To: File periodogram_var.wpd

From: Jon Hagen

Date: May 3, 2004

Subject: Variance in Spectral Points from Periodogram

Summary: Considering the case of a rectangular spectrum, sampled at the Nyquist rate, we find that the relative variance of each spectral point is 2. If we average the spectrum over M non-
overlapping periodograms, the relative variance of the average spectral value is therefore 2/M.

If, instead of non-overlapped periodograms, we use an equivalent bandpass filter of width fs/N sampled at its Nyquist rate, 2 fs /N, the total number of independent samples in the time for M groups of N would be 2 fs /N x MN / fs = 2M and the variance of a square-law power estimate would be 2/(2M) = 1/M. Non-overlapping periodograms require twice the integration time.

Calculation of the Variance of a single periodogram

Since we are considering a uniform spectrum, the variance will be equal for all channels. We will consider the first (zeroth) channel, where all the W factors are unity. The value of the periodogram for this channel is

The expectation of S0 is just N<V2 > = N 2. To find the variance of S0, we calculate the mean square minus the square mean:

In the sum over four indices, we first can pull out the N terms where i = j = k = m. The total contribution of these terms is N<V4 > = 3N 4 . The other terms that will be non-zero are those where two indices have one value and the remaining two indices have another value. Each of these terms will have the value <V2>2 = 4. To find how many such terms exist, consider that there are N choices for i. Then there are 3 ways to pick the other index that will have the same value as i. Finally there are N-1 ways to pick the second index, for a total N x 3 x (N-1) = 3N2 -
3N terms. Thus

Dividing by <S0>2, we get the relative variance,