Aufgaben:Exercise 1.3Z: Thermal Noise - Revision history

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	[[File:P_ID953__Mod_Z_1_3.png\|right\|frame\|Examplary signals for low-pass and band-pass noise]]		[[File:P_ID953__Mod_Z_1_3.png\|right\|frame\|Examplary signals for low-pass and band-pass noise]]
	A fundamental disturbance and one that occurs in any communication system is  "thermal noise",  since any resistance  $R$  with an absolute temperature of  $θ$  (in "degrees Kelvin")  produces a noise signal  $n(t)$  with a  (one-sided)  noise power-spectral density		A fundamental disturbance and one that occurs in any communication system is  "thermal noise",  since any resistance  $R$  with an absolute temperature of  $θ$  (in "degrees Kelvin")  produces a noise signal  $n(t)$  with a  (one-sided)  noise power-spectral density
	:$${N_{\rm 0, \hspace{0.05cm}min}}= k_{\rm B} \cdot \theta		:$${N_{\rm 0, \hspace{0.05cm}min}}= k_{\rm B} \cdot \theta\hspace{0.3cm}\left(k_{\rm B} = 1.38 \cdot 10^{-23}\hspace{0.05cm}{\rm Ws}/{\rm K}\right)$$
	\hspace{0.3cm}\left(k_{\rm B} = 1.38 \cdot 10^{-23}
	\hspace{0.05cm}{\rm Ws}/{\rm K}\right)$$
	where $k_{\rm B}$  denotes the  "Boltzmann constant" (from German  "Konstante").		where $k_{\rm B}$  denotes the  "Boltzmann constant" (from German  "Konstante").

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	'''(1)'''  Using the Boltzmann constant  $k_{\rm B}$  it holds that:		'''(1)'''  Using the Boltzmann constant  $k_{\rm B}$  it holds that:
	:$$N_0 = F \cdot k_{\rm B} \cdot \theta = 10 \~~cdot~~		:$$N_0 = F \cdot k_{\rm B} \cdot \theta = 10 \cdot1.38\hspace{0.05cm}\cdot 10^{-23} \hspace{0.05cm}\frac{\rmWs}{\rm K}\cdot 290\,{\rm K} \hspace{0.15cm}\underline {\approx 4\hspace{0.05cm}\cdot10^{-20} \hspace{0.05cm}{\rm W}/{\rm Hz}}\hspace{0.05cm}.$$
	1.38\hspace{0.05cm}\cdot 10^{-23} \hspace{0.05cm}\frac{\rm
	Ws}{\rm K}\cdot 290\,{\rm K} \hspace{0.15cm}\underline {\approx 4\hspace{0.05cm}\~~cdot~~
	10^{-20} \hspace{0.05cm}{\rm W}/{\rm Hz}}\hspace{0.05cm}.$$


	'''(2)'''  The specified noise power density  $N_0$  is physically limited to  $6$  THz.  Thus the maximum noise power is:		'''(2)'''  The specified noise power density  $N_0$  is physically limited to  $6$  THz.  Thus the maximum noise power is:
	:$$N_{\rm max} = 4\hspace{0.05cm}\cdot 10^{-20}		:$$N_{\rm max} = 4\hspace{0.05cm}\cdot 10^{-20}\hspace{0.08cm}\frac{\rm W}{\rm Hz}\cdot 6 \cdot10^{12}\hspace{0.08cm}{\rm Hz}\hspace{0.15cm}\underline {= 0.24\hspace{0.08cm}\cdot 10^{-6}\;{\rmW}}\hspace{0.05cm}.$$
	\hspace{0.08cm}\frac{\rm W}{\rm Hz}\cdot 6 \cdot10^{12}
	\hspace{0.08cm}{\rm Hz}\hspace{0.15cm}\underline {= 0.24\hspace{0.08cm}\cdot 10^{-6}\;{\rm
	W}}\hspace{0.05cm}.$$


