Bandwidth of The Sinc Function

FvM said:

In the first case, half of the sinc signal is cut. The steep rise is causing the out-of-band components.

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Mustaine said:

hello my friends currently i am studying the Problem-based learning communication Systems Using Matlab and Simulink by kwonge choi and huaping lıu.
First i transformed the sinc function i'm expecting to see a square function but the function i get is so much different than the square one but when i applied a dely to the sinc transformed function becomes square. Can you explain why is that
Also I'm having a problem about the bandwidth of the sinc function. When i transform the sinc function below i'm expecting to see a square wave which has a B bandwidth but no, the bandwidth is much bigger than B. Can you explain why is that. thanks

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kaz1 said:

It is all too easy to get mixed up with maths, tools and normalisation. If it is me I will do this:

B = 2000;
t = -.02:1/8192:.02;
h = 2*B*sinc(2*B*t); %cutoff = B/8192
freqz(h,1,-.5:.001:.5,1);

so I see your normalised cutoff is B/8192. i.e. If your sampling frequency is unity (=1).

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Mustaine said:

i'm sorry but i didn't get your reply
magnitude must be 1 i guess but its 80 in the figure

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