Aufgaben:Exercise 3.3Z: Optimization of a Coaxial Cable System - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:54:48Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:54, 16 March 2026
Line 63:		Line 63:

	'''(2)'''  Using the result from  '''(1)''',  we further obtain:		'''(2)'''  Using the result from  '''(1)''',  we further obtain:
	:$$\rho_{\rm U} = \left ( {3.73}/{2} \right )^2 \approx 3.48 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\~~rmlg~~}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline { = 5.41\,{\rm dB}}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q}\left ({3.73}/{2} \right) \hspace{0.15cm}\underline {\approx 0.031} \hspace{0.05cm}.$$		:$$\rho_{\rm U} = \left ( {3.73}/{2} \right )^2 \approx 3.48 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline { = 5.41\,{\rm dB}}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q}\left ({3.73}/{2} \right) \hspace{0.15cm}\underline {\approx 0.031} \hspace{0.05cm}.$$


Line 69:		Line 69:
	*However,  for the worst-case error probability  $p_{\rm U} = 10^{\rm -6}$   ⇒   $10 \cdot {\rm lg} \, \rho_{\rm U} > 13.55 \, {\rm dB}$  must be obtained.		*However,  for the worst-case error probability  $p_{\rm U} = 10^{\rm -6}$   ⇒   $10 \cdot {\rm lg} \, \rho_{\rm U} > 13.55 \, {\rm dB}$  must be obtained.
	*This is achieved by increasing the quotient  $E_{\rm B}/N_0$  accordingly:		*This is achieved by increasing the quotient  $E_{\rm B}/N_0$  accordingly:
	:$$10 \cdot {\~~rmlg~~}\hspace{0.1cm}{E_{\rm B}}/{N_0} = 40\,{\rm dB}\hspace{0.1cm}+\hspace{0.1cm}13.55\,{\rm dB}\hspace{0.1cm}-\hspace{0.1cm}5.41\,{\rm dB}= 48.14\,{\rm dB}\hspace{0.3cm}\Rightarrow\hspace{0.3cm} {E_{\rm B}}/{N_0} = 10^{4.814}\approx 65163\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {N_0}/{E_{\rm B}}\hspace{0.15cm}\underline { =1.53 \cdot 10^{-5}} \hspace{0.05cm}.$$		:$$10 \cdot {\rm lg}\hspace{0.1cm}{E_{\rm B}}/{N_0} = 40\,{\rm dB}\hspace{0.1cm}+\hspace{0.1cm}13.55\,{\rm dB}\hspace{0.1cm}-\hspace{0.1cm}5.41\,{\rm dB}= 48.14\,{\rm dB}\hspace{0.3cm}\Rightarrow\hspace{0.3cm} {E_{\rm B}}/{N_0} = 10^{4.814}\approx 65163\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {N_0}/{E_{\rm B}}\hspace{0.15cm}\underline { =1.53 \cdot 10^{-5}} \hspace{0.05cm}.$$


Line 80:		Line 80:

	[[Category:Digital Signal Transmission: Exercises\|^3.3 Channel Distortion and Equalization^]]		[[Category:Digital Signal Transmission: Exercises\|^3.3 Channel Distortion and Equalization^]]
	[[de:3.~~3Z_Koaxialkabelsystem_-_Optimierung~~]]		[[de:Aufgaben:Aufgabe 3.3Z: Optimierung eines Koaxialkabelsystems]]

