Aufgaben:Exercise 2.6: Free Space Attenuation - Revision history

Guenter at 16:38, 8 December 2021

2021-12-08T16:38:19Z

Guenter at 16:23, 8 December 2021

2021-12-08T16:23:33Z

Reed at 16:37, 3 December 2021

2021-12-03T16:37:44Z

Reed: /* Solution */

2021-12-03T16:37:22Z

Solution

Reed at 16:23, 3 December 2021

2021-12-03T16:23:34Z

Reed at 16:02, 3 December 2021

2021-12-03T16:02:30Z

Reed at 16:01, 3 December 2021

2021-12-03T16:01:45Z

Guenter at 14:52, 15 November 2021

2021-11-15T14:52:51Z

Guenter: Guenter moved page Aufgaben:Aufgabe 2.6: Freiraumdämpfung to Aufgaben:Exercise 2.6: Free Space Attenuation

2021-11-15T14:44:42Z

Guenter moved page Aufgabe 2.6: Freiraumdämpfung to Exercise 2.6: Free Space Attenuation

Javier: Text replacement - "Category:Aufgaben zu Modulationsverfahren" to "Category:Modulation Methods: Exercises"

2021-03-23T13:01:39Z

Text replacement - "Category:Aufgaben zu Modulationsverfahren" to "Category:Modulation Methods: Exercises"

@@ Line 63: / Line 63: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; According to the equation for free space attenuation, when &nbsp; $d = 10\ \rm  km$&nbsp; and&nbsp; $f_{\rm T} = 20 \ \rm  MHz$, then:
+'''(1)'''&nbsp; According to the equation for free space attenuation,&nbsp; when &nbsp; $d = 10\ \rm  km$&nbsp; and&nbsp; $f_{\rm T} = 20 \ \rm  MHz$,&nbsp; then:
 :$$\frac{a_{\rm K}(d, f_{\rm T})}{\rm dB}  =  34 + 20 \cdot {\rm lg }\hspace{0.1cm}\frac{d}{\rm km} + 20 \cdot {\rm lg }\hspace{0.1cm}\frac{f_{\rm T}}{\rm MHz}=  34 + 20 \cdot {\rm lg }\hspace{0.1cm}(10) + 20 \cdot {\rm lg }\hspace{0.1cm}(20)\approx 80\hspace{0.1cm}{\rm dB} \hspace{0.05cm}.$$
-*This corresponds to a power reduction by a factor of &nbsp; $10^{8}$:
+*This corresponds to a power reduction by a factor of&nbsp; $10^{8}$:
 :$$P_{\rm E}= 10^{-8} \cdot P_{\rm S}= 10^{-8} \cdot 100\,{\rm kW}\hspace{0.15cm}\underline {= 1\, {\rm mW} \hspace{0.05cm}}.$$
-'''(2)'''&nbsp; From&nbsp; $P_{\rm S} = 10^5 \ \rm  W$, $P_{\rm E} = 10{^–4}\ \rm   W$&nbsp; follows a free space attenuation of&nbsp; $90 \ \rm  dB$.&nbsp; From this, we further obtain:
+'''(2)'''&nbsp; From&nbsp; $P_{\rm S} = 10^5 \ \rm  W$,&nbsp; $P_{\rm E} = 10{^–4}\ \rm   W$&nbsp; follows a free space attenuation of&nbsp; $90 \ \rm  dB$.&nbsp; From this,&nbsp; we further obtain:
 :$$20 \cdot {\rm lg }\hspace{0.1cm}\frac{d}{\rm km} = ( 90-34 - 26)\hspace{0.1cm}{\rm dB}= 30\,{\rm dB}\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} d = 10^{1.5}\,{\rm km}\hspace{0.15cm}\underline { = 31.6\,{\rm km}\hspace{0.05cm}}.$$
@@ Line 76: / Line 76: @@
