- #1
tworitdash
- 104
- 25
I found a paper by Brethorst where he developed a periodogram that is a generalized version of the Lomb-Scargle periodogram. You can find it here [1].
I tried to implement (22) from this paper to make a periodogram for an aperiodically sampled complex data that is stochastic. I observed that it is the same as a Schuster periodogram. I want to verify what I did. Please let me know if something is wrong.
In the paper, they added a decay factor in the model ## Z ##, which I set to [itex ] 0 [/itex ].
Second, they also have different lengths for the real and imaginary parts of the signal. However, for me, they are collected at the same time. ## N_R = N_I = N_d ##, and ## t_i = t_j ##.
I choose the $H$ as the basis ## H = 2 \pi f t ## as they do in (23).
If I go by these assumptions, the following quantities become:
$$\theta = \frac{1}{2} \tan^{-1}\left(\frac{0}{0}\right) = 0$$ From (20)
$$ C = N_d $$ from (17)
$$ S = N_d $$ from (18)
$$ R = \sum_{i = 1}^{N_d} d_R(t_i) \cos{(H(t_i))} - d_I(t_i) \sin{(H(t_i))} $$
$$ I = \sum_{i = 1}^{N_d} d_R(t_i) \sin{(H(t_i))} + d_I(t_i) \cos{(H(t_i))} $$
Here, ## d_R = \Re({z}) ##, and ## d_I = \Im({z}) ##. So, the final expression (22) becomes:
$$ \bar{h}^2 = \frac{1}{N_d} \times (R^2 + I^2) $$
I think this is the same as the Schuster periodogram. Am I correct? In that case, which periodogram should I use with lower side-lobe levels than the Schuster periodogram for the aperiodically sampled complex signal?
[1]: https://bayes.wustl.edu/glb/general.pdf
I tried to implement (22) from this paper to make a periodogram for an aperiodically sampled complex data that is stochastic. I observed that it is the same as a Schuster periodogram. I want to verify what I did. Please let me know if something is wrong.
In the paper, they added a decay factor in the model ## Z ##, which I set to [itex ] 0 [/itex ].
Second, they also have different lengths for the real and imaginary parts of the signal. However, for me, they are collected at the same time. ## N_R = N_I = N_d ##, and ## t_i = t_j ##.
I choose the $H$ as the basis ## H = 2 \pi f t ## as they do in (23).
If I go by these assumptions, the following quantities become:
$$\theta = \frac{1}{2} \tan^{-1}\left(\frac{0}{0}\right) = 0$$ From (20)
$$ C = N_d $$ from (17)
$$ S = N_d $$ from (18)
$$ R = \sum_{i = 1}^{N_d} d_R(t_i) \cos{(H(t_i))} - d_I(t_i) \sin{(H(t_i))} $$
$$ I = \sum_{i = 1}^{N_d} d_R(t_i) \sin{(H(t_i))} + d_I(t_i) \cos{(H(t_i))} $$
Here, ## d_R = \Re({z}) ##, and ## d_I = \Im({z}) ##. So, the final expression (22) becomes:
$$ \bar{h}^2 = \frac{1}{N_d} \times (R^2 + I^2) $$
I think this is the same as the Schuster periodogram. Am I correct? In that case, which periodogram should I use with lower side-lobe levels than the Schuster periodogram for the aperiodically sampled complex signal?
[1]: https://bayes.wustl.edu/glb/general.pdf