Maintenance script: Join 1 split formula blocks

2026-03-16T14:16:30Z

Join 1 split formula blocks

← Older revision		Revision as of 16:16, 16 March 2026
Line 69:		Line 69:
	*However,  for the worst-case error probability  $p_{\rm U} = 10^{\rm -6}$   ⇒   $10 \cdot {\rm lg} \, \rho_{\rm U} > 13.55 \, {\rm dB}$  must be obtained.		*However,  for the worst-case error probability  $p_{\rm U} = 10^{\rm -6}$   ⇒   $10 \cdot {\rm lg} \, \rho_{\rm U} > 13.55 \, {\rm dB}$  must be obtained.
	*This is achieved by increasing the quotient  $E_{\rm B}/N_0$  accordingly:		*This is achieved by increasing the quotient  $E_{\rm B}/N_0$  accordingly:
	:$$10 \cdot {\rm		:$$10 \cdot {\rmlg}\hspace{0.1cm}{E_{\rm B}}/{N_0} = 40\,{\rm dB}\hspace{0.1cm}+\hspace{0.1cm}13.55\,{\rm dB}\hspace{0.1cm}-\hspace{0.1cm}5.41\,{\rm dB}= 48.14\,{\rm dB}\hspace{0.3cm}\Rightarrow\hspace{0.3cm} {E_{\rm B}}/{N_0} = 10^{4.814}\approx 65163\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {N_0}/{E_{\rm B}}\hspace{0.15cm}\underline { =1.53 \cdot 10^{-5}} \hspace{0.05cm}.$$
	lg}\hspace{0.1cm}{E_{\rm B}}/{N_0} = 40\,{\rm dB}
	\hspace{0.1cm}+\hspace{0.1cm}13.55\,{\rm dB}
	\hspace{0.1cm}-\hspace{0.1cm}5.41\,{\rm dB}= 48.14\,{\rm dB}\hspace{0.3cm}
	\Rightarrow
	\hspace{0.3cm} {E_{\rm B}}/{N_0} = 10^{4.814}\approx 65163
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm} {N_0}/{E_{\rm B}}\hspace{0.15cm}\underline { =
	1.53 \cdot 10^{-5}} \hspace{0.05cm}.$$

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:30:51Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:30, 16 March 2026
Line 15:		Line 15:

	The table shows the eye opening   $\ddot{o}(T_{\rm D})$   as well as the detection noise rms value   $\sigma_{\rm d}$   – each normalized to the transmitted amplitude  $s_0$  – for different cutoff frequencies  $f_{\rm G}$.  The cutoff frequency is to be chosen such that the worst-case error probability is as small as possible,  with the following definition:		The table shows the eye opening   $\ddot{o}(T_{\rm D})$   as well as the detection noise rms value   $\sigma_{\rm d}$   – each normalized to the transmitted amplitude  $s_0$  – for different cutoff frequencies  $f_{\rm G}$.  The cutoff frequency is to be chosen such that the worst-case error probability is as small as possible,  with the following definition:
	:$$p_{\rm U} = {\rm Q} \left( \frac{\ddot{o}(T_{\rm D})/2}{ \sigma_d}		:$$p_{\rm U} = {\rm Q} \left( \frac{\ddot{o}(T_{\rm D})/2}{ \sigma_d}\right) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}}\right)$$
	\right) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}}\right)$$

	*This quantity represents an upper bound for the mean error probability   $p_{\rm S} \le p_{\rm U}$.		*This quantity represents an upper bound for the mean error probability   $p_{\rm S} \le p_{\rm U}$.
Line 64:		Line 63:

	'''(2)'''  Using the result from  '''(1)''',  we further obtain:		'''(2)'''  Using the result from  '''(1)''',  we further obtain:
	:$$\rho_{\rm U} = \left ( {3.73}/{2} \right )^2 \approx 3.48 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$\rho_{\rm U} = \left ( {3.73}/{2} \right )^2 \approx 3.48 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rmlg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline { = 5.41\,{\rm dB}}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q}\left ({3.73}/{2} \right) \hspace{0.15cm}\underline {\approx 0.031} \hspace{0.05cm}.$$
	10 \cdot {\rm
	lg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline { = 5.41\,{\rm dB}}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q}\left (
	{3.73}/{2} \right) \hspace{0.15cm}\underline {\approx 0.031} \hspace{0.05cm}.$$

Maintenance script: Add German interlanguage link

2026-03-16T12:25:45Z

Add German interlanguage link

← Older revision		Revision as of 14:25, 16 March 2026
Line 91:		Line 91:

	[[Category:Digital Signal Transmission: Exercises\|^3.3 Channel Distortion and Equalization^]]		[[Category:Digital Signal Transmission: Exercises\|^3.3 Channel Distortion and Equalization^]]
			[[de:3.3Z_Koaxialkabelsystem_-_Optimierung]]

Hwang at 13:06, 28 June 2022

2022-06-28T13:06:14Z

← Older revision		Revision as of 15:06, 28 June 2022
Line 3:		Line 3:
	}}		}}

	[[File:P_ID1409__Dig_Z_3_3.png\|right\|frame\|Normalized system parameters for different ~~cut-off~~ frequencies]]		[[File:P_ID1409__Dig_Z_3_3.png\|right\|frame\|Normalized system parameters for different cutoff frequencies]]
	We consider a redundancy-free binary transmission system with the following specifications:		We consider a redundancy-free binary transmission system with the following specifications:
	* The transmission pulses are NRZ rectangular and have energy  $E_{\rm B} = s_0^2 \cdot T$.		* The transmission pulses are NRZ rectangular and have energy  $E_{\rm B} = s_0^2 \cdot T$.

