Aufgaben:Exercise 3.6: Noisy DC Signal - Revision history

Guenter at 15:02, 17 February 2022

2022-02-17T15:02:39Z

Bene: Text replacement - "rms value" to "standard deviation"

2022-02-17T11:11:39Z

Text replacement - "rms value" to "standard deviation"

Bene: Text replacement - "root mean square" to "standard deviation"

2022-02-17T11:00:38Z

Text replacement - "root mean square" to "standard deviation"

Guenter at 14:35, 1 February 2022

2022-02-01T14:35:32Z

Guenter at 14:09, 1 February 2022

2022-02-01T14:09:36Z

Guenter at 14:08, 1 February 2022

2022-02-01T14:08:42Z

Guenter at 14:08, 1 February 2022

2022-02-01T14:08:14Z

Guenter at 12:39, 14 January 2022

2022-01-14T12:39:10Z

Noah at 23:34, 31 December 2021

2021-12-31T23:34:31Z

Noah at 23:31, 31 December 2021

2021-12-31T23:31:09Z

@@ Line 34: / Line 34: @@
-{Calculate the standard deviation&nbsp; (standard deviation)&nbsp; of the signal&nbsp; $x(t)$.
+{Calculate the standard deviation of the signal&nbsp; $x(t)$.
 |type="{}"}
 $\sigma_x \ = \ $ { 1 3% } $\ \rm V$
@@ Line 68: / Line 68: @@
 :$$\sigma_{x}^{\rm 2}=m_{\rm 2 \it x}-m_{x}^{\rm 2}. $$
-*The quadratic mean&nbsp; $m_{2x}$&nbsp; is equal to the&nbsp; $($referred to&nbsp; $1\hspace{0.05cm} \Omega)$&nbsp; total power&nbsp; $P_x = 5\hspace{0.05cm}\rm V^2$.
+*The second order moment&nbsp; $m_{2x}$&nbsp; is equal to the&nbsp; $($referred to&nbsp; $1\hspace{0.05cm} \Omega)$&nbsp; total power&nbsp; $P_x = 5\hspace{0.05cm}\rm V^2$.
 *With the mean&nbsp; $m_x = 2\hspace{0.05cm}\rm V$&nbsp; it follows for the standard deviation: &nbsp;
 :$$\sigma_{x} = \sqrt{5\hspace{0.05cm}\rm V^2 - (2\hspace{0.05cm}\rm V)^2} \hspace{0.15cm}\underline{= 1\hspace{0.05cm}\rm V}.$$
@@ Line 74: / Line 74: @@
-'''(3)'''&nbsp; The CDF of a Gaussian random variable&nbsp; $($mean&nbsp; $m_x$,&nbsp; rms&nbsp; $\sigma_x)$&nbsp; is with the Gaussian error integral:
+'''(3)'''&nbsp; The CDF of a Gaussian random variable&nbsp; $($mean&nbsp; $m_x$,&nbsp; standard deviation&nbsp; $\sigma_x)$&nbsp; is with the Gaussian error integral:
 :$$F_x(r)=\rm\phi(\it\frac{r-m_x}{\sigma_x}\rm ).$$

@@ Line 69: / Line 69: @@
 *The quadratic mean&nbsp; $m_{2x}$&nbsp; is equal to the&nbsp; $($referred to&nbsp; $1\hspace{0.05cm} \Omega)$&nbsp; total power&nbsp; $P_x = 5\hspace{0.05cm}\rm V^2$.
-*With the mean&nbsp; $m_x = 2\hspace{0.05cm}\rm V$&nbsp; it follows for the rms value: &nbsp;
+*With the mean&nbsp; $m_x = 2\hspace{0.05cm}\rm V$&nbsp; it follows for the standard deviation: &nbsp;
 :$$\sigma_{x} = \sqrt{5\hspace{0.05cm}\rm V^2 - (2\hspace{0.05cm}\rm V)^2} \hspace{0.15cm}\underline{= 1\hspace{0.05cm}\rm V}.$$

@@ Line 34: / Line 34: @@
-{Calculate the standard deviation&nbsp; (root mean square)&nbsp; of the signal&nbsp; $x(t)$.
+{Calculate the standard deviation&nbsp; (standard deviation)&nbsp; of the signal&nbsp; $x(t)$.
 |type="{}"}
 $\sigma_x \ = \ $ { 1 3% } $\ \rm V$