-'''(3)'''&nbsp; For DSB–AM without a carrier, that is, for a modulation depth&nbsp; $m → ∞$, the following would hold:
+'''(3)'''&nbsp; For DSB–AM without carrier,&nbsp; that is,&nbsp; for a modulation depth&nbsp; $m → ∞$,&nbsp; the following would hold:
 :$$ \rho_{v } = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{{N_0} \cdot B_{\rm NF}} = \frac{ P_{\rm E}}{{N_0} \cdot B_{\rm NF}}= \frac{10^{-4}\,{\rm W}}{10^{-14}\,{\rm W/Hz}\cdot 8 \cdot 10^{3}\,{\rm Hz} } = 1.25 \cdot 10^6\hspace{0.3cm}
   \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg }\hspace{0.1cm}\rho_{v } \approx 61\,{\rm dB}\hspace{0.05cm}.$$
-*With modulation depth&nbsp; $m = 0.5$&nbsp; the sink SNR becomes smaller by a factor of&nbsp; $[1 +{2}/{m^2}]^{-1} = {1}/{9}$&nbsp;.&nbsp; Thus, the signal-to-noise ratio at the sink is also smaller:
+*With modulation depth&nbsp; $m = 0.5$&nbsp; the sink SNR becomes smaller by a factor of&nbsp; $[1 +{2}/{m^2}]^{-1} = {1}/{9}$&nbsp;.&nbsp; Thus,&nbsp; the signal-to-noise ratio at the sink is also smaller:
 :$$ 10 \cdot {\rm lg }\hspace{0.1cm}\rho_{v } = 61\,{\rm dB}- 10 \cdot {\rm lg }\hspace{0.1cm}(9) \hspace{0.15cm}\underline {\approx 51.5\,{\rm dB}\hspace{0.05cm}}.$$
-'''(4)'''&nbsp; According to the calculations in subtask&nbsp; '''(3)'''&nbsp;, the following condition must be satisfied:
+'''(4)'''&nbsp; According to the calculations in subtask&nbsp; '''(3)''',&nbsp; the following condition must be satisfied:
 :$$ 10 \cdot {\rm lg }\hspace{0.1cm}\left({1 + {2}/{m^2}}\right) < 1\,{\rm dB}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} 1 +{2}/{m^2} < 10^{0.1}=1.259
 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{2}/{m^2} < 0.259 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} m > \sqrt{8}\approx 2.83 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} m_{\rm min} \hspace{0.15cm}\underline {= 2.83} \hspace{0.05cm}.$$
@@ Line 90: / Line 90: @@
-'''(5)'''&nbsp; <u>Answers 1 and 3</u> are correct:
+'''(5)'''&nbsp; <u>Answers 1 and 3</u>&nbsp; are correct:
 *When using a synchronous demodulator, the addition of the carrier makes no sense unless the former is useful for the required carrier recovery.
-*Since the carrier cannot be used for demodulation, only a fraction of the transmit power is available for demodulation &nbsp; $($one third for &nbsp; $m = 1$, one ninth for&nbsp;  $m = 0.5)$.
+*Since the carrier cannot be used for demodulation,&nbsp; only a fraction of the transmit power is available for demodulation &nbsp; $($one third for &nbsp; $m = 1$,&nbsp; one ninth for&nbsp;  $m = 0.5)$.
 {{ML-Fuß}}