	'''(3)'''  Now the resulting noise power is:		'''(3)'''  Now the resulting noise power is:
	:$$N = N_0 \cdot B = 4\hspace{0.08cm}\cdot 10^{-20}		:$$N = N_0 \cdot B = 4\hspace{0.08cm}\cdot 10^{-20}\hspace{0.08cm}\frac{\rm W}{\rm Hz}\cdot 3 \cdot10^{4}\hspace{0.08cm}{\rm Hz}\hspace{0.15cm}\underline {= 12\hspace{0.05cm}\cdot 10^{-16}\;{\rmW}}\hspace{0.05cm}.$$
	\hspace{0.08cm}\frac{\rm W}{\rm Hz}\cdot 3 \cdot10^{4}
	\hspace{0.08cm}{\rm Hz}\hspace{0.15cm}\underline {= 12\hspace{0.05cm}\cdot 10^{-16}\;{\rm
	W}}\hspace{0.05cm}.$$

	* Converting to the reference resistance  $R = 1 \ Ω$:		* Converting to the reference resistance  $R = 1 \ Ω$:
	:$$N = N_0 \cdot B \cdot R = 12\hspace{0.05cm}\cdot 10^{-16}\;{\rm		:$$N = N_0 \cdot B \cdot R = 12\hspace{0.05cm}\cdot 10^{-16}\;{\rmW}\hspace{0.05cm} \cdot 50 \; {\rm \Omega}= 6\hspace{0.05cm}\cdot10^{-14}\;{\rm V^2}\hspace{0.05cm}.$$
	W}\hspace{0.05cm} \cdot 50 \; {\rm \Omega}= 6\hspace{0.05cm}\~~cdot~~
	10^{-14}\;{\rm V^2}\hspace{0.05cm}.$$
	[[File:EN_Mod_Z_1_3_e.png\|rechts\|frame\|Power-spectral densities with band-limited noise]]		[[File:EN_Mod_Z_1_3_e.png\|rechts\|frame\|Power-spectral densities with band-limited noise]]
	*The noise standard deviation $σ_n$  is the square root of this:		*The noise standard deviation $σ_n$  is the square root of this:
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	'''(5)'''  The noise power density of the random signal  $n_1(t)$  is constant in the frequency range  $\|f\| < 30$  kHz:		'''(5)'''  The noise power density of the random signal  $n_1(t)$  is constant in the frequency range  $\|f\| < 30$  kHz:
	:$${\it \Phi}_{n,\hspace{0.05cm}{ \rm LP} }(f) \hspace{-0.05cm}=\hspace{-0.05cm} \frac{N_0}{2} \hspace{0.15cm}\underline {=2\hspace{0.05cm}\hspace{-0.05cm}\cdot \hspace{-0.05cm}		:$${\it \Phi}_{n,\hspace{0.05cm}{ \rm LP} }(f) \hspace{-0.05cm}=\hspace{-0.05cm} \frac{N_0}{2} \hspace{0.15cm}\underline {=2\hspace{0.05cm}\hspace{-0.05cm}\cdot \hspace{-0.05cm}10^{-12} \hspace{0.05cm}{\rm W}/{\rm Hz}}\hspace{0.05cm}.$$
	10^{-12} \hspace{0.05cm}{\rm W}/{\rm Hz}}\hspace{0.05cm}.$$
	*Thus,  this value is also valid for the frequency  $f = 20$  kHz.		*Thus,  this value is also valid for the frequency  $f = 20$  kHz.

