); 
ylabel('Amplitude ');
title('Impulse Response'); 
%Compute and plot the zero-state response to a sinusoidal input. 
%figure; 
dt = 0.002; 
tf = 4; 
t = 0 : dt : tf;
u = sin(t); 
[ysine, t] = lsim(sys, u, t);
subplot(2,2,3)
plot(t, ysine); 
grid 
xlabel('Time (sec)'); 
ylabel('Amplitude ');
title('Sine Response');
%bode response
hold on
figure
%R=1;L=2e-3;C=3e-6;
G = tf([1/(L*C)],[1 R/L 1/(L*C)])
bode(G), grid
end

[Walter Roberson]

Which line does the problem occur on?

[Elia]

G = tf([1/(L*C)],[1 R/L 1/(L*C)]) was the line where the error came from.

After looking at it again, I saw tf is mentioned twice, once in G and once in t=0:dt:tf;

I changed that latter line to t=0:dt:tff and it ran. I'm not sure why it did since the tf is a transfer function call while tf is just something I made up. Anybody have any thoughts? Thanks.

Answer 1

[Matt J]

I saw tf is mentioned twice

No, it's mentioned 3 times. Presumably you also changed the line

 tf=4;

to

 tff=4;

[Elia]

Yes that is correct. Any thoughts as to why changing that affects the tf function itself? When you name a variable as an ordinary function like I just did even without arguments, the variable replaces the function? It seemed like it does.

[Matt J]

Yes. It does.