@@ Line 4: / Line 4: @@
 [[File:P_ID1016__Mod_A_2_6.jpg|right|frame|Photo of a transmitter]]
-A shortwave transmitter operated according to the modulation method "DSB-AM with carrier" works with carrier frequency &nbsp;$f_{\rm T} = 20 \ \rm MHz$&nbsp; and transmit power&nbsp;$P_{\rm S} = 100\ \rm  kW$.&nbsp; It is designed for a bandwidth of &nbsp;$B_{\rm NF} = 8 \ \rm kHz$&nbsp;.
+A shortwave transmitter operated according to the modulation method&nbsp; "DSB-AM with carrier" works with carrier frequency &nbsp;$f_{\rm T} = 20 \ \rm MHz$&nbsp; and transmit power&nbsp;$P_{\rm S} = 100\ \rm  kW$.&nbsp; It is designed for a low-frequency bandwidth of &nbsp;$B_{\rm NF} = 8 \ \rm kHz$.
+For test operation,&nbsp; a mobile receiver is used, which operates with a synchronous demodulator.&nbsp;  If this is located at distance &nbsp; $d$&nbsp; from the transmitter,&nbsp; the attenuation function of the transmission channel can be approximated as follows:
 :$$\frac{a_{\rm K}(d, f)}{\rm dB} = 34 + 20 \cdot {\rm lg }\hspace{0.2cm}\frac{d}{\rm km} + 20 \cdot {\rm lg }\hspace{0.2cm}\frac{f}{\rm MHz}
   \hspace{0.05cm}.$$
-This equation describes so-called&nbsp; '''free space attenuation''', which also depends on the (carrier) frequency.
+This equation describes so-called&nbsp; '''free space attenuation''',&nbsp; which also depends on the (carrier) frequency.
+It can be assumed that the entire DSB-AM spectrum is attenuated like the carrier frequency.&nbsp;  This means that
 *the slightly larger attenuation of the upper sideband (USB), and
 *the slightly smaller attenuation of the lower sideband (LSB)
@@ Line 23: / Line 21: @@
-For the first two subtasks, it is assumed that the transmitter transmits only the carrier, which is equivalent to the modulation depth being &nbsp;$m = 0$&nbsp;.
+For the first two subtasks,&nbsp; it is assumed that the transmitter transmits only the carrier,&nbsp; which is equivalent to the modulation depth being &nbsp;$m = 0$.
@@ Line 33: / Line 28: @@
-''Hints:''
+Hints:
 *This exercise belongs to the chapter &nbsp; [[Modulation_Methods/Synchronous_Demodulation|Synchronous Demodulation]].
 *Particular reference is made to the page&nbsp;  [[Modulation_Methods/Synchronous_Demodulation#Sink_SNR_and_the_performance_parameter|Sink SNR and the performance parameter]].
@@ Line 50: / Line 45: @@
 $d \ = \ $ { 31.6 3% } $\ \rm km$
-{Which sink SNR results from the distance &nbsp;$d$ calculated in subtask&nbsp; '''(2)'''&nbsp; when the modulation depth is &nbsp;$m = 0.5$&nbsp;?
+{Which sink SNR results from the distance &nbsp;$d$&nbsp; calculated in subtask&nbsp; '''(2)'''&nbsp; when the modulation depth is &nbsp;$m = 0.5$&nbsp;?
 |type="{}"}
 $10 · \lg ρ_v \ = \ $  { 51.5 3% } $\ \text{dB}$
@@ Line 60: / Line 55: @@
 {Which of the following statements are true?
 |type="[]"}
-+ DSB–AM with carrier does not make sense for energy reasons if a synchronous demodulator is used.
++ "DSB–AM with carrier"&nbsp; does not make sense for energy reasons if a synchronous demodulator is used.
-- DSB–AM without carrier does not make sense for energy reasons if a synchronous demodulator is used.
+- "DSB–AM without carrier"&nbsp; does not make sense for energy reasons if a synchronous demodulator is used.
 + A small carrier component can be helpful for the required frequency and phase synchronization.

