Aufgaben:Exercise 2.6Z: Signal-to-Noise Ratio - Revision history

Hwang at 14:19, 18 January 2023

2023-01-18T14:19:09Z

Bene: Text replacement - "power spectral density" to "power-spectral density"

2022-02-17T11:41:20Z

Text replacement - "power spectral density" to "power-spectral density"

Bene: Text replacement - "power density spectrum" to "power spectral density"

2022-02-17T11:20:27Z

Text replacement - "power density spectrum" to "power spectral density"

Guenter at 13:40, 9 December 2021

2021-12-09T13:40:25Z

Guenter at 13:30, 9 December 2021

2021-12-09T13:30:36Z

Guenter at 13:21, 9 December 2021

2021-12-09T13:21:17Z

Reed at 17:05, 3 December 2021

2021-12-03T17:05:05Z

Reed: /* Musterlösung */

2021-12-03T17:04:48Z

Musterlösung

Reed: /* Musterlösung */

2021-12-03T17:02:31Z

Musterlösung

Reed: /* Questions */

2021-12-03T16:54:32Z

Questions

@@ Line 15: / Line 15: @@
-In the graph,&nbsp; these specifications are shown in the spectral domain.&nbsp;  It should be explicitly mentioned that the power-spectral density &nbsp;${\it Φ}_z(f)$&nbsp; of the cosine oscillation &nbsp;$z(t)$&nbsp;  is composed of two Dirac lines at &nbsp;$±f_{\rm T}$,&nbsp; as in the amplitude spectrum &nbsp; $Z(f)$,&nbsp; but with weight &nbsp;$A^2/4$&nbsp; instead of &nbsp;$A/2$.&nbsp; The amplitude should always be set to &nbsp;$A=1$&nbsp; in this exercise.
+In the graph,&nbsp; these specifications are shown in the spectral domain.&nbsp;  It should be explicitly mentioned that the power-spectral density &nbsp;${\it Φ}_z(f)$&nbsp; of the cosine oscillation &nbsp;$z(t)$&nbsp;  is composed of two Dirac delta lines at &nbsp;$±f_{\rm T}$,&nbsp; as in the amplitude spectrum &nbsp; $Z(f)$,&nbsp; but with weight &nbsp;$A^2/4$&nbsp; instead of &nbsp;$A/2$.&nbsp; The amplitude should always be set to &nbsp;$A=1$&nbsp; in this exercise.
 The sink signal &nbsp;$v(t)$&nbsp; is composed of the useful component &nbsp;$α · q(t)$&nbsp; and the noise component &nbsp;$ε(t)$&nbsp;.&nbsp;  Thus, the general rule for the signal-to-noise power ratio to be determined is:
@@ Line 69: / Line 69: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The power-spectral density of a cosine signal with amplitude &nbsp; $A$&nbsp; consists of two Dirac lines,&nbsp; each with weight &nbsp; $A^2/4$.
+'''(1)'''&nbsp; The power-spectral density of a cosine signal with amplitude &nbsp; $A$&nbsp; consists of two Dirac delta lines,&nbsp; each with weight &nbsp; $A^2/4$.
 *The power is obtained from the integral over the PDS and is thus equal to the sum of the two Dirac weights.&nbsp;
 *Thus,&nbsp; when &nbsp; $A = 4 \ \rm V$,&nbsp; we obtain the power of the source signal:

