Why the FFT of the conjugate doesn't coincide exactly at the negative frequency of interest?

I am trying to understand why the conjugate of a signal in the time domain doesn't produce an exact flip of the frequency domain spectrum. There is always a one-pixel shift in the result.

The MATLAB code is shown below. I use a flip for the conjugate spectrum to show that it doesn't match the original spectrum and there is a delay of one resolution cell. I want to understand why this happens. It shouldn't as it is the perfect conjugate.

v_max = 7.5;
v = 5;
N = 128;
PRT = 1e-3;
lambda = 0.03;
vel_axis = linspace(-v_max, v_max, N);

x = exp(1j .* 4 .* pi .* v ./ lambda .* (1:N) .* PRT);

x_conj = conj(exp(1j .* 4 .* pi .* v ./ lambda .* (1:N) .* PRT));

x_fft = 1/sqrt(N) .* fftshift(fft(x, N));
x_conj_fft = 1/sqrt(N) .* fftshift(fft(x_conj, N));
figure; plot(vel_axis, db(abs(x_fft)), 'LineWidth', 2); hold on; plot(vel_axis, flip(db(abs(x_conj_fft))), 'LineWidth', 2);grid on;

Answer 1

[Matt J]

Because the frequency origin shouldn't change positions when you do a frequency reversal. But with an even length array N, the flip() command will always relocate all elements in the array.

[Matt J]

Here is how I would code it. I made up a value for lambda, since you didn't post one.

v_max = 7.5;
v = 5;
N = 128;
PRT = 1e-3;
lambda=1;

Axis=((0:N-1) -ceil((N-1)/2))/N;  %Normalized axis

t_axis= Axis;
vel_axis =2*v_max*Axis; 

x = ifftshift( exp(1j .* 4 .* pi .* v ./ lambda .* t_axis .* PRT) );

x_conj = ifftshift( conj(exp(1j .* 4 .* pi .* v ./ lambda .* t_axis .* PRT)) );

flipit=@(z) z([1,end:-1:2]) ; 

x_fft = 1/sqrt(N) .* fft(x, N);
x_conj_fft = 1/sqrt(N) .* flipit( fft(x_conj, N)); 



figure; plot(vel_axis, fftshift(db(abs(x_fft))), 'LineWidth', 2); 
hold on; 
        plot(vel_axis, fftshift(db(abs(x_conj_fft))), 'LineWidth', 2);
hold off        
grid on;

[Tworit Dash]

@Matt J Thank you for the response. Why ifftshift on the signal before doing a fft? And with your example, the peak is not at v which should be 5. It is at 0. And lambda = 0.03 which I forgot to mention.

[Matt J]

Why ifftshift on the signal before doing a fft?

Because in my modification, I constructed the signals in the fftshifted position.

And with your example, the peak is not at v which should be 5. It is at 0

You can revert t_axis to 1:N, if you wish. I don't really understand what this signal is supposed to represent.

[Tworit Dash]

The signal is a radar monochromatic signal. The radar beam is rotating. Due to the rotation I want to have a cos(theta) term in the phase. For example, here I assume theta to be pi therefore it becomes the conjugate of the original signal. However, the moment I use cos(theta) in the time domain signal and make a fft, I alwayts encounter a positive one pixel shift like in this case. I use the flip on the conjugate fft just to show that it should coincide with the original signal. However, I actually want to see red plot exactly on the negative velocity axis and symmetric to the blue one. However, there is alwasys a shift of one pixel. I am pasting a screenshot without the flip. The red plot peak should be exactly as the blue plot peak but negative and they should cross each other at 0 velocity, but they don't . You can see a shift towards positive velocity for the conjugate. I don't want to manipulate the flip. The original problem is this. I suspect that the dc component of FFT is causing some issue as it is always included in the positive velocity axis. I don't know really how to avoid the shift in the positive velocity axis.

