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To: file acf-to-spectrum.wpd From: Jon Hagen Date: April 2, 2004 Subject: Transforming correlation functions to spectra I. Recovering a Power Spectrum from a Sampled Autocorrelation Function The power spectrum, by definition, is the Fourier Transform of the autocorrelation function (ACF). The inverse transformation is written as
4

R() '
&4

S () e

j

d.

101)

For a band-limited spectrum, the integral becomes

B

R() '
&

S () e
B

j

d.

102)

In Equation 102, S() can be replaced by Sp(), a periodic function created by repeating S(), with a period of 2P, i.e.

B

R() '
&

S P () e
B

j

d,

103)

since S() and Sp() are identical in the range of integration. Now the function Sp(), because it is periodic, can be represented by the Fourier Series

4

&j

SP() '
n'&4

an e

2n 2B

104)

where the Fourier coefficient an is found, in the usual way, to be


1 an ' 2

B

j

S p() e
B &
B

2n 2B

d

105)

1


Comparing Equation 105 with Equation 103, we see that 1 2 n R ' , 2B 2B

an '

106)

showing that discrete samples of R() are sufficient to calculate the periodic function SP(), and hence S(), for any value of . Substituting Equation 106 into Equation 104, we have

1 SP() ' 2

4

B n'&4

2 n R ' e 2B

&j

2 n 2B

107)

Of course we need only calculate SP() for values of such that -B<
1 SP() ' 2

N &1

B n'&(N&1)

2 n R ' e 2B

&j

2 n 2B

108)

Again, since R() = R(-), we can arrange the sum to run over just the positive values of n:
N &1 2 n 2B

1 SP() ' 2

&R(0) % 2 Re
B n '0

2 n R ' e 2B

&j

109)

Normally we want to evaluate SP() at N points, from = 0 to = (N-1)B/N. Letting m = m B/N, we have
N &1 2 nm 2N

1 SP(m) ' 2

&R0 % 2 Re
B n '0

Rn e 2

&j

.

110) If we want


to calculate the sum in Equation 110 via an FFT, we can zero-extend the ACF by defining Rn = 0 for N-1 < n< 2N. Equation 110 then becomes
2 N &1

1 SP(m) ' 2

&R0 % 2 Re
B n '0

Rn e

&j

2 nm 2N

.

111)

The sum within the brackets now has the desired form. The transform will produce 2N points, the first N of which are the desired spectral points. Note that the N+1 term is also valid, providing a spectral point at B. The last N-1 points can be ignored. II. Recovering a Cross Spectrum from Sampled Autocorrelation Functions Let us denote the two signals as VX (t) and VY(t). These could correspond to xpolarization and y-polarization. Normally we produce the N positive tau (including zero) cross correlation values in one correlator unit and N negative tau (including zero) cross correlation values in another correlation unit, i.e. R+ RXY

(n) = estimate of < VX (t)VY (t + 2 n/(2 B)) > (n) = estimate of < VX (t)VY (t - 2 n/(2 B)) >

XY

The zero lag is included in both sets. Note that the two zero lags are identical, as they are formed from sums of identical products. Together, these two sets of lagged products give us the two-sided function, RXY (n) for -N < n < N). Let us denote the cross spectrum as U() + jV(). The cross spectrum is, by definition, the Fourier Transform of the cross-correlation function. Writing the inverse transform, we have
4 j t

RXY() '
&4

U()%jV() e

d

201)

where U and V are the real and imaginary parts of the cross spectrum. When VX (t) and VY(t) are band limited, their cross spectrum will also be band limited and we can write

B

RXY() '
&
B

U()%jV() e

j t

d

202)

Using the previous arguments, this band limited function can be represented as a Fourier series: 3


4

&j

U()%jV() '
n'&4

an e

2n 2B

204)

where the coefficients, an, are given by


1 an ' 2

B

j

U()%jV() e
B &
B

2n 2B

d

205)

Comparing Equation 205 with Equation 202, we see that 1 R 2B 2 n , 2B

an '

XY

'

206)

showing that discrete samples of RXY() are sufficient to calculate the function U() + jV() for any value of . Substituting Equation 206 into Equation 204, we have 1 U()%jV() ' 2
4

R
B n'&4

XY

2 n ' e 2B

&j

2 n 2B

207)

Normally we want to evaluate the cross spectrum for N points, starting with =0 and separated by a spacing B/N. The mth cross spectral point is given by
N &1

1 U(m) % jV(m) ' 2

RXY (n) e
B n'&(N&1)

&j

2 nm 2N

208)

1 ' 2

N &1

R
B n '0

%

XY

(n) e

&j

2 nm 2N

N &1

%
n '0

R

&

XY

(n) e

j

2 nm 2N

&RXY(0)

,

209)

where RXY(0) is either R+XY(0) or R-XY(0), since they are identical. To put this into the standard 4


form for evaluation via the FFT algorithm, we can add zeros to R+XY and R-XY to extend both to length 2N, i.e. R+XY(n) = R-XY(n) = 0 for N-1 < n < 2N. The cross spectrum becomes
2 nm 2N 2 nm 2N

1 U(m) % jV(m) ' 2

2 N &1

R
B n '0

%

XY

(n) e

&j

2 N &1

%
n '0

R

&

XY

(n) e

j

&RXY(0)

210)

As before, the first N+1 terms provide cross spectrum values for equally spaced values of from zero to B.

5