@@ Line 15: / Line 15: @@
-In the graph,&nbsp; these specifications are shown in the spectral domain.&nbsp;  It should be explicitly mentioned that the power spectral density &nbsp;${\it Φ}_z(f)$&nbsp; of the cosine oscillation &nbsp;$z(t)$&nbsp;  is composed of two Dirac lines at &nbsp;$±f_{\rm T}$,&nbsp; as in the amplitude spectrum &nbsp; $Z(f)$,&nbsp; but with weight &nbsp;$A^2/4$&nbsp; instead of &nbsp;$A/2$.&nbsp; The amplitude should always be set to &nbsp;$A=1$&nbsp; in this exercise.
+In the graph,&nbsp; these specifications are shown in the spectral domain.&nbsp;  It should be explicitly mentioned that the power-spectral density &nbsp;${\it Φ}_z(f)$&nbsp; of the cosine oscillation &nbsp;$z(t)$&nbsp;  is composed of two Dirac lines at &nbsp;$±f_{\rm T}$,&nbsp; as in the amplitude spectrum &nbsp; $Z(f)$,&nbsp; but with weight &nbsp;$A^2/4$&nbsp; instead of &nbsp;$A/2$.&nbsp; The amplitude should always be set to &nbsp;$A=1$&nbsp; in this exercise.
 The sink signal &nbsp;$v(t)$&nbsp; is composed of the useful component &nbsp;$α · q(t)$&nbsp; and the noise component &nbsp;$ε(t)$&nbsp;.&nbsp;  Thus, the general rule for the signal-to-noise power ratio to be determined is:
@@ Line 69: / Line 69: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The power spectral density of a cosine signal with amplitude &nbsp; $A$&nbsp; consists of two Dirac lines,&nbsp; each with weight &nbsp; $A^2/4$.
+'''(1)'''&nbsp; The power-spectral density of a cosine signal with amplitude &nbsp; $A$&nbsp; consists of two Dirac lines,&nbsp; each with weight &nbsp; $A^2/4$.
 *The power is obtained from the integral over the PDS and is thus equal to the sum of the two Dirac weights.&nbsp;
 *Thus,&nbsp; when &nbsp; $A = 4 \ \rm V$,&nbsp; we obtain the power of the source signal:
@@ Line 90: / Line 90: @@
-'''(4)'''&nbsp; The power spectral density of the product &nbsp; $n(t) · z(t)$&nbsp; is obtained by convolving the two power density spectra of &nbsp; $n(t)$&nbsp; and&nbsp; $z(t)$:
+'''(4)'''&nbsp; The power-spectral density of the product &nbsp; $n(t) · z(t)$&nbsp; is obtained by convolving the two power density spectra of &nbsp; $n(t)$&nbsp; and&nbsp; $z(t)$:
 :$$ {\it \Phi}_\varepsilon \hspace{0.01cm} '(f) = {\it \Phi}_n (f) \star {\it \Phi}_{z }(f)= \frac{N_0}{2} \star \left[\delta(f - f_{\rm T}) + \delta(f + f_{\rm T}) \right]= N_0 \hspace{0.05cm}.$$
-*For the power spectral density of the signal &nbsp; $ε(t)$&nbsp; after the low-pass filter,&nbsp; we obtain a rectangular shape with the same value at &nbsp; $f = 0$:
+*For the power-spectral density of the signal &nbsp; $ε(t)$&nbsp; after the low-pass filter,&nbsp; we obtain a rectangular shape with the same value at &nbsp; $f = 0$:
 :$${\it \Phi}_\varepsilon (f) = {\it \Phi}_\varepsilon \hspace{0.01cm} '(f) \cdot |H_{\rm E}(f)|^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\it \Phi}_\varepsilon (f=0)= N_0\hspace{0.15cm}\underline {= 4 \cdot 10^{-19}\,{\rm W/Hz}} \hspace{0.05cm}.$$

