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# Bandwidth of The Sinc Function

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#### Mustaine

##### Member level 1
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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In the first case, half of the sinc signal is cut. The steep rise is causing the out-of-band components.

### Mustaine

Points: 2
In the first case, half of the sinc signal is cut. The steep rise is causing the out-of-band components.
thanks for your reply i get the issue on the first problem but still i do not get why the bandwidth is different from the B value.

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
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).

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).
magnitude must be 1 i guess but its 80 in the figure

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magnitude must be 1 i guess but its 80 in the figure
That is another case of normalising response:

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

you will get zero dB

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