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	:$$N = N_0 \cdot B\cdot R = \sigma_n^2 \hspace{0.01cm}.$$		:$$N = N_0 \cdot B\cdot R = \sigma_n^2 \hspace{0.01cm}.$$
	*According to the first equation,  the result is the actual,  physical power in "watts"  $\rm (W)$.		*According to the first equation,  the result is the actual,  physical power in "watts"  $\rm (W)$.
	*According to the second equation,  the result has the unit   "$\rm V^{ 2 }$".		*According to the second equation,  the result has the unit   "$\rm V^{2}$".
	*This means that the power is here converted to the reference resistance  $R = 1\ Ω$  – as is often the case in Communications Engineering.		*This means that the power is here converted to the reference resistance  $R = 1\ Ω$  – as is often the case in Communications Engineering.
	*This equation must also be used to calculate the standard deviation  $σ_n$  of the noise signal  $n(t)$ .		*This equation must also be used to calculate the standard deviation  $σ_n$  of the noise signal  $n(t)$ .
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	{Calculate the noise power-spectral density  $N_0$  with a noise factor of  $F = 10$  and  $θ = 290^\circ$  Kelvin.		{Calculate the noise power-spectral density  $N_0$  with a noise factor of  $F = 10$  and  $θ = 290^\circ$  Kelvin.
	\|type="{}"}		\|type="{}"}
	$N_0 \ = \ $ { 4 3% } $\ \cdot 10^{ -20 }\ \text{W/Hz}$		$N_0 \ = \ $ { 4 3% } $\ \cdot 10^{-20}\ \text{W/Hz}$

	{What is the maximum noise power  (without bandwidth limits)?		{What is the maximum noise power  (without bandwidth limits)?
	\|type="{}"}		\|type="{}"}
	$N_{\rm max} \ = \ $ { 0.24 3% } $\ \cdot 10^{ -6 }\ \text{W/Hz}$		$N_{\rm max} \ = \ $ { 0.24 3% } $\ \cdot 10^{-6}\ \text{W/Hz}$


	{What is the noise power  $N$  with bandwidth  $B = 30\text{ kHz}$?  What is the standard deviation $σ_n$?		{What is the noise power  $N$  with bandwidth  $B = 30\text{ kHz}$?  What is the standard deviation $σ_n$?
	\|type="{}"}		\|type="{}"}
	$N \ = \ $ { 12 3% } $\ \cdot 10^{ -16 }\ \text{W/Hz}$		$N \ = \ $ { 12 3% } $\ \cdot 10^{-16}\ \text{W/Hz}$
	$σ_n \ = \ ${ 0.245 3% } $\ \cdot 10^{ -6 }\ \text{V}$		$σ_n \ = \ ${ 0.245 3% } $\ \cdot 10^{-6}\ \text{V}$

	{Which of the signals –  $n_1(t)$  or  $n_2(t)$  – shows low-pass noise and which shows band-pass noise?		{Which of the signals –  $n_1(t)$  or  $n_2(t)$  – shows low-pass noise and which shows band-pass noise?
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	{What is the value of the noise power-spectral density of the low-pass  $\rm (LP)$  noise at frequency  $f = 20\text{ kHz}$?  Let  $B = 30\text{ kHz}$.		{What is the value of the noise power-spectral density of the low-pass  $\rm (LP)$  noise at frequency  $f = 20\text{ kHz}$?  Let  $B = 30\text{ kHz}$.
	\|type="{}"}		\|type="{}"}
	${\it Φ}_{n, \hspace{0.05cm}\rm LP}(f = 20 \ \rm kHz) \ = \ $ { 2 3% } $\ \cdot 10^{ -12 }\ \text{W/Hz}$		${\it Φ}_{n, \hspace{0.05cm}\rm LP}(f = 20 \ \rm kHz) \ = \ $ { 2 3% } $\ \cdot 10^{-12}\ \text{W/Hz}$

	{What is the value of the noise power-spectral density of the band-pass  $\rm (BP)$  noise at  $f = 120\text{ kHz}$?  Let  $f_{\rm M} = 100\text{ kHz}$  and  $B = 30\text{ kHz}$.		{What is the value of the noise power-spectral density of the band-pass  $\rm (BP)$  noise at  $f = 120\text{ kHz}$?  Let  $f_{\rm M} = 100\text{ kHz}$  and  $B = 30\text{ kHz}$.
	\|type="{}"}		\|type="{}"}
	${\it Φ}_{n, \hspace{0.05cm}\rm BP}(f = 120 \ \rm kHz) \ = \ $ { 0. } $\ \cdot 10^{ -12 }\ \text{W/Hz}$		${\it Φ}_{n, \hspace{0.05cm}\rm BP}(f = 120 \ \rm kHz) \ = \ $ { 0. } $\ \cdot 10^{-12}\ \text{W/Hz}$

