# Frequency estimation of N FFTs of one sine signal in AWGN

Assume N receivers, detecting the same sine signal in AWGN with FFT, then we have N peaks for the same frequency corrupted by noise.

How to use the N FFT observations of the same signal to get a more accurate estimate of the frequency.

My first intuition is averaging the frequency values at each peak, but the noise in frequency domain is not gaussian, so I guess averaging is not the right way to do it

thanks

• Why do you say the noise is not Gaussian? The FFT coefficients of noise of just about any distribution will be as close to Gaussian as you are ever likely to see IRL.
– Peter K.
Apr 17 '16 at 15:08
• It will be Gaussian. But correlated. Anyway, you can idft them all and then take the mean and perform FFT. Apr 17 '16 at 19:08
• @NirRegev Since DFT(x+y) = DFT(x) + DFT(y), how will that make a difference? Apr 17 '16 at 20:55
• i mistaken uniformly with identically @PeterK. The white noise is uniformly distributed in the frequency domain, therefore, average is not even an unbiased estimator for a constant in uniformly distributed noise
– Amro
Apr 18 '16 at 3:06
• The noise is not uniformly distributed in the frequency domain. It will be a Gaussian noise in both domains, as @PeterK. Wrote. Apr 18 '16 at 4:44

Let's calculate: Suppose we have a $\underline{x} \sim \mathcal{N}\left ( \underline{0}, \sigma_x^2 I_N \right )$, where $I_N$ is the $N \times N$ identity matrix.
Denote by $\bf{W}$ the unitary DFT matrix, and by $\bf{W}^H$ its Hermitian transpose. Note that $\bf{W}\bf{W}^H=\bf{W}^H\bf{W}=I_N$, thus, $\underline{X} = DFT\left ( \underline{x}\right ) = \bf{W} \underline{x}$ and \begin{equation} E\left \{ \underline{X} \right \} = E\left \{ \bf{W} \underline{x} \right \} = \bf{W} E\left \{ \underline{x} \right \} = \underline{0}, \end{equation}
so in the frequency domain, $\underline{X} \sim \mathcal{N}\left ( \underline{0}, \sigma_x^2 I_N \right )$, thus, you can take the mean in the frequency domain as well as in the time domain and the estimation will be minimum variance unbiased.
• your calculations will be true if I'm to estimate the magnitude at a specific frequency, then the Noise magnitude has a gaussian distribution. I'm having N frequency values crossing some threshold, so what I have is $f_1, f_2,..f_N$, not $X(f_1), X(f_2),...$.
• Absolutely not. The magnitude of the signal is $chi^2$ distributed. I totally lost you. Sorry. Apr 18 '16 at 19:45