Difference between revisions of "Aufgaben:Exercise 1.3Z: Thermal Noise"

Latest revision as of 17:27, 10 April 2022

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

$${N_{\rm 0, \hspace{0.05cm}min}}= k_{\rm B} \cdot \theta \hspace{0.3cm}\left(k_{\rm B} = 1.38 \cdot 10^{-23} \hspace{0.05cm}{\rm Ws}/{\rm K}\right)$$

where $k_{\rm B}$ denotes the "Boltzmann constant" (from German "Konstante").

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-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:

$$N_0 = F \cdot {N_{\rm 0, \hspace{0.05cm}min}}= F \cdot k_{\rm B} \cdot \theta \hspace{0.05cm}.$$

With a bandwidth $B$, the effective noise power is characterized by:

$$N = N_0 \cdot B \hspace{0.1cm} \hspace{0.01cm}.$$

$$N = N_0 \cdot B\cdot R = \sigma_n^2 \hspace{0.01cm}.$$

According to the first equation, the result is the actual, physical power in "watts" $\rm (W)$.
According to the 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 equation must also be used to calculate the standard deviation $σ_n$ of the noise signal $n(t)$ .

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-spectral density of band-limited low-pass noise $n_{\rm TP}(t)$ is:

$$ {\it \Phi}_{n, {\hspace{0.05cm}\rm LP}}(f) = \left\{ \begin{array}{c} N_0/2 \\ 0 \\ \end{array} \right. \begin{array}{*{10}c} {\rm{for}} \\ \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < B,} \\ {\rm otherwise.} \\ \end{array}$$

In contrast, for band-pass noise $n_{\rm BP}(t)$ with center frequency $f_{\rm M}$, it holds that:

$${\it \Phi}_{n, {\hspace{0.05cm}\rm BP}}(f) = \left\{ \begin{array}{c} N_0/2 \\ 0 \\ \end{array} \right. \begin{array}{*{10}c} {\rm{for}} \\ \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} f - f_{\rm M}\hspace{0.05cm} \right| < B/2,} \\ {\rm otherwise.} \\ \end{array}.$$

For all subsequent numerical calculations it is assumed:

$$ F = 10, \hspace{0.2cm}\theta = 290\,{\rm K},\hspace{0.2cm}R = 50\,{\rm \Omega},\hspace{0.2cm}B = 30\,{\rm kHz},\hspace{0.2cm}f_{\rm M} = 0 \hspace{0.5cm}{\rm or}\hspace{0.5cm}f_{\rm M} =100\,{\rm kHz}\hspace{0.05cm}.$$

Hints:

This exercise belongs to the chapter Quality Criteria.
Particular reference is made to the page Some remarks on the AWGN channel model.
By specifying the powers in $\rm W$atts , they are independent of the reference resistance $R$, while power with the unit $\rm V^2$ can only be evaluated directly for $R = 1\ \Omega$.

Questions

$N_0 \ = \ $

$\ \cdot 10^{ -20 }\ \text{W/Hz}$

$N_{\rm max} \ = \ $

$\ \cdot 10^{ -6 }\ \text{W/Hz}$

$N \ = \ $

$\ \cdot 10^{ -16 }\ \text{W/Hz}$

$σ_n \ = \ $

$\ \cdot 10^{ -6 }\ \text{V}$

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

${\it Φ}_{n, \hspace{0.05cm}\rm LP}(f = 20 \ \rm kHz) \ = \ $

$\ \cdot 10^{ -12 }\ \text{W/Hz}$

${\it Φ}_{n, \hspace{0.05cm}\rm BP}(f = 120 \ \rm kHz) \ = \ $

$\ \cdot 10^{ -12 }\ \text{W/Hz}$

Solution

(1) Using the Boltzmann constant $k_{\rm B}$ it holds that:

$$N_0 = F \cdot k_{\rm B} \cdot \theta = 10 \cdot 1.38\hspace{0.05cm}\cdot 10^{-23} \hspace{0.05cm}\frac{\rm Ws}{\rm K}\cdot 290\,{\rm K} \hspace{0.15cm}\underline {\approx 4\hspace{0.05cm}\cdot 10^{-20} \hspace{0.05cm}{\rm W}/{\rm Hz}}\hspace{0.05cm}.$$