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	[[Category:Modulation Methods: Exercises\|^1.2 Quality Criteria^]]		[[Category:Modulation Methods: Exercises\|^1.2 Quality Criteria^]]
			[[de:Aufgaben:1.3Z_Thermisches_Rauschen]]

Guenter at 16:27, 10 April 2022

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	All equations apply regardless of whether the noise is low-pass or band-pass. The diagram shows two noise signals  $n_1(t)$  and  $n_2(t)$  of equal bandwidth.  Question   '''(4)'''  asks which of these signals will appear at the output of a low-pass and a band-pass, respectively.		All equations apply regardless of whether the noise is low-pass or band-pass. The diagram shows two noise signals  $n_1(t)$  and  $n_2(t)$  of equal bandwidth.  Question   '''(4)'''  asks which of these signals will appear at the output of a low-pass and a band-pass, respectively.

	The two-sided noise power density of band-limited low-pass noise  $n_{\rm TP}(t)$  is:		The two-sided noise power-spectral density of band-limited low-pass noise  $n_{\rm TP}(t)$  is:
	:$$ {\it \Phi}_{n, {\hspace{0.05cm}\rm LP}}(f) = \left\{ \begin{array}{c} N_0/2 \\ 0 \\ \end{array} \right. \begin{array}{{10}c} {\rm{for}} \\ \\ \end{array}\begin{array}{{20}c} {\left\| \hspace{0.005cm} f\hspace{0.05cm} \right\| < B,} \\ {\rm otherwise.} \\ \end{array}$$		:$$ {\it \Phi}_{n, {\hspace{0.05cm}\rm LP}}(f) = \left\{ \begin{array}{c} N_0/2 \\ 0 \\ \end{array} \right. \begin{array}{{10}c} {\rm{for}} \\ \\ \end{array}\begin{array}{{20}c} {\left\| \hspace{0.005cm} f\hspace{0.05cm} \right\| < B,} \\ {\rm otherwise.} \\ \end{array}$$
	In contrast,  for band-pass noise  $n_{\rm BP}(t)$  with center frequency  $f_{\rm M}$, it holds that:		In contrast,  for band-pass noise  $n_{\rm BP}(t)$  with center frequency  $f_{\rm M}$, it holds that:

Guenter at 16:26, 10 April 2022

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	- The noise signal  $n_1(t)$  is characteristically band-pass.		- The noise signal  $n_1(t)$  is characteristically band-pass.

	{What is the value of the noise power density of the low-pass  $\rm (LP)$  noise at frequency  $f = 20\text{ kHz}$?  Let  $B = 30\text{ kHz}$.		{What is the value of the noise power-spectral density of the low-pass  $\rm (LP)$  noise at frequency  $f = 20\text{ kHz}$?  Let  $B = 30\text{ kHz}$.
	\|type="{}"}		\|type="{}"}
	${\it Φ}_{n, \hspace{0.05cm}\rm LP}(f = 20 \ \rm kHz) \ = \ $ { 2 3% } $\ \cdot 10^{ -12 }\ \text{W/Hz}$		${\it Φ}_{n, \hspace{0.05cm}\rm LP}(f = 20 \ \rm kHz) \ = \ $ { 2 3% } $\ \cdot 10^{ -12 }\ \text{W/Hz}$

