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If I have:

$$ y = x_r+jx_i + n_r +j n_i$$

with $n_r$ and $n_i$ Gaussian with mean 0 and variance $\sigma^2$, what is the pdf of the envelope |y| and phase(y)? Is it still Rayleigh-distributed and uniform, respectively?

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  • $\begingroup$ um, something's wrong with your equation: it says $y=jy+\text{other stuff}$, and I don't think that's what you meant. $\endgroup$ May 5 '20 at 7:07
  • $\begingroup$ Am I right to assume that you define $x$ is a deterministic thing, and you only consider $n$ to be random? $\endgroup$ May 5 '20 at 7:46
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No, it will be Rician Distribution.

Rayleigh distribution is a special case of Rician Distribution when the normal random variables involved are of zero mean and equal variances.

I have assumed you meant :$$y=x_r + j \cdot x_i + n_r + j \cdot n_i$$ where $x_r, x_i$ are deterministic variables.

Basically, you now have $y_r = (n_r + x_r),\ \sim \mathcal N(x_r, \sigma^2) $ and $y_i = (n_i + x_i), \ \sim \mathcal N(x_i, \sigma^2)$

Now, you can write $y = y_r +j \cdot y_i = r \cdot e^{j\phi}$, where $r,\phi$ are random variables and, $y_r = r \cdot \cos(\phi)$ and $y_i = r \cdot \sin(\phi)$.

Using method of transformations, Jacobian matrix can be given as follows: $$\begin{pmatrix} y_r \\ y_i \end{pmatrix} = \begin{pmatrix} r \cdot \cos(\phi) \\ r \cdot \sin(\phi)\end{pmatrix}$$ $$\mathbf J = \begin{pmatrix} \cos(\phi)&&-r \cdot \sin(\phi) \\ \sin(\phi)&& r \cdot \cos(\phi)\end{pmatrix}$$ $$|\mathbf J| = r$$ Hence, you see that Jacoboan does not change by changing the mean of the normal random variables from 0 to non-zero.

Joint PDF of derived RVs $r, \phi$ will be given by the following : $$f_{R,\Phi}(r, \phi) = f_{Y_r, Y_i}(r\cos(\phi), r\sin(\phi)) \cdot |\mathbf J |$$ $$f_{R,\Phi}(r, \phi) = \frac{r}{2\pi \sigma^2} \cdot \exp\left[-\frac{(r\cos(\phi) - x_r)^2+(r\sin(\phi)-x_i)^2)}{2\sigma^2}\right]$$ $$\forall r\in [0,\infty) \\ \phi \in[-\pi, \pi]$$

Solve for Marginal PDF of $R$ and $\Phi$ to see that it is not Rayleigh anymore. But a more generalized Rician Distribution.

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    $\begingroup$ Also, $\Phi$ is not uniform but is peaked in the vicinity of the mean location, and $R$ and $\Phi$ are not independent. $\endgroup$ May 5 '20 at 21:14
  • $\begingroup$ Hay, DSP Rookie, this is a friendly request. Please use \cdot instead of . And try to remember the backslash `\` preceding function names. $\endgroup$ May 19 '20 at 2:08
  • $\begingroup$ and you don't need to use so many dots for multiplication. we know when two expressions sit side-by-side, they're meant to be multiplied. i would suggest leaving the dot out except for dot product and maybe some expression of inner product (but that would be crappy notation). $\endgroup$ May 19 '20 at 2:12
  • $\begingroup$ @robertbristow-johnson Sure, will keep in mind. Thanks. $\endgroup$
    – DSP Rookie
    May 19 '20 at 10:13
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The envelope |y| follows

The phase of y follows the PDF given in On the envelope and phase distributions for correlated gaussian quadratures.

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