Hwang at 08:57, 27 June 2022

2022-06-27T08:57:17Z

← Older revision		Revision as of 10:57, 27 June 2022
Line 27:		Line 27:
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization\|"Consideration of Channel Distortion and Equalization"]].		*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization\|"Consideration of Channel Distortion and Equalization"]].

	* Use the interaction module  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|"Complementary Gaussian Error Functions"]]  for numerical evaluation of the Q-function..		* Use the interaction module  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|"Complementary Gaussian Error Functions"]]  for numerical evaluation of the Q-function.

Guenter at 12:43, 19 June 2022

2022-06-19T12:43:40Z

← Older revision		Revision as of 14:43, 19 June 2022
Line 55:		Line 55:
	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  For the optimization it is sufficient to maximize the quotient $\ddot{o}(T_{\rm D})/\sigma_d$:		'''(1)'''  For the optimization it is sufficient to maximize the quotient  $\ddot{o}(T_{\rm D})/\sigma_d$:
	*This is maximized from the values given in the table for the cutoff frequency  $f_{\rm G, opt} \cdot T \underline {= 0.4}$  with $0.735/0.197 \approx 3.73$.		*This is maximized from the values given in the table for the cutoff frequency  $f_{\rm G, opt} \cdot T \underline {= 0.4}$  with  $0.735/0.197 \approx 3.73$.
	*As a comparison:   For  $f_{\rm G} \cdot T = 0.3$  the result is $0.192/0.094 \approx 2.04$ due to the smaller eye opening ~~and for~~  $f_{\rm G} \cdot T = 0.5$  the quotient is also smaller than for the optimum: $1.159/0.379 \approx 3.05$.		*As a comparison:   For  $f_{\rm G} \cdot T = 0.3$  the result is  $0.192/0.094 \approx 2.04$  due to the smaller eye opening.
			*For  $f_{\rm G} \cdot T = 0.5$  the quotient is also smaller than for the optimum:  $1.159/0.379 \approx 3.05$.
	*An even larger cutoff frequency leads to a very large noise rms value without simultaneously increasing the vertical eye opening in the same way.		*An even larger cutoff frequency leads to a very large noise rms value without simultaneously increasing the vertical eye opening in the same way.



	'''(2)'''  Using the result from '''(1)''', we further obtain:		'''(2)'''  Using the result from  '''(1)''',  we further obtain:
	:$$\rho_{\rm U} = \left ( {3.73}/{2} \right )^2 \approx 3.48 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$\rho_{\rm U} = \left ( {3.73}/{2} \right )^2 \approx 3.48 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	10 \cdot {\rm		10 \cdot {\rm
Line 69:		Line 70:


	'''(3)'''		'''(3)'''  With the given  $10 \cdot {\rm lg} \, E_{\rm B}/N_0 = 40 \ \rm dB$,  i.e. $E_{\rm B}/N_0 = 10^4$,  the worst-case signal-to-noise ratio has been found to be  $10 \cdot {\rm lg} \, \rho_{\rm U} \approx 5.41 \, {\rm dB}$.
	*With the given $10 \cdot {\rm lg} \, E_{\rm B}/N_0 = 40 \ \rm dB$, i.e. $E_{\rm B}/N_0 = 10^4$, the worst-case signal-to-noise ratio has been found to be $10 \cdot {\rm lg} \, \rho_{\rm U} \approx 5.41 \, {\rm dB}$.		*However,  for the worst-case error probability  $p_{\rm U} = 10^{\rm -6}$   ⇒   $10 \cdot {\rm lg} \, \rho_{\rm U} > 13.55 \, {\rm dB}$  must be obtained.
	*However, for the worst-case error probability $p_{\rm U} = 10^{\rm -6}$, $10 \cdot {\rm lg} \, \rho_{\rm U} > 13.55 \, {\rm dB}$ must be obtained.		*This is achieved by increasing the quotient  $E_{\rm B}/N_0$  accordingly:
	*This is achieved by increasing the quotient $E_{\rm B}/N_0$ accordingly:
	:$$10 \cdot {\rm		:$$10 \cdot {\rm
	lg}\hspace{0.1cm}{E_{\rm B}}/{N_0} = 40\,{\rm dB}		lg}\hspace{0.1cm}{E_{\rm B}}/{N_0} = 40\,{\rm dB}
Line 84:		Line 84:

