The improvement of signal-noise

1. Apr 5, 2012

Jamipat

On page 26 of http://www.maths.qmul.ac.uk/~rpn/ASTM735/Week7.pdf [Broken], it claims that the signal-noise can be improved by integrating over long time periods since the noise amplitude scales with $\frac{1}{\sqrt{t}}$ for Gaussian statistics.

Can someone explain that to me?

Last edited by a moderator: May 5, 2017
2. Apr 5, 2012

marcusl

The scaling is proportional to sqrt(t), not its inverse. This comes from the classic random walk, see Gaussian Random Walk heading here
http://en.wikipedia.org/wiki/Random_walk
The signal adds coherently (in your case you are integrating so I assume that the signal is a constant ), so its integrated value scales as t. The overall increase in amplitude SNR is sqrt(t), the power SNR increases as t.

Last edited: Apr 5, 2012
3. Apr 5, 2012

RoshanBBQ

The idea behind this statement is that Gaussian noise with zero mean will in the long run spend as much time above 0 as it does below 0 in such a way that its total area contribution in an integral is zero if you integrate enough of it. The more you integrate, the surer you can be about it. The signal, which would be a constant in this example, always contributes the same amount of area per unit time in the integral. So

$$\frac{1}{t_0}\int \limits_{0}^{t_0} S + N\, dt$$

N is the noise. S is the constant signal.

$$\frac{\int \limits_{0}^{t_0} S\, dt}{t_0} + \frac{\int \limits_{0}^{t_0}N\, dt}{t_0}$$

The term on the right approaches zero for large t_0. The term on the left is a constant, so as t_0 approaches infinity, we get

$$\frac{t_0S}{t_0} = S$$
Which has the best signal to noise ratio possible.

So I just presented an intuitive argument for bettering the SNR. Is that for what you were looking?