@@ Line 97: / Line 97: @@
 '''(5)'''&nbsp; <u>Answers 1 and 3</u> are correct:
 *When using a synchronous demodulator, the addition of the carrier makes no sense unless the former is useful for the required carrier recovery.
-*Since the carrier cannot be used for demodulation, only a fraction of the transmit power is available for demodulation &nbsp; $($one third for &nbsp; $m = 1$, one ninth for&nbsp;  $m = 0.5$)$.
+*Since the carrier cannot be used for demodulation, only a fraction of the transmit power is available for demodulation &nbsp; $($one third for &nbsp; $m = 1$, one ninth for&nbsp;  $m = 0.5)$.
 {{ML-Fuß}}

@@ Line 70: / Line 70: @@
 '''(1)'''&nbsp; According to the equation for free space attenuation, when &nbsp; $d = 10\ \rm  km$&nbsp; and&nbsp; $f_{\rm T} = 20 \ \rm  MHz$, then:
 :$$\frac{a_{\rm K}(d, f_{\rm T})}{\rm dB}  =  34 + 20 \cdot {\rm lg }\hspace{0.1cm}\frac{d}{\rm km} + 20 \cdot {\rm lg }\hspace{0.1cm}\frac{f_{\rm T}}{\rm MHz}=  34 + 20 \cdot {\rm lg }\hspace{0.1cm}(10) + 20 \cdot {\rm lg }\hspace{0.1cm}(20)\approx 80\hspace{0.1cm}{\rm dB} \hspace{0.05cm}.$$
-*Dies entspricht einer Leistungsverminderung um den Faktor&nbsp; $10^{8}$:
+*This corresponds to a power reduction by a factor of &nbsp; $10^{8}$:
 :$$P_{\rm E}= 10^{-8} \cdot P_{\rm S}= 10^{-8} \cdot 100\,{\rm kW}\hspace{0.15cm}\underline {= 1\, {\rm mW} \hspace{0.05cm}}.$$
-'''(2)'''&nbsp; Aus&nbsp; $P_{\rm S} = 10^5 \ \rm  W$, $P_{\rm E} = 10{^–4}\ \rm   W$&nbsp; folgt eine Freiraumdämpfung von&nbsp; $90 \ \rm  dB$.&nbsp; Daraus erhält man weiter:
+'''(2)'''&nbsp; From&nbsp; $P_{\rm S} = 10^5 \ \rm  W$, $P_{\rm E} = 10{^–4}\ \rm   W$&nbsp; follows a free space attenuation of&nbsp; $90 \ \rm  dB$.&nbsp; From this, we further obtain:
 :$$20 \cdot {\rm lg }\hspace{0.1cm}\frac{d}{\rm km} = ( 90-34 - 26)\hspace{0.1cm}{\rm dB}= 30\,{\rm dB}\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} d = 10^{1.5}\,{\rm km}\hspace{0.15cm}\underline { = 31.6\,{\rm km}\hspace{0.05cm}}.$$
@@ Line 81: / Line 81: @@
-'''(3)'''&nbsp; Bei ZSB–AM ohne Träger, das heißt für den Modulationsgrad&nbsp; $m → ∞$, würde gelten:
+'''(3)'''&nbsp; For DSB–AM without a carrier, that is, for a modulation depth&nbsp; $m → ∞$, the following would hold:
 :$$ \rho_{v } = \frac{\alpha_{\rm K}^2 \cdot P_{\rm S}}{{N_0} \cdot B_{\rm NF}} = \frac{ P_{\rm E}}{{N_0} \cdot B_{\rm NF}}= \frac{10^{-4}\,{\rm W}}{10^{-14}\,{\rm W/Hz}\cdot 8 \cdot 10^{3}\,{\rm Hz} } = 1.25 \cdot 10^6\hspace{0.3cm}
   \Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg }\hspace{0.1cm}\rho_{v } \approx 61\,{\rm dB}\hspace{0.05cm}.$$
-*Mit dem Modulationsgrad&nbsp; $m = 0.5$&nbsp; wird das Sinken–SNR um den Faktor&nbsp; $[1 +{2}/{m^2}]^{-1} = {1}/{9}$&nbsp; kleiner.&nbsp; Der Sinken–Störabstand ist somit ebenfalls geringer:
+*With modulation depth&nbsp; $m = 0.5$&nbsp; the sink SNR becomes smaller by a factor of&nbsp; $[1 +{2}/{m^2}]^{-1} = {1}/{9}$&nbsp;.&nbsp; Thus, the signal-to-noise ratio at the sink is also smaller:
 :$$ 10 \cdot {\rm lg }\hspace{0.1cm}\rho_{v } = 61\,{\rm dB}- 10 \cdot {\rm lg }\hspace{0.1cm}(9) \hspace{0.15cm}\underline {\approx 51.5\,{\rm dB}\hspace{0.05cm}}.$$
-'''(4)'''&nbsp; Entsprechend den Berechnungen zur Teilaufgabe&nbsp; '''(3)'''&nbsp; muss nun folgende Bedingung erfüllt sein:
+'''(4)'''&nbsp; According to the calculations in subtask&nbsp; '''(3)'''&nbsp;, the following condition must be satisfied:
 :$$ 10 \cdot {\rm lg }\hspace{0.1cm}\left({1 + {2}/{m^2}}\right) < 1\,{\rm dB}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} 1 +{2}/{m^2} < 10^{0.1}=1.259
 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{2}/{m^2} < 0.259 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} m > \sqrt{8}\approx 2.83 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} m_{\rm min} \hspace{0.15cm}\underline {= 2.83} \hspace{0.05cm}.$$
@@ Line 95: / Line 95: @@
-'''(5)'''&nbsp; Richtig sind die <u>Vorschläge 1 und 3</u>:
+'''(5)'''&nbsp; <u>Answers 1 and 3</u> are correct:
-*Bei Verwendung eines Synchrondemodulators macht die Zusetzung des Trägers keinen Sinn, außer, dieser ist für die erforderliche Trägerrückgewinnung nützlich.
+*When using a synchronous demodulator, the addition of the carrier makes no sense unless the former is useful for the required carrier recovery.
-*Da der Träger zur Demodulation nicht genutzt werden kann, steht nur ein Bruchteil der Sendeleistung für die Demodulation zur Verfügung&nbsp; $($ein Drittel bei&nbsp; $m = 1$, ein Neuntel bei&nbsp;  $m = 0.5)$.
+*Since the carrier cannot be used for demodulation, only a fraction of the transmit power is available for demodulation &nbsp; $($one third for &nbsp; $m = 1$, one ninth for&nbsp;  $m = 0.5$)$.
 {{ML-Fuß}}

