Fourier transform of impulse function

I calculated the Fourier transform of a pulse function(figure 1) Using the fft function. However The fft result if kind of weird. Can anyone check if my code is right. //Thanks

clc
clear all
close all

t1=7.0e-08;
sigma=1e-08;
t=linspace(0,4.0000e-7,1000);

P=exp(-(t-t1).^2./sigma.^2);

P_FT=fft(P);                  %fourier transform of P

figure(1)                                     
plot(t*10^6,P);
grid on
xlabel('time[\mus]')
ylabel('amplitude[a.u]')

figure(2)                                     
plot(P_FT);
grid on

[Hira Asghar]

Can u explain your signal 'p'?

Answer 1

[Star Strider]

The Fourier transform of an impulse function is uniformly 1 over all frequencies from -Inf to +Inf. You did not calculate an impulse function.

You calculated some sort of exponential function that will appear as an exponential function in the Fourier transform.

Your slightly modified code:

t1=7.0e-08;
sigma=1e-08;
L = 1000;
t=linspace(0,4.0000e-7,L);
Ts = mean(diff(t));
Fs = 1/Ts;
Fn = Fs/2;

P=exp(-(t-t1).^2./sigma.^2);

P_FT=fft(P)/L;                  %fourier transform of P
Fv = linspace(0, 1, fix(L/2)+1)*Fn; % Frequency Vector
Iv = 1:length(Fv);                  % Index Vector

figure(1)                                     
plot(t*1E+6, P);
grid on
xlabel('time[\mus]')
ylabel('amplitude[a.u]')

figure(2)                                     
plot(Fv, abs(P_FT(Iv))*2);
grid on

The correct representation of the Fourier transform of your signal is in figure(2).

[friet]

Hi Star Strider! Thanks for your answer. I understand it now. I have further question. I have a function which is frequency and space dependent and I post my question a week ago here https://www.mathworks.com/matlabcentral/answers/326466-fourier-transform-space-dependent-function however, didnt get any answer. With your help today I get closer to solve my problem however, can't makes it to the final answer. Can u please help me to compute the space and frequency dependent function. I have tried the following code.

clc
  clear all
  close all

    t1=7.0e-08;
    sigma=1e-08;
    L = 1000;
  t=linspace(0,4.0000e-7,L);
  Ts = mean(diff(t));

Fs = 1/Ts;
Fn = Fs/2;
P=exp(-(t-t1).^2./sigma.^2);
P_FT=fft(P)/L;                         %fourier transform of P
Fv = linspace(0, 1, fix(L/2)+1)*Fn;    % Frequency Vector
Iv = 1:length(Fv);                     % Index Vector

figure(1)                                     
plot(Fv, abs(P_FT(Iv))*2);
grid on
xlabel('f[Hz]')
ylabel('amplitude[a.u]')

%................................................

z=linspace(0,2*10^-3,1000);     %[m]

v=2000;
alpha= 5;                      %[Neper/cm]

beta=(2*pi*Fv/v);              %[Hz]

figure(3)
plot(Fv*10^-6,beta*0.01)
xlabel('f[MHz]')
ylabel('alpha[N/cm]')

pfz=zeros(length(z),length(Fv));
pft=zeros(length(z),length(Fv));

for j_f=1:length(Fv)
    for i_z=length(z)
        pfz(i_z,j_f)=P_FT(j_f).*exp(-alpha.*z(i_z)).*exp(1i.*beta(j_f).*z(i_z));
    end
end

for i=length(z)
        pft(i,:)=ifft(pfz(i,:));
end
figure(4)
plot(z,abs(pft))

[Star Strider]

I answered you in the email you sent. The essence of that being that you can use Laplace transforms to solve partial differential equations in time-domain and space-domain by converting them to ordinary differential equations in s-domain and space-domain. That is relatively straightforward with Laplace transforms but much more difficult with Fourier transforms, since they can involve complex frequency variables. Those are much more difficult to work with.

I do not understand the problem you want to solve.

Answer 2

[Walter Roberson]

plot() with one argument that is complex-valued (hint!) plots real() of the parameter against imag() of the parameter.