why the plot figure is empty?
6 vues (au cours des 30 derniers jours)
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claudia francis
le 24 Juin 2022
Réponse apportée : Yash Srivastava
le 28 Juin 2022
i'm trying to plot funcion of t that has sigma of n in it.
it's not showing any errors
Ts = 4;
f = @(t) sinc((t-n*Ts)/Ts);
x1n = sinc(n*4);
x2n = (sinc(n*4/12)).^2;
x3n = cos(pi*n*4/12);
syms n
xr1 = @(t) symsum(x1n.*f,n,-13,13);
syms n
xr2 = @(t) symsum(x2n.*f,n,-13,13);
syms n
xr3 = @(t) symsum(x3n.*f,n,-13,13);
figure(1)
fplot(xr1,[-13 13])
title("x_1[n]")
figure(2)
fplot(xr2,[-13 13])
title("x_2[n]")
figure(3)
fplot(xr3,[-13 13])
title("x_3[n]")
1 commentaire
Dyuman Joshi
le 24 Juin 2022
sinc(0) throws off an error in symbolic usage
syms x1n(n)
x1n(n) = sinc(n*4)
x1n(0)
Réponse acceptée
Yash Srivastava
le 28 Juin 2022
Hi Claudia
I have modified your code a little bit. Instead of using function handle, we can directly use the expression treating n and t of "syms" type.
Also, sinc(x) = sin(pi*x) / pi*x, will result into "division by 0" error. Therefore, in xr1, xr2 and xr3 I have changed the range of n from [-13 13] to [1 13] to avoid that error.
clear;
syms n t
Ts = 4;
f = sinc((t-n*Ts)/Ts);
x1n = sinc(n*4);
x2n = (sinc(n*4/12)).^2;
x3n = cos(pi*n*4/12);
xr1 = symsum(x1n.*f,n,1,13);
xr2 = symsum(x2n.*f,n,1,13);
xr3 = symsum(x3n.*f,n,1,13);
figure(1)
fplot(xr1,[-13 13])
title("x_1[n]")
figure(2)
fplot(xr2,[-13 13])
title("x_2[n]")
figure(3)
fplot(xr3,[-13 13])
title("x_3[n]")
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