why fft of an even and real signal in time domain gives imaginary part?

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2NOR_Kh 2022-12-2

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https://ww2.mathworks.cn/matlabcentral/answers/1869647-why-fft-of-an-even-and-real-signal-in-time-domain-gives-imaginary-part

评论： 2NOR_Kh 2022-12-3

This is the question:

a. Create a real and even-symmetric signal for use as a test signal. Since this is easiest to do when the length is odd, take N to be odd (e.g., N = 21). Verify that the DFT of this signal is purely real, and that it is also even-symmetric.

This is my attempt:

I have a coine function which is real and even and I expecy its fourier transform to be real and conjugate symmetry. I applied f1=fftshift(abs(fft(x1))); for taking the fft and the imaginary part is zero. But for part b it asks:

b. Now take the FFT of the same signal, but with the length three times the original (i.e., N = 63). Verify that the transform is no longer purely real (unless, of course, your test signal were the impulse).

still with the same procedure the imaginary part is zero. Where did I make a mistake?

This is my code for a:

N = 21;

tmin=-1;

tmax=1;

t = linspace(tmin,tmax,N+1);

t(end) = [] ;

fs=1/abs(t(1)-t(2));

a = 2;

x1 = cos(a*pi*t);

figure,plot(t,x1);xlabel 'x axis', ylabel 'y axis', title 'cos(2*pi*t)'

k = ( (-(N-1)/2) : ((N-1)/2) )/N*fs;

f1=fftshift(abs(fft(x1)));

figure,plot(k,f1);xlabel 'x axis', ylabel 'y axis',

title 'fourier transform of cos(2*pi*t)'

figure,plot(k,imag(f1)),

title 'imaginary part of the fourier transform of cos(2*pi*t)'

xlabel 'x axis', ylabel 'y axis'

code for part b:

%% the same signal with three times more points

N2 = 63;

tmin=-1;

tmax=1;

t2 = linspace(tmin,tmax,N2+1);

t2(end) = [] ;

fs2=1/abs(t2(1)-t2(2));

a = 2;

x2 = cos(a*pi*t2);

figure,plot(t2,x2);xlabel 'x axis', ylabel 'y axis', title 'cos(2*pi*t)'

k2 = ( (-(N2-1)/2) : ((N2-1)/2) )/N2*fs2;

f2=fftshift(abs(fft(x2)));

figure,plot(k2,f2);xlabel 'x axis', ylabel 'y axis',

title 'fourier transform of cos(2*pi*t)'

figure,plot(k2,imag(f2)),

title 'imaginary part of the fourier transform of cos(2*pi*t)'

xlabel 'x axis', ylabel 'y axis';

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Answer 1

Walter Roberson 2022-12-2

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https://ww2.mathworks.cn/matlabcentral/answers/1869647-why-fft-of-an-even-and-real-signal-in-time-domain-gives-imaginary-part#answer_1118632

在 MATLAB Online 中打开

t = linspace(tmin,tmax,N+1);
t(end) = [] ;

Your signal must not return to origin in the final sample. You need a signal such that repmat of the signal would be multiple periods. For example if the signal was triangle 0 1 then if you input 0 1 0 then fft treats it as if it were an indefinite chain of 0 1 0 0 1 0 0 1 0 rather than as 0 1 0 1 0 1...

Also if you are going to use that k then you need to fftshift to plot properly. The negative frequencies are after the positive.

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2NOR_Kh 2022-12-2

编辑：2NOR_Kh 2022-12-2

在 MATLAB Online 中打开

Besides of the changes you mentioned I took the fftshift and fft of the abs value of the signal and now it is what I expect in theory. Why?

Also, your suggestion of not including the final sample was like a magic. ;-)

In the part b of the question it asks:

b. Now take the FFT of the same signal, but with the length three times the original (i.e., N = 63). Verify that the transform is no longer purely real (unless, of course, your test signal were the impulse).

With the same procedure I took the fft of this signal with 63 points and still the imaginary part is zero. I am confused right now. Maybe I am doing something wrong.

N2 = 63;

tmin=-1;

tmax=1;

t2 = linspace(tmin,tmax,N2+1);

t2(end) = [] ;

fs2=1/abs(t2(1)-t2(2));

a = 2;

x2 = cos(a*pi*t2);

figure,plot(t2,x2);xlabel 'x axis', ylabel 'y axis', title 'cos(2*pi*t)'

k2 = ( (-(N2-1)/2) : ((N2-1)/2) )/N2*fs2;

f2=fftshift(abs(fft(x2)));

figure,plot(k2,f2);xlabel 'x axis', ylabel 'y axis',

title 'fourier transform of cos(2*pi*t)'

figure,plot(k2,imag(f2)),

title 'imaginary part of the fourier transform of cos(2*pi*t)'

xlabel 'x axis', ylabel 'y axis';

Walter Roberson 2022-12-3

在 MATLAB Online 中打开

In both cases the non-zero imaginary component of the fft is due to round-off error.

