0
$\begingroup$

let us suppose that we have non stationary signal,whose value is given by

56.69
69.22
-2.19
21.80
9.85
-80.51
-56.28
-18.79
-14.32
12.26
74.71
67.37
24.43
-9.05
-47.30
-62.60
-47.64
-26.36
20.78
67.58
49.83
28.24
25.14
-43.10
-72.81
-17.82
-12.38
-34.82
58.96
69.83
-5.10
14.65
28.29
-64.48
-61.23
26.43
1.90
-47.52
51.01
41.97
-41.54
-0.62
60.86
-52.44
-37.61
61.99
-10.36
-60.20
30.45
22.95
-55.33
34.50
85.71
-13.38
-12.63
54.70
-37.86
-77.38
4.65
9.28
-24.83
68.53
85.37
-15.80
-7.33
4.31
-72.06
-73.52
1.21
17.90
16.23
63.14
63.68
-14.40
-22.79
-46.29
-73.40
-35.51
13.92
27.94
65.99
50.76
6.08
-5.44
-33.80
-73.27
-14.35
14.32
-10.47
41.59
65.77
-9.65
-17.11
25.02
-34.94
-57.50
32.16
35.22
-47.44
14.10
68.68
-35.63
-23.43
52.82
-18.17
-45.41
35.20
13.02
-77.34
9.09
54.62
-41.86
5.54
100.61
-2.50
-47.15
21.43
-30.39
-105.88
10.12
54.47
-15.25
51.19
85.27
-13.40
-63.08
-3.18
-55.73
-85.04
18.21
56.47
24.51
54.11
55.81
-22.11
-48.48
-52.59
-52.83
-32.17
21.34
52.16
57.12
32.52
3.33
-26.44
-45.33
-51.45
-26.23
22.70
7.25
30.30
49.79
10.05
-47.82
7.81
5.60
-52.47
12.12
49.13
-27.13
-11.40
53.12
-14.12
-37.25
64.87
24.10
-64.07
16.83
25.67
-84.88
-23.01
76.72
2.82
-18.01
81.46
19.46
-69.46
-4.69
-15.00
-95.40
-14.27
81.06
10.99
12.24
77.50
-13.87
-80.90
-39.37
-33.24
-55.32
27.60
87.91
46.31
35.38
34.22
-36.01
-70.78
-43.57
-18.78
-5.34
48.43
53.05
39.49
20.91
-11.03
-39.82
-42.87
-34.33
-21.77
33.55
22.34
11.90
27.41
7.81
-40.11
1.82
9.00
-45.68
-10.45
44.91
-7.91
-47.04
41.03
24.43
-45.47
28.51
54.19
-51.30
-33.24
27.62
-63.64
-53.94
84.27
31.39
-19.69
67.12
39.14
-85.28
-46.64
3.16
-58.10
-30.32
96.07
56.72
2.62
46.12
4.95
-114.65
-43.85
0.83
-27.53
16.15
99.72
40.78
-9.38
15.97
-32.67
-78.52
-36.29
8.67
3.76
36.16
61.85
31.10
-9.09
-14.47
-51.55
-35.03
-16.66
0.81
39.57
42.17
-8.22
13.39
20.67
-36.92
-21.53
25.32
-13.28
-27.59
41.31
-3.05
-56.63
21.87
41.42
-24.58
19.23
65.43
-55.32
-57.21
28.64
-12.84
-50.87
75.03

i have tried to apply autoregressive model to this data,by using following matlab code

[Pxx,f]=pyulear(B,20,1024,100);

plot(f,Pxx)

and i have got following picture

enter image description here

in this case i had not any information about order,just i have took randomly,now one point is that this signal is represented by following form

$y[t]=A_1(sin(\omega_1*t+\phi_1)+A_2*sin(\omega_2*t+\phi_2)+....+A_p*sin(\omega_p*t+\phi_p)$+$z(t)$

where $\phi$ are some constants and also frequencies and amplitudes,$z(t)$ is white noise,because this signal is nonstationary,can we apply standard autoregressive models(AR,ARMA) to this model?which methods are suitable for better spectral resolution?i am not interested estimation of parameters,just i am concerned about spectral resolution,i want to distinguish from each other closed spaced frequencies,or when $\omega_p\approx \omega_{p+1}$thanks in advance

$\endgroup$
  • $\begingroup$ Why do you think the signal is non-stationary? $\endgroup$ – Peter K. Mar 28 '14 at 13:25
  • $\begingroup$ amplitude,phases and frequencies are constants,are not they non stationary? $\endgroup$ – dato datuashvili Mar 28 '14 at 13:31
  • $\begingroup$ If $A_k$, $\omega_k$ and $\phi_k$ are all constants (for $k=1\ldots p$) then the signal is stationary. If $A_k = A_k(t)$ (i.e. is time-varying) then the signal may be non-stationary. $\endgroup$ – Peter K. Mar 28 '14 at 13:34
  • $\begingroup$ but i have hear that for example in such model if phases are uniformly distributed,then signal is stationary right?but there they are constants $\endgroup$ – dato datuashvili Mar 28 '14 at 13:36
  • $\begingroup$ Both can be true: the phases can be uniformly distributed, but will generally be constant for a single realization of the signal. The idea is that we don't know what the $A_k$, $\omega_k$ and $\phi_k$ are, so we have to model that statistically (with a distribution). Once we start measuring a particular signal, though, all of those values are now constants. $\endgroup$ – Peter K. Mar 28 '14 at 13:38
1
$\begingroup$

As I've said in the (copious) comments, the signal modeled using

$$ y[t]=A_1(\sin(\omega_1 t+\phi_1)+A_2 \sin(\omega_2 t+\phi_2)+....+A_p \sin(\omega_p t+\phi_p) + z(t) $$

is stationary.

Your statement i am not interested estimation of parameters,just i am concerned about spectral resolution seems like an oxymoron. I think what you mean is that you're not concerned with estimating $A_k$ and $\phi_k$, but you are interested in $\omega_k$ for the case where $\omega_k \approx \omega_{k+1}$.

Can you please refine your question, based on this information and the discussion in the comments above?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.