Given below is the code for a radiated electric field as a function of time. I need to take its Fourier transform.

clc;clear all;

tau=35*10^-15

mu=200*(10^(-4))

E_b=30

e0=8.854*(10^(-12))

pi=3.14

w0=20*10^-6

A=pi*(w0^(2))

lambda=800*10^-9

c=3*10^8

I0=1.59*(10^20)

t=linspace(-0.3*10^-12,0.3*10^-12,500)

I=I0*exp(-((t./tau).^2))

plot(t,I)

xlabel('time(ps)');

ylabel('I(t)');

e=1.602*10^-19;

eta=377;

F_opt=10

F_opt_2=1.3

F_opt_3=1.0

b=5*10^-6;

n=3.94;

m=200*(10^(-4));

E_b=30;

pi=3.14;

R=0.359

c=3*10^8;

I0=1.59*(10^20)

h=6.626*10^-34;

h_cross=h./(2*(pi));

w=2.3561*10^15;

E_p=h_cross*w

r=20*10^-6;

A=pi*(r^(2))

B=(A.*e.*(1-R).*m)./(4.*pi.*e0.*c.^(-2).*b.*h_cross.*w.*sqrt(pi))

D=(eta.*e.*(1-R).*m)./((n+1).*h_cross.*w.*sqrt(pi))

syms x

f_1=tau.*exp(-x.^(2));

t=linspace(-0.3*10^-12,0.3*10^-12,500);

fun=matlabFunction(f_1,'Vars',x)

for k = 1:length(t)

z_1(k)=integral(fun,-Inf,(t(k)/tau),'ArrayValued',true);

end

E_rad=-B.*E_b.*(F_opt/tau).*exp(-((t./tau).^2)).*((1 + (D.*F_opt.*z_1(k))).^(-2))

Q=-E_rad./(B.*E_b)

plot(t,Q)

xlabel('time(ps)');

ylabel('-E_ rad/BE_b');

Answer 1

[Christopher McCausland]

Hi Jasmine,

You can use the inbuilt fft() function. Documentation is here.

Christopher