%21

N1 = 21;

tmin=-1;

tmax=1;

t1 = linspace(tmin,tmax,N1+1);

t1(end) = [] ;

fs1=1/abs(t1(1)-t1(2));

a = 2;

x1 = cos(a*pi*t1);

figure

plot(t1,x1);

xlabel('x axis');

ylabel('y axis');

title('cos(2*pi*t) 21')

k1 = ( (-(N1-1)/2) : ((N1-1)/2) )/N1*fs1;

f1 = fftshift((fft(x1)));

figure

subplot(2,1,1);

plot(k1,real(f1)); xlabel('x axis'); ylabel('y axis'); title('fft real part 21')

subplot(2,1,2);

plot(k1,imag(f1)); xlabel('x axis'); ylabel('y axis'); title('fft imag part 21')

%63

N2 = 63;

tmin=-1;

tmax=1;

t2 = linspace(tmin,tmax,N2+1);

t2(end) = [] ;

fs2=1/abs(t2(1)-t2(2));

a = 2;

x2 = cos(a*pi*t2);

figure

plot(t2,x2);

xlabel('x axis');

ylabel('y axis');

title('cos(2*pi*t) 63')

k2 = ( (-(N2-1)/2) : ((N2-1)/2) )/N2*fs2;

f2 = fftshift((fft(x2)));

figure

subplot(2,1,1);

plot(k2,real(f2)); xlabel('x axis'); ylabel('y axis'); title('fft real part 63')

subplot(2,1,2);

plot(k2,imag(f2)); xlabel('x axis'); ylabel('y axis'); title('fft imag part 63')

2NOR_Kh 2022-12-3

But how can I show what part a and b asks for?

That for N=21 the imaginary is zero but for n=63 non_zero. This is a weird question and makes me confused.

This is the refrence of this question and the whole question in case I am missing anything in my solution or explanation:

The problem is adopted from Exercise 5.2 in “Computer-Based Exercises for Signal Processing Using MATLAB,” C. S. Burrus, J. H. McClellan, A. V. Oppenheim, T. W. Parks, and R. W. Schaefer, Pearson, 1994.

One difficulty with zero padding is that it destroys symmetry, in other words, if the original signal were such that its DFT were purely real, then after zero padding the resultant DFT is probably not real. There is a method of zero padding that will preserve such symmetries – this is padding “in the middle.”

a. Create a real and even-symmetric signal for use as a test signal. Since this is easiest to do when the length is odd, take N to be odd (e.g., N = 21). Verify that the DFT of this signal is purely real, and that it is also even-symmetric.

b. Now take the FFT of the same signal, but with the length three times the original (i.e., N = 63). Verify that the transform is no longer purely real (unless, of course, your test signal were the impulse).

c. Now create a new signal that is also three times as long as the original, but do the padding in the following way:

1. In the first 0.5* (N + 1) points of the new signal, place the first 0.5* (N + 1) points of the original signal. (Remember that N is assumed to be odd.)

2. In the next 2N points of the new signal, place zeros.

3. In the last 0.5(N −1) points of new signal, place the (remaining) last 0.5 (N −1) points of the original signal. Write a Matlab program that does this operation, and then verify that it preserves symmetry by testing with simple inputs and their DFTs. Also check that the interpolation property still holds

Walter Roberson 2022-12-3

fft of "the same signal" "with length 63" is probably fft(x1, 63) rather than creating a new signal of length 63

2NOR_Kh 2022-12-3

在 MATLAB Online 中打开

Thanks for your attention. I fixed it and these are the results:

%%part a:

N = 21;

tmin=-1;

tmax=1;

t = linspace(tmin,tmax,N+1);

t(end) = [] ;

fs=1/abs(t(1)-t(2));

a = 2;

x1 = cos(a*pi*t);

figure,plot(t,x1);xlabel 'x axis', ylabel 'y axis', title 'cos(2*pi*t)'

k = ( (-(N-1)/2) : ((N-1)/2) )/N*fs;

f1=fftshift((fft(x1)));

figure,plot(k,f1);xlabel 'x axis', ylabel 'y axis',

Warning: Imaginary parts of complex X and/or Y arguments ignored.

title 'fourier transform of cos(2*pi*t)'

figure,plot(k,imag(f1)),

title 'imaginary part of the fourier transform of cos(2*pi*t)'

xlabel 'x axis', ylabel 'y axis';

%% partb:the same signal wit

h three times more points

N2=63;

k2 = ( (-(N2-1)/2) : ((N2-1)/2) )/N2*fs;

f2=fftshift((fft(x1,63)));

figure,plot(k2,f2);xlabel 'x axis', ylabel 'y axis',

Warning: Imaginary parts of complex X and/or Y arguments ignored.

title 'fourier transform of cos(2*pi*t)'

figure,plot(k2,imag(f2)),

title 'imaginary part of the fourier transform of cos(2*pi*t)'

xlabel 'x axis', ylabel 'y axis';

So, based on what we discuused so far, I've got to this conclusion that for N=21 the imaginary part is negligible since it in the range of power of -16. Ans as you said this id due to the round-off error.

I learned alot from you in this post and I appreciate it so much.

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Answer 2

Image Analyst 2022-12-2

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1869647-why-fft-of-an-even-and-real-signal-in-time-domain-gives-imaginary-part#answer_1118767

You might like to look at Steve's blog on the topic.

https://blogs.mathworks.com/steve/2021/02/26/fft-spectral-leakage/

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2NOR_Kh 2022-12-2

It was a great link, I appreciate your help.

However, I couldnt find the answer for my question. In theory we expect the fourier transform of a real and even function to be real and conjugate symmetric but I cant see that in MATLAB. I chose cosine function from -1 to 1. So, the imaginary part of its fourier transform must be zero but it is not. Why?

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