	{What is the value of the noise power density of the band-pass  $\rm (BP)$  noise at  $f = 120\text{ kHz}$?  Let  $f_{\rm M} = 100\text{ kHz}$  and  $B = 30\text{ kHz}$.		{What is the value of the noise power-spectral density of the band-pass  $\rm (BP)$  noise at  $f = 120\text{ kHz}$?  Let  $f_{\rm M} = 100\text{ kHz}$  and  $B = 30\text{ kHz}$.
	\|type="{}"}		\|type="{}"}
	${\it Φ}_{n, \hspace{0.05cm}\rm BP}(f = 120 \ \rm kHz) \ = \ $ { 0. } $\ \cdot 10^{ -12 }\ \text{W/Hz}$		${\it Φ}_{n, \hspace{0.05cm}\rm BP}(f = 120 \ \rm kHz) \ = \ $ { 0. } $\ \cdot 10^{ -12 }\ \text{W/Hz}$

Guenter at 16:25, 10 April 2022

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	However,  this noise is limited to  $6\text{ THz}$  for physical reasons.  Furthermore,  it can be observed that this minimum value can only be achieved with exact impedance matching.		However,  this noise is limited to  $6\text{ THz}$  for physical reasons.  Furthermore,  it can be observed that this minimum value can only be achieved with exact impedance matching.

	In the realization of a circuit unit - for example, an amplifier - the effective noise power density is usually significantly greater,  since several noise sources add up,  and mismatches also play a role.  This effect is captured by the noise factor  $F \ge 1$  .  It holds that:		In the realization of a circuit unit - for example, an amplifier - the effective noise power-spectral density is usually significantly greater,  since several noise sources add up,  and mismatches also play a role.  This effect is captured by the noise factor  $F \ge 1$  .  It holds that:
	:$$N_0 = F \cdot {N_{\rm 0, \hspace{0.05cm}min}}= F \cdot k_{\rm B} \cdot \theta \hspace{0.05cm}.$$		:$$N_0 = F \cdot {N_{\rm 0, \hspace{0.05cm}min}}= F \cdot k_{\rm B} \cdot \theta \hspace{0.05cm}.$$
	With a bandwidth  $B$,  the effective noise power is characterized by:		With a bandwidth  $B$,  the effective noise power is characterized by:
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	<quiz display=simple>		<quiz display=simple>
	{Calculate the noise power density  $N_0$  with a noise factor of  $F = 10$  and  $θ = 290^\circ$  Kelvin.		{Calculate the noise power-spectral density  $N_0$  with a noise factor of  $F = 10$  and  $θ = 290^\circ$  Kelvin.
	\|type="{}"}		\|type="{}"}
	$N_0 \ = \ $ { 4 3% } $\ \cdot 10^{ -20 }\ \text{W/Hz}$		$N_0 \ = \ $ { 4 3% } $\ \cdot 10^{ -20 }\ \text{W/Hz}$

Guenter at 16:24, 10 April 2022

2022-04-10T16:24:16Z

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	}}		}}

	[[File:P_ID953__Mod_Z_1_3.png\|right\|frame\|~~Example~~ signals for low-pass and band-pass noise]]		[[File:P_ID953__Mod_Z_1_3.png\|right\|frame\|Examplary signals for low-pass and band-pass noise]]
	A fundamental disturbance and one that occurs in any communication system is  "thermal noise",  since any resistance  $R$  with an absolute temperature of  $θ$  (in "degrees Kelvin")  produces a noise signal  $n(t)$  with a  (one-sided)  noise power density		A fundamental disturbance and one that occurs in any communication system is  "thermal noise",  since any resistance  $R$  with an absolute temperature of  $θ$  (in "degrees Kelvin")  produces a noise signal  $n(t)$  with a  (one-sided)  noise power-spectral density
	:$${N_{\rm 0, \hspace{0.05cm}min}}= k_{\rm B} \cdot \theta		:$${N_{\rm 0, \hspace{0.05cm}min}}= k_{\rm B} \cdot \theta
	\hspace{0.3cm}\left(k_{\rm B} = 1.38 \cdot 10^{-23}		\hspace{0.3cm}\left(k_{\rm B} = 1.38 \cdot 10^{-23}
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	W}\hspace{0.05cm} \cdot 50 \; {\rm \Omega}= 6\hspace{0.05cm}\cdot		W}\hspace{0.05cm} \cdot 50 \; {\rm \Omega}= 6\hspace{0.05cm}\cdot
	10^{-14}\;{\rm V^2}\hspace{0.05cm}.$$		10^{-14}\;{\rm V^2}\hspace{0.05cm}.$$
	[[File:EN_Mod_Z_1_3_e.png\|rechts\|frame\|Power ~~density spectra~~ with band-limited noise]]		[[File:EN_Mod_Z_1_3_e.png\|rechts\|frame\|Power-spectral densities with band-limited noise]]
	*The noise standard deviation $σ_n$  is the square root of this:		*The noise standard deviation $σ_n$  is the square root of this:
	:$$\sigma_n= \sqrt{6\hspace{0.05cm}\cdot 10^{-14}\;{\rm V^2}} \hspace{0.15cm}\underline {= 0.245 \hspace{0.05cm}\cdot 10^{-6}\;{\rm V}}\hspace{0.05cm}.$$		:$$\sigma_n= \sqrt{6\hspace{0.05cm}\cdot 10^{-14}\;{\rm V^2}} \hspace{0.15cm}\underline {= 0.245 \hspace{0.05cm}\cdot 10^{-6}\;{\rm V}}\hspace{0.05cm}.$$