@@ Line 57: / Line 57: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Correct are <u>solutions 2 and 4</u>:
+'''(1)'''&nbsp; Correct are&nbsp; <u>solutions 2 and 4</u>:
-*The uniform signal&nbsp; $s(t)$&nbsp; is not uniformly distributed, rather its PDF consists of only one Dirac function at&nbsp; $m_x = 2\hspace{0.05cm}\rm V$&nbsp; with weight&nbsp; $1$.
+*The uniform signal&nbsp; $s(t)$&nbsp; is not uniformly distributed,&nbsp; rather its PDF consists of only one Dirac delta function at&nbsp; $m_x = 2\hspace{0.05cm}\rm V$&nbsp; with weight&nbsp; $1$.
-*The signal&nbsp; $n(t)$&nbsp; is gaussian and mean-free &nbsp; &rArr; &nbsp; $m_n = 0$.
+*The signal&nbsp; $n(t)$&nbsp; is Gaussian and mean-free &nbsp; &rArr; &nbsp; $m_n = 0$.
-*Therefore the sum signal&nbsp; $x(t)$&nbsp; is also Gaussian, but now with mean&nbsp; $m_x = 2\hspace{0.05cm}\rm V$.
+*Therefore the sum signal&nbsp; $x(t)$&nbsp; is also Gaussian,&nbsp; but now with mean&nbsp; $m_x = 2\hspace{0.05cm}\rm V$.
-*This is due to the DC signal alone&nbsp; $s(t) = 2\hspace{0.05cm}\rm V$&nbsp;.
+*This is due to the DC signal alone&nbsp; $s(t) = 2\hspace{0.05cm}\rm V$.
@@ Line 68: / Line 68: @@
 :$$\sigma_{x}^{\rm 2}=m_{\rm 2 \it x}-m_{x}^{\rm 2}. $$
-*The quadratic mean&nbsp; $m_{2x}$&nbsp; is equal to the&nbsp; $($referred$to&nbsp; $1\hspace{0.05cm} \Omega$&nbsp;$) total power&nbsp; $P_x = 5\hspace{0.05cm}\rm V^2$.
+*The quadratic mean&nbsp; $m_{2x}$&nbsp; is equal to the&nbsp; $($referred to&nbsp; $1\hspace{0.05cm} \Omega)$&nbsp; total power&nbsp; $P_x = 5\hspace{0.05cm}\rm V^2$.
-*With the mean&nbsp; $m_x = 2\hspace{0.05cm}\rm V$&nbsp; it follows for the rms: &nbsp;
+*With the mean&nbsp; $m_x = 2\hspace{0.05cm}\rm V$&nbsp; it follows for the rms value: &nbsp;
 :$$\sigma_{x} = \sqrt{5\hspace{0.05cm}\rm V^2 - (2\hspace{0.05cm}\rm V)^2} \hspace{0.15cm}\underline{= 1\hspace{0.05cm}\rm V}.$$
-'''(3)'''&nbsp; The CDF of a Gaussian random variable&nbsp; $($mean&nbsp; $m_x$,&nbsp; rms $\sigma_x)$&nbsp; is with the Gaussian error integral:
+'''(3)'''&nbsp; The CDF of a Gaussian random variable&nbsp; $($mean&nbsp; $m_x$,&nbsp; rms&nbsp; $\sigma_x)$&nbsp; is with the Gaussian error integral:
 :$$F_x(r)=\rm\phi(\it\frac{r-m_x}{\sigma_x}\rm ).$$
-*The distribution function at the point&nbsp; $r = 0\hspace{0.05cm}\rm V$&nbsp; is equal to the probability that&nbsp; $x$&nbsp; is less than or equal to&nbsp; $0\hspace{0.05cm}\rm V$&nbsp;.
+*The cumulative distribution function at the point&nbsp; $r = 0\hspace{0.05cm}\rm V$&nbsp; is equal to the probability that&nbsp; $x$&nbsp; is less than or equal to&nbsp; $0\hspace{0.05cm}\rm V$&nbsp;.
-*But for continuous random variables, ${\rm Pr}(x \le r) = {\rm Pr}(x < r)$ also holds&nbsp;.
+*But for continuous random variables,&nbsp; ${\rm Pr}(x \le r) = {\rm Pr}(x < r)$&nbsp; also holds&nbsp;.
-*Using the complementary Gaussian error integral, we obtain:
+*Using the complementary Gaussian error integral,&nbsp; 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)'''&nbsp; Because of the symmetry around the mean&nbsp; $m_x = 2\hspace{0.05cm}\rm V$&nbsp; this gives the same probability, viz.
+'''(4)'''&nbsp; Because of the symmetry around the mean&nbsp; $m_x = 2\hspace{0.05cm}\rm V$&nbsp; this gives the same probability,&nbsp; viz.
 :$$\rm Pr(\it x > \rm 4\,V)\hspace{0.15cm}\underline{=\rm 2.27\%}.$$