@@ Line 15: / Line 15: @@
-In the graph,&nbsp; these specifications are shown in the spectral domain.&nbsp;  It should be explicitly mentioned that the power density spectrum &nbsp;${\it Φ}_z(f)$&nbsp; of the cosine oscillation &nbsp;$z(t)$&nbsp;  is composed of two Dirac lines at &nbsp;$±f_{\rm T}$,&nbsp; as in the amplitude spectrum &nbsp; $Z(f)$,&nbsp; but with weight &nbsp;$A^2/4$&nbsp; instead of &nbsp;$A/2$.&nbsp; The amplitude should always be set to &nbsp;$A=1$&nbsp; in this exercise.
+In the graph,&nbsp; these specifications are shown in the spectral domain.&nbsp;  It should be explicitly mentioned that the power spectral density &nbsp;${\it Φ}_z(f)$&nbsp; of the cosine oscillation &nbsp;$z(t)$&nbsp;  is composed of two Dirac lines at &nbsp;$±f_{\rm T}$,&nbsp; as in the amplitude spectrum &nbsp; $Z(f)$,&nbsp; but with weight &nbsp;$A^2/4$&nbsp; instead of &nbsp;$A/2$.&nbsp; The amplitude should always be set to &nbsp;$A=1$&nbsp; in this exercise.
 The sink signal &nbsp;$v(t)$&nbsp; is composed of the useful component &nbsp;$α · q(t)$&nbsp; and the noise component &nbsp;$ε(t)$&nbsp;.&nbsp;  Thus, the general rule for the signal-to-noise power ratio to be determined is:
@@ Line 69: / Line 69: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The power density spectrum of a cosine signal with amplitude &nbsp; $A$&nbsp; consists of two Dirac lines,&nbsp; each with weight &nbsp; $A^2/4$.
+'''(1)'''&nbsp; The power spectral density of a cosine signal with amplitude &nbsp; $A$&nbsp; consists of two Dirac lines,&nbsp; each with weight &nbsp; $A^2/4$.
 *The power is obtained from the integral over the PDS and is thus equal to the sum of the two Dirac weights.&nbsp;
 *Thus,&nbsp; when &nbsp; $A = 4 \ \rm V$,&nbsp; we obtain the power of the source signal:
@@ Line 90: / Line 90: @@
-'''(4)'''&nbsp; The power density spectrum of the product &nbsp; $n(t) · z(t)$&nbsp; is obtained by convolving the two power density spectra of &nbsp; $n(t)$&nbsp; and&nbsp; $z(t)$:
+'''(4)'''&nbsp; The power spectral density of the product &nbsp; $n(t) · z(t)$&nbsp; is obtained by convolving the two power density spectra of &nbsp; $n(t)$&nbsp; and&nbsp; $z(t)$:
 :$$ {\it \Phi}_\varepsilon \hspace{0.01cm} '(f) = {\it \Phi}_n (f) \star {\it \Phi}_{z }(f)= \frac{N_0}{2} \star \left[\delta(f - f_{\rm T}) + \delta(f + f_{\rm T}) \right]= N_0 \hspace{0.05cm}.$$
-*For the power density spectrum of the signal &nbsp; $ε(t)$&nbsp; after the low-pass filter,&nbsp; we obtain a rectangular shape with the same value at &nbsp; $f = 0$:
+*For the power spectral density of the signal &nbsp; $ε(t)$&nbsp; after the low-pass filter,&nbsp; we obtain a rectangular shape with the same value at &nbsp; $f = 0$:
 :$${\it \Phi}_\varepsilon (f) = {\it \Phi}_\varepsilon \hspace{0.01cm} '(f) \cdot |H_{\rm E}(f)|^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\it \Phi}_\varepsilon (f=0)= N_0\hspace{0.15cm}\underline {= 4 \cdot 10^{-19}\,{\rm W/Hz}} \hspace{0.05cm}.$$

@@ Line 28: / Line 28: @@
 *Particular reference is made to the pages &nbsp;[[Modulation_Methods/Synchronous_Demodulation#Calculating_noise_power|Calculating noise power]]&nbsp; and &nbsp;[[Modulation_Methods/Synchronous_Demodulation#Relationship_between_the_powers_from_source_signal_and_transmitted signal|Relationship between the powers from source and_transmitted signal]].
 *Please note that the variables &nbsp;$α$&nbsp; and &nbsp;$α_{\rm K}$&nbsp; need not be the same.
-*All powers refer to a resistance of &nbsp;$R = 50 \ \rm Ω$&nbsp; with the exception of subtask &nbsp; '''(2)'''.
+*All powers refer to a resistance of &nbsp;$R = 50 \ \rm Ω$&nbsp; with the exception of subtask &nbsp; '''(1)'''.
 *For DSB-AM without carrier,&nbsp; $P_q$&nbsp; also represents the transmit power &nbsp;$P_{\rm S}$.