(2) The specified noise power density $N_0$ is physically limited to $6$ THz. Thus the maximum noise power is:

$$N_{\rm max} = 4\hspace{0.05cm}\cdot 10^{-20} \hspace{0.08cm}\frac{\rm W}{\rm Hz}\cdot 6 \cdot10^{12} \hspace{0.08cm}{\rm Hz}\hspace{0.15cm}\underline {= 0.24\hspace{0.08cm}\cdot 10^{-6}\;{\rm W}}\hspace{0.05cm}.$$

(3) Now the resulting noise power is:

$$N = N_0 \cdot B = 4\hspace{0.08cm}\cdot 10^{-20} \hspace{0.08cm}\frac{\rm W}{\rm Hz}\cdot 3 \cdot10^{4} \hspace{0.08cm}{\rm Hz}\hspace{0.15cm}\underline {= 12\hspace{0.05cm}\cdot 10^{-16}\;{\rm W}}\hspace{0.05cm}.$$

Converting to the reference resistance $R = 1 \ Ω$:

$$N = N_0 \cdot B \cdot R = 12\hspace{0.05cm}\cdot 10^{-16}\;{\rm W}\hspace{0.05cm} \cdot 50 \; {\rm \Omega}= 6\hspace{0.05cm}\cdot 10^{-14}\;{\rm V^2}\hspace{0.05cm}.$$

Power-spectral densities with band-limited noise

The noise standard deviation $σ_n$ is the square root of this:

$$\sigma_n= \sqrt{6\hspace{0.05cm}\cdot 10^{-14}\;{\rm V^2}} \hspace{0.15cm}\underline {= 0.245 \hspace{0.05cm}\cdot 10^{-6}\;{\rm V}}\hspace{0.05cm}.$$

(4) Answer 1 is correct:

In the random signal $n_2(t)$ one can recognize certain regularities similar to a harmonic oscillation ⇒ it is bandpass noise.
In contrast, the signal $n_1(t)$ is low-pass noise.

(5) The noise power density of the random signal $n_1(t)$ is constant in the frequency range $|f| < 30$ kHz:

$${\it \Phi}_{n,\hspace{0.05cm}{ \rm LP} }(f) \hspace{-0.05cm}=\hspace{-0.05cm} \frac{N_0}{2} \hspace{0.15cm}\underline {=2\hspace{0.05cm}\hspace{-0.05cm}\cdot \hspace{-0.05cm} 10^{-12} \hspace{0.05cm}{\rm W}/{\rm Hz}}\hspace{0.05cm}.$$

Thus, this value is also valid for the frequency $f = 20$ kHz.

(6) As can be seen from the diagram, ${\it Φ}_{n, \hspace{0.05cm}\rm BP}(f)$ is non-zero only in the range between $85$ kHz and $115$ kHz, when the bandwidth is $B = 30$ kHz.

Thus, at the frequency $f = 120$ kHz, the noise power density is zero:

$${\it Φ}_{n, \hspace{0.05cm}\rm BP}(f = 120 \ \rm kHz)\hspace{0.15cm}\underline{=0}.$$

