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I Truncated Fourier transform and power spectral density

  1. Mar 31, 2017 #1
    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. jcsd
  3. Apr 1, 2017 #2

    blue_leaf77

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    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?
     
  4. Apr 3, 2017 #3

    jasonRF

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    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
     
  5. Apr 3, 2017 #4

    jasonRF

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    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
     
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