Fourier series and transform of Sinc Function

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)));

Answer 1

[Paul]

Hello @cikalekli,

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 ....

[David Goodmanson]

Hello cikalekli,

It appears that no one is clear on the meaning of the question yet. You may need to rephrase it. All part of the process. Perhaps you are asking about the parts of the bars protruding outside of the red curve, perhaps not. If so, don't forget that, since bars are centered at the x value they represent, a corresponding line y value appears at the center of the top of the bar. Anyway, yes, same sinc envelope for both.

bar(1:10,1:10),hold on,plot(1:10,1:10),hold off

[cikalekli]

Dear David,

Thank you for understanding me. It means a lot to me. Literally I asked that approx. 1 week ago and I could not understand why someone did not help even I was clear with my question and my codes.

I literally even showed my progress with my code outputs as well.

Finally you did understood my question and elaborate it smoothly.

Thanks a lot and happy new year...

[Paul]

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)));

Answer 2

[Kshitij Chhabra]

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!

[Rik]

@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.

[cikalekli]

I apologize for my inappropriate behavior. But I asked everything clearly in my question. I never expected an answer from anyone. I just wanted to ask about the correctness and relevance of my code. That's why I reacted to such a response that came days later. Where did I go wrong? I wonder if I should have asked differently when asking a question? What is the error in my question while posting here please? Again pardon me and I am not going to do the same mistake ever again. Thank you for your understanding as well.

Edit: Typo correction.