[Matt J]

Looks pretty symmetric to me,

v_max = 7.5;
v = 5;
N = 128;
PRT = 1e-3;
lambda=0.03;

Axis=((0:N-1) -ceil((N-1)/2))/N;

t_axis=1:N;
vel_axis =2*v_max*Axis; %<--- Changed

x = ifftshift( exp(1j .* 4 .* pi .* v ./ lambda .* t_axis .* PRT) );

x_conj =  conj(x) ;

flipit=@(z) z([1,end:-1:2]) ; %<--- Added

x_fft = 1/sqrt(N) .* fft(x, N);
x_conj_fft = 1/sqrt(N) .* fft(x_conj, N); %<--- Changed


[x_max,idx]=max(db(abs(x_fft)));

  x_max

x_max = 19.4222

  vel_max=vel_axis(idx)

vel_max = -2.4609

 [xconj_max,idx]=max(db(abs(x_conj_fft)));

  xconj_max

xconj_max = 19.4222

  velconj_max=vel_axis(idx) 

velconj_max = 2.4609

  
  
figure; plot(vel_axis, fftshift(db(abs(x_fft))), 'LineWidth', 2); 
hold on; 
        plot(vel_axis, fftshift(db(abs(x_conj_fft))), 'LineWidth', 2);
hold off        
grid on;

[Tworit Dash]

Nice solution. I suppose you just removed one pixel from the right side and included the zero frequency in a nice way. I really want to know if this is the real reason why you did this or is there anything related to the zero frequency (DC) component because that always makes me confused :P

However, try the same example with fewer points. Like instead of 128 just N = 8. In that case, we lose a lot of spectrum of the original signal (blue plot).

P.S. I also sometimes in my radar application have so few samples like 8 or even 4. Now, I try the same principle as yours and zero padded it and I can reproduce the same signal. Is that the way I should do things when the samples are too few?

[Matt J]

I really want to know if this is the real reason why you did this or is there anything related to the zero frequency (DC)

I set up vel_axis to be consistent with the conventions of fftshift. Note that for an even-sized array, the DC position is not at the center of the array after fftshift:

fftshift(0:7)

ans = 1×8

4 5 6 7 0 1 2 3

Here, the DC component is the element with 0 in it. It starts at location 1 and moves to location 5, which is not the center.

Is that the way I should do things when the samples are too few?

The design decisions to be made are the choices for N, the time sampling inderval dT, and the frequency sampling interval dF. They must be chosen to satisfy these constraints,

N*dT*dF=1

N*dT>=TimeDuration

N*dF=1/dT>=FrequencyDuration %Nyquist criterion

[Tworit Dash]

Ok. I have a very silly doubt. When we write the signal like the following, how did you already know that it is a fftshiftEd version and applied ifftshift to it. Is it because we only deal with the velocity v of only one sign?

exp(1j .* 4 .* pi .* v ./ lambda .* t_axis .* PRT)

[Matt J]

I created t_axis, so I know that t=0 is near the middle and not near the beginning.

[Tworit Dash]

The t_axis is 1 from N multiplied by PRT. So, the in time_axis t=0 is the first element t_axis(1) = 1 * PRT. I didn't understand why it is at the center.

[Matt J]

Here is an alternative way you could set up the sampling so that time is at the fftshift center, but still give you the same spectra. Since your code only involves the amplitude spectrum, it doesn't ultimately matter in this case where your time origin starts - the amplitude spectrum is invariant to time shifts.

N=128;
v_max = 7.5;
PRT = 1e-3;
v=5;
lambda=0.03;

dt=2*PRT/lambda; %time sampling interval
dv=1/N/dt;       %fourier sampling interval

Axis=(0:N-1)-ceil((N-1)/2);
t_axis=Axis*dt;
vel_axis =Axis*dv; 

x =  exp(1j .* 2 .* pi .* v .* t_axis) ;

Fourier=@(z) 1/sqrt(N).* fftshift(fft(ifftshift(z)));

x_fft = Fourier(x);
x_conj_fft = Fourier(conj(x));

  
figure; plot(vel_axis, db(abs(x_fft)), 'LineWidth', 2); 
hold on; 
        plot(vel_axis, db(abs(x_conj_fft)), 'LineWidth', 2);
hold off        
grid on;

[Tworit Dash]

Thank you for this exxplanation as well. I used to make the axis like this before like one year ago. I just didn't pay more attention this time. :)