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Empirical Orthogonal Functions (EOF) homework.--Background reading: Hsieh book chapter 2Empirical Orthogonal Functions (EOF) homework.--Background reading: Hsieh book chapter 2Background reading: Hsieh book chapter 2
Empirical Orthogonal Functions (EOF) homework.--1. First, a question about the sense of EOF's:Empirical Orthogonal Functions (EOF) homework.--1. First, a question about the sense of EOF's:1. First, a question about the sense of EOF's:
You have some data(x,t) with space-time structure: 144 space bins (in this case, just longitude), by 240 time bins (months).
You want to decompose it into a set of orthogonal terms that add together to give the total.
Since they are orthogonal, each term represents some variance: cross terms disappear when you average the square of the sum.
If you keep enough terms you will get back all the variance (and more importantly, you can reconstruct the data in all its detail).
In the case of EOF (also known as Principal Components (PC)) analysis, you express your data as:
[[image:space/math/3a322530c7d160da73e6a98417605451.gif caption=" {EOF}_1(bold x) {PC}_1(t) + {EOF}_2(bold x) {PC}_2(t) + {EOF}_3(bold x) {PC}_3(t) + ... = mydata(x,t)"]]
{EOF}_1(bold x) {PC}_1(t) + {EOF}_2(bold x) {PC}_2(t) + {EOF}_3(bold x) {PC}_3(t) + ... = mydata(x,t)
How many values (numbers) are in your input data array? 144 x 240 (lon vs t)
How many values (numbers) are needed to build each term on the left? 144 for EOFs, 240 for PCs
{m4.jpg} {m5.jpg}
{m6.jpg} {m7.jpg}
Empirical Orthogonal Functions (EOF) homework.--Extra credit/ teach us something new:Empirical Orthogonal Functions (EOF) homework.--Extra credit/ teach us something new:Extra credit/ teach us something new:
Do an EOF analysis of your second field. Display and interpret. Compare and contrast with your first field.
Remove the mean, or don't, or remove a different mean. How are the results affected? Explain the sense of the results.
Standardize the time series at each longitude: this gives eigenvectors and eigenvalues of the correlation matrix rather than the covariance matrix (recall HW3 where you plotted slices of these).
Try doing the computation with x and t transposed. Now the "coefficients" or "eigenvectors" are in time (240) and the "scores" are in space (144).
There is a part of the total spacetime variance that EOF's can't reach if you remove the TIME mean, but then use SPACE as the statistical dimension over which you sum to compute covariances. (Or, for that matter, if you remove the SPACE mean to define anomalies but then perform a TIME covariance analysis). What is that unreachable part of the spacetime variance? (Just look at the difference between the input data and the reconstruction and you will see what I am getting at.)
Do a "Combined EOF" analysis of a vector that combines the two fields (each field must be standardized, since the units are different).
you just make a (240x288) array where the 288 values at each time are the 144 field1 (standardized) and then the 144 field2.
Run princomp() in the usual way
Unpack the results at plotting time: the first 144 values are your field1, the others your field2. Rescale with physical units for a better plot.
CEOFs here maximized the variance of the combined data, so they indicate related variations between the 2 fields.
examples: [[file/view/CEOF.sst.slp.ps|CEOF.sst.slp.ps]] [[file/view/CEOF.sst.precip.ps|CEOF.sst.precip.ps]] from code [[file/view/HW5_CEOF_BEM.pro|HW5_CEOF_BEM.pro]]
Read rotatefactors() documentation (Matlab) and learn and teach us all about "rotated EOFs".
Background: you first truncate to the first few EOF/PC pairs, then relax one or more of the orthogonality conditions.
(background of background: EOFs are orthogonal in space, AND PCs are orthogonal in time. Heavy constraint! Either one would allow us to still speak of variance as being cleanly partitioned among modes. Rotated modes may or might or could (?) thus be more physical, since a purely mathematical constraint has been relaxed).
The double orthogonality condition:
[[image:space/math/6ae5df9b48e5c6b2bf39479f51e7b41f.gif caption=" int EOF_i(vec x) EOF_j(vec x) d vec x = 0andint PC_i(t) PC_j(t) dt = 0"]]
int EOF_i(vec x) EOF_j(vec x) d vec x = 0andint PC_i(t) PC_j(t) dt = 0
Empirical Orthogonal Functions (EOF) homework.-IDL results for OLR fieldEmpirical Orthogonal Functions (EOF) homework.-IDL results for OLR fieldIDL results for OLR field
[[file/view/HW5_EOF_BEM.ps|HW5_EOF_BEM.ps]]
[[file/view/HW5_EOF_BEM.pro|IDL code HW5_EOF_BEM.pro]]
{HW5_EOF_OLRmodes_BEM.png} HW5_EOF_OLRmodes_BEM.png
HW5_EOF_OLRmodes_BEM.png
{HW5_EOF_OLRreconstructions_BEM.png} HW5_EOF_OLRreconstructions_BEM.png
HW5_EOF_OLRreconstructions_BEM.png
Empirical Orthogonal Functions (EOF) homework.-IDL results for OLR field-Showing that an EOF is an eigenvector of the covariance matrixEmpirical Orthogonal Functions (EOF) homework.-IDL results for OLR field-Showing that an EOF is an eigenvector of the covariance matrixShowing that an EOF is an eigenvector of the covariance matrix
(so that the results here are related back to those of Question 6.1 of HW3).
Recall that an eigenvector of a matrix is a vector that, when multiplied by the matrix, returns the same vector (times a constant, the eigenvalue):
[[image:space/math/13db31ddd1f603b4c1c102d17c1165b7.gif caption=" matrix M vec x = lambda vec x"]]
matrix M vec x = lambda vec x
{EOF1_is_eigenvector_check_olrBEM.png} EOF1_is_eigenvector_check_olrBEM.png
EOF1_is_eigenvector_check_olrBEM.png
{EOF2_is_eigenvector_check_olrBEM.png} EOF2_is_eigenvector_check_olrBEM.png
EOF2_is_eigenvector_check_olrBEM.png

