Fourier transform of impulse function

2017 2 28
2 个回答
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更新时间：2018 2 25
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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

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 采纳的回答

Star Strider
Star Strider 2017-2-28
编辑：Star Strider 2017-2-28

1 个投票

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).

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friet
friet 2017-2-28
编辑：friet 2017-2-28
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
Star Strider 2017-2-28
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.

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更多回答（1 个）

Walter Roberson
Walter Roberson 2017-2-28

0 个投票

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

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