# How do you find the null to null bandwidth for the signal below?

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

• 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. Sep 21 at 15:40

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')


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


$$5000$$

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


this matches with half the result of

abs(2*x0)


$$5000$$

• 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.
– Jdip
Sep 22 at 19:00
• might want to modify your code a little, you’re doing 4.10e-5 instead of 4.10e-4 ;)
– Jdip
Sep 23 at 0:26
• 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? Sep 25 at 3:48
• Also thanks for the visualization mate. Appreciate it Sep 25 at 3:49
• Yes 2500hz is right if you only consider the positive frequency part…
– Jdip
Sep 25 at 4:14

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 $$\text{5000 Hz}$$.

• 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? Sep 25 at 3:47

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 $$2 /(4*10^{-4}) = 5000 Hz$$

• 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? Sep 25 at 3:47