Effect of sampling interval length and Gaussian noise on the frequency-spectrum

Question

Suppose you have a signal $s(t)$, corrupted by Gaussian noise of $\eta(t)$ with zero mean and standard deviation $\sigma$. So that the measured signal is $f(t)=s(t)+\eta(t)$.

I now want to calculate the spectrum of the signal using FFT. For my particular case, I have found that for wavenumbers $q$ above a given cuttoff value $q_c$ (e.g. for short wavelengths), my spectrum of $s(t)$ will run into the noise floor of $\eta$. The noise floor due to the (ideal) Gaussian noise $\eta(t)$ in Fourier space is given by $\int \sigma^2 dt$ and thus depends on the length of sampling interval.

My question is:

• Is it possible to modify the noise-floor of the frequency spectrum in Fourier space by reducing the length of the sampling interval in the time-domain and thus gain SNR for a certain $q$-range of interest? In other words, could I do something like high-pass filter the whole signal and then take the FFT of a short region to reduce the noise. Or perhaps do something similar.

• If this is conceptually flawed, why?

• If it is not flawed, how would I do this and what are the limits/tradeoffs to doing this (obviously there has to be a limit). Also, what are the implications/requirements for the stationarity vs. non-stationarity of signal $s(t)$ in order to justify this?

Answer 1

Yes, you're on the right track. The key thing that would help you would not involve decreasing your sampling interval, but increasing your observation duration. Here's a qualitative explanation:

Recall the filterbank interpretation of the DFT:

$$ X[k] = \sum_{n=0}^{N-1} x[n] e^{\frac{-j2\pi k n}{N}} $$

In this view of the transform, the DFT implements a uniformly-spaced bank $N$ of critically-sampled filters. The frequency response of each filter in normalized frequency $\omega$ has the shape of a Dirichlet kernel, whose width is inversely proportional to $N$. So, as you increase the observation duration, you implicitly increase $N$, meaning that each bandpass filter covers a smaller portion of the frequency band. That means that proportionally less noise power is passed by each bin's corresponding filter.

This can help you if you're analyzing a narrowband signal such as a sinusoidal tone. Again, think of that situation as placing a bunch of bandpass filters in the hope that one will catch your signal of interest. You will maximize the SNR in the resulting observations by making those bandpass filters as narrow as possible, i.e. by making your DFT as long as possible. Thus, you can increase your DFT length in order to "push down the noise floor." To those who are familiar with spectrum analyzers, this is analogous to decreasing the resolution bandwidth.

You can also make a theoretical argument for why this is the case. Assume the unitary definition of the DFT to avoid having scaling factors get in the way:

$$ X[k] = \frac{1}{\sqrt{N}}\sum_{n=0}^{N-1} x[n] e^{\frac{-j2\pi k n}{N}} $$

And recall Parseval's theorem for the DFT:

$$ \sum_{n=0}^{N-1} |x[n]|^2 = \sum_{k=0}^{N-1} |X[k]|^2 $$

Assume that $x[n]$ is a zero-mean white noise process with variance $\sigma^2$ and take the expectation of both sides:

$$ \sum_{n=0}^{N-1} \mathbb{E}\left(|x[n]|^2\right) = \sum_{k=0}^{N-1} \mathbb{E}\left(|X[k]|^2\right) $$ $$ N \sigma^2 = N \mathbb{E}\left(|X[k]|^2\right) $$ $$ \sigma^2 = \mathbb{E}\left(|X[k]|^2\right) $$

where I have made the assumption that $\mathbb{E}\left(|X[k]|^2\right)$ is a constant independent of $k$ (which is true, but I haven't shown it here; you can see a proof in this analysis of white noise in the frequency domain).

The takeaway from this is that as you increase the DFT size $N$, the expected value of the noise spectrum's squared magnitude doesn't change under the unitary definition of the transform. However, let's look at what happens to the same quantity for the case of $x[n]$ being a sinusoidal tone of interest. Assume without loss of generality that the tone is at zero frequency (i.e. it is a constant; I will assume $x[n] = 1$):

$$ X[k] = \frac{1}{\sqrt{N}}\sum_{n=0}^{N-1} x[n] e^{\frac{-j2\pi k n}{N}} $$ $$ X[k] = \frac{1}{\sqrt{N}}\sum_{n=0}^{N-1} e^{\frac{-j2\pi k n}{N}} $$ $$ X[k] = \frac{1}{\sqrt{N}} N \delta[k] $$ $$ X[k] = \sqrt{N} \delta[k] $$ $$ |X[k]|^2 = N \delta[k] $$

This illustrates the payoff:

As you increase the DFT size $N$, the expected squared magnitude of the noise process in the frequency domain stays constant. However, the expected frequency-domain squared magnitude for a narrowband signal is proportional to $N$. Therefore, you can effectively increase the frequency-domain SNR by increasing the transform length $N$.

A bonus illustration in MATLAB:

% tone at zero frequency plus noise
x = 1+randn(1e5,1);  
% calculate DFTs of various lengths
X1 = 1/sqrt(100)*fft(x(1:1e2));
X2 = 1/sqrt(1e3)*fft(x(1:1e3));
X3 = 1/sqrt(1e4)*fft(x(1:1e4));
X4 = 1/sqrt(1e5)*fft(x);
% plot them all
figure;
hold all;
plot(10*log10(abs(X1).^2));
plot(10*log10(abs(X2).^2));
plot(10*log10(abs(X3).^2));
plot(10*log10(abs(X4).^2));
% zoom in on the first part of the plot
a = axis;
a(1:2) = [1 10];
axis(a);
legend('N = 100', 'N = 1000', 'N = 10000', 'N = 100000');
grid on;

Note the logarithmic scaling on the Y axis, showing the proportionality to $N$.

A final note to address the other part of your question: there are of course limits to what you can accomplish here. One key one is the uncertainty principle; namely, when you increase the transform length in this way to gain resolution on the frequency axis, you lose a corresponding amount of time resolution. Secondly, your results may vary if your signal of interest or your background noise is not stationary over your observation interval; increasing the transform duration obviously increases the chances of this occurring.