how to plot a sequence over a given interval

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Connor Wright
Connor Wright 2020 年 11 月 28 日
コメント済み: Connor Wright 2020 年 12 月 3 日
Hello,
After hours upon hours of trying to work this out and a couple of questions here I am having no luck.
I need to plot a seqeunce over a given interval, I have gotten to the point where the sequence is there however I can't find a way around 'vectors must be the same length'
Here is my code...
n = -20:12
x = repmat([5,4,3,2,1],3 15)
linespace(n)
plot(n,x)
I would expect a saw tooth wave looking graph but I get the above mentioned error.
What am I doing wrong? I don't have enough time to play about with the code to try to figure this out as I need to submit a report tomorrow by 11pmand after trying to figure this out for two or three days I am out of ideas.
Any help will be greatly appriciated.
  4 件のコメント
Image Analyst
Image Analyst 2020 年 11 月 28 日
So, n is the x-axis and you want one triangle going from 1 to 5, peaking in the middle, and you want to replicate it 5 more times, for a total of 6 or 198 elements with the final element being n=178?
Connor Wright
Connor Wright 2020 年 11 月 28 日
Yes, as long as that gives me a graph that looks like a saw tooth wave. Its for a Signal Processing lab report based on MATLAB exercises. I can provide the original data from it if that helps.

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採用された回答

Roshan Hingnekar
Roshan Hingnekar 2020 年 12 月 1 日
In the code mentioned above the length of vector 'n' is 33, the length of vector 'x' is 75, Hence the plot function is giving error.
Use the following code:
n = linspace(-20,12,75);
x = repmat([5,4,3,2,1],3 ,15);
plot(n,x)
xlim([-20 12])
The aboce code should give the saw-tooth output.
  1 件のコメント
Connor Wright
Connor Wright 2020 年 12 月 3 日
Thanks for the help, I ended up using
n = linspace(-20,12,length(x))
Whether or not that is the 'correct' way of doing things is another matter.

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