How to workaround this problem with angle?

Question

SIMONE 2024-3-4

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2090331-how-to-workaround-this-problem-with-angle

编辑： SIMONE 2024-3-5

(I started with MATLAB today, so I'm new). I created a script to calculate the fourier transform of a function, the amplitude spectrum and the phase spectrum. For testing purposes i tried calculating using as the input function the rectangular impulse. The fourier transform and the amplitude spectrum are right, the phase spectrum is not. When i use the angle function or the atan2 to calculate the angle i get as output alternating +pi and -pi where the output should be +pi. Watching the inputs and outputs of the angle function everything seems fine, the angle function works as intended. I think that the problem is related to the sign of the imaginary part, and maybe to the fact that they are really small number (e^-18) . I know that the code is all based on an approximation so it could be normal but i want to know if there is a workaround. (I know there is already a function(fft))

%total range of omega

delta_omega=20;

%intervals of omega that are used to approximate

omega_precision=0.01;

%initializatoin of the results vector

res=0:omega_precision:delta_omega;

%all the values of omega

val_omega=-delta_omega/2:omega_precision:delta_omega/2;

%for every omega calculate it's fourier transform (from -delta_omega/2 to delta_omega/2)

for n=1:delta_omega/omega_precision+1

res(n)=X((n-(delta_omega/omega_precision/2))*omega_precision);

end

%plot of the real value of the fourier tranform

plot(val_omega,real(res),"red");

hold on;

%plot og the immaginary values of the fourier tranform

plot(val_omega,imag(res),"cyan");

%amplitude spectrum values

V=abs(res)/pi;

hold on;

%plot of the amplitude spectrum

plot(val_omega,V,"green");

xlabel("omega");

%phase spectrum values

Fi=-angle(res);

%plot of the phase spectrumm values

plot(val_omega,Fi,"blue")

%fourier transform of function x(t)

function X=X(omega)

handle= @(t) x(t).*exp(-1i*omega*t);

X=integral(handle,-1100,1100);

end

%function x(t) (rectangular impulse)

function x = x(t)

%amplitude

A=1;

tao=pi;

x=1:length(tao);

for j=1:length(t)

if(t(j)<=tao/2 && t(j)>=-tao/2)

x(j)=A;

else

x(j)=0;

end

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Answer 1

Chunru 2024-3-5

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2090331-how-to-workaround-this-problem-with-angle#answer_1421021

在 MATLAB Online 中打开

You can try unwrap function.

%total range of omega

delta_omega=20;

%intervals of omega that are used to approximate

omega_precision=0.01;

%initializatoin of the results vector

res=0:omega_precision:delta_omega;

%all the values of omega

val_omega=-delta_omega/2:omega_precision:delta_omega/2;

%for every omega calculate it's fourier transform (from -delta_omega/2 to delta_omega/2)

for n=1:delta_omega/omega_precision+1

res(n)=X((n-(delta_omega/omega_precision/2))*omega_precision);

end

%plot of the real value of the fourier tranform

plot(val_omega,real(res),"red");

hold on;

%plot og the immaginary values of the fourier tranform

plot(val_omega,imag(res),"cyan");

%amplitude spectrum values

V=abs(res)/pi;

hold on;

%plot of the amplitude spectrum

plot(val_omega,V,"green");

xlabel("omega");

%phase spectrum values

Fi=-angle(res);

%plot of the phase spectrumm values

% plot(val_omega,Fi,"blue")

plot(val_omega,unwrap(Fi),"blue")

%fourier transform of function x(t)

function X=X(omega)

handle= @(t) x(t).*exp(-1i*omega*t);

X=integral(handle,-1100,1100);

end

%function x(t) (rectangular impulse)

function x = x(t)

%amplitude

A=1;

tao=pi;

x=1:length(tao);

for j=1:length(t)

if(t(j)<=tao/2 && t(j)>=-tao/2)

x(j)=A;

else

x(j)=0;

end

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SIMONE 2024-3-5

编辑：SIMONE 2024-3-5

I have 2 problems:

1) I already tried to unse unwrap and my output was different from yours, does some know why it looke like this when i run it?