@@ Line 69: / Line 69: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The power density spectrum of a cosine signal with amplitude &nbsp; $A$&nbsp; consists of two Dirac lines, each with weight &nbsp; $A^2/4$.
+'''(1)'''&nbsp; The power density spectrum of a cosine signal with amplitude &nbsp; $A$&nbsp; consists of two Dirac lines,&nbsp; each with weight &nbsp; $A^2/4$.
-*The power is obtained from the integral over the PDS and is thus equal to the sum of the two Dirac weights.  Thus, when nbsp; $A = 4 \ \rm V$&nbsp;, we obtain the power of the source signal:
+*The power is obtained from the integral over the PDS and is thus equal to the sum of the two Dirac weights.&nbsp;
 :$$ P_q = \frac{A^2}{2} \hspace{0.15cm}\underline {= 8\,{\rm V^2}} \hspace{0.05cm}.$$
-*For the modulation method "DSB-AM without a carrier", this is also the transmit power $P_{\rm S}$ in reference to the unit resistance&nbsp; $1\ \rm  Ω$&nbsp;.
+*For the modulation method&nbsp; "DSB-AM without a carrier",&nbsp; this is also the transmit power $P_{\rm S}$ in reference to the unit resistance&nbsp; $1\ \rm  Ω$.
-'''(2)'''&nbsp; According to the elementary laws of electrical engineering:
+'''(2)'''&nbsp; According to the elementary laws of Electrical Engineering:
 :$$P_q = \frac{8\,{\rm V^2}}{50\,{\Omega}} \hspace{0.15cm}\underline {= 0.16\,{\rm W}} \hspace{0.05cm}.$$
-'''(3)'''&nbsp; In the theory section, it is shown that &nbsp; $v(t) = q(t)$&nbsp; holds under ideal conditions. However, the following should be taken into account:
+'''(3)'''&nbsp; In the theory section,&nbsp; it is shown that&nbsp; $v(t) = q(t)$&nbsp; holds under ideal conditions.&nbsp; However,&nbsp; the following should be taken into account:
-*From the graph, it can be seen that&nbsp; $Z_{\rm E}(f) = Z(f)$&nbsp; holds.&nbsp; Thus, the receiver-side carrier signal &nbsp; $z_{\rm E}(t)$&nbsp;, like &nbsp; $z(t)$&nbsp;, has amplitude &nbsp; $1$.
+*From the graph,&nbsp; it can be seen that&nbsp; $Z_{\rm E}(f) = Z(f)$&nbsp; holds.&nbsp; Thus,&nbsp; the receiver-side carrier signal &nbsp; $z_{\rm E}(t)$&nbsp; has like &nbsp; $z(t)$&nbsp; the amplitude &nbsp; $1$.
-*Ideally, however, the receiver-side carrier signal &nbsp; $z_{\rm E}(t)$&nbsp; should have amplitude &nbsp;  $2$&nbsp;.
+*Ideally,&nbsp; however,&nbsp; the receiver-side carrier signal &nbsp; $z_{\rm E}(t)$&nbsp; should have amplitude &nbsp;  $2$.
-*Therefore, $υ(t) = q(t)/2$ applies here.
+*Therefore,&nbsp; $υ(t) = q(t)/2$&nbsp; applies here.
-*If we further consider the channel attenuation &nbsp; $α_{\rm K} = 10^{–4}$, we obtain the final result:
+*If we further consider the channel attenuation &nbsp; $α_{\rm K} = 10^{–4}$,&nbsp; we obtain the final result:&nbsp; $α\hspace{0.15cm}\underline { = 0.5 · 10^{–4}}.$
-'''(4)'''&nbsp; The power density spectrum of the product &nbsp; $n(t) · z(t)$&nbsp; is obtained by convolving the two power density spectra of nbsp; $n(t)$&nbsp; and&nbsp; $z(t)$:
+'''(4)'''&nbsp; The power density spectrum of the product &nbsp; $n(t) · z(t)$&nbsp; is obtained by convolving the two power density spectra of &nbsp; $n(t)$&nbsp; and&nbsp; $z(t)$:
 :$$ {\it \Phi}_\varepsilon \hspace{0.01cm} '(f) = {\it \Phi}_n (f) \star {\it \Phi}_{z }(f)= \frac{N_0}{2} \star \left[\delta(f - f_{\rm T}) + \delta(f + f_{\rm T}) \right]= N_0 \hspace{0.05cm}.$$
-*For the power density spectrum of the signal &nbsp; $ε(t)$&nbsp; after the low-pass filter, we obtain a rectangular shape with the same value at &nbsp; $f = 0$:
+*For the power density spectrum of the signal &nbsp; $ε(t)$&nbsp; after the low-pass filter,&nbsp; we obtain a rectangular shape with the same value at &nbsp; $f = 0$:
 :$${\it \Phi}_\varepsilon (f) = {\it \Phi}_\varepsilon \hspace{0.01cm} '(f) \cdot |H_{\rm E}(f)|^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\it \Phi}_\varepsilon (f=0)= N_0\hspace{0.15cm}\underline {= 4 \cdot 10^{-19}\,{\rm W/Hz}} \hspace{0.05cm}.$$

