4.5.2 Clipping correction for 4-level quantization

The following determination of the clipping correction is due to [Hagen et al 1973]:
Given two jointly normal random variables x and y with covariance $\mu$, and given some arbitrary function g(x,y), Price's theorem states that

$$\frac{\partial^m\langle g(x,y) \rangle}{\partial \mu^m} = \la... ...\partial^{2m} g(x,y)}{\partial x^m \partial y^m} p(x,y) dx}dy}$$ (4.24)

For random signals of zero mean, the covariance $\mu$ is identical with the cross-correlation function $R_{ij}({\tau})$ defined in Eq.4.1. As shown by Eq.4.1, we need to accumulate products of the voltage outputs of two antennas (i,j), but using the quantized signals rather than the continuous ones. Thus, with the identification x = v_i(t) and $y = v_j(t+\tau)$, and using $\tilde{x}$and $\tilde{y}$ for the quantized signals, we can apply Price's theorem to the 4-level cross-correlation function $R_4 = \langle \tilde{x} \tilde{y} \rangle$ such that

$$\frac{dR_4}{d\rho} = \sigma^2 \frac{dR_4}{dR} = \sigma^2\frac... ...ial x}\cdot \frac{\partial\tilde{y}}{\partial y} p(x,y) dx}dy}$$ (4.25)

( $R = \rho\sigma^2$ denotes the continuous cross correlation function, for the sake of simplicity, antenna indices are omitted). The partial derivatives in the integrand are easily found by using the transfer function shown in Fig.4.5:

$$\tilde{x} = \Theta(x)+(n-1)\left[\Theta(x-v_0)-\Theta(-x-v_0) \right] \Theta(-x)$$ (4.26)

where $\Theta(x) = 1$ for x>0, and 0 else. Thus,

$$\frac{\partial \tilde{x}}{\partial x} = 2\delta(x)+(n-1)\left[ \delta(x-v_0)+\delta(x+v_0)\right]$$ (4.27)

Re-writing Price's theorem, we find

$$\frac{dR_4}{d\rho} \sigma^2 \int_{-\infty}^{\infty} {\int_{-\infty}^{\infty}{ \left(2\delta(x)+(n-1)\left[\delta(x-v_0)+\delta(x+v_0)\right] \right)}}$$ =

$${{\cdot\left(2\delta(y)+(n-1)\left[\delta(y-v_0)+\delta(y+v_0)\right] \right)p(x,y)dx}dy}\,.$$ (4.28)

Inserting the jointly normal distribution p(x,y), and evaluating the integral yields

$$\frac{dR_4}{d\rho} = \frac{\sigma^2}{\pi} \frac{1}{\sqrt{1-r^2}}\biggl\{(n-1)^2 \left[\ex... ...a^2(1+\rho)}\right)}+ \exp{\left(\frac{-v_0^2}{\sigma^2(1-\rho)}\right)}\right]$$

$$+4(n-1)\exp{\left(\frac{-v_0^2}{2\sigma^2(1-\rho^2)}\right)}+2\biggr\}\,,$$ (4.29)

or, alternatively, the integral form given in Eq.4.11.