Exercise 3.6: Noisy DC Signal

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Noisy DC signal and PDF

A DC signal  $s(t) = 2\hspace{0.05cm}\rm V$  is additively overlaid by a noise signal  $n(t)$.

  • In the upper picture you can see a section of the sum signal   $x(t)=s(t)+n(t).$
  • The probability density function  $\rm (PDF)$  of the signal  $x(t)$  is shown below.
  • The  $($related to the resistor  $1\hspace{0.05cm} \Omega)$  total power of this signal is  $P_x = 5\hspace{0.05cm}\rm V^2$.


$$\rm Q(0) = 0.5,\hspace{0.5cm} Q(1) = 0.1587, \hspace{0.5cm}\rm Q(2) = 0.0227, \hspace{0.5cm} Q(3) = 0.0013. $$



Which of the following statements are true?

The signal $s(t)$  is uniformly distributed.
The noise signal  $n(t)$  is Gaussian distributed.
The noise signal  $n(t)$  has a mean value  $m_n \ne 0$.
The total signal  $x(t)$  is Gaussian distributed with mean  $m_x = 2\hspace{0.05cm}\rm V$.


Calculate the standard deviation of the signal  $x(t)$.

$\sigma_x \ = \ $

$\ \rm V$


What is the probability that  $x(t) < 0\hspace{0.05cm}\rm V$ ?

${\rm Pr}(x < 0\hspace{0.05cm}\rm V)\ = \ $

$\ \%$


What is the probability that  $x(t) > 4\hspace{0.05cm}\rm V$?

${\rm Pr}(x > 4\hspace{0.05cm}\rm V)\ = \ $

$\ \%$


What is the probability that  $x(t)$  is between  $3\hspace{0.05cm}\rm V$  and  $4\hspace{0.05cm}\rm V$?

${\rm Pr}(+3\hspace{0.05cm}{\rm V} < x < +4\hspace{0.05cm}{\rm V}) \ = \ $

$\ \%$


(1)  Correct are  solutions 2 and 4:

  • The uniform signal  $s(t)$  is not uniformly distributed,  rather its PDF consists of only one Dirac delta function at  $m_x = 2\hspace{0.05cm}\rm V$  with weight  $1$.
  • The signal  $n(t)$  is Gaussian and mean-free   ⇒   $m_n = 0$.
  • Therefore the sum signal  $x(t)$  is also Gaussian,  but now with mean  $m_x = 2\hspace{0.05cm}\rm V$.
  • This is due to the DC signal alone  $s(t) = 2\hspace{0.05cm}\rm V$.

(2)  According to Steiner's theorem:

$$\sigma_{x}^{\rm 2}=m_{\rm 2 \it x}-m_{x}^{\rm 2}. $$
  • The second order moment  $m_{2x}$  is equal to the  $($referred to  $1\hspace{0.05cm} \Omega)$  total power  $P_x = 5\hspace{0.05cm}\rm V^2$.
  • With the mean  $m_x = 2\hspace{0.05cm}\rm V$  it follows for the standard deviation:  
$$\sigma_{x} = \sqrt{5\hspace{0.05cm}\rm V^2 - (2\hspace{0.05cm}\rm V)^2} \hspace{0.15cm}\underline{= 1\hspace{0.05cm}\rm V}.$$

(3)  The CDF of a Gaussian random variable  $($mean  $m_x$,  standard deviation  $\sigma_x)$  is with the Gaussian error integral:

$$F_x(r)=\rm\phi(\it\frac{r-m_x}{\sigma_x}\rm ).$$
  • The cumulative distribution function at the point  $r = 0\hspace{0.05cm}\rm V$  is equal to the probability that  $x$  is less than or equal to  $0\hspace{0.05cm}\rm V$ .
  • But for continuous random variables,  ${\rm Pr}(x \le r) = {\rm Pr}(x < r)$  also holds .
  • Using the complementary Gaussian error integral,  we obtain:
$$\rm Pr(\it x < \rm 0\,V)=\rm \phi(\rm \frac{-2\,V}{1\,V})=\rm Q(\rm 2)\hspace{0.15cm}\underline{=\rm 2.27\%}.$$

(4)  Because of the symmetry around the mean  $m_x = 2\hspace{0.05cm}\rm V$  this gives the same probability,  viz.

$$\rm Pr(\it x > \rm 4\,V)\hspace{0.15cm}\underline{=\rm 2.27\%}.$$

(5)  The probability that  $x$  is greater than  $3\hspace{0.05cm}\rm V$ is given by.

$${\rm Pr}( x > 3\text{ V}) = 1- F_x(\frac{3\text{ V}-2\text{ V}}{1\text{ V}})=\rm Q(\rm 1)=\rm 0.1587.$$
  • For the sought probability one obtains from it:
$$\rm Pr(3\,V\le \it x \le \rm 4\,V)= \rm Pr(\it x > \rm 3\,V)- \rm Pr(\it x > \rm 4\,V) = 0.1587 - 0.0227\hspace{0.15cm}\underline{=\rm 13.6\%}. $$