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Time Series: Frequency Domain We Will Discuss Here First Fourier Transform and Spectra, and Then Time Filters. Fourier Series Fo 58 Time series: Frequency domain We will discuss here first Fourier transform and spectra, and then time filters. Fourier series for continuous data Assume we have data f (t) in a time interval !0,T # . Then the data can be " $ expressed as a Fourier series . A0 ( " 2! % " 2! % + f (t) = + * Ak cos k t + Bk sin k t - (1), 2 / #$ T &' #$ T &' k =1 ) , or, using an exponential form, ) " 2! % f (t) = C exp ik t . * k $ T ' k =() # & 2" Here !0 = is the fundamental frequency (corresponding to a period T). T 2"k !k = = !0k are the higher frequencies (harmonics). T The Fourier transforms of f (t) can be obtained by multiplying (1) by " 2! % " 2! % cos k t or sin k t , integrating over !0,T # and using #$ T &' #$ T &' " $ T T T 2 2! 1 # 2 2! 2 2! & 1 T " cos (k t)dt = " %cos (k t) + sin (k t)(dt = " dt = . We then get 0 T 2 0 $ T T ' 2 0 2 2 T # 2" & A = f (t)cos k t dt k T ! $% T '( 0 , 2 T # 2" & B = f (t)sin k t dt k T ! % T ( 0 $ ' or, in exponential form 1 T $ 2# ' C = f (t)exp "ik t dt k T ! & T ) 0 % ( 59 The sine/cosine and exponential coefficients are related by #1 ' ( A ! iB ) for k " 0 %2 k k % C = $ ( . k 1 % ( A + iB ) for k < 0% % !k !k % &2 ) T A0 1 = f (t)dt = f (t) corresponds to the frequency ! = 0. 2 T ! 0 The power spectrum of a time series is the square of the amplitude of each harmonic, and it provides the contribution of each harmonic to the total energy of the time series f '2 (t): 2 2 2 Pk = Ak + Bk Now, if we have a discrete time series of length N, with equal time intervals !t , the formulas are similar, witht = n!t; T = N!t , and N+1 n distinct Fourier coefficients (N+1 is the number of discrete points in the series, since n=0,1,…,N). n=0 shortest detectable wave n=N t=0 tn = n!t t=T N N 2 # 2! & 2 # 2!k & f f (t ) C exp ik n t C exp i n n = n = * k % " ( = * k % ( N $ N"t ' N $ N ' k =) k =) 2 2 Or, in terms of sines and cosines, 60 N 2 A0 ( " 2!k % " 2!k % + f (tn ) = + * Ak cos n + Bk sin n - 2 . #$ N &' #$ N &' k =1 ) , The coefficients are obtained, as before, from the discrete Fourier transform that gives the amplitude of the signal due to each wave number or harmonic: 2 N ( " 2!k % + A f cos n k = .* n $ ' - N n=1 ) # N & , 2 N ( " 2!k % + B f sin n k = .* n $ ' - N n=1 ) # N & , or, in complex exponential form 1 N ( ( " 2!k % ++ Ck = * fn exp*/i n -- N . #$ N &' n=0 )* ) ,,- with power spectrum 2 2 2 2 Pk = Ak + Bk = Ck Parseval’s theorem: the average energy is the same in physical or Fourier space: N 1 T A2 2 f (t )2dt = 0 + " A2 + B2 $ T ! n 2 &# k k % 0 k =1 61 Example: Annual cycle Assume we have monthly means, and we average all years to obtain f , n monthly averages corresponding to each month of the year. !t is then one month, and N=12. Typically, we can represent the annual cycle with at least two harmonics, to allow for a lack of symmetry between winter and summer: A " 2! % " 2! % " 2! % " 2! % f = 0 + A cos n + B sin n + A cos n + B sin n ACn 2 1 #$ 12 / 1 &' 1 #$ 12 / 1 &' 2 #$ 12 / 2 &' 2 #$ 12 / 2 &' The A0 term represents the annual average (with zero frequency), the A1 and B1 terms represent the periodic component with period 12 months (fundamental frequency), and the A2 and B2 terms represent a periodic component with period 6 months (first harmonic). The coefficients A , A , A , B , B can be obtained as before, e.g., 0 1 2 1 2 2 12 " 2! % B = f sin n , where the bar represents the monthly average over 2 12 ( n $ 12 / 2 ' n=1 # & several years. Once the coefficients are obtained, the annual cycle can be subtracted from the time series in order to deal with anomalies. .
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