2) This would be wrong, maybe it's my fault and i did something else wrong, as the phase spectrum should look like this from what i found online:

Thanks for rsponding and trying ti help me

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Answer 2

VBBV 2024-3-5

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2090331-how-to-workaround-this-problem-with-angle#answer_1421031

在 MATLAB Online 中打开

You have real and imaginary componens for variable Fi, so use only real angles

%total range of omega

delta_omega=20;

%intervals of omega that are used to approximate

omega_precision=0.01;

%initializatoin of the results vector

res=0:omega_precision:delta_omega;

%all the values of omega

val_omega=-delta_omega/2:omega_precision:delta_omega/2;

%for every omega calculate it's fourier transform (from -delta_omega/2 to delta_omega/2)

for n=1:delta_omega/omega_precision+1

res(n)=X((n-(delta_omega/omega_precision/2))*omega_precision);

end

%plot of the real value of the fourier tranform

plot(val_omega,real(res),"red");

hold on;

%plot og the immaginary values of the fourier tranform

plot(val_omega,imag(res),"cyan");

%amplitude spectrum values

V=abs(res)/pi;

hold on;

%plot of the amplitude spectrum

plot(val_omega,V,"green");

xlabel("omega");

%phase spectrum values

Fi=angle(real(res));

%plot of the phase spectrumm values

plot(val_omega,Fi,"blue")

%fourier transform of function x(t)

function X=X(omega)

handle= @(t) x(t).*exp(-1i*omega*t);

X=integral(handle,-1100,1100);

end

%function x(t) (rectangular impulse)

function x = x(t)

%amplitude

A=1;

tao=pi;

%x=1:length(tao);

for j=1:length(t)

if(t(j)<=tao/2 & t(j)>=-tao/2)

x(j)=A;

else

x(j)=0;

end

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SIMONE 2024-3-5

在 MATLAB Online 中打开

This would work for this specific function x(t), another solution for that would be abs(angle(res)). The problem is that it is not generalized, if i try using a function x(t) that has the imaginary part in the fourier transform then the phase spectrum is wrong. An example is using x(t)={A*exp(-abs(t)/t_0) for t>=0 , 0 for t<0}. With this funtion -angle(res) gives the right phase spectrum and angle(real(res)) equals 0. Thanks for your response and your time.

function x = x(t)
%amplitude
A=1;
t_0=1;
for j=1:length(t)
    if t(j)>=0 
        x(j)=A*exp(-abs(t(j))/t_0);
    else
        x(j)=0;
    end
end
end

VBBV 2024-3-5

在 MATLAB Online 中打开

If you consider the imaginary part of Fourier spectrum, then use the imag(res) . For a real periodic signal, the phase information of imaginary component of signal has no significance

%total range of omega

delta_omega=20;

%intervals of omega that are used to approximate

omega_precision=0.01;

%initializatoin of the results vector

res=0:omega_precision:delta_omega;

%all the values of omega

val_omega=-delta_omega/2:omega_precision:delta_omega/2;

%for every omega calculate it's fourier transform (from -delta_omega/2 to delta_omega/2)

for n=1:delta_omega/omega_precision+1

res(n)=X((n-(delta_omega/omega_precision/2))*omega_precision);

end

%plot of the real value of the fourier tranform

plot(val_omega,real(res),"red");

hold on;

%plot og the immaginary values of the fourier tranform

plot(val_omega,imag(res),"cyan");

%amplitude spectrum values

V=abs(res)/pi;

hold on;

%plot of the amplitude spectrum

plot(val_omega,V,"green");

xlabel("omega");

%phase spectrum values

Fi= angle(imag(res));

%plot of the phase spectrumm values

plot(val_omega,Fi,"blue")

%fourier transform of function x(t)

function X=X(omega)

handle= @(t) x(t).*exp(-1i*omega*t);

X=integral(handle,-1100,1100);

end

%function x(t) (rectangular impulse)

function x = x(t)

%amplitude

A=1;

t_0=1;

for j=1:length(t)

if t(j)>=0

x(j)=A*exp(-abs(t(j))/t_0);

else

x(j)=0;

end

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