@@ Line 42: / Line 42: @@
 <quiz display=simple>
-{ Welche Leistung wird im Abstand &nbsp;$d = 10 \ \rm km$&nbsp; vom Sender empfangen, wenn nur der Träger abgestrahlt wird &nbsp;$(m = 0)$?
+{ What power is received at a distance &nbsp;$d = 10 \ \rm km$&nbsp; from the transmitter when only the carrier is transmitted &nbsp;$(m = 0)$?
 |type="{}"}
 $P_{\rm E} \ = \ $ { 1 3% } $\ \rm mW$
-{ In welcher Entfernung &nbsp;$d$&nbsp; vom Sender befindet sich der Empfänger, wenn die empfangene Leistung &nbsp;$P_{\rm E} = 100 \ \rm &micro; W$ beträgt??
+{ At what distance &nbsp;$d$&nbsp; from the transmitter is the receiver located when the received power is &nbsp;$P_{\rm E} = 100 \ \rm &micro; W$??
 |type="{}"}
 $d \ = \ $ { 31.6 3% } $\ \rm km$
-{Welches Sinken–SNR ergibt sich bei der unter&nbsp; '''(2)'''&nbsp; berechneten Distanz &nbsp;$d$, wenn der Modulationsgrad &nbsp;$m = 0.5$&nbsp; beträgt?
+{Which sink SNR results from the distance &nbsp;$d$ calculated in subtask&nbsp; '''(2)'''&nbsp; when the modulation depth is &nbsp;$m = 0.5$&nbsp;?
 |type="{}"}
 $10 · \lg ρ_v \ = \ $  { 51.5 3% } $\ \text{dB}$
-{Wie groß muss der Modulationsgrad &nbsp;$m$&nbsp; mindestens gewählt werden, damit sich ein Sinken–Störabstand von &nbsp;$60  \ \rm dB$&nbsp; ergibt?
+{What is the minimum modulation depth &nbsp;$m$&nbsp; that can be chosen for a resulting sink-to-noise ratio of &nbsp;$60  \ \rm dB$&nbsp;?
 |type="{}"}
 $m_{\min} \ = \ $ { 2.83 5% }
-{Welche der folgenden Aussagen sind zutreffend?
+{Which of the following statements are true?
 |type="[]"}
-+ ZSB–AM mit Träger macht aus energetischen Gründen keinen Sinn, wenn ein Synchrondemodulator verwendet wird.
++ DSB–AM with carrier does not make sense for energy reasons if a synchronous demodulator is used.
-- ZSB–AM ohne Träger macht aus energetischen Gründen keinen Sinn, wenn ein Synchrondemodulator verwendet wird.
+- DSB–AM without carrier does not make sense for energy reasons if a synchronous demodulator is used.
-+ Ein kleiner Trägeranteil kann für die erforderliche Frequenz– und Phasensynchronisation hilfreich sein.
++ A small carrier component can be helpful for the required frequency and phase synchronization.
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Entsprechend der Gleichung für die Freiraumdämpfung gilt mit&nbsp; $d = 10\ \rm  km$&nbsp; und&nbsp; $f_{\rm T} = 20 \ \rm  MHz$:
+'''(1)'''&nbsp; According to the equation for free space attenuation, when &nbsp; $d = 10\ \rm  km$&nbsp; and&nbsp; $f_{\rm T} = 20 \ \rm  MHz$, then:
 :$$\frac{a_{\rm K}(d, f_{\rm T})}{\rm dB}  =  34 + 20 \cdot {\rm lg }\hspace{0.1cm}\frac{d}{\rm km} + 20 \cdot {\rm lg }\hspace{0.1cm}\frac{f_{\rm T}}{\rm MHz}=  34 + 20 \cdot {\rm lg }\hspace{0.1cm}(10) + 20 \cdot {\rm lg }\hspace{0.1cm}(20)\approx 80\hspace{0.1cm}{\rm dB} \hspace{0.05cm}.$$
 *Dies entspricht einer Leistungsverminderung um den Faktor&nbsp; $10^{8}$:

