unable to plot the graph

Hey, I am trying this code to generate a graph between voltage (x axis) and resistance(y axis), but i get an error saying the index aren't of the same length. below is the code, please advice..

close all; 
clear all; 
un=2000*10^-4;
cox=10*10^-3;
vtn=0.4;
wl=1;
kn=un*cox*wl;
vgs=4;
vds=0:5;
m=length(vds);
for i=1:m
    if vgs>vtn
        if vds(i)<vgs-vtn
            rdsl(i)=1./kn*(vgs-vtn-vds(i));
        elseif vds(i)>vgs-vtn
            rdss(i)=0;
        end
    end
end

plot(vds,rdsl(i));
xlabel('voltage vds');
ylabel('resistance rds');
title('qualitative graph between voltage and resistance');

Answer 1

[Geoff Hayes]

Apurva - when I run your code, I observe the error

 Index exceeds matrix dimensions.

because of the line

 plot(vds,rdsl(i));

vds is an array of six elements whereas rdsl is an array of four elements of which you are tying to access the ith element. There are two problems with this - the first is the error which is telling you that you are trying to access an element outside of the dimensions of the array. i will be six yet there are only four elements in the array. I suspect that since this line of code is outside of the for loop, you don't actually need the index into rdsl. (If you had been plotting this on each iteration of the for loop, then you would need to do something like

 plot(vds(i),rdsl(i));
 hold on;
 % etc.

Outside of the for loop, you would just need to do

 plot(ids,rdsl);

The second problem is that ids is six element whereas rdsl is four. Why the discrepancy? Look closely at your elseif body

 rdss(i)=0;

This seems to be a typo and should read

 rdsl(i)=0;

A couple other things - MATLAB uses i and j to represent the imaginary number so it is good practice to avoid naming indexing variables as i or j. Secondly, pre-allocate your rdsl array outside of the for loop since you know that there are exactly six elements. You can do

 m=length(vds);
 rdsl = zeros(m,1);

This will ensure that rdsl and vds are of the same dimension. Finally, examine your elseif - is this necessary? Can you just use an else instead to capture the case where

 vds(i)>=vgs-vtn

as

        if vds(i)<vgs-vtn
            rdsl(i)=1./kn*(vgs-vtn-vds(i));
        else
            rdsl(i)=0;
        end

(Technically, the else isn't necessary since rdsl has been initialized to an array of all zeros.)

Try implementing the above changes and see what happens!

[Apurva Kulkarni]

Thank You Geoff! Preallocating rdsl worked!

Answer 2

[Walter Roberson]

You assign to rdsl in one place and to rdss in another. As a result, the two arrays might not end up the same length.