Exercise 3.8: Amplification and Limitation

Amplification and limitation
of the random variable $x$

We consider a random signal $x(t)$ with symmetric probability density function:

$$f_x(x)=A\cdot \rm e^{\rm -2 \hspace{0.05cm}\cdot \hspace{0.05cm}|\it x|}.$$

This signal is applied to the input of a nonlinearity with the characteristic curve (see lower figure):

$$y=\left\{\begin{array}{*{4}{c}}0 &\rm for\hspace{0.2cm} \it x <\rm 0, \\\rm2\it x & \rm for\hspace{0.2cm} \rm 0\le \it x\le \rm 0.5, \\1 & \rm for\hspace{0.2cm}\it x > \rm 0.5\\\end{array}\right.$$

The characteristic sketched below limits the variable $x(t)$ at the input asymmetrically and amplifies it in the linear range.

Hints:

The exercise belongs to the chapter Exponentially Distributed Random Variable.

Given the following definite integral:

$$\int_{0}^{\infty}\it x^n\cdot\rm e^{-\it a \hspace{0.03cm}\cdot \hspace{0.03cm}x}\, d{\it x} =\frac{\it n{\rm !}}{\it a^{n}}.$$

Questions

$A \ = \ $

$\sigma_x \ = \ $

$K_x \ = \ $

${\rm Pr}(x > 0.5) \ = \ $

$\ \%$

	The PDF contains a Dirac delta function at $y = 0$.
	The PDF contains a Dirac delta function at $y = 0.5$.
	The PDF contains a Dirac delta function at $y = 1$.

$f_y(y = 0.5) \ = \ $

$m_y \ = \ $

Solution

(1) The area under the probability density function yields

$$\it F=\rm 2\cdot \it A\int_{\rm 0}^{\infty}\hspace{-0.15cm}\rm e^{\rm -2\it x}\, \rm d \it x=\frac{\rm 2\cdot \it A}{\rm -2}\cdot \rm e^{\rm -2\it x}\Big|_{\rm 0}^{\infty}=\it A.$$

Since this area must be equal by definition $F = 1$ $\underline{A = 1}$.

(2) All moments with odd index $k$ are equal to zero due to symmetric PDF.

For even $k$ the left part of the PDF can be mirrored into the right one and we get:

$$\it m_k=\rm 2 \cdot \int_{\rm 0}^{\infty}\hspace{-0. 15cm}\it x^{k}\cdot \rm e^{-\rm 2\it x}\,\rm d \it x=\frac{\rm 2\cdot\rm\Gamma(\it k{\rm +}\rm 1)}{\rm 2^{\it k{\rm +}\rm 1}}=\frac{\it k{\rm !}}{\rm 2^{\it k}}.$$

From this it follows with $k = 2$ considering the mean $m_1 = 0$:

$$m_{\rm 2}=\frac{\rm 2!}{\rm 2^2}={\rm 0.5\hspace{0.5cm}bzw.\hspace{0.5cm} }\sigma_x=\sqrt{ m_{\rm 2}}\hspace{0.15cm}\underline{=\rm 0.707}.$$

PDF after reinforcement and boundary

(3) The fourth-order central moment is $\mu_4 = m_4 = 4!/2^4 = 1.5$.

From this follows for the kurtosis:

$$K_{x}=\frac{ \mu_{\rm 4}}{ \sigma_{\it x}^{4}}=\frac{1.5}{0.25}\hspace{0.15cm}\underline{=\rm 6}.$$

(4) Using the result from (1) we get:

$${\rm Pr}( x> 0.5)=\int_{0.5}^{\infty}{\rm e}^{- 2 x}\,{\rm d}x=\frac{\rm 1}{\rm 2\rm e}\hspace{0.15cm}\underline{=\rm 18.4\%}.$$

(5) Correct are solutions 1 and 3:

The PDF $f_y(y)$ involves a Dirac delta function at the point $y= 0$ with weight ${\rm Pr}(x < 0) = 0.5$.
In addition, another Dirac delta function at $y= 1$ with weight ${\rm Pr}(x > 0.5) = 0.184$.

(6) The signal range $0 \le x \le 0.5$ is linearly mapped to the range $0 \le y \le 1$ at the output.

The derivative of the characteristic curve is constantly equal to $2$ (amplification). From this one obtains:

$$f_y(y)=\frac{f_x(x)}{|g'(x)|}\Bigg|_{x=h(y)}=\frac{\rm e^{-\rm 2\it x}{\rm 2}\Bigg|_{\it x={\it y}/{\rm 2}}=0.5 \cdot {\rm e^{\it -y}} .$$

For $y= 0.5$ accordingly, the continuous PDF portion is

$$f_y(y = 0.5)\hspace{0.15cm}\underline{\approx 0.304}.$$

(7) For the mean value of the random grö&aerospace;e $y$ holds:

$$m_y=\frac{1}{\rm 2\rm e} \cdot 1 +\int_{\rm 0}^{\rm 1}\frac{\it y}{\rm 2}\cdot \rm e^{\it -y}\, \rm d \it y=\frac{\rm 1}{\rm 2\rm e}{\rm +}\frac{\rm 1}{\rm 2}-\frac{\rm 1}{\rm e}=\frac{\rm 1}{\rm 2}-\frac{\rm 1}{\rm 2 e}\hspace{0. 15cm}\underline{=\rm 0.316}.$$

The first term is from the Dirac delta at $y= 1$, the second from the continuous PDF–fraction.