question about sinc function
显示 更早的评论
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));
2 个评论
Dimitris Kalogiros
2019-9-2
Hi Adnan
What exactly do want to plot ?
A sync(t) function ? Or function u(t) which you gave ?
ADNAN KIRAL
2019-9-2
采纳的回答
KALYAN ACHARJYA
2019-9-2
编辑：KALYAN ACHARJYA
2019-9-2
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);
10 个评论
ADNAN KIRAL
2019-9-2
KALYAN ACHARJYA
2019-9-2
编辑：KALYAN ACHARJYA
2019-9-2
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
2019-9-2
Torsten
2019-9-2
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
2019-9-2
编辑：KALYAN ACHARJYA
2019-9-2
Its my pleasure Adnan!
KALYAN ACHARJYA
2019-9-2
编辑：KALYAN ACHARJYA
2019-9-2
sinc(x)=sin(x)/x for otherwise case.
Hence sin(x)=x*sinc(x)
KALYAN ACHARJYA
2019-9-2
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
2019-9-2
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
2019-9-2
@Torsten OK sir, thanks for rectifying me. I will update it.
Gratitude!
更多回答（1 个）
t = -511*0.005:0.005:512*0.005;
u_1 = 3/2*(110*sinc(110*t)-60*sinc(60*t));
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