RLC circuit equation problem

Can some one tell me where I'm going wrong with this code?
I need to write a program that plots 2d graphics of the current (i) against time (t) with time changing from 0:0.001:0.4 seconds.
This is what I have so far:

1 Commento

Stephen23
Stephen23 il 20 Gen 2015
Not that you should never use i or j as your loop indices, as these are the names of the inbuilt imaginary unit .

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 Risposta accettata

Star Strider
Star Strider il 21 Gen 2015
If you haven’t already solved it by now, you need to incorporate ‘t’ into your equation. Note that including it in ‘omega’ will make your code work:
Vm=240; f=50; R=5; L=0.004; C=0.0015; t=0:0.001:0.04;
for k1=1:length(t)
omega=2*pi*f*t(k1);
z(k1)=sqrt(R^2)+(omega*L-(1./(omega*C)).^2);
i(k1)=Vm/z(k1);
end;
plot(t, i)

2 Commenti

steve
steve il 21 Gen 2015
You are a legend. It needed some tweaking, but you got me there!

My pleasure!

I appreciate your compliment.

I remember my own struggles with ESCI 101 (‘Basic Circuits and Systems’).

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Più risposte (1)

John
John il 20 Gen 2015
You must use positive integers to index into an array. z(j) doesn't work, because you have defined j as a number between 0 and .04. Add an index such as
i = 0;
for j ...
i=i+1;
z(i) = ...
...

6 Commenti

Stephen23
Stephen23 il 20 Gen 2015
Except you should never use i or j as your loop indices, as these are the names of the inbuilt imaginary unit .
John
John il 21 Gen 2015
It is true that i and j are initially defined as sqrt(-1), but imaginary numbers can be defined as any variable you want. And if you don't ever use i or j as an imaginary number, then it doesn't really matter.
John
Bravo, correct,
actually you can use i and j as indices as often as you want, as long as not conflicting, because when using i or j as (-1).^.5 MATLAB already requires a different syntax: to put the imaginary root right after the numeral, without any multiplication operator.
3+i*4
this is 4 times i, plus whatever real value i has, or MATLAB will return 'unknown variable' if i had nothing assigned previously.
On the other side
3+4i
or
3+4j
this is a complex number.
So, John, feel free to use i and j as loop indices, or use other more compact functions than loops.
Cobeldick has a really high credit ranking in this forum but when it comes to complex variables, he really needs to refresh class notes.
To read why I say this just check my answer and his answer to an asin related question:
crack on John
John BG
James Tursa
James Tursa il 26 Ago 2016
Modificato: James Tursa il 26 Ago 2016
3+i*4
"this is 4 times i, plus whatever real value i has, or MATLAB will return 'unknown variable' if i had nothing assigned previously."
I don't follow your comment. If i was not assigned a value as a variable previously then wouldn't MATLAB just call the toolbox i function?
@James Tursa —
You’re absolutely correct.
It would do exactly that:
z = 3+i*4
z =
3.0000 + 4.0000i
Stephen23
Stephen23 il 27 Ago 2016
Modificato: Stephen23 il 27 Ago 2016
@John BG: Apparently I "really needs to refresh class notes."
Okay, lets have a look at my sources of information:
  • The imaginary unit documentation itself currently states: "it is best to avoid using i and j for variable names if you intend to use them in complex arithmetic."
  • The accepted answer written by an TMW employee (TMW, who write MATLAB in case you were wondering), who states regarding avoiding using i and j that "Stephen, I'd say it's a best practice".
  • The official MATLAB documentation Programming Patterns: Some Common MATLAB Programming Pitfalls and How to Avoid Them: "to avoid this problem, either use different variable names or clear the variables...".
  • The many comments by other MATLAB experts, both on this forum and elsewhere, who essentially comment that "using i or j works, but I would not do it myself..."
Can you please tell me exactly what your references are, that are so much better than these official MATLAB references, and comments by expert users and MATLAB employees?
@John: the attitude of "then it doesn't really matter" would get you fired where I work. Standards and best practice are there to make code easier to write, easier to read, easier to debug, easier to fix, easier to integrate with other code. Contrary to what many beginners think, good code practice make your own life easier too!

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