1. Lecture 18 advanced topics is spectral analysis Parsival’s Theorem Multi-taper spectral analysis Auto- and Cross- Correlation Phase spectra

2. Parsival’s Theorem Total power in the time-series Total power its spectrum   i T i 2   i |C i | 2 Discrete Version    T(t) 2 dt     |C(  )| 2 d  Integral Version

3. Write the inverse DFT as Gm=d, where the data d k =T k are the timeseries, and the unknowns, m k =C k, are the Fourier coefficients, and G is N -1 times the matrix of complex exponentials Recall that G H G=N -1 I Note d H d = (Gm) H (Gm) = m H (G H G)m = N -1 m H m Or  i T i 2 = N -1  i |C i | 2 ….. Parsival’s Theorem

4. Multi-taper spectral analysis

5. A criticism of the Hamming Taper is that you’re throwing away ‘hard-earned’ data at the ends of the interval, because the taper is near-zero there … Multi-taper spectral analysis offers a way to ‘save’ that data

6. Let’s try to design a good taper A good taper … has a fourier transform with … a lot of energy in an interval near  =0 compared to the intervals far from  =0  

7. So try the formal maximization problem: find taper coefficients, T i i=0, N-1 with Fourier coefficents C(  ) that maximize the ratio of the energy near  =0, say -W<  <W (green interval) to the total energy (green plus red intervals) R =  -W W |C i | 2 d  /  -  ny  ny |C i | 2 d  Here 2W is the width of the near  =0 (green) interval   WW WW

8. R =  -W W |C i | 2 d  /  -  ny  ny |C i | 2 d  Note that the denominator is proportional to  i=0 N T i 2 by Parsival’s Theorem. So we can rewrite the problem Maximize R =  -W W |C i | 2 d  Subject to the constraint that the power in the timeseries is a constant, say unity  i=0 N T i 2 = 1

9. R =  -W W |C| 2 d  =  -W W C i * C i d  But C(  ) =  n=-N/2 N/2 T n exp(-i n  t  ) and C * (  ) =  m=-N/2 N/2 T m exp(+i m  t  ) R =  -W W C k * C k d  =  n  m T n T m  - W W exp{ i(m-n)  t  } d   n  m T n T m M nm with M nm =  - W W exp{ i(m-n)  t  } d 

10. M nm =  - W W exp{ i(m-n)  t  } d    0 W cos( (m-n)  t  ) d   2 sin{ (m-n)  t W } / { (m-n)  t}  2W sin{ (m-n)  t W } / { W (m-n)  t} = 2W sinc{ (m-n)  t W } Note M is a symmetric matrix … We can do the integral analytically …

11. Does anyone remember how to do a constrained maximization ? Maximize: R =  n  m T n T m M nm with constraint C =  n T n 2 – 1= 0

12. Method of Lagrange Multipliers Maximize R with constraint C=0 equivalent to Maximize  =R- C with no constraint is new unknown, the Lagrange multiplier

13. Minimize  =  n  m T n T m M nm -  n T n 2 + With respect to T q d  /dT q = 0 =  n  m dT n /dT q T m M nm +  n  m T n dT m /dT q M nm - 2  n T n dT n /dT q =  n  m  nq T m M nm +  m T n  mq M nm - 2  n T n  nq =   m T m M qm - 2 T q 

14. Minimize  =  n  m T n T m M nm -   n T n 2 – 1) with respect to T q Yields this equation for T q :  m M qm T m - T q  Does anyone recognize it?

15. M T = T The algebraic eigenvalue problem With symmetric N  N matrix, M has N solutions, say T (n) each with a different eigenvalue, say (n) a distinct pair of solutions are mutually orthogonal if their eigenvalues are different

16. Interpretation of eigenvalues Start with M T = T and premultiply by T T T T M T = T T T and rearrange  T T M T / T T T = R = amount of power near  =0 The value of the quantity that we sought to maximize.

