1 / 39

Lecture 15 Orthogonal Functions Fourier Series

Lecture 15 Orthogonal Functions Fourier Series. LGA mean daily temperature time series is there a global warming signal?. Model that includes annual variability.

dinesh
Télécharger la présentation

Lecture 15 Orthogonal Functions Fourier Series

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 15Orthogonal FunctionsFourier Series

  2. LGA mean daily temperature time seriesis there a global warming signal?

  3. Model that includes annual variability T(t) = a + bt + A1 cos(2pf1t) + B1 sin(2pf1t) + A2 cos(2pf2t) + B2 sin(2pf2t) + A3 cos(2pf3t) + B3 sin(2pf3t) + …with f1 = 1 cycle per year f2 = 2 cycles per year etc

  4. Why both sines and cosines? cos{2pf1(t-t0)}

  5. Why both sines and cosines? cos{2pf1(t-t0)} Cosine does not ‘start’ at t=0 But remember cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

  6. cos(a+b)=cos(a) cos(b) - sin(a) sin(b) cos{2pf1(t-t0)} = cos(2pf1t0) cos(2pf1t) – sin(2pf1t0) sin(2pf1t) = A cos(2pf1t) + B sin(2pf1t)

  7. cos(a+b)=cos(a) cos(b) - sin(a) sin(b) So using both sines and cosines moves the delay, t0, out of the cosine, and into the coefficients of the sines and cosines. This trick ‘linearizes’ the unknown, t0. cos{2pf1(t-t0)} = cos(2pf1t0) cos(2pf1t) – sin(2pf1t0) sin(2pf1t) = A cos(2pf1t) + B sin(2pf1t)

  8. Why more than one frequency? f1 = 1 cycle per year f2 = 2 cycles per year etcAllows us to represent non-sinusoidal shape of annual cycle.

  9. cos(ft) 0.3cos(2ft) sum: cos(ft)+0.3cos(2ft) exactly periodic, but shape not exactly sinusoidal

  10. Least-squares fit to LGA data (up to f8) data fit constant term, a error of fit, e linear term, bt

  11. Statistics of linear term, bt b = 0.31 degrees F per decade sd = [ eTe / N ]1/2 = 7 deg F Cm = sd2[GTG]-1 sb = [ sd2Cmb,b ]1/2 = 0.05 degrees F per decade 95% confidence b = 0.31±0.1 degrees F per decade So LGA is warming

  12. sines and cosines are“orthogonal” functions T(t) = A0 + A1 cos(2pf1t) + B1 sin(2pf1t) + A2 cos(2pf2t) + B2 sin(2pf2t) + A3 cos(2pf3t) + B3 sin(2pf3t) + …with f2=2f1, f3=3f1, etc Called a “Fourier Series”

  13. Standard least-squares G matrix 1cos(2pf1t1) sin(2pf1t1) cos(2pf2t1) sin(2pf2t1) … 1cos(2pf1t2) sin(2pf1t2) cos(2pf2t2) sin(2pf2t2) … 1cos(2pf1t3) sin(2pf1t3) cos(2pf2t3) sin(2pf2t3) … 1cos(2pf1t4) sin(2pf1t4) cos(2pf2t4) sin(2pf2t4) … 1cos(2pf1t5) sin(2pf1t5) cos(2pf2t5) sin(2pf2t5) … 1cos(2pf1t6) sin(2pf1t6) cos(2pf2t6) sin(2pf2t6) … … G=

  14. With the proper choice of f1the matrix GTG is diagonaldot product of any pair of columns of G is zerocolumns of G are orthogonal

  15. The proper choice of f1 Suppose the time-series is N data points long, with spacing Dt. Then the lowest frequency must be f1 = 1 / (NDt) one oscillation over the length of the time-series And the highest frequency must be fN/2 = 1 / (2Dt) one-half oscillation per sampling interval