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1017__Mod_Z_2_6.png|right|frame|Diverse Power Density Spectra]]
+[[File:P_ID1017__Mod_Z_2_6.png|right|frame|Spectra and power density spectra]]
-In the following exercise, we assume:
+In the following exercise,&nbsp; we assume:
 *a cosine source signal:
 :$$ q(t) = 4 \,{\rm V} \cdot \cos(2 \pi \cdot 5\,{\rm kHz} \cdot t )\hspace{0.05cm},$$
@@ Line 10: / Line 10: @@
 :$$z(t) = 1 \cdot \cos(2 \pi \cdot 20\,{\rm kHz} \cdot t )\hspace{0.05cm},$$
 * a frequency-independent attenuation on the channel corresponding to &nbsp;$α_{\rm K} = 10^{–4}$,
-* additive white input noise with noise power density &nbsp;$N_0 = 4 · 10^{–19} \ \rm W/Hz$,
+* additive white input noise with power density &nbsp;$N_0 = 4 · 10^{–19} \ \rm W/Hz$,
-* phase- and frequency-synchronous demodulation by multiplication with the same &nbsp;$z(t)$&nbsp; as at the transmitter,
+* phase-synchronous and frequency-synchronous demodulation by multiplication the same &nbsp;$z(t)$&nbsp; as at the transmitter,
 * a rectangular low-pass at the synchronous demodulator with cutoff frequency &nbsp;$f_{\rm E} = 5 \ \rm kHz$.
-In the graph, these specifications are shown in the spectral domain.  It should be explicitly mentioned that the power density spectrum &nbsp;${\it Φ}_z(f)$&nbsp; of the cosine oscillation &nbsp;$z(t)$&nbsp;  is composed of two Dirac lines at &nbsp;$±f_{\rm T}$&nbsp; zusammensetzt, as is the amplitude spectrum &nbsp; $Z(f)$&nbsp;, but with weight &nbsp;$A^2/4$&nbsp; instead of &nbsp;$A/2$.&nbsp; The amplitude should always be set to &nbsp;$A=1$&nbsp; in this exercise.
+In the graph,&nbsp; these specifications are shown in the spectral domain.&nbsp;  It should be explicitly mentioned that the power density spectrum &nbsp;${\it Φ}_z(f)$&nbsp; of the cosine oscillation &nbsp;$z(t)$&nbsp;  is composed of two Dirac lines at &nbsp;$±f_{\rm T}$,&nbsp; as in the amplitude spectrum &nbsp; $Z(f)$,&nbsp; but with weight &nbsp;$A^2/4$&nbsp; instead of &nbsp;$A/2$.&nbsp; The amplitude should always be set to &nbsp;$A=1$&nbsp; in this exercise.
 The sink signal &nbsp;$v(t)$&nbsp; is composed of the useful component &nbsp;$α · q(t)$&nbsp; and the noise component &nbsp;$ε(t)$&nbsp;.&nbsp;  Thus, the general rule for the signal-to-noise power ratio to be determined is:
 :$$ \rho_{v } = \frac{\alpha^2 \cdot P_q}{P_\varepsilon}\hspace{0.05cm}.$$
-This important quality criterion is often abbreviated to&nbsp; '''SNR'''&nbsp; (&nbsp; ''Signal–to–Noise–Ratio'').
+This important quality criterion is often abbreviated to&nbsp; $\rm SNR$&nbsp; ("signal–to–noise power ratio").
 Hints:
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Synchronous_Demodulation|Synchronous Demodulation]].
-*Particular reference is made to the pages &nbsp;[[Modulation_Methods/Synchronous_Demodulation#Calculating_noise_power|Calculating noise power]]&nbsp; and &nbsp;[[Modulation_Methods/Synchronous_Demodulation#Relationship_between_.7F.27.22.60UNIQ-MathJax167-QINU.60.22.27.7F_and_.7F.27.22.60UNIQ-MathJax168-QINU.60.22.27.7F|Relationship between $P_q$ &nbsp;and&nbsp; $P_{\rm S}$]].
+*Particular reference is made to the pages &nbsp;[[Modulation_Methods/Synchronous_Demodulation#Calculating_noise_power|Calculating noise power]]&nbsp; and &nbsp;[[Modulation_Methods/Synchronous_Demodulation#Relationship_between_the_powers_from_source_signal_and_transmitted signal|Relationship between the powers from source and_transmitted signal]].
 *Please note that the variables &nbsp;$α$&nbsp; and &nbsp;$α_{\rm K}$&nbsp; need not be the same.
-*All powers refer to a resistance of &nbsp;$R = 50 \ \rm Ω$ with the exception of subtask &nbsp; '''(1)'''&nbsp;.
+*All powers refer to a resistance of &nbsp;$R = 50 \ \rm Ω$&nbsp; with the exception of subtask &nbsp; '''(2)'''.
-*For DSB-AM without a carrier, $P_q$&nbsp; also represents the transmit power &nbsp;$P_{\rm S}$&nbsp;.
+*For DSB-AM without carrier,&nbsp; $P_q$&nbsp; also represents the transmit power &nbsp;$P_{\rm S}$.
@@ Line 49: / Line 47: @@
-{Which damping factor &nbsp;$α$&nbsp; results for the whole system?
+{Which attenuation factor &nbsp;$α$&nbsp; results for the whole system?
 |type="{}"}
 $α  \ = \ $ { 0.5 3% } $\ \cdot 10^{-4}$