@@ Line 16: / Line 16: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables|Gaussian Distributed Random Variables]].
-*Use the&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceeding_probabilityy|complementary Gaussian error integral]]&nbsp; ${\rm Q}(x)$ to solve.
+*Use the&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability|complementary Gaussian error integral]]&nbsp; ${\rm Q}(x)$ to solve.
 *The following are some values of this monotonically decreasing function:
 :$$\rm Q(0) = 0.5,\hspace{0.5cm} Q(1) = 0.1587, \hspace{0.5cm}\rm Q(2) = 0.0227, \hspace{0.5cm} Q(3) = 0.0013. $$

@@ Line 4: / Line 4: @@
 [[File:P_ID126__Sto_A_3_6.png|right|frame|Noisy DC signal and PDF]]
-A DC signal&nbsp; $s(t) = 2\hspace{0.05cm}\rm V$&nbsp; is additively overlaid by a noise signal&nbsp; $n(t)$&nbsp; .
+A DC signal&nbsp; $s(t) = 2\hspace{0.05cm}\rm V$&nbsp; is additively overlaid by a noise signal&nbsp; $n(t)$.
 *In the upper picture you can see a section of the sum signal &nbsp; $x(t)=s(t)+n(t).$
-*The probability density function (PDF) of the signal&nbsp; $x(t)$&nbsp; is shown below.
+*The probability density function&nbsp; $\rm (PDF)$&nbsp; of the signal&nbsp; $x(t)$&nbsp; is shown below.
-*The&nbsp; (related to the resistor&nbsp; $1\hspace{0.05cm} \Omega$&nbsp; total power of this signal is&nbsp; $P_x = 5\hspace{0.05cm}\rm V^2$.
+*The&nbsp; $($related to the resistor&nbsp; $1\hspace{0.05cm} \Omega)$&nbsp; total power of this signal is&nbsp; $P_x = 5\hspace{0.05cm}\rm V^2$.
@@ Line 34: / Line 34: @@
-{Calculate the standard deviation (rms) of the signal&nbsp; $x(t)$.
+{Calculate the standard deviation&nbsp; (root mean square)&nbsp; of the signal&nbsp; $x(t)$.
 |type="{}"}
 $\sigma_x \ = \ $ { 1 3% } $\ \rm V$
-{What is the probability that $x(t) < 0\hspace{0.05cm}\rm V$&nbsp; ?
+{What is the probability that&nbsp; $x(t) < 0\hspace{0.05cm}\rm V$ ?
 |type="{}"}
 ${\rm Pr}(x < 0\hspace{0.05cm}\rm V)\ = \ $ { 2.27 3% } $\ \%$
-{What is the probability that&nbsp; $x(t) > 4\hspace{0.05cm}\rm V$&nbsp; ?
+{What is the probability that&nbsp; $x(t) > 4\hspace{0.05cm}\rm V$?
 |type="{}"}
 ${\rm Pr}(x > 4\hspace{0.05cm}\rm V)\ = \ $ { 2.27 3% } $\ \%$
-{What is the probability&nbsp; $x(t)$&nbsp; between&nbsp; $3\hspace{0.05cm}\rm V$&nbsp; and&nbsp; $4\hspace{0.05cm}\rm V$?
+{What is the probability that&nbsp; $x(t)$&nbsp; is between&nbsp; $3\hspace{0.05cm}\rm V$&nbsp; and&nbsp; $4\hspace{0.05cm}\rm V$?
 |type="{}"}
 ${\rm Pr}(+3\hspace{0.05cm}{\rm V} < x < +4\hspace{0.05cm}{\rm V}) \ = \ $ { 13.6 3% } $\ \%$

@@ Line 15: / Line 15: @@
 Hints:
-*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variable|Gaussian Distributed Random Variable]].
+*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables|Gaussian Distributed Random Variable]].
-*Use the&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_distributed_random_sizes#Exceeding_probability|complementary Gaussian error integral]]&nbsp; ${\rm Q}(x)$ to solve.
+*Use the&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceeding_probabilityy|complementary Gaussian error integral]]&nbsp; ${\rm Q}(x)$ to solve.
 *The following are some values of this monotonically decreasing function:
 :$$\rm Q(0) = 0.5,\hspace{0.5cm} Q(1) = 0.1587, \hspace{0.5cm}\rm Q(2) = 0.0227, \hspace{0.5cm} Q(3) = 0.0013. $$