question about sinc function

hi how can i get this figure ?

i have used the following eq.

by coding the following type;

but it did not work can u please help me with that ? i cannot get negative values. thanks.

    t=-511*0.005:0.01:512*0.005;
u_1=3/2*(110*sinc(110*pi*t)-60*sinc(60*pi*t));

[Dimitris Kalogiros]

Hi Adnan

What exactly do want to plot ?

A sync(t) function ? Or function u(t) which you gave ?

[ADNAN KIRAL]

thanks Dimitris Kalogiros

Mr KALYAN ACHARJYA replied.

Answer 1

[KALYAN ACHARJYA]

t=-linspace(-511*0.005,512*0.005,1000)
u_t=3*(sin(2*55*pi*t)-sin(2*30*pi*t))./(2*pi*t);
plot(u_t);

[ADNAN KIRAL]

hi KALYAN ACHARJYA

thanks for quick feedback.

would you please tell me the equivalent of this function in sinc ?

As this function in my code goes infinite form so that i need its sinc formation.

[KALYAN ACHARJYA]

Using sinc

t=-linspace(-511*0.005,512*0.005,1000)
u_t=3*((2*55*pi*t).*sinc(2*55*pi*t)-(2*30*pi*t).*sinc(2*30*pi*t))./(2*pi*t);
plot(t,u_t);

Will give the same results.

I am sure, you can do basic multiplication operation to simply the code.

[ADNAN KIRAL]

many thanks

[Torsten]

Note that the definition for sinc in MATLAB is

sinc(t) = sin(pi*t)/(pi*t) 

So your u_t written in sinc is not correct.

[KALYAN ACHARJYA]

Its my pleasure Adnan!

[KALYAN ACHARJYA]

sinc(x)=sin(x)/x for otherwise case.

Hence sin(x)=x*sinc(x)

[Torsten]

No.

sin(x) = x*sinc(x/pi)

https://de.mathworks.com/help/signal/ref/sinc.html

[KALYAN ACHARJYA]

Now this one sir?

t=-linspace(-511*0.005,512*0.005,1000)
u_t=3*((2*55*pi*t).*sinc(2*55*t)-(2*30*pi*t).*sinc(2*30*t))./(2*pi*t);
plot(t,u_t);

[Torsten]

This doesn't help since you still divide by 0 at t=0.

If you divide out, you get what I wrote in my answer.

[KALYAN ACHARJYA]

@Torsten OK sir, thanks for rectifying me. I will update it.

Gratitude!

Answer 2

[Torsten]

t = -511*0.005:0.005:512*0.005;
u_1 = 3/2*(110*sinc(110*t)-60*sinc(60*t));

[ADNAN KIRAL]

thanks for your reply.