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1016__Mod_A_2_6.jpg|right|frame|Foto einer Sendeanlage]]
+[[File:P_ID1016__Mod_A_2_6.jpg|right|frame|Photo of a transmitter]]
-Ein gemäß dem Modulationsverfahren&nbsp; „ZSB–AM mit Träger”&nbsp; betriebener Kurzwellensender arbeitet mit der Trägerfrequenz &nbsp;$f_{\rm T} = 20 \ \rm MHz$&nbsp; und der Sendeleistung &nbsp;$P_{\rm S} = 100\ \rm  kW$.&nbsp; Er ist für eine Bandbreite von &nbsp;$B_{\rm NF} = 8 \ \rm kHz$&nbsp; ausgelegt.
+A shortwave transmitter operated according to the modulation method "DSB-AM with carrier" works with carrier frequency &nbsp;$f_{\rm T} = 20 \ \rm MHz$&nbsp; and transmit power&nbsp;$P_{\rm S} = 100\ \rm  kW$.&nbsp; It is designed for a bandwidth of &nbsp;$B_{\rm NF} = 8 \ \rm kHz$&nbsp;.
-Zum Testbetrieb wird ein mobiler Empfänger eingesetzt, der mit einem Synchrondemodulator arbeitet.&nbsp; Befindet sich dieser in der Distanz&nbsp; $d$&nbsp; zum Sender, so kann die Dämpfungsfunktion des Übertragungskanals wie folgt angenähert werden:
+For test operation, a mobile receiver is used, which operates with a synchronous demodulator.  If this is located at distance &nbsp; $d$&nbsp; from the transmitter, the attenuation function of the transmission channel can be approximated as follows:
 :$$\frac{a_{\rm K}(d, f)}{\rm dB} = 34 + 20 \cdot {\rm lg }\hspace{0.2cm}\frac{d}{\rm km} + 20 \cdot {\rm lg }\hspace{0.2cm}\frac{f}{\rm MHz}
   \hspace{0.05cm}.$$
-Die Gleichung beschreibt die so genannte&nbsp; '''Freiraumdämpfung''', die auch von der (Träger-)Frequenz abhängt.
+This equation describes so-called&nbsp; '''free space attenuation''', which also depends on the (carrier) frequency.
-Man kann davon ausgehen, dass das gesamte ZSB–AM–Spektrum wie die Trägerfrequenz gedämpft wird.&nbsp; Das bedeutet, dass
+It can be assumed that the entire DSB-AM spectrum is attenuated like the carrier frequency.  This means that
-*die etwas größere Dämpfung des oberen Seitenbandes (OSB), bzw.
+*the slightly larger attenuation of the upper sideband (USB), and
-*die geringfügig kleinere Dämpfung des unteren Seitenbandes (USB)
+*the slightly smaller attenuation of the lower sideband (LSB)
-durch eine entsprechende Vorverzerrung beim Sender ausgeglichen wird.
+are compensated for by a corresponding pre-distortion at the transmitter.
-Die am Empfänger wirksame Rauschleistungsdichte sei &nbsp;$N_0 = 10^{–14}  \ \rm W/Hz.$
+Let the effective noise power density at the receiver be &nbsp;$N_0 = 10^{–14}  \ \rm W/Hz.$
-Für die beiden ersten Teilaufgaben wird vorausgesetzt, dass der Sender nur den Träger überträgt, was gleichbedeutend damit ist, dass der Modulationsgrad &nbsp;$m = 0$&nbsp; ist.
+For the first two subtasks, it is assumed that the transmitter transmits only the carrier, which is equivalent to the modulation depth being &nbsp;$m = 0$&nbsp;.
@@ Line 33: / Line 33: @@
-''Hinweise:''
+''Hints:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Modulation_Methods/Synchrondemodulation|Synchrondemodulation]].
+*This exercise belongs to the chapter &nbsp; [[Modulation_Methods/Synchronous_Demodulation|Synchronous Demodulation]].
-*Bezug genommen wird insbesondere auf die Seite&nbsp;  [[Modulation_Methods/Synchrondemodulation#Sinken-SNR_und_Leistungskenngr.C3.B6.C3.9Fe|Sinken-SNR und Leistungskenngröße]].
+*Bezug genommen wird insbesondere auf die Seite&nbsp;  [[Modulation_Methods/Synchronous_Demodulation#Sink_SNR_and_the_performance_parameter|Sink SNR and the performance paramter]].
-===Fragebogen===
+===Questions===
 <quiz display=simple>

← Older revision		Revision as of 14:52, 15 November 2021
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← Older revision		Revision as of 13:01, 23 March 2021
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−	[[Category:~~Aufgaben zu Modulationsverfahren~~\|^2.2 Synchrondemodulation^]]	+	[[Category:Modulation Methods: Exercises\|^2.2 Synchrondemodulation^]]