Quantifying goodness of amplitude estimation

Assume you have a single sinusoid in bandlimited Gaussian noise with unknown amplitude $A$, known frequency $f_0$, and known noise spectral density $S(f)$ in $\frac{\mathrm{units}^2}{\mathrm{Hz}}$:

$$x(t) = A\sin(2\pi f_0t) + n(t)$$

The signal is sampled for a known finite duration $T$ such that the frequency component of the sinusoid alone would have finite magnitude $$\bigg\lvert\mathscr{F}\left\{A\sin(2\pi f_0t)\cdot \mathrm{rect}\left(\frac{t}{T}\right)\right\}(f_0)\bigg\rvert = AT$$.

How would one then quantify the 'goodness' of amplitude estimation $\hat{A} = \frac{|X(f_0)|}{T}$? I assume the SNR would be a distribution with the following shape? (Since the signal amplitude would be divided by an integral of zero-mean noise, which is also zero-mean) Edit: Actually, using the definition of SNR as $\frac{E_s}{\sigma^2}$, SNR is constant since we would know both $E_s (=\frac{A^2T}{2})$ and $\sigma^2 (= S(f) \cdot BW)$. However, now I'm a bit confused. This would make it seem that reducing the BW to an infinitesimal amount would make the SNR very high. On the other hand, I thought the fourier transform was essentially applying a very selective BP filter at each frequency but still results in noise corrupting the frequency magnitude response because there is also a finite amount of energy per bandwidth in the 'smeared' (time-windowed) sinusoid.

Edit2: I believe the main sub-problem is Frequency magnitude distribution of noise

Your understanding of the FT as several extremely narrow bandpass filters is fine. However, you mix up the terms SNR and noise variance. SNR is defined by

$$SNR=\frac{E(s(t)^2)}{E(n(t)^2)}$$

where $E[n(t)^2]=\int_\mathbb{R}S(f)df$ due to the Wiener-Khintschin Theorem ($S(f)$ is the PSD of the noise). So, if you have a measured signal $x(t)=s(t)+n(t)$ with $s(t)=A\sin(2\pi f_0t)$ and $n(t)$ is bandlimited AWGN with bandwidth $B$ and density $N_0$, the SNR decreases when the noise bandwidth increases. So, in order to have a good SNR, you should keep the bandwidth as small as possible (as narrow, as your $s(t)$ permits).

However, your problem is to estimate the amplitude $A$ of $s(t)=A\sin(2\pi f_0t)$ where $f_0$ is known to you. Optimally, you take the Fourier transform of your signal and see what is the value of $X(f)$ at the frequency $f_0$. What will you see there? Does this measurement depend on SNR? Does this measurement actually depend on the noise bandwidth? No, the measurement $X(f_0)$ is independent of the noise bandwidth (as long as the noise has contribution at $f_0$). So, by performing bandpass filtering, increasing/decreasing bandwidth you will not get any better estimates.

The only thing which you can do is to increase the duration of the measurement window. If you have a finite length measurement of duration $T$, your frequency domain essentially has resolution $1/T$. Hence, on each frequency bin (assuming the DFT) the noise variance is $N_0/T$ (assuming AWGN). So, the longer the measurement, the better the estimate.

You can also understand this from a different point: What you are interested in, is the value

$$\rho=\frac{\int_{f_0-\Delta_f}^{f_0+\Delta_f}S(f)df}{\int_{f_0-\Delta_f}^{f_0+\Delta_f}N(f)df}$$

where $2\Delta_f$ describes the frequency domain resolution, $S(f)$ is the signal spectrum and $N(f)$ is the noise spectrum. The numerator is always the same (independent of $\Delta_f$) since $S(f)$ is just a (weighted) Dirac at $f_0$. But, the denominator depends on $\Delta_f$.

• Okay, this clears up some misunderstanding. However, I don't get why you can say "If you have a finite length measurement of duration T, your frequency domain essentially has resolution 1/T. Hence, on each frequency bin (assuming the DFT) the noise variance is N0/T (assuming AWGN). " Maybe it's easier to just talk continuous Fourier transform (since a DFT of sufficient sampling rate and zero padding could reproduce it). I would think that the noise at $f_0$ would be affected by the 'area under the Fourier transform' of the window function, and you want that to be small relative to the peak. – abc Apr 11 '17 at 14:58