Phase amplitude coupling correct?

Discussions
1

Hi!

I wanted to test if the PAC is working as expected. Therefore I generated a test signal in Matlab and imported the signal as raw file:


    f_l=5;    % low frequency
    f_h=50; % high frequency
   
    t = (0:0.001:360);
    x=(sin(2*pi*f_l*t).*3)+(sin(2*pi*f_l*t)).*(0.5*cos(2*pi*f_h*t));

If I understand the PAC right, there should be a peak in the comodulogram around f_P=5 Hz and f_A=50 Hz. However, the result I get with brainstorm is:

PAC_x.jpg684×512 27.3 KB

No coupling is visible at the frequencies 5 Hz and 50 Hz...

Am I missing something, do I misunderstand the PAC, or is something wrong with the function?

Thanks in advance.

Best Regards,
Manuel

2

Hi Manuel,

the code you provided generates a signal, which has a burst of 50Hz oscillations both at the peak and the trough of the low-frequency (5Hz). In the computation of the PAC they should cancel each other out (Since in the PAC estimation we’re averaging vectors in complex space). That way it’s not surprising that you don’t see anything in the comodulogram.

There is a process in Brainstorm to simulate PAC signals (process_pac_simulate.m). You can use it to evaluate the method.

Chee

3

Hi Manuel,

Peter gave me some more explanations about your example signal. He suggests you replace your signal:
sin(f_pt) * sin(f_at), where f_p is 5Hz and f_a is 50 Hz
with the following, so that the modulating sine wave (5Hz) is strictly positive:
(1+sin(f_pt)) * sin(f_at)
In this way you get the burst of 50Hz only at the peak of the modulating 5Hz.

For the FFT: If you multiply two sine waves of frequencies a and b you get a sum of two other sine waves (the sidebands at 50+/-5Hz).
sin(at)sin(bt) = [cos(a-b)t - cos(a+b)t]/2
In his signal this is what happens, you see only the sidebands. If we do it the right way (1+...), then we get an extra term sin(f_at) in the sum, we see the 50Hz peak in the FFT and the PAC estimation works.

Your signal: 3.sin(2pif_lt) + sin(2pif_lt) . 0.5 .* cos(2pif_h*t)

The modified signal: 3.sin(2pif_lt) + (1 + sin(2pif_lt)) . 0.5 .* cos(2pif_h*t)

Francois & Peter

4

Thank you for your explanations!

Best regards,
Manuel