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Modulationsverfahren/Qualitätskriterien
+{{quiz-Header|Buchseite=Modulation_Methods/Quality_Criteria
 }}
-[[File:P_ID953__Mod_Z_1_3.png|right|]]
+[[File:P_ID953__Mod_Z_1_3.png|right|frame|Examplary signals for low-pass and band-pass noise]]
-Eine fundamentale und bei jedem Nachrichtensystem auftretende Störung ist das thermische Rauschen, da jeder Widerstand $R$ mit der absoluten Temperatur $θ$ (in „Grad Kelvin”) ein Rauschsignal $n(t)$ mit der (einseitigen) Rauschleistungsdichte
+A fundamental disturbance and one that occurs in any communication system is&nbsp; "thermal noise",&nbsp; since any resistance &nbsp;$R$&nbsp; with an absolute temperature of &nbsp;$θ$&nbsp; (in "degrees Kelvin")&nbsp; produces a noise signal &nbsp;$n(t)$&nbsp; with a&nbsp; (one-sided)&nbsp; noise power-spectral density
-$$N_{0,min} = k_B \cdot \theta ( k_B = 1.38 \cdot 10^{ -23 } Ws/K)$$
+:$${N_{\rm 0, \hspace{0.05cm}min}}= k_{\rm B} \cdot \theta
-abgibt. $k_B$ bezeichnet man als die Boltzmann–Konstante.
+\hspace{0.3cm}\left(k_{\rm B} = 1.38 \cdot 10^{-23}
+\hspace{0.05cm}{\rm Ws}/{\rm K}\right)$$
+where $k_{\rm B}$&nbsp; denotes the&nbsp; "Boltzmann constant" (from German&nbsp; "Konstante").
-Allerdings ist diese aus physikalischen Gründen auf $6$ $\text{THz}$ begrenzt. Weiterhin ist zu beobachten, dass dieser minimale Wert nur bei exakter Widerstandsanpassung erreicht werden kann.
+However,&nbsp; this noise is limited to &nbsp;$6\text{ THz}$&nbsp; for physical reasons.&nbsp; Furthermore,&nbsp; it can be observed that this minimum value can only be achieved with exact impedance matching.
-Bei der Realisierung einer Schaltungseinheit – zum Beispiel eines Verstärkers – ist die wirksame Rauschleistungsdichte meist deutlich größer, da sich mehrere Rauschquellen addieren und zudem Fehlanpassungen eine Rolle spielen. Dieser Effekt wird durch die Rauschzahl F erfasst, und es gilt:
+In the realization of a circuit unit - for example, an amplifier - the effective noise power-spectral density is usually significantly greater,&nbsp; since several noise sources add up,&nbsp; and mismatches also play a role.&nbsp;  This effect is captured by the noise factor &nbsp;$F \ge 1$&nbsp; .&nbsp; It holds that:
-$$N_0 = F \cdot {N_{\rm 0, \hspace{0.05cm}min}}= F \cdot k_{\rm B} \cdot \theta \hspace{0.05cm}.$$
+:$$N_0 = F \cdot {N_{\rm 0, \hspace{0.05cm}min}}= F \cdot k_{\rm B} \cdot \theta \hspace{0.05cm}.$$
-Für die wirksame Rauschleistung gilt mit der Bandbreite $B$:
+With a bandwidth &nbsp;$B$,&nbsp; the effective noise power is characterized by:
-$$N = N_0 \cdot B \hspace{0.1cm}\left(= N_0 \cdot B\cdot R = \sigma_n^2\right) \hspace{0.01cm}.$$
+:$$N = N_0 \cdot B \hspace{0.1cm} \hspace{0.01cm}.$$
-Nach der ersten Gleichung ergibt sich die tatsächliche, physikalische Leistung in „W”. Nach der zweiten, in Klammern angegebenen Gleichung hat das Ergebnis die Einheit „$V^{ 2 }$”. Das heißt: Hier ist die Leistung – wie in der Nachrichtentechnik allgemein üblich – auf den Bezugswiderstand $\text{R = 1 Ω}$ umgerechnet. Diese Gleichung muss auch herangezogen werden, um den Effektivwert (die Streuung) σn des Rauschsignals $n(t)$ zu berechnen.
+:$$N = N_0 \cdot B\cdot R = \sigma_n^2 \hspace{0.01cm}.$$
+*According to the first equation,&nbsp; the result is the actual,&nbsp; physical power in "watts"&nbsp; $\rm (W)$.