@@ Line 92: / Line 92: @@
-'''(4)'''&nbsp; Das Leistungsdichtespektrum des Produktes&nbsp; $n(t) · z(t)$&nbsp; ergibt sich aus der Faltung der beiden Leistungsdichtespektren von&nbsp; $n(t)$&nbsp; und&nbsp; $z(t)$:
+'''(4)'''&nbsp; The power density spectrum of the product &nbsp; $n(t) · z(t)$&nbsp; is obtained by convolving the two power density spectra of nbsp; $n(t)$&nbsp; and&nbsp; $z(t)$:
 :$$ {\it \Phi}_\varepsilon \hspace{0.01cm} '(f) = {\it \Phi}_n (f) \star {\it \Phi}_{z }(f)= \frac{N_0}{2} \star \left[\delta(f - f_{\rm T}) + \delta(f + f_{\rm T}) \right]= N_0 \hspace{0.05cm}.$$
-*Für das Leistungsdichtespektrum des Signals&nbsp; $ε(t)$&nbsp; nach dem Tiefpass erhält man eine Rechteckform mit dem gleichen Wert bei&nbsp; $f = 0$:
+*For the power density spectrum of the signal &nbsp; $ε(t)$&nbsp; after the low-pass filter, we obtain a rectangular shape with the same value at &nbsp; $f = 0$:
 :$${\it \Phi}_\varepsilon (f) = {\it \Phi}_\varepsilon \hspace{0.01cm} '(f) \cdot |H_{\rm E}(f)|^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\it \Phi}_\varepsilon (f=0)= N_0\hspace{0.15cm}\underline {= 4 \cdot 10^{-19}\,{\rm W/Hz}} \hspace{0.05cm}.$$
-'''(5)'''&nbsp;  Die Rauschleistung ist das Integral über die Rauschleistungsdichte:
+'''(5)'''&nbsp; The noise power is the integral over the noise power density:
 :$$ P_{\varepsilon} = \int_{-f_{\rm E}}^{ + f_{\rm E}} {{\it \Phi}_\varepsilon (f)}\hspace{0.1cm}{\rm d}f = N_0 \cdot 2 f_{\rm E} = 4 \cdot 10^{-19}\,\frac{ \rm W}{\rm  Hz} \cdot 10^{4}\,{\rm Hz} \hspace{0.15cm}\underline {= 4 \cdot 10^{-15}\,{\rm W}}\hspace{0.05cm}.$$
-'''(6)'''&nbsp; Aus den Ergebnissen der Teilaufgaben&nbsp; '''(2)''',&nbsp; '''(3)'''&nbsp; und&nbsp; '''(5)'''&nbsp; folgt:
+'''(6)'''&nbsp; From the results of subtasks&nbsp; '''(2)''',&nbsp; '''(3)'''&nbsp; and&nbsp; '''(5)'''&nbsp; it follows that:
 :$$\rho_{v } = \frac{\alpha^2 \cdot P_q}{P_\varepsilon} = \frac{(0.5 \cdot 10^{-4})^2 \cdot 0.16\,{\rm W}}{4 \cdot 10^{-15}\,{\rm W}} \hspace{0.15cm}\underline {= 100000} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg }\hspace{0.1cm}\rho_{v } \hspace{0.15cm}\underline {= 50\,{\rm dB}}\hspace{0.05cm}.$$