	'''(4)'''		'''(4)'''
	*The upper bound for $p_{\rm S}$ is equal to the worst-case error probability $p_{\rm U} = \underline {10^{\rm -6}}$.		*The upper bound  for $p_{\rm S}$  is equal to the worst-case error probability  $p_{\rm U} = \underline {10^{\rm -6}}$.
	*The lower bound is $\underline {0.25 \cdot 10^{\rm -6}}$, which is smaller by a factor of 4.
			*The lower bound is  $\underline {0.25 \cdot 10^{\rm -6}}$,  which is smaller by a factor of  $4$.
	{{ML-Fuß}}		{{ML-Fuß}}


	[[Category:Digital Signal Transmission: Exercises\|^3.3 Channel Distortion and Equalization^]]		[[Category:Digital Signal Transmission: Exercises\|^3.3 Channel Distortion and Equalization^]]

Guenter at 11:18, 19 June 2022

2022-06-19T11:18:00Z

← Older revision		Revision as of 13:18, 19 June 2022
Line 18:		Line 18:
	\right) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}}\right)$$		\right) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}}\right)$$

	*This quantity represents an upper bound for the mean error probability ~~ $p_{\rm S}$: ~~   $p_{\rm S} \le p_{\rm U}$.		*This quantity represents an upper bound for the mean error probability   $p_{\rm S} \le p_{\rm U}$.

	*For  $f_{\rm G} \cdot T ≥ 0.4$,  a lower bound can also be given:   $p_{\rm S} \ge p_{\rm U}/4$.		*For  $f_{\rm G} \cdot T ≥ 0.4$,  a lower bound can also be given:   $p_{\rm S} \ge p_{\rm U}/4$.

Guenter at 11:17, 19 June 2022

2022-06-19T11:17:16Z

← Older revision		Revision as of 13:17, 19 June 2022
Line 18:		Line 18:
	\right) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}}\right)$$		\right) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}}\right)$$

	*This quantity represents an upper bound for the mean error probability  $p_{\rm S}$:    $p_{\rm S} \le p_{\rm U}$.		*This quantity represents an upper bound for the mean error probability  $p_{\rm S}$:    $p_{\rm S} \le p_{\rm U}$.

	*For  $f_{\rm G} \cdot T ≥ 0.4$,  a lower bound can also be given:   $p_{\rm S} \ge p_{\rm U}/4$.		*For  $f_{\rm G} \cdot T ≥ 0.4$,  a lower bound can also be given:   $p_{\rm S} \ge p_{\rm U}/4$.



			Notes:
			*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization\|"Consideration of Channel Distortion and Equalization"]].

			* Use the interaction module  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|"Complementary Gaussian Error Functions"]]  for numerical evaluation of the Q-function..
	~~''Notes:''~~
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization\|"Consideration of Channel Distortion and Equalization"]].
	* Use the interaction module  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|"Complementary Gaussian Error Functions"]] for numerical evaluation of the Q-function..


Line 34:		Line 34:
	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{Within the given grid, determine the optimal cutoff frequency with respect to the "worst-case error probability" criterion.		{Within the given grid,  determine the optimal cutoff frequency with respect to the  "worst-case error probability"  criterion.
	\|type="{}"}		\|type="{}"}
	$f_\text{G, opt} \cdot T \ = \ $ { 0.4 3% }		$f_\text{G, opt} \cdot T \ = \ $ { 0.4 3% }

	{What values does this give for the worst-case signal-to-noise ratio and the worst-case error probability?		{What values does this give for the  "worst-case signal-to-noise ratio"  and the worst-case error probability?
	\|type="{}"}		\|type="{}"}
	$f_\text{G} = \text{G, opt:}\hspace{0.4cm} 10 \cdot {\rm lg} \, \rho_{\rm U} \ = \ $ { 5.41 3% } ${\ \rm dB}$		$f_\text{G} = \text{G, opt:}\hspace{0.4cm} 10 \cdot {\rm lg} \, \rho_{\rm U} \ = \ $ { 5.41 3% } ${\ \rm dB}$
	$\hspace{4.07cm}p_{\rm U} \ = \ $ { 3.1 3% } $\ \rm \%$		$\hspace{4.07cm}p_{\rm U} \ = \ $ { 3.1 3% } $\ \rm \%$