  16. f1 = 1 / (2NDt) f1 = 1 / (2Dt) note sine is zero

  17. Count of unknowns The constant term, one unknown plus 2 coefficients per frequency, N/2 frequencies so N unknowns minus One unknown since the fN/2 term, which has no sine term equals N unknowns, same as number of data

  18. MatLab Code N = 100; % times vector dt = 0.5; tmin = 0.0; t = tmin + dt*[0:N-1]'; tmax = tmin + dt*(N-1); df = 1/(N*dt); % frequency spacing M = N; % number of unknowns same as data G = zeros(N,M); % set up least-squares G matrix G(:,1)=ones(N,1); for p = 2*[1:M/2-1] G(:,p) = cos(pi*p*df*t); G(:,p+1) = sin(pi*p*df*t); end p=M/2; G(:,M) = cos(2*pi*p*df*t);

  19. GTG for N=100 [GTG]11= [GTG]NN=N Other diagonal elements [GTG]ii=N/2 Off diagonal elements are zero

  20. So least-squares solution is m = [GTG]-1GTd = = diag( N-1, 2/N, … 2/N, N-1 ) GTd NO matrix inversion required!

  21. GTG for N=100 Example: Neuse River Hydrograph (100 days) data, d d=Gm with m=[GTG]-1GTd d=Gm with m=DGTd where D=diag( N-1, 2/N, … 2/N, N-1 )

  22. “spectrum”amount of power at different frequencies si2 = Ai2 + Bi2 time-series has a lot of energy at frequency fp si2 fi fp

  23. Spectrum of Neuse data set for N=4380

  24. Close up of low frequencies 6 mo 12 mo 4 mo 3 mo 2 mo Big annual cycle in Neuse hydrograph

  25. Error Estimates for Fourier Series Assume uncorrelated, normally-distributed data, d, with variance sd2 The problem Gm=d is linear, so the unknowns, m, (the coefficients of the cosines and sines, Ai and Bi) are also normally-distributed. Since sines and cosines are orthogonal, GTG is diagonal and Cm= sd2 [GTG]-1 is diagonal, too So that m’s have uncorrelated errors. All but the first and last have variance sm2= 2sd2/N. The spectrum si2=Ai2+Bi2 is the sum of two uncorrelated, normally distributed random variables and is thus c22-distributed. The c22-distribution has a variance of 4, so that ss2= 8sd2/N

  26. Switching to complex numbersnothing different in principlebut calculations become easier

  27. But first Lets switch to angular frequency measured in radians per second wi = 2p fi Beats writing all those 2p’s !

  28. Remember Euler’s formula exp( iwt ) = cos( wt ) + i sin( wt ) ?

  29. exp( iwt ) = cos( wt ) + i sin( wt ) exp( -iwt ) = cos( wt ) - i sin( wt ) cos( wt ) = (1/2) [exp( iwt ) + exp( -iwt )] sin( wt ) = (1/2i) [exp( iwt ) - exp( -iwt )]

  30. =1 =0 Let’s compareT(t) = A0 cos(w0t) + B0 sin(w0t) + A1 cos(w1t) + B1 sin(w1t) + A2 cos(w2t) + B2 sin(w2t) + A3 cos(w3t) + B3 sin(w3t) + …withT(t) = ... + C-2 exp(-iw2t) + C-1 exp(-w1t) + C0 exp(iw0t) + C1 exp(iw1t) + C2 exp(iw2t) + C3 exp(iw3t) + … with wp=pDw w-p= -wp First, if T is real, then we must have C-p = Cp* Then C-p exp(-wpt) + Cp exp(wpt) = (Cpr-iCpi) [cos(wpt) - i sin(wpt)] + (Cpr+iCpi) [cos(wpt) + i sin(wpt)] = 2Cpr cos(wpt) - 2Cpi sin(-w-pt)] So Ap= 2Cpr and Bp= -2Cpi So these two representations are equivalent

