# I Truncated Fourier transform and power spectral density

1. Mar 31, 2017

### Mishra

Hello,

I am trying to find an expression for the signal-to-noise ratio of an oscillating signal on top some white noise. In particular I would like to know how the SNR scales with the integration time. It is well known that during some integration time $T$, the SNR increases as $T^{1/2}$ (because the noise increases as $T^{1/2}$ and the signal increases at $T$). I am trying to prove this with, my limited skills in maths, using the power spectral density formalism.

The integration time is $T$. I assume some signal: $x(t)=A \sin(2 \pi \omega_0 t)$ on top of some white noise of power spectral density $S$. I am guessing that I should compare the power spectral density of the noise vs signal:
$$P(f)=X(f)^2$$
where $X(f)$ is the truncated Fourier transform:
$$X(f)=\frac{1}{\sqrt{T}}\int_0 ^{T} x(t) e^{-i \omega t} dt$$
In the limit where $T$ is large, the integral is a Fourier tranform, yielding a delta function:
$$X(f)=\frac{A}{\sqrt{T}} \delta (\omega - \omega_0)$$
Ence:
$$P(f)=\frac{A^2}{ T } \delta (\omega - \omega_0)$$
$$P(f_0)=\frac{A^2}{ T }$$

Here I have that the power spectral density at the oscillation frequency decreases linearly in time. Can someone explain me where my misunderstanding is please ?

2. Apr 1, 2017

### blue_leaf77

It can be that my memory is playing trick in my head but as I recall the white noise variance in frequency domain increases as $T^3$. So, do you have any reference that states what you wrote there?

3. Apr 3, 2017

### jasonRF

I think the issue is that you are playing fast and loose with delta functions. The integral is simple enough that you may as well just do it. If I do a simplified case of $x(t) = A e^{i\omega_0 t}$ and let t run from -T/2 to T/2, then
$$X(\omega) = \frac{A}{\sqrt{T}} \int_{-T/2}^{T/2} dt \, e^{i(\omega_0-\omega)t} = A \sqrt{T} \left[\frac{\sin\left(\frac{1}{2} T (\omega-\omega_0) \right)}{\frac{1}{2} T (\omega-\omega_0)} \right]$$
The term in square brackets is a very nice number when $\omega\rightarrow\omega_0$.

jason

4. Apr 3, 2017

### jasonRF

By the way, for the noise I get $\langle |X(\omega)|^2 \rangle$ to be independent of time, so the SNR $\propto A^2 T$, as I would expect. You should do the calculation.

Jason