17. Tapering strategy Pick a W Find N tapers by solving the eigenvalue problem MT= T Sort the tapers, smallest eigenvalue first Choose the last few*, they have the best R values Compute spectra of time-series times each of these tapers Average the results   WW WW It turns out that the number of dominant eigenvalues is about NW/  ny, which is called the Shannon number.

18. Example based on 128-point time-series W/  ny = 1/(N/2); tapers computed by MatLab [V,D] = eig( M ) function (in practice you would not do this, but rather use a custom multi-taper code that computes the tapers for you in a more efficient way)

19. Eigenvalues, smallest to biggest ’s. 0.0035 0.0430 0.2746 0.7218 0.9594 0.9976 0.9999 use

20. Tapers, T i time, t Similar to Hanning Taper T1T1 T2T2 T3T3 T4T4 T5T5 T6T6 Gives more emphasis to ends of time-series

21. time-series, x time, t cosine T 1 x T 2 x T 3 x T 4 x T 5 x T 6 x

22. frequency,  spectrum of T 1 spectrum of T 2 spectrum of T 3 spectrum of T 4 spectrum of T 5 spectrum of T 6

23. frequency,  spectrum of T 1 x spectrum of T 2 x spectrum of T 3 x spectrum of T 4 x spectrum of T 5 x spectrum of T 6 x sum of first 4 spectrum of x Results

24. Autocorrelation A(  ) =  -  +  T(t) T(t-  ) dt  called the lag T(t) t t  Offset by , multiply and add to get A(  )

25. Autocorrelation A(  ) =  -  +  T(t) T(t-  ) dt  called the lag A(  )  A(  ) =  -  +  T(t) 2 dt = power in time-series A(  )=A(-  ) symmetrical function A(  ) = T(  ) * T(-  )  Signal convolved with backwards-in-time signal

26. Example T(t) A(  )   t

27. Another Example T(t) A(  )   t Decay of tales due to finite length of time-series 2-point running average of random number sequence

28. MatLab Code t=dt*[0:N-1]'; x = whatever y=xcorr(x); tau=dt*[1-N:N-1]';

29. Fourier Autocorrelation Theorem FT of Autocorrelation is spectrum of T(t) A(  ) = |T(  )| 2

30. A(  ) =  -  +   -  +  T(t) T(t-  ) dt exp(-  ) d   -  +  T(t)  -  +  T(t-  ) exp(-  ) d  dt = -  -  +  T(t)  +  -  T(t’) exp{-  (t-t’)} dt’  dt =  -  +  T(t) exp(-  t) dt  -  +  T(t’) exp(+  t’) dt’ = T(  ) T(  )* = |T(  )| 2 t’= t-  so  =t-t’ and dt’=-d  and  t’  - 

31. Stationary time-series has a well-defined autocorrelation to arbitrarily large lag length-N section cut from time-series autocorrelation non-zero only to lag N and furthermore more you lag it, the more you ‘lose the ends’

32. [T 0 T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8 ]  =0 [T 0 T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8 ] multiply and add 9 terms [T 0 T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8 ]  =4 [T 0 T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8 ] multiply and add only 5 terms … not as good an approximation …

33. So another way of understanding the problem of estimating the spectrum of a stationary time series is Autocorrelation is poorly estimated at large lags …

34. Crosscorrelation Auto-correlation A(  ) =  -  +  T(t) T(t-  ) dt Cross-correlation X(  ) =  -  +  T 1 (t) T 2 (t-  ) dt Generalization of the autocorrelation to two different time- series. In MatLab xcorr(x) is the autocorrelation and xcorr(x1,x2) is the crosscorrelation

35. Filter design and autocorrelation

36. y0y1y2…yNy0y1y2…yN f0f1…fNf0f1…fN x 0 0 0 0 0 0 x 1 x 0 0 0 0 0 x 2 x 1 x 0 0 0 0 … x N … x 3 x 2 x 1 x 0 = In matrix form this is the equation y = G f the least-squares estimate of y involves the two matrices G T G and G T y Recall this formulation of the convolution equation y = f * x