  31. T(t) = ... + C-2 exp(-iw2t) + C-1 exp(-w1t) + C0 exp(iw0t) + C1 exp(iw1t) + C2 exp(iw2t) + C3 exp(iw3t) + … Implies a simple form of the equation d=Gm … C-2C-1C0C1C2 … T0T1T2T3T4 … … exp(-iw2t0) exp(-iw1t0) exp(iw0t0) exp( iw1t0) exp(iw2t0) … … exp(-iw2t1) exp(-iw1t1) exp(iw0t1) exp( iw1t1) exp( iw2t1) … … exp(-iw2t2) exp(-iw1t2) exp(iw0t2) exp( iw1t2) exp( iw2t2) … … exp(-iw2t3) exp(-iw1t3) exp(iw0t3) exp( iw1t3) exp( iw2t3) … … exp(-iw2t4) exp(-iw1t4) exp(iw0t4) exp( iw1t4) exp( iw2t4) … =

  32. Least-squares with complex numbers real numbers: given Gm =d minimize E=eTe implies m=[GTG]-1GTd complex nos: given Gm =d minimize E=eHe where eH = e*T implies m=[GHG]-1GHd The Hermitian transpose, that is, the transpose of the complex conjugate. The formula m=[GHG]-1GHd is not hard to work out using the standard minimization procedure, but we don’t have time to work it out in class.

  33. d=Gm … C-2C-1C0C1C2 … T0T1T2T3T4 … … exp(-iw2t0) exp(-iw1t0) exp(iw0t0) exp(iw1t0) exp(iw2t0) … … exp(-iw2t1) exp(-iw1t1) exp(iw0t1) exp(iw1t1) exp(iw2t1) … … exp(-iw2t2) exp(-iw1t2) exp(iw0t2) exp(iw1t2) exp(iw2t2) … … exp(-iw2t3) exp(-iw1t3) exp(iw0t3) exp(iw1t3) exp(iw2t3) … … exp(-iw2t4) exp(-iw1t4) exp(iw0t4) exp(iw1t4) exp(iw2t4) … = Note T2Si Ci exp(+iwit2) m=N-1GHm T0T1T2T3T4 … … C-2C-1C0C1C2 … … exp(iw2t0) exp(iw2t1) exp(iw2t2) exp(iw2t3) exp(iw2t4) … … exp(iw1t0) exp(iw1t1) exp(iw1t2) exp(iw1t3) exp(iw1t4) … … exp(iw0t0) exp(iw0t1) exp(iw0t2) exp(iw0t3) exp(iw0t4) … … exp(-iw1t0) exp(-iw1t1) exp(-iw1t2) exp(-iw1t3) exp(-iw1t4) …… exp(-iw2t0) exp(-iw2t1) exp(-iw2t2) exp(-iw2t3) exp(-iw2t4) … =N-1 Note C2Si Ti exp(-iw2ti)

  34. d=Gm … C-2C-1C0C1C2 … T0T1T2T3T4 … … exp(-iw2t0) exp(-iw1t0) exp(iw0t0) exp(iw1t0) exp(iw2t0) … … exp(-iw2t1) exp(-iw1t1) exp(iw0t1) exp(iw1t1) exp(iw2t1) … … exp(-iw2t2) exp(-iw1t2) exp(iw0t2) exp(iw1t2) exp(iw2t2) … … exp(-iw2t3) exp(-iw1t3) exp(iw0t3) exp(iw1t3) exp(iw2t3) … … exp(-iw2t4) exp(-iw1t4) exp(iw0t4) exp(iw1t4) exp(iw2t4) … = Note T2Si Ci exp(+iwit2) Opposite signs M=N-1GHm T0T1T2T3T4 … … C-2C-1C0C1C2 … … exp(iw2t0) exp(iw2t1) exp(iw2t2) exp(iw2t3) exp(iw2t4) … … exp(iw1t0) exp(iw1t1) exp(iw1t2) exp(iw1t3) exp(iw1t4) … … exp(iw0t0) exp(iw0t1) exp(iw0t2) exp(iw0t3) exp(iw0t4) … … exp(-iw1t0) exp(-iw1t1) exp(-iw1t2) exp(-iw1t3) exp(-iw1t4) …… exp(-iw2t0) exp(-iw2t1) exp(-iw2t2) exp(-iw2t3) exp(-iw2t4) … =N-1 Note C2Si Ti exp(-iw2ti)