Reed at 16:59, 23 March 2022

2022-03-23T16:59:11Z

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	:$$N = N_0 \cdot B\cdot R = \sigma_n^2 \hspace{0.01cm}.$$		:$$N = N_0 \cdot B\cdot R = \sigma_n^2 \hspace{0.01cm}.$$
	*According to the first equation,  the result is the actual,  physical power in "watts"  $\rm (W)$.		*According to the first equation,  the result is the actual,  physical power in "watts"  $\rm (W)$.
	*According to the second equation,  the result has the unit   „$\rm V^{ 2 }$”.		*According to the second equation,  the result has the unit   "$\rm V^{ 2 }$".
	*This means that the power is here converted to the reference resistance  $R = 1\ Ω$  – as is often the case in Communications Engineering.		*This means that the power is here converted to the reference resistance  $R = 1\ Ω$  – as is often the case in Communications Engineering.
	*This equation must also be used to calculate the standard deviation  $σ_n$  of the noise signal  $n(t)$ .		*This equation must also be used to calculate the standard deviation  $σ_n$  of the noise signal  $n(t)$ .

Guenter at 15:16, 17 February 2022

2022-02-17T15:16:59Z

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	*According to the second equation,  the result has the unit   „$\rm V^{ 2 }$”.		*According to the second equation,  the result has the unit   „$\rm V^{ 2 }$”.
	*This means that the power is here converted to the reference resistance  $R = 1\ Ω$  – as is often the case in Communications Engineering.		*This means that the power is here converted to the reference resistance  $R = 1\ Ω$  – as is often the case in Communications Engineering.
	*This equation must also be used to calculate the standard deviation~~  ("standard deviation")~~  $σ_n$  of the noise signal  $n(t)$ .		*This equation must also be used to calculate the standard deviation  $σ_n$  of the noise signal  $n(t)$ .

← Older revision		Revision as of 17:57, 16 March 2026
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	'''(1)'''  Using the Boltzmann constant  $k_{\rm B}$  it holds that:		'''(1)'''  Using the Boltzmann constant  $k_{\rm B}$  it holds that:
	:$$N_0 = F \cdot k_{\rm B} \cdot \theta = 10 \cdot1.38\hspace{0.05cm}\cdot 10^{-23} \hspace{0.05cm}\frac{\~~rmWs~~}{\rm K}\cdot 290\,{\rm K} \hspace{0.15cm}\underline {\approx 4\hspace{0.05cm}\cdot10^{-20} \hspace{0.05cm}{\rm W}/{\rm Hz}}\hspace{0.05cm}.$$		:$$N_0 = F \cdot k_{\rm B} \cdot \theta = 10 \cdot1.38\hspace{0.05cm}\cdot 10^{-23} \hspace{0.05cm}\frac{\rm Ws}{\rm K}\cdot 290\,{\rm K} \hspace{0.15cm}\underline {\approx 4\hspace{0.05cm}\cdot10^{-20} \hspace{0.05cm}{\rm W}/{\rm Hz}}\hspace{0.05cm}.$$