	{To what value would we need to reduce the noise power density  $N_0$  (with respect to signal energy) so that  $p_{\rm U}$  is no greater than  $10^{\rm -6}$?		{To what value would we need to reduce the noise power density  $N_0$  (with respect to signal energy)  so that  $p_{\rm U}$  is not greater than  $10^{\rm -6}$?
	\|type="{}"}		\|type="{}"}
	$N_0/E_{\rm B} \ = \ $ { 1.53 3% } $\ \cdot 10^{\rm -5}$		$N_0/E_{\rm B} \ = \ $ { 1.53 3% } $\ \cdot 10^{\rm -5}$

	{For the assumptions made in '''(3)''', give a lower ~~bound~~ and an upper bound for the average error probability  $p_{\rm S}$.		{For the assumptions made in  '''(3)''',  give a lower and an upper bound for the  "average error probability"   $p_{\rm S}$.
	\|type="{}"}		\|type="{}"}
	$p_\text{ S, min}\hspace{0.02cm} \ = \ $ { 0.25 3% } $\ \cdot 10^{\rm -6}$		$p_\text{ S, min}\hspace{0.02cm} \ = \ $ { 0.25 3% } $\ \cdot 10^{\rm -6}$

Guenter at 11:13, 19 June 2022

2022-06-19T11:13:06Z

← Older revision		Revision as of 13:13, 19 June 2022
Line 3:		Line 3:
	}}		}}

	[[File:P_ID1409__Dig_Z_3_3.png\|right\|frame\|Normalized system ~~sizes~~ for different cut-off frequencies]]		[[File:P_ID1409__Dig_Z_3_3.png\|right\|frame\|Normalized system parameters for different cut-off frequencies]]
	We consider a redundancy-free binary transmission system with the following specifications:		We consider a redundancy-free binary transmission system with the following specifications:
	* The transmission pulses are NRZ rectangular and have energy  $E_{\rm B} = s_0^2 \cdot T$.		* The transmission pulses are NRZ rectangular and have energy  $E_{\rm B} = s_0^2 \cdot T$.

	* The channel is a coaxial cable with characteristic cable attenuation  $a_* = 40 \, {\rm dB}$.		* The channel is a coaxial cable with characteristic cable attenuation  $a_* = 40 \, {\rm dB}$.
	* AWGN noise with noise power density  $N_0 = 0.0001 \cdot E_{\rm B}$  is present.
			* AWGN noise with (one-sided) noise power density  $N_0 = 0.0001 \cdot E_{\rm B}$  is present.

	* The receiver frequency response  $H_{\rm E}(f)$  includes an ideal channel equalizer  $H_{\rm K}^{\rm -1}(f)$  and a Gaussian low-pass filter  $H_{\rm G}(f)$  with cutoff frequency  $f_{\rm G}$  for noise power limitation.		* The receiver frequency response  $H_{\rm E}(f)$  includes an ideal channel equalizer  $H_{\rm K}^{\rm -1}(f)$  and a Gaussian low-pass filter  $H_{\rm G}(f)$  with cutoff frequency  $f_{\rm G}$  for noise power limitation.


	The table shows the eye opening  $\ddot{o}(T_{\rm D})$  as well as the detection noise rms value  $\sigma_{\rm d}$  – each normalized to the transmitted amplitude  $s_0$  – for different cutoff frequencies  $f_{\rm G}$. The cutoff frequency is to be chosen such that the worst-case error probability is as small as possible, with the following definition:		The table shows the eye opening   $\ddot{o}(T_{\rm D})$   as well as the detection noise rms value   $\sigma_{\rm d}$   – each normalized to the transmitted amplitude  $s_0$  – for different cutoff frequencies  $f_{\rm G}$.  The cutoff frequency is to be chosen such that the worst-case error probability is as small as possible,  with the following definition:
	:$$p_{\rm U} = {\rm Q} \left( \frac{\ddot{o}(T_{\rm D})/2}{ \sigma_d}		:$$p_{\rm U} = {\rm Q} \left( \frac{\ddot{o}(T_{\rm D})/2}{ \sigma_d}
	\right) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}}\right)$$		\right) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U} = {\rm Q} \left( \sqrt{\rho_{\rm U}}\right)$$