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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)

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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

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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 $\Delta_f$.

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  • $\begingroup$ 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. $\endgroup$ – abc Apr 11 '17 at 14:58

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