Fourier series and transform of Sinc Function
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Does the line spectrum acquired in 2nd have a sinc envelope like the one obtained in 3rd?
Here is my code below:
x = [-5:0.001:5];
y = sinc(x);
plot(x,y);
% 2nd sinc graph:
duty = 0.2;
n = [0:1:15];
cn = 5 * duty * abs(sinc(n*duty));
bar(n*duty,cn);
hold on
% 3rd sinc graph:
n = [0:0.001:3];
plot(x,abs(sinc(x)));
0 个评论
采纳的回答
Paul
2022-1-1
编辑:Paul
2022-1-1
The second and third graph are both plots of the magnitude of the sinc() function, so it would appear that the third must evelop the first.
Perhaps I've misunderstood the question ....
3 个评论
Paul
2022-1-2
Perhaps a stem plot would be more appropriate for this type of plot
x = [-5:0.001:5];
duty = 0.2;
n = [0:1:15];
cn = 5 * duty * abs(sinc(n*duty));
stem(n*duty,cn); % changed bar to stem
hold on
% 3rd sinc graph:
% n = [0:0.001:3];
plot(x,abs(sinc(x)));
更多回答(1 个)
Kshitij Chhabra
2021-12-31
Hi,
You can try leveraging the Curve Fitting Toolbox offering from MATLAB to get insights on how to plot a curve over the bar chart.
You can check the various examples to get a clearer insight. Once the curve is optained, you can compare the values between the two plots.
Hope it helps!
2 个评论
Rik
2022-1-1
@cikalekli, Why don't you post a comment explaining what would be wrong with this answer? If you just reject the answer flagging it as misinformation, that will not encourage anyone to help you.
You have a question. Apparently Kshitij didn't provide an answer that you were happy with. Remember that this process has many steps: first you have a problem, then you tried to explain it, then Kshitij read it, tried to understand it, found two links for you, and wrote up an answer. Then you read it and tried to understand it.
ALL those steps must go well for you to be happy with the result. This is the internet, so you can't assume to have the same linguistic and cultural background as anyone responding. Kshitij took the time to attempt to help you. Now you can repay that kindness by explaining why this isn't what you meant. Flagging accurate (though apparently not helpful) information as 'misinformation' isn't going to help you.
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