How can I calculate the inverse fourier transform in matlab?

Hi. I have to calculate the inverse fourier transform of the function F in may code and compare with the original function f. The code is

w0=15; %frequencia da fonte
w=linspace(1,150,150); %vetor da frequencia
deltaw=w(2)-w(1);
t=linspace(0,30,150); %vetor do tempo
alpha=0.1; %coeficiente de atenuação de sinal sísmico

for nt=1:length(t)
  if t(nt)>=0.0
      H(nt)=1.0; %H(t) é a função de Heaviside
  end
  if t(nt)<0.0
      H(nt)=0.0;
  end
  f(nt)=H(nt)*(exp(1)^(-alpha*t(nt)))*sin((pi/180)*w0*t(nt));
end

for nw=1:length(w)
  F(nw)=alpha/(alpha^(2)+(w(nw)+(pi/180)*w0)^(2));
end

A=(i*t'*w);
B=exp(A);
Ftempo=(1/(2*pi))*deltaw*(F*B.');
plot(t,f,t,real(Ftempo))

But the graphic is not good. What I have to do?

[Youssef Khmou]

i tried to change your code , but i do not understand the lines : "A=i*t'*w);......"

Answer 1

[Youssef Khmou]

hi Marcia,

Are you trying to compute the inverse Fourier Transform of the Heaviside function and compare it with Heaviside function?

you can try this code in which Fast Fourier Transform is computed using loops not built in fft function :

 w0=15; %frequencia da fonte
w=linspace(1,150,150); %vetor da frequencia
deltaw=w(2)-w(1);
t=linspace(-10,30,150); %vetor do tempo
alpha=0.1; %coeficiente de atenuação de sinal sísmico

 % HEAVIVISE FUNCTION H,  F(H)=z  and F^1(z)=ZZ

 for nt=1:length(t)

    if t(nt)<0.0
      H(nt)=0.00; %H(t) é a função de Heaviside
    elseif t(nt)==0
        H(nt)=0.5;
    elseif t(nt)>0.0
      H(nt)=1.00;
    end
end

 figure, plot(t,H,'LineWidth',1.9), axis([-10 30 -1 2]), title(' Heaviside function H(x)'), grid on ,

 % Fourier Transform,  without using FFT

 N=length(H);
 nfft=N;
 z=zeros(1,nfft);
 Sum=0;
 for k=1:nfft
     for jj=1:N
         Sum=Sum+H(jj)*exp(-2*pi*j*(jj-1)*(k-1)/nfft);
     end
 z(k)=Sum;
 Sum=0;% Reset
 end

 figure, plot(w, abs(z),'LineWidth',1.9), title(' Fourier Transform of F[H(x)]'), grid on

 % Inverse FOURIER Transform
  ZZ=zeros(1,nfft);
 Sum=0;
 for k=1:nfft
     for jj=1:N
         Sum=Sum+z(jj)*exp(2*pi*j*(jj-1)*(k-1)/nfft);
     end
 ZZ(k)=Sum;
 Sum=0;% Reset
 end
 ZZ=ZZ/N;

 figure, plot(t, real(ZZ),'LineWidth',1.9), title(' Inverse Fourier Transform F^{-1}[F[H(x)]]=H(x)'), grid on
 axis([-10 30 -1 2])

[Marcia Azeredo]

Yes. This is what I would like to do. Thanks. I'l test this code.