+*According to the second equation,&nbsp; the result has the unit &nbsp; "$\rm V^{ 2 }$".
+*This means that the power is here converted to the reference resistance &nbsp;$R = 1\ Ω$&nbsp; – as is often the case in Communications Engineering.
+*This equation must also be used to calculate the standard deviation&nbsp; $σ_n$&nbsp; of the noise signal &nbsp;$n(t)$&nbsp;.
-Alle Gleichungen gelten unabhängig davon, ob es sich um Tiefpass– oder Bandpass–Rauschen handelt. Die Grafik zeigt zwei Rauschsignale $n_1(t)$ und $n_2(t)$ bei gleicher Bandbreite. In der Teilaufgabe (d) ist gefragt, welches dieser Signale am Ausgang eines Tiefpasses bzw. eines Bandpasses auftreten wird.
-Die zweiseitige Rauschleistungsdichte von bandbegrenztem Tiefpass–Rauschen $n_{TP}(t)$ lautet:
+All equations apply regardless of whether the noise is low-pass or band-pass.  The diagram shows two noise signals &nbsp;$n_1(t)$&nbsp; and &nbsp;$n_2(t)$&nbsp; of equal bandwidth.&nbsp; Question &nbsp; '''(4)'''&nbsp; asks which of these signals will appear at the output of a low-pass and a band-pass, respectively.
-$$ {\it \Phi}_{n, {\hspace{0.05cm}\rm TP}}(f) = \left\{ \begin{array}{c} N_0/2 \\ 0 \\ \end{array} \right. \begin{array}{*{10}c} {\rm{f\ddot{u}r}} \\ \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < B,} \\ {\rm sonst.} \\ \end{array}$$
-Dagegen gilt bei bandpassartigem Rauschen $n_{BP}(t)$ mit der Mittenfrequenz $f_M$:
+The two-sided noise power-spectral density of band-limited low-pass noise &nbsp;$n_{\rm TP}(t)$&nbsp; is:
-$${\it \Phi}_{n, {\hspace{0.05cm}\rm BP}}(f) = \left\{ \begin{array}{c} N_0/2 \\ 0 \\ \end{array} \right. \begin{array}{*{10}c} {\rm{f\ddot{u}r}} \\ \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} f - f_{\rm M}\hspace{0.05cm} \right| < B/2,} \\ {\rm sonst.} \\ \end{array}.$$
+:$$ {\it \Phi}_{n, {\hspace{0.05cm}\rm LP}}(f) = \left\{ \begin{array}{c} N_0/2 \\ 0 \\ \end{array} \right. \begin{array}{*{10}c} {\rm{for}} \\ \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < B,} \\ {\rm otherwise.} \\ \end{array}$$
-Für alle nachfolgenden numerischen Berechnungen wird vorausgesetzt:
+In contrast,&nbsp; for band-pass noise &nbsp;$n_{\rm BP}(t)$&nbsp; with center frequency &nbsp;$f_{\rm M}$, it holds that:
-$$ F = 10, \hspace{0.2cm}\theta = 290\,{\rm K},\hspace{0.2cm}R = 50\,{\rm \Omega},\hspace{0.2cm}B = 30\,{\rm kHz},\hspace{0.2cm}f_{\rm M} = 0 \hspace{0.1cm}{\rm bzw.}\hspace{0.1cm}100\,{\rm kHz}\hspace{0.05cm}.$$
+:$${\it \Phi}_{n, {\hspace{0.05cm}\rm BP}}(f) = \left\{ \begin{array}{c} N_0/2 \\ 0 \\ \end{array} \right. \begin{array}{*{10}c} {\rm{for}} \\ \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} f - f_{\rm M}\hspace{0.05cm} \right| < B/2,} \\ {\rm otherwise.} \\ \end{array}.$$
-'''Hinweis:''' Die Aufgabe bezieht sich auf die theoretischen Grundlagen von [http://en.lntwww.de/Modulationsverfahren/Qualit%C3%A4tskriterien Kapitel 1.2].
+For all subsequent numerical calculations it is assumed:
-===Fragebogen===
+:$$ F = 10, \hspace{0.2cm}\theta = 290\,{\rm K},\hspace{0.2cm}R = 50\,{\rm \Omega},\hspace{0.2cm}B = 30\,{\rm kHz},\hspace{0.2cm}f_{\rm M} = 0 \hspace{0.5cm}{\rm or}\hspace{0.5cm}f_{\rm M} =100\,{\rm kHz}\hspace{0.05cm}.$$
+Hints:
+*This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Quality_Criteria|Quality Criteria]].