	'''(2)'''  The specified noise power density  $N_0$  is physically limited to  $6$  THz.  Thus the maximum noise power is:		'''(2)'''  The specified noise power density  $N_0$  is physically limited to  $6$  THz.  Thus the maximum noise power is:
	:$$N_{\rm max} = 4\hspace{0.05cm}\cdot 10^{-20}\hspace{0.08cm}\frac{\rm W}{\rm Hz}\cdot 6 \cdot10^{12}\hspace{0.08cm}{\rm Hz}\hspace{0.15cm}\underline {= 0.24\hspace{0.08cm}\cdot 10^{-6}\;{\~~rmW~~}}\hspace{0.05cm}.$$		:$$N_{\rm max} = 4\hspace{0.05cm}\cdot 10^{-20}\hspace{0.08cm}\frac{\rm W}{\rm Hz}\cdot 6 \cdot10^{12}\hspace{0.08cm}{\rm Hz}\hspace{0.15cm}\underline {= 0.24\hspace{0.08cm}\cdot 10^{-6}\;{\rm W}}\hspace{0.05cm}.$$


	'''(3)'''  Now the resulting noise power is:		'''(3)'''  Now the resulting noise power is:
	:$$N = N_0 \cdot B = 4\hspace{0.08cm}\cdot 10^{-20}\hspace{0.08cm}\frac{\rm W}{\rm Hz}\cdot 3 \cdot10^{4}\hspace{0.08cm}{\rm Hz}\hspace{0.15cm}\underline {= 12\hspace{0.05cm}\cdot 10^{-16}\;{\~~rmW~~}}\hspace{0.05cm}.$$		:$$N = N_0 \cdot B = 4\hspace{0.08cm}\cdot 10^{-20}\hspace{0.08cm}\frac{\rm W}{\rm Hz}\cdot 3 \cdot10^{4}\hspace{0.08cm}{\rm Hz}\hspace{0.15cm}\underline {= 12\hspace{0.05cm}\cdot 10^{-16}\;{\rm W}}\hspace{0.05cm}.$$

	* Converting to the reference resistance  $R = 1 \ Ω$:		* Converting to the reference resistance  $R = 1 \ Ω$:
	:$$N = N_0 \cdot B \cdot R = 12\hspace{0.05cm}\cdot 10^{-16}\;{\~~rmW~~}\hspace{0.05cm} \cdot 50 \; {\rm \Omega}= 6\hspace{0.05cm}\cdot10^{-14}\;{\rm V^2}\hspace{0.05cm}.$$		:$$N = N_0 \cdot B \cdot R = 12\hspace{0.05cm}\cdot 10^{-16}\;{\rm W}\hspace{0.05cm} \cdot 50 \; {\rm \Omega}= 6\hspace{0.05cm}\cdot10^{-14}\;{\rm V^2}\hspace{0.05cm}.$$
	[[File:EN_Mod_Z_1_3_e.png\|rechts\|frame\|Power-spectral densities with band-limited noise]]		[[File:EN_Mod_Z_1_3_e.png\|rechts\|frame\|Power-spectral densities with band-limited noise]]
	*The noise standard deviation $σ_n$  is the square root of this:		*The noise standard deviation $σ_n$  is the square root of this:
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	[[Category:Modulation Methods: Exercises\|^1.2 Quality Criteria^]]		[[Category:Modulation Methods: Exercises\|^1.2 Quality Criteria^]]
	[[de:Aufgaben:1.~~3Z_Thermisches_Rauschen~~]]		[[de:Aufgaben:Aufgabe 1.3Z: Thermisches Rauschen]]