The mooring is located at the origin of the ˜mean

part. Studies pursuing this idea are listed by Horel

state™ vector bundle. The EOF vector bundle is

[181]. In general, though, such an approach is

drawn at another point for convenience.

unable to detect propagating signals.

Adapted from Brink and Muench [67].

16.3.5 EOFs of the Complexi¬ed Process. We

towards the southeast on average. Consistent now consider the complexi¬ed process

with Ekman theory, the near-surface currents are

southerly (i.e., to the right of the mean wind stress) Y = X + i X

H

(16.36)

and deeper currents, between 20 m and the bottom where XH is the Hilbert transform (16.2) of X.

at about 60 m, are northerly. Without loss of generality, we can assume that

Brink and Muench [67] performed separate the process has zero mean, and we ¬nd that

complex EOF analyses for the horizontal velocities (cf. (16.30))

at ¬ve depths and for the wind stress at two

locations. If U j and V j are the zonal and Σ yy = E (Xt + i XH )(Xt + i XH )†t t

meridional velocities at depth j, then the complex

= 2Σx x + i(Σx H x ’ Σx x H )

random vector is X = U j + iV j , where

= 2 Σx x + iΣx H x . (16.37)

U = (U1 , . . . , U5 )T and V = (V1 , . . . , V5 )T .

The two-dimensional complex wind stress vector This is a Hermitian matrix and therefore has a set

is constructed similarly from the zonal and of orthogonal complex eigenvectors e k with real

meridional components of the wind stress at the non-negative eigenvalues »k . These eigenvectors

two locations. are said to be the Hilbert EOFs of the process X.

The ¬rst complex EOFs of the velocity vectors The principal components (or EOF coef¬cients,

and of the two wind stress vectors are also shown cf. (16.32)) of the Hilbert EOFs have special

in Figure 16.5 as a vector bundle in the upper right properties. If we write the Hilbert EOFs as

16: Complex Eigentechniques

360

e k = e R + i e Ik , then the EOF coef¬cient may be

k 16.3.6 The Spectral Matrix of the Complexi¬ed

Process. Equation (16.37), together with (16.28)

expanded as

and (16.30), tell us that the covariance matrix of

±k (t) = (Xt )T e R + (XH )T e Ik

k (16.38) the complexi¬ed process equals the integral of the

t

+ i (Xt )T e Ik ’ (XH )T e R . spectral matrix of X over all positive frequencies:

k

t

1

2

Then, if we take the Hilbert transform of the EOF Σ yy = 4 “x x (ω) dω (16.45)

coef¬cients themselves, we see that 0

(16.39) where Y is the complexi¬ed process (16.36). Thus,

(±k (t))H = ’i±k (t)

the Hilbert EOFs are not only the eigenvectors

(a proof is given in Appendix M). Thus, the of the covariance matrix of the complexi¬ed

Hilbert transform of the EOF coef¬cient is just process, but also the eigenvectors of the frequency

the untransformed coef¬cient rotated 90—¦ in the integrated spectral matrix of process X.

5

complex domain. It therefore follows that the

real and imaginary parts of the complex EOF 16.3.7 Frequency Domain EOFs. The Hilbert

coef¬cients are related through their Hilbert EOFs can be interpreted as characteristic patterns

transforms by of the spectral matrix of X when the variability of

the process is con¬ned to a narrow frequency band

H

Re(±k (t)) = ’ Im(±k (t)) (16.40) ω0 ± δω. In that case

H

Im(±k (t)) = Re(±k (t)) , (16.41) Σ yy ∝ “x x (ω0 ) (16.46)

and that their variances are equal: and the Hilbert EOFs are the eigenvectors of the

spectral matrix at frequency ω0 . It is therefore

Var(Im(±k )) = Var(Re(±k )). (16.42) natural to extend the Hilbert EOF analysis to the

frequency domain by applying it to the spectral

The EOF expansion (16.31) of the complexi¬ed matrix “ (ω) so that the characteristic modes of

xx

process (16.36) also has special properties. variation can be identi¬ed for arbitrary time scales

ω’1 where ω ∈ [0, 1/2].

Expanding Yt as

Yt = Yk ,

t

16.3.8 Equivalence of Complex and Real

k

Eigenproblems. The real and imaginary parts

k = ± (t)e k , and equating with the real of the Hilbert EOFs are related to the cross-

where Yt k

covariances between the components of the input

and imaginary parts of (16.36), we ¬nd that

vector and its Hilbert transform. This relationship

Xt = Re(Yk ) (16.43) is easier to see when the eigenproblem is

t

k expressed in real terms. We therefore describe

kH

Xt = Im(Yt ) = Re(Yt ) .

H k the corresponding real eigenproblem here, and

then return to the role of the cross-covariances in

k k

[16.3.9].

Thus the real and imaginary parts of the It is easily shown that e k = e R + i e Ik is

k

complexi¬ed process (16.36) have the same an eigenvector of the complex Hermitian matrix

Hilbert EOF expansion. This is easily con¬rmed Σ = Σ + i Σ with eigenvalue » if and only

yy R I k

with (16.40) and (16.41) by noting that if e k satis¬es the real eigen-equation

Re(Yk ) = Re ±k (t) e R ’ Im ±k (t) e Ik

k k k

Σ R ’Σ I eR eR

t

= »k . (16.47)

(16.44) e Ik e Ik

ΣI ΣR

Im(Yk ) = Re ±k (t) e Ik + Im ±k (t) e R

k

t 5 Note that “ (ω) is only integrated over positive

xx

H

= Re ±k (t) ’ Im ±k (t)

k e Ik frequencies. When “x x (ω) is integrated over both positive and

eR

negative frequencies, the contribution from the anti-symmetric

H

= Re(Yk ) quadrature spectrum is cancelled and we arrive at the real

t

covariance matrix and the conventional EOFs since

It follows, therefore, that the Hilbert EOF 1 1

2 2

represents equal amounts of variance in the input Σx x = “x x (ω) dω = 2 Λx x (ω) dω.

’1 0

time series and its Hilbert transform. 2

16.3: Complex and Hilbert EOFs 361

The eigenvectors of Σ yy are orthogonal and are 16.3.10 Example: Several Uncorrelated Proces-

ses. What are the Hilbert EOFs of a stationary

ordinarily chosen with unit length so that

process Xt = (X1t , . . . , Xmt )T such that the cross-

e k , e j = (e k )† e j = δkl . (16.48) covariance function (and thus quadrature spectra)

between any two components is zero? Under these

In real terms, equation (16.48) reads circumstances the complex covariance matrix Σ yy

equals twice the real covariance matrix Σx x . Thus

j j

(e R )T e R + (e Ik )T e I = δkl

k

6

(16.49) the conventional EOFs are also the Hilbert EOFs.

j j

(e R )T e I ’ (e Ik )T e R = 0.

k What are the coef¬cients of the Hilbert EOFs

in this case? If we assume that the Hilbert EOFs

k = » e k has

e c have been normalized so that they equal