Aufgaben:Exercise 3.8: Amplification and Limitation

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Amplification and limitation
of the random variable  $x$

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

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

• 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

1

Calculate the function value  $A= f_x(0)$  of the PDF at the location  $x = 0$.

$A \ = \ $

2

Calculate the moments  $m_k$  of the random variable  $x$.  Reason that all moments with odd index are zero.  How big is the rms value?

$\sigma_x \ = \ $

3

What is the value of the kurtosis of the random variable  $x$?

$K_x \ = \ $

4

What is the probability that  $x$  exceeds  $0.5$ ?

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

5

Which of the following statements are true regarding the PDF  $f_y(y)$ ?

	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  $y = 1$.

6

What is the continuous part of the PDF  $f_y(y)$?  What value results for  $y = 0.5$ ?

$f_y(y = 0.5) \ = \ $

7

What is the mean of the bounded and amplified random variable  $y$?

$m_y \ = \ $

Solution

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 component is

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

(7)  For the mean value of the random variable  $y$  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 component.