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Diffstat (limited to 'content/blog')
| -rw-r--r-- | content/blog/fourier-noise.md | 8 |
1 files changed, 4 insertions, 4 deletions
diff --git a/content/blog/fourier-noise.md b/content/blog/fourier-noise.md index e17c119..8e1a114 100644 --- a/content/blog/fourier-noise.md +++ b/content/blog/fourier-noise.md @@ -83,11 +83,11 @@ $$\begin{aligned}dx_rdx_i &= \|{J_{x_r, x_i}(\alpha, \theta)}\| d\alpha d\theta Therefore, the integral now becomes the following. Note that for the distribution of $A$ we allow $\theta$ across its entire range. The value of $a$ also has to remain positive, therefore, the lower limit is changed from $- \infty$ to $0$. -$$\begin{equation}P_A(a) = \int_0^{2\pi} \int_0^a \frac{1}{2\pi\sigma_n^2} exp(\frac{-\alpha^2}{2\sigma_n^2}) \alpha d\alpha d\theta\end{equation}$$ +$$P_A(a) = \int_0^{2\pi} \int_0^a \frac{1}{2\pi\sigma_n^2} exp(\frac{-\alpha^2}{2\sigma_n^2}) \alpha d\alpha d\theta$$ I leave the simplification of this to the reader but this results in the following distribution: -$$\begin{equation}P_A(a) = 1 - exp(\frac{-a^2}{2\sigma_n^2}) \end{equation}$$ +$$P_A(a) = 1 - exp(\frac{-a^2}{2\sigma_n^2})$$ This makes the Probability Distribution Function (PDF) of $A$ to be: @@ -99,10 +99,10 @@ This is a well know distribution called the [Rayleigh Distribution](https://en.w The similar derivation for the phase can be done and shows that the Cummulative Distribution Function (CDF) is: -$$\begin{equation}P_\Phi(\phi) = \frac{\phi}{2\pi}\end{equation}$$ +$$P_\Phi(\phi) = \frac{\phi}{2\pi}$$ which makes the Probability Distribution Function (PDF): -$$\begin{equation}p_\Phi(\phi) = \frac{1}{2\pi}\end{equation}$$ +$$p_\Phi(\phi) = \frac{1}{2\pi}$$ This means that the phase is uniformly distributed across the the interval $(0, 2\pi)$. |