explained = latent/sum(latent(:)) .*100;
explained(1:10)
way to compute EOF and PC;do it; here is
X = [2 6 1 5 2;
9 4 0 5 4];

{chen_EOF1.jpg} {chen_EOF2.jpg}
{SSTAOr.jpg} {SSTARe.jpg} {Truncation1.jpg} {Truncation2.jpg}
The following is the main part of my code for computing EOF and PC:
%% EOF=COEFF; PC=SCORE
sst = nc_varget('data.nc','sst');
sstm = mean(sst);
sstm = repmat(sstm,240,1);
ssta = sst-sstm;
[COEFF,SCORE,latent,tsquare] = princomp(ssta); % n x m; SCORE=PC, COEFF=EOF!!!
recon_ssta = SCORE*transpose(COEFF);
explained = latent/sum(latent(:)) .*100;
explained(1:10)
There is another way to compute EOF and PC; here is a simple demonstration:
X = [2 6 1 5 2;
9 4 0 5 4];
X(1,:) = X(1,:)-mean(X(1,:)); X(2,:)=X(2,:)-mean(X(2,:));
% co-variance matrix
C=X*X'/5;
[EOF,E] = eig(C); % EOF: eigenvectors; E:eigenvalues
PC = EOF*X;
% reverse the order
E = fliplr(flipud(E));
lambda = diag(E); % retain eigenvalues only; exactly the same as variable 'latent' in the foregoing code.
EOF = fliplr(EOF);
PC = flipud(PC);
%% check
EOF*EOF' % = I
PC*PC'/5 % = lambda
EOF*PC % = X

{http://mpo581-hw3-eofs.wikispaces.com/space/math/3a322530c7d160da73e6a98417605451.gif}
How many values (numbers) are in your input data array? 144*240
left? 144*240 No! 144 + 240, much smaller.
If 5 set? (144-5)*144*240 No, 5*(144+240) versus 144x240.
Empirical Orthogonal Functions (EOF) homework.--2. Read in your field1 (let's call it x again). Use the same data from HW3 data source here.2. Read in your field1 (let's call it x again). Use the same data from HW3 data source here.
Perform and display an EOF analysis of your first field.

ExtraCredits
1. First, a question about the sense of EOF's:
You have some data(x,t) with space-time structure: 144 space bins (in this case, just longitude), by 240 time bins (months).
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{5.jpg}
ExtraExtraCreditsExtra credit/ teach
Do an EOF analysis of your second field. Display and interpret. Compare and contrast with your first field.
The first several EOF analysis of SST explained much more variability than Precip. This indicates that there’s more noise for the precip data. For the first mode, they both represent ENSO, but the strongest signal appears more east for the SST.

ExtraCredits
1. First, a question about the sense of EOF's:
You have some data(x,t) with space-time structure: 144 space bins (in this case, just longitude), by 240 time bins (months).

{4.jpg}
{5.jpg}
Extra credit/ teach us something new:
Do an EOF analysis of your second field. Display and interpret. Compare and contrast with your first field.
The first several EOF analysis of SST explained much more variability than Precip. This indicates that there’s more noise for the precip data. For the first mode, they both represent ENSO, but the strongest signal appears more east for the SST.
{b1.jpg}
{b2.jpg}
{b3.jpg} {b4.jpg}
{b5.jpg}
Remove the mean, or don't, or remove a different mean. How are the results affected? Explain the sense of the results.
It doesn’t seem to affect the results.
{c1.jpg}
Standardize the time series at each longitude: this gives eigenvectors and eigenvalues of the correlation matrix rather than the covariance matrix (recall HW3 where you plotted slices of these).
This change slightly reduces the variability explained by the first several EOF modes. For the first mode, the maximum signal is slightly shifted eastward.
{d1.jpg}
{d5.jpg}
Try doing the computation with x and t transposed. Now the "coefficients" or "eigenvectors" are in time (240) and the "scores" are in space (144).
There is a part of the total spacetime variance that EOF's can't reach if you remove the TIME mean, but then use SPACE as the statistical dimension over which you sum to compute covariances. (Or, for that matter, if you remove the SPACE mean to define anomalies but then perform a TIME covariance analysis). What is that unreachable part of the spacetime variance? (Just look at the difference between the input data and the reconstruction and you will see what I am getting at.)
EOF with x and t transposed:
{e1.jpg}
{e2.jpg}
{e5.jpg}
EOF with spatial mean removed.
{f1.jpg}
{f5.jpg}