Aufgaben:Exercise 3.3: Moments for Cosine-square PDF

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Squared cosine PDF
and a similar PDF

As in  Exercise 3.1  and  Exercise 3. 2  we consider the random variable restricted to the range of values from  $-2$  to  $+2$  with the following PDF in this section:

$$f_x(x)= {1}/{2}\cdot \cos^2({\pi}/{4}\cdot { x}).$$

Next to this,  we consider a second random variable  $y$  that returns only values between  $0$  and  $2$  with the following PDF:

$$f_y(y)=\sin^2({\pi}/{2}\cdot y).$$

• Both probability density functions are shown in the graph.
• Outside the ranges  $-2 < x < +2$  resp.  $0 < x < +2$,  the followings holds:   $f_x(x) = 0$  resp.  $f_y(y) = 0$.
• Both random variables can be taken as  (normalized)  instantaneous values of the associated random signals  $x(t)$  or  $y(t)$  resp.

Hints:

• This exercise belongs to the chapter  Expected values and moments.

• To solve this problem,  you can use the following indefinite integral:

$$\int x^{2}\cdot {\cos}(ax)\,{\rm d}x=\frac{2 x}{ a^{ 2}}\cdot \cos(ax)+ \left [\frac{x^{\rm 2}}{\it a} - \frac{\rm 2}{\it a^{\rm 3}} \right ]\cdot \rm sin(\it ax \rm ) .$$

Questions

1

Which of the following statements are true for any given PDF  $f_x(x)$ ?  Used are the following quantities:   linear mean  $m_1$,  second moment  $m_2$,  variance  $\sigma^2$.

	$m_2 = 0$,  if   $m_1 \ne 0$.
	$m_2 = 0$,  if   $m_1 = 0$.
	$m_1 = 0$,  if   $m_2 = 0$.
	$m_2 > \sigma^2$,  if   $m_1 \ne 0$.
	$m_1 = 0$,   if   $f_x(-x) = f_x(x)$.
	$f_x(-x) = f_x(x)$,   if   $m_1 = 0$.

2

What is the DC component  ("linear mean")  of the signal  $x(t)$?

$m_x \ = \ $

3

What is the standard deviation of the signal  $x(t)$?

$\sigma_x \ = \ $

4

The random variable  $y$  can be derived from  $x$.  Which assignment is valid?

	$y = 1+x/2.$
	$y = 2x.$
	$y = x/2-1.$

5

What is the DC component of the signal  $y(t)$?

$m_y\ = $

6

What is the standard deviation of the signal  $y(t)$?

$\sigma_y\ = \ $

Solution

Solution

(1)  Under all circumstances,  statements 3, 4, and 5  are correct:

• The first statement is never true,  as is evident from the  "Steiner's theorem".
• The second statement is only valid in the  (trivial)  special case  $x = 0$.

However, there are also zero-mean random variables with asymmetric PDF.

• This means:  Statement 6 is not always true.

(2)  Because of the PDF symmetry with respect to  $x = 0$  it follows that for the linear mean  $m_x \hspace{0.15cm}\underline{= 0}$.

(3)  The standard deviation of the signal  $x(t)$  is equal to the standard deviation   $\sigma_x$  or equal to the root of the variance  $\sigma_x^2$.

• Since the random variable  $x$  has mean  $m_x {= 0}$,  the variance is equal to the second moment according to Steiner's theorem.
• This is also referred to as the power  $($with respect to  $1 \ \rm \Omega)$  in the context of signals. Thus:

$$\sigma_x^{\rm 2}=\int_{-\infty}^{+\infty}x^{\rm 2}\cdot f_x(x)\hspace{0.1cm}{\rm d}x=2 \cdot \int_{\rm 0}^{\rm 2} x^2/2 \cdot \cos^2({\pi}/4\cdot\it x)\hspace{0.1cm} {\rm d}x.$$

• With the relation  $\cos^2(\alpha) = 0.5 \cdot \big[1 + \cos(2\alpha)\big]$  it follows:

$$\sigma_x^2=\int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{\rm 2} \hspace{0.1cm}{\rm d}x + \int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{2}\cdot \cos({\pi}/{\rm 2}\cdot\it x) \hspace{0.1cm} {\rm d}x.$$

• These two standard integrals can be found in tables. One obtains with  $a = \pi/2$:

$$\sigma_x^{\rm 2}=\left[\frac{x^{\rm 3}}{\rm 6} + \frac{x}{a^2}\cdot {\cos}(a x) + \left( \frac{x^{\rm2}}{{\rm2}a} - \frac{1}{a^3} \right) \cdot \sin(a \cdot x)\right]_{x=0}^{x=2} \hspace{0.5cm} \rightarrow \hspace{0.5cm} \sigma_{x}^{\rm 2}=\frac{\rm 4}{\rm 3}-\frac{\rm 8}{\rm \pi^2}\approx 0.524\hspace{0.5cm} \Rightarrow \hspace{0.5cm}\sigma_x \hspace{0.15cm}\underline{\approx 0.722}.$$

(4)  Correct is the  first mentioned suggestion:

• The variant  $y = 2x$  would yield a random variable distributed between  $-4$  and  $+4$.
• In the last proposition  $y = x/2-1$  the mean would be  $m_y = -1$.

(5)  From the graph on the specification sheet it is already obvious that  $m_y \hspace{0.15cm}\underline{=+1}$  must hold.

(6)  The mean value does not change the variance and rms.

• By compressing by the factor  $2$  the standard deviation becomes smaller compared to subtask  (3)  also by this factor:

$$\sigma_y=\sigma_x/\rm 2\hspace{0.15cm}\underline{\approx 0.361}.$$