Borrar filtros
Borrar filtros

fouriertransform with piecewise function

26 visualizaciones (últimos 30 días)
Elisa
Elisa el 11 de Dic. de 2022
Editada: Paul el 13 de Dic. de 2022
I am trying to apply a fft (fourier transform) into a piecewise function, but i receive the following message: "Error using fft
Invalid data type. First argument must be double, single, int8, uint8, int16, uint16, int32, uint32, or logical."
Code:
figure(1);
syms t
x = piecewise(-1/2<=t<=1/2, 1-abs(t), t<-1/2, 0 , t>1/2, 0);
fplot(x,[-2,2])
grid
xlabel("Time")
ylabel("Amplitude")
fa = 15000;
t = 0:1/fa:2;
y=fft(x);
f = -fa:1:fa;
figure(2);
subplot(2,1,1);
plot(f,abs(y_ajustado)/fa);
xlabel("Frequency (Hz)")
ylabel("Amplitude")
xlim([0 400])
subplot(2,1,2)
plot(f,angle(y_ajustado));
xlabel("Frequency (Hz)")
ylabel("Fase (Graus)")
xlim([0 400])

Respuesta aceptada

VBBV
VBBV el 11 de Dic. de 2022
use double and subs() the result of x
figure(1);
syms t
x = piecewise(-1/2<=t<=1/2, 1-abs(t), t<-1/2, 0 , t>1/2, 0)
x = 
fplot(x,[-2,2])
grid
xlabel("Time")
ylabel("Amplitude")
fa = 15000;
t = 0:1/fa:2;
y=fft(double(subs(x,t)));
plot(t,real(y),t,imag(y))
  2 comentarios
Elisa
Elisa el 12 de Dic. de 2022
thank you!!!!
Paul
Paul el 12 de Dic. de 2022
Editada: Paul el 13 de Dic. de 2022
Why is y plotted agains t? Shouldn't y be plotted against frequency? And the fft is only taken for the portion of the signal for t>=0, but it's non-zero for t < 0.

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Fourier Analysis and Filtering en Help Center y File Exchange.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by