+*Particular reference is made to the page&nbsp;  [[Modulation_Methods/Quality_Criteria#Some_remarks_on_the_AWGN_channel_model|Some remarks on the AWGN channel model]].
+*By specifying the powers in &nbsp;$\rm W$atts&nbsp;, they are independent of the reference resistance &nbsp;$R$,&nbsp; while power with the unit  &nbsp;$\rm V^2$&nbsp; can only be evaluated directly for &nbsp;$R = 1\ \Omega$.
+===Questions===
 <quiz display=simple>
-{Berechnen Sie die Rauschleistungsdichte mit $F = 10$ und $θ = 290$$\text{ K}$.
+{Calculate the noise power-spectral density &nbsp;$N_0$&nbsp; with a noise factor of  &nbsp;$F = 10$&nbsp; and &nbsp;$θ = 290^\circ$&nbsp; Kelvin.
 |type="{}"}
-$N_0$ = { 4 3% }  $10^{ -20 }$ $\text{W/Hz}$
+$N_0 \ = \ $ { 4 3% }  $\ \cdot 10^{ -20 }\ \text{W/Hz}$
-{Wie groß ist die maximale Rauschleistung (ohne Bandbegrenzung)?
+{What is the maximum noise power&nbsp; (without bandwidth limits)?
 |type="{}"}
-$N_{max}$= { 2.4 3% } $10^{ -7 }$ $\text{W}$
+$N_{\rm max} \ = \ $ { 0.24 3% } $\ \cdot 10^{ -6 }\ \text{W/Hz}$
-{Welche Rauschleistung N ergibt sich mit der Bandbreite $B = 30$ $\text{kHz}$? Wie groß ist der Rauscheffektivwert $σ_n$?
+{What is the noise power &nbsp;$N$&nbsp;  with bandwidth &nbsp;$B = 30\text{ kHz}$?&nbsp; What is the standard deviation $σ_n$?
 |type="{}"}
-$N$={ -12 3% }  $10^{ -16 }$ $\text{W}$
+$N \ = \ $ { 12 3% }  $\ \cdot 10^{ -16 }\ \text{W/Hz}$
-$σ_n$={ 0.245 3% } $10^{ -6 }$ $\text{V}$
+$σ_n \ = \ ${ 0.245 3% } $\ \cdot 10^{ -6 }\ \text{V}$
-{Welches der Signale $n_1(t)$ und $n_2(t)$ zeigt TP– und welches BP–Rauschen?
+{Which of the signals &ndash; &nbsp;$n_1(t)$&nbsp; or &nbsp;$n_2(t)$&nbsp; &ndash;  shows low-pass noise and which shows band-pass noise?
-|type="[]"}
+|type="()"}
-+ Das Rauschsignal $n_1(t)$ hat Tiefpass–Charakter.
++ The noise signal &nbsp;$n_1(t)$&nbsp; is characteristically low-pass.
-- Das Rauschsignal $n_1(t)$ hat Bandpass–Charakter.
+- The noise signal &nbsp;$n_1(t)$&nbsp; is characteristically band-pass.
-{Welchen Wert hat die Rauschleistungsdichte des Tiefpass–Rauschens bei der Frequenz $f = 20$ $\text{kHz}$? Es gelte $B = 30$ $\text{kHz}$.
+{What is the value of the noise power-spectral density of the low-pass&nbsp; $\rm (LP)$&nbsp; noise at frequency &nbsp;$f = 20\text{ kHz}$?&nbsp; Let &nbsp;$B = 30\text{ kHz}$.
 |type="{}"}
-$Φ_{n, TP}(f = 20 kHz)$={ -12 3% } $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}$
-{Welchen Wert besitzt die Rauschleistungsdichte des Bandpass–Rauschens bei der Frequenz $f = 120$ $\text{kHz}$? Es gelte $f_M = 100 kHz$ und $B = 30 kHz$.
+{What is the value of the noise power-spectral density of the band-pass&nbsp; $\rm (BP)$&nbsp; noise at  &nbsp;$f = 120\text{ kHz}$?&nbsp; Let &nbsp;$f_{\rm M} = 100\text{ kHz}$&nbsp; and &nbsp;$B = 30\text{ kHz}$.
 |type="{}"}
-$Φ_{n, BP}(f = 120 kHz)$= { 0 3% } $\text{W\Hz}$
+${\it Φ}_{n, \hspace{0.05cm}\rm BP}(f = 120 \ \rm kHz) \ = \ $ { 0. } $\ \cdot 10^{ -12 }\ \text{W/Hz}$
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''1.'''Mit der Boltzmann–Konstante $k_B$ gilt:
+'''(1)'''&nbsp; Using the Boltzmann constant&nbsp; $k_{\rm B}$&nbsp; it holds that:
+:$$N_0 = F \cdot k_{\rm B} \cdot \theta = 10 \cdot
+.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
+^{-20} \hspace{0.05cm}{\rm W}/{\rm Hz}}\hspace{0.05cm}.$$
+'''(2)'''&nbsp; The specified noise power density&nbsp; $N_0$&nbsp; is physically limited to&nbsp; $6$&nbsp; THz.&nbsp;  Thus the maximum noise power is:
