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enter image description here

My tutor does not explain it very well. Can someone please explain to me the reasoning and what null to null bandwidth actually is?

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    $\begingroup$ It's asking to find where the first zeroes occur in the spectrum. You already have the expression for the spectrum so set it to zero and solve. $\endgroup$
    – Envidia
    Sep 21 at 15:40

3 Answers 3

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The following MATLAB script is a solution to your question

clear all;close all;clc

1. fzero only returns 1 zero

syms f
y1=@(f) 10*sinc(4e-4*f);

x0=fzero(y1,0)

$x_0=-2500$

$\text{sinc}$ has more than 1 zero but now we know where to look, because fzero found the 1st zero after $f = 0$ and this is half the interval you have been asked for.

2.

x0=100000;
x=[-x0:.01:x0];
y=10*sinc(.00004*x);
figure
plot(x,y)
grid on
xlabel('x')
axis([-x0 x0 -2.2 10])

y2=abs(y);
n1=find(y2<.000001)

hold on
plot(x(n1),zeros(1,numel(n1)),'or')

3. the answer is

BW=x(n1(5))-x(n1(4)) % answer

$5000$

plot([x(n1(4)) x(n1(5))],[0 0],'LineWidth',2,'Color','r')

enter image description here

this matches with half the result of

abs(2*x0)

$5000$

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    $\begingroup$ Not a math-based answer, which I'm guessing is what the OP was looking for, however this is a nice visualization that will help the OP. $\endgroup$
    – Jdip
    Sep 22 at 19:00
  • $\begingroup$ might want to modify your code a little, you’re doing 4.10e-5 instead of 4.10e-4 ;) $\endgroup$
    – Jdip
    Sep 23 at 0:26
  • $\begingroup$ The answer in the tute is given as 2500hz. And when I asked why its not 5000 the lecturer said only to consider the positive frequency part. Is he right? $\endgroup$
    – MALLU
    Sep 25 at 3:48
  • $\begingroup$ Also thanks for the visualization mate. Appreciate it $\endgroup$
    – MALLU
    Sep 25 at 3:49
  • $\begingroup$ Yes 2500hz is right if you only consider the positive frequency part… $\endgroup$
    – Jdip
    Sep 25 at 4:14
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The key observation is that $\text{sinc}(x)$ is zero for all arguments $x$ that are nonzero integers, so the problem reduces to "what values of $f$ yield the nonzero integers closest to zero, that is, $1$ and $-1$?"

The answer to that question is $f = \pm\text{2500 Hz}$, and the distance between them is $\text{5000 Hz}$.

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  • $\begingroup$ The answer in the tute is given as 2500hz. And when I asked why its not 5000 the lecturer said only to consider the positive frequency part. Is he right? $\endgroup$
    – MALLU
    Sep 25 at 3:47
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The $\text{sinc}$ function is the Fourier Transform of a Rectangular pulse.
Its zeros are located at non-zero integers of $x$.

Therefore, if your spectrum can be expressed as $\text{sinc}(Bx)$, the first zero occurs at $1/B$, the second at $2/B$, etc, and the corresponding time-domain signal is a Rectangular Pulse with Bandwidth $2 * 1/B$.

In your case, the bandwidth is $2 /(4*10^{-4}) = 5000 Hz$

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  • $\begingroup$ The answer in the tutorial is given as 2500hz. And when I asked why its not 5000 the lecturer said only to consider the positive frequency part. Is he right? $\endgroup$
    – MALLU
    Sep 25 at 3:47

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