How to plot a 2D graph using for loop values?

Hello, Could anyone tell me what the problem is in the following code?
Many thanks.
% code:
lambda= [1.5:0.005:1.6];
llambda=length(lambda);
for di=1:llambda
k(di)=2*3.14/lambda(di)
end
lambda= [1.5:0.005:1.6];
plot(lambda,k);

4 Kommentare

Joseph Cheng
Joseph Cheng am 3 Jun. 2015
Bearbeitet: Joseph Cheng am 3 Jun. 2015
Whats wrong with it and why is the plot not what you're looking for? the code matches what the plot should be.
also no need for the for loop as
if true
lambda= [1.5:0.005:1.6];
k=2*3.14./lambda;
plot(lambda,k);
end
accomplishes the same thing
sha
sha am 3 Jun. 2015
Thanks, Cheng. I would like to plot the same thing, however. using for loop. Just curious how to plot the data using for loop? Could you help?
Joseph Cheng
Joseph Cheng am 3 Jun. 2015
I am trying. The main question is why is the plot in your sample code not what you're trying to get. If it is then what is the question or is the question to plot within the for loop?
sha
sha am 3 Jun. 2015
Bearbeitet: sha am 3 Jun. 2015
If I run the code, the following errors appear:
Error using plot Vectors must be the same lengths.
Error in Untitled7 (line 7) plot(lambda,k);
How to fix this error?

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 Akzeptierte Antwort

sha
sha am 3 Jun. 2015

0 Stimmen

Thanks, Cheng. I would like to plot the same thing, however. using for loop. Just curious how to plot the data using for loop? Could you help?

Weitere Antworten (2)

Image Analyst
Image Analyst am 3 Jun. 2015

1 Stimme

The code works fine:
lambda= [1.5:0.005:1.6];
llambda = length(lambda);
for di = 1 : llambda
k(di) = 2 * 3.14 / lambda(di)
end
plot(lambda,k);
It plots with absolutely no warnings or errors about different lengths. What is the actual code you ran (it's not that , that's for sure)?
jyoti mishra
jyoti mishra am 20 Feb. 2018

0 Stimmen

how to plot a vector and its delayed version in matlab?

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sha
am 3 Jun. 2015

Kommentiert:

am 20 Feb. 2018

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