  35. Note normalization factor of N-1 has been omitted Discrete Fourier Transform Find the coefficients C given the data, T Equivalent to m = GHd Ck = Sn=-N/2N/2 Tn exp(±2pikn/N ) with k=-½N, …, ½N Discrete Inverse Fourier Transform Find the data T given the coefficients, C Equivalent to d = N-1Gm Tn = N-1Sk=-N/2N/2 Ck exp(2pikn/N ) with n=-½N, …, ½N Warnings: 1) no one can agree on signs 2) no one can agree on normalizations Note normalization factor of N-1 has been added ±

  36. Counting unknowns frequencies from –(N/2)Dw to (N/2)Dw in steps of Dw So N+1 complex numbers, Cp So 2N+2 real and imaginary parts, Cpr and Cpi But C-p = Cp*, so really only N/2+1 unknown complex numbers So N+2 real and imaginary parts , Cpr and Cpi (p0) But C0i=0 and CN/2i=0 (always) So N unknowns, matching N data

  37. % standard fft setup. The standard implementation of the digital fourier % transform is VERY INFLEXIBLE. Learn these rules: N=256; % you can choose the length N of the time series % in some implementations N can be any positive % integer, but in others it MUST be a % power-or-two. I set it here to 256, which % is two-to-the-eigth-power. dt=1.0; % and you can choose the sampling interval dt % but then the following variables are set tmax=dt*(N-1); % we presume the time series starts at t=0, so % the maximum time is tmax t=dt*[0:N-1]; % time then goes from 0 to (N-1)*dt fmax=1/(2.0*dt); % the maximum frequency in the fft calculation is % called the Nyquist frequency. It is % determined by the two-points-per wavelength % rule df=fmax/(N/2); % the frequency spacing, df, assumes that a N-point % time series is reperesented by an N-point fourier % transform f=df*[0:N/2,-N/2+1:-1]'; % The fourier transform has N values, from a negative % frequency of -(fmax-df) through zero freqency, to % positive frequency of fmax. But note the weird order. The % zero and positive frequencies are put in the first % half of the array and the negative frequencies are % put in the second half.

  38. % p is the timeseries whose transform is being computed w0 = 2*pi*fmax/10; % sample p, a simple sinusoid of frequency w0 p = sin(w0*t); % fourier transform using MatLab's fft function. The help function % says that it uses the NEGATIVE sign in the exponential. pt=fft(p); % these are the coefficients, C, of the complex exponential % presumably one would do something with the fourier transform % at this point - apply a filter, for example. But I do nothing. % Inverse fourier transform using the Matlab function ifft. % Help says it uses the POSITIVE sign in the exponential, and that % it has the right normalization that ifft(fft(x))=x. But % BE WARNED, that doesn't mean that the normalization on fft % is 1 and that the normalization on ifft is (1/N), like % I had it in class. You can put any constant, b, in front % of the fft integral, as long as you put 1/b in front of % the IFFT integral. But judging by the Help, I think that % Matlab used b=1. pr=ifft(pt); % this reconstructs the function from the coefficients

  39. The fast Fourier transform algorithm The Fourier Transform equation m = [GTG]-1GTd = diag(N-1,2/N, … 2/N,N-1) GTd has N multiplications for each of N unknowns, So N2 in total. For example, for N=1024, N2=1,048,576 But in the special case of N being a power of two, there is an algorithm – the fast fourier transform algorithm - that can compute m in only Nlog2N multiplications For example, N=1024, Nlog2N=10,240 A substantial savings! MatLab implements it. Use it!

More Related