@@ Line 39: / Line 39: @@
 <quiz display=simple>
-{Berechnen Sie die Sendeleistung bezogen auf den Einheitswiderstand &nbsp;$R = 1 \ \rm Ω$.
+{Calculate the transmit power with respect to the unit resistance &nbsp;$R = 1 \ \rm Ω$.
 |type="{}"}
 $P_q \ = \ $ { 8 3% } $\ \rm V^2$
-{Wie groß ist die Leistung &nbsp;$P_q$&nbsp; in „Watt” für den Widerstand &nbsp;$R = 50  \ \rm  Ω$?
+{What is the power &nbsp;$P_q$&nbsp; in watts for the resistor &nbsp;$R = 50  \ \rm  Ω$?
 |type="{}"}
 $P_q \ = \ $ { 0.16 3% } $\ \rm W$
-{Welcher Dämpfungsfaktor &nbsp;$α$&nbsp; ergibt sich für das Gesamtsystem?
+{Which damping factor &nbsp;$α$&nbsp; results for the whole system?
 |type="{}"}
 $α  \ = \ $ { 0.5 3% } $\ \cdot 10^{-4}$
-{Berechnen Sie die Leistungsdichte der Rauschkomponente &nbsp;$ε(t)$&nbsp; am Ausgang.&nbsp; Wie groß ist der Wert bei &nbsp;$f = 0$?&nbsp; Es gelte &nbsp;$H_{\rm E}(f = 0) = 1$.
+{Calculate the power density of the noise component &nbsp;$ε(t)$&nbsp; at the output.&nbsp; What is the value when &nbsp;$f = 0$?&nbsp; Let &nbsp;$H_{\rm E}(f = 0) = 1$.
 |type="{}"}
 ${\it Φ}_ε(f = 0) \ = \ $ { 4 3% } $\ \cdot 10^{-19} \ \rm   W/Hz$
-{Wie groß ist die Rauschleistung im Sinkensignal?
+{What is the noise power of the sink signal?
 |type="{}"}
 $P_ε \ = \ $ { 4 3% } $\ \cdot 10^{-15} \ \rm   W$
-{Wie groß ist das Signal–zu–Rausch–Leistungsverhältnis (SNR) an der Sinke?&nbsp; Welcher dB–Wert ergibt sich daraus?
+{What is the signal-to-noise power ratio (SNR) at the sink?  What is the resulting dB value?
 |type="{}"}
 $ρ_v \ = \ $ { 100000 3% }