+:$$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}\;{\rm
+W}}\hspace{0.05cm}.$$
-$$N_0= F \cdot K_B \cdot \theta = 10 \cdot  1.38 \cdot 10^{ -23 } \frac{Ws}{K} \cdot 290K \approx  4 \cdot 10^{ -20 } W/Hz.$$
+'''(3)'''&nbsp; Now the resulting noise power is:
+:$$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}\;{\rm
+W}}\hspace{0.05cm}.$$
-'''2.'''Die angegebene Rauschleistungsdichte $N_0$ ist physikalisch auf $6 \text{THz}$ begrenzt. Damit beträgt die maximale Rauschleistung:
+* Converting to the reference resistance&nbsp; $R = 1 \ Ω$:
-$$N_{max} = 4 \cdot  10^{ -20 }\frac{ W}{Hz} \cdot 6 \cdot 10^{ 12 } Hz = 0.24 \cdot 10^{-6} W.$$
+:$$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}\cdot
+^{-14}\;{\rm V^2}\hspace{0.05cm}.$$
+[[File:EN_Mod_Z_1_3_e.png|rechts|frame|Power-spectral densities with band-limited noise]]
+*The noise standard deviation $σ_n$&nbsp; is the square root of this:
+:$$\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}.$$
-'''3.'''  Nun ergibt sich für die Rauschleistung:
+'''(4)'''&nbsp; <u>Answer 1</u>&nbsp; is correct:
-$$N=N_0 \cdot B = 4 \cdot 10^{ -20 } \frac{W}{Hz} \cdot 3 \cdot 10^{ 4 } Hz = 12 \cdot 10^{ -16  }W,$$
+*In the random signal&nbsp; $n_2(t)$&nbsp;one can recognize certain regularities similar to a harmonic oscillation &nbsp; &rArr; &nbsp;  it is bandpass noise.
+*In contrast, the signal&nbsp; $n_1(t)$&nbsp; is low-pass noise.
-bzw. umgerechnet auf den Bezugswiderstand $R = 1 Ω$:
-$$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}\cdot 10^{-14}\;{\rm V^2}\hspace{0.05cm}.$$
-Der Rauscheffektivwert $σ_n$ ist die Quadratwurzel hieraus:
-$$\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}.$$
-'''4.'''[[File:P ID954 Mod Z 1 3e neu.png|mini|rechts]]
+'''(5)'''&nbsp; The noise power density of the random signal&nbsp; $n_1(t)$&nbsp; is constant in the frequency range&nbsp; $|f| < 30$&nbsp; kHz:
-Im Zufallssignal $n_2(t)$ erkennt man gewisse Regelmäßigkeiten ähnlich einer harmonischen Schwingung – es ist Bandpass–Rauschen. Dagegen handelt es sich beim Signal $n_1(t)$ um Tiefpass–Rauschen.
+:$${\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}
+^{-12} \hspace{0.05cm}{\rm W}/{\rm Hz}}\hspace{0.05cm}.$$
+*Thus,&nbsp; this value is also valid for the frequency&nbsp; $f = 20$&nbsp; kHz.
-'''5.'''Die Rauschleistungsdichte des Zufallssignals $n_1(t)$ ist im Frequenzbereich $|f| < 30 kHz$ konstant gleich
-$$\phi_{n,TP}(f)= \frac{N_0}{2}= 2 \cdot 10^{-12} W/Hz$$
-Dieser Wert gilt somit auch für die Frequenz $f = 20$ $\text{kHz}$.
-'''6.'''Wie aus der Grafik hervorgeht, ist $Φ_{n,BP}(f)$ nur zwischen $85 kHz$ und $115 kHz$ ungleich 0, wenn die Bandbreite $B = 30 kHz$ beträgt. Bei der Frequenz $f = 120 kHz$ ist die Rauschleistungsdichte somit Null.
+'''(6)'''&nbsp; As can be seen from the diagram, &nbsp; ${\it Φ}_{n, \hspace{0.05cm}\rm BP}(f)$&nbsp; is non-zero only in the range between&nbsp; $85$&nbsp; kHz and&nbsp; $115$&nbsp; kHz,&nbsp; when the bandwidth is&nbsp; $B = 30$&nbsp; kHz.
+*Thus,&nbsp; at the frequency &nbsp; $f = 120$&nbsp; kHz, the noise power density is zero:
+:$${\it Φ}_{n, \hspace{0.05cm}\rm BP}(f = 120 \ \rm kHz)\hspace{0.15cm}\underline{=0}.$$
 {{ML-Fuß}}
@@ Line 89: / Line 126: @@
-[[Category:Aufgaben zu Modulationsverfahren|^1.2 Qualitätskriterien^]]
+[[Category:Modulation Methods: Exercises|^1.2 Quality Criteria^]]