37. y0y1y2…yNy0y1y2…yN f0f1…fNf0f1…fN x 0 0 0 0 0 0 x 1 x 0 0 0 0 0 x 2 x 1 x 0 0 0 0 … x N … x 3 x 2 x 1 x 0 = but G T G are just the columns of G dotted with each other and G T y is just the columns of G dotted with y [GTy]1[GTy]1 [G T G] 3,5 By inspection: [G T G] i,j autocorrelation of x at lag i-j and [G T y] i crosscorrelation of x and y at lag i

38. X(0) X(1) X(2) … X(N) f0f1…fNf0f1…fN A(0) A(1) A(2) … A(1) A(0) A(1) … A(2) A(1) A(0) … … A(N) A(N-1) A(N-2) … = [G T G] f = G T y So least-squares filter design relies very heavilty on knowledge of the autocorrelation function of a time-series To find a filter of length N, you need to know the first N-1 lags of the autocorrelation function

39. Now back to spectral analysis Suppose that we want to compute the spectrum of a stationary time-series x(t) We think we know the only the first N-1 autocorrelation coefficients well Strategy: compute a filter of length N that tells you something about the spectrum of x(t)

40. The Autoregressive (AR) Spectral Method Assume that a stationary time-series x(t) is created by a filter, S(t) acting on uncorrelated random noise n(t) x(t) = S(t) * n(t) Now in the Fourier domain, x(  )=S(  ) n(  ) and |x(  )| 2 = |S(  )| 2 |n(  )| 2

41. But noise has a ‘white spectrum, that is ‘fluctuating but on-average constant’ So |x(  )| 2  |S(  )| 2 The AR method chooses S(t) so that S inv (t) is a short filter, f.

42. x(t) = S(t) * n(t) = f inv (t) * n(t) or x(t) * f(t) = n(t) The least-squares equations are  j A ij f j = X i = 0 With A ij the autocorrelation of x at lag (i-j) and With X i the crosscorrelation of x and n at lag i But the cross-correlation of anything with random noise is zero

43. In order to be able to solve this equation, you need to impose some constraint on f, say f 0 =1. So the least-squares equation is:  j A ij f j = 0 with constraint f 0 =1 One you get f, then the spectrum of x(t) is computed as |x(  )| 2 = |f(  )| -2

44. Example LGA Temperature Timeseries LGA time-series Hanning Taper Tapered time-series

45. AR spectrum, filter-length = 256 Standard Spectrum

46. Phase spectra If we write the Fourier Transform C(  ) = |C(  )| exp{ -i  (  ) } The quantity  (  ) is called the phase spectrum  (  ) = -tan -1 ( C imag / C real )

47. shifting a spike We’ve seen something like this before … If we have a time-series p(t), and want to shift it in time by t 0, then we multiply its FT by exp(-  t 0 )

48. p(t) ifft(exp(-i  t 0 )  fft(p(t))) with t 0 =50 t t

49. shifting a spike Suppose we take p(t) to be spike at the origin The FT of a spike at the origin is just unity |C(  )|=1 So the FT of a shifted spike is exp(-  t 0 ) So a shifted spike as a phase spectrum of  (  )=  t 0 … this is called a phase ramp

50. But suppose we wanted to shift each frequency by a different amount? We might imagine that at position x=0, the time-series is a spike, but that at a position x>0 each frequency has shifted by t 0 (  )=x/v 0 (  ) where v(  ) is a frequency- dependent velocity

51. Example v(  )    ()() ()() x=100 x=200 v(  )=1+exp(-|  |/c) with c=2  ny

52. Time-series p(t,x=0)    x=100 p(t,x=100) p(t,x=200) dispersed wave train

53. Attempt to recover phase spectrum from timeseries     =  x/v with x=100 Phase wrapping problem   =atan2(imag(r),imag(r)) with r=fft(p100)./fft(p0)

54. Recovering phase spectrum is tricky Problem is that arc tangent cycles between –  and +  Whereas phase  (  ) and increase out of the –  to +  range Even the best phase unwrapping algorithms are unreliable