Aufgaben:Exercise 3.5: PM and FM for Rectangular Signals - Revision history

Hwang at 15:19, 23 January 2023

2023-01-23T15:19:55Z

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Reed at 12:20, 17 March 2022

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Reed at 12:20, 17 March 2022

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Reed at 12:19, 17 March 2022

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 Assume a bipolar and rectangular source signal&nbsp;$q(t)$&nbsp;, as shown in the upper diagram.&nbsp; This signal can only take on the two signal values &nbsp;$±A = ±2 \ \rm V$&nbsp; and the duration of the positive and negative rectangles are each&nbsp;$T = 1 \ \rm ms$.&nbsp; The period of &nbsp;$q(t)$&nbsp; is therefore  &nbsp;$T_0 = 2 \ \rm ms$.
-The signals&nbsp;$s_1(t)$&nbsp; and &nbsp;$s_2(t)$&nbsp; display two transmit signals with angle modulation&nbsp; $\rm (WM)$, each of which can be represented as
+The signals&nbsp;$s_1(t)$&nbsp; and &nbsp;$s_2(t)$&nbsp; display two transmitted signals with angle modulation&nbsp; $\rm (WM)$, each of which can be represented as
 :$$s(t) = A_{\rm T} \cdot \cos \hspace{-0.05cm}\big [\psi (t) \big ]$$
 Here, we distinguish between phase modulation&nbsp; $\rm (PM)$&nbsp; with the angular function

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Modulationsverfahren/Frequenzmodulation (FM)
+{{quiz-Header|Buchseite=Modulation_Methods/Frequency_Modulation_(FM)
 }}
-[[File:P_ID1099__Mod_A_3_5.png|right|frame|Zwei Signalverläufe bei Winkelmodulation ]]
+[[File:P_ID1099__Mod_A_3_5.png|right|frame|Two signal waveforms in angle modulation]]
 Assume a bipolar and rectangular source signal&nbsp;$q(t)$&nbsp;, as shown in the upper diagram.&nbsp; This signal can only take on the two signal values &nbsp;$±A = ±2 \ \rm V$&nbsp; and the duration of the positive and negative rectangles are each&nbsp;$T = 1 \ \rm ms$.&nbsp; The period of &nbsp;$q(t)$&nbsp; is therefore  &nbsp;$T_0 = 2 \ \rm ms$.

@@ Line 93: / Line 93: @@
-'''(6)'''&nbsp;  Der Frequency deviation can also be represented as follows:
+'''(6)'''&nbsp;  The frequency deviation can also be represented as follows:
 :$$\Delta f_{\rm A} = \frac {K_{\rm FM}}{2\pi}\cdot A \hspace{0.05cm}.$$
-*Mit&nbsp; $Δf_{\rm A} · {\rm A}  = 2$&nbsp; erhält man somit:
+*With &nbsp; $Δf_{\rm A} · {\rm A}  = 2$&nbsp; we thus get:
 :$$K_{\rm FM} = \frac {2 \cdot 2\pi}{A \cdot T}= \frac {4\pi}{2\,{\rm V} \cdot 1\,{\rm ms}}\hspace{0.15cm}\underline {= 6283 \,{\rm V}^{-1}{\rm s}^{-1}} \hspace{0.05cm}.$$

@@ Line 77: / Line 77: @@
 *By counting the oscillations of&nbsp; $s_2(t)$&nbsp; in the time interval&nbsp; $T$&nbsp;, it can be seen that&nbsp; $f_{\rm T} · T\hspace{0.15cm}\underline{ = 6}$&nbsp; was used.
 *When frequency modulating a bipolar source signal, &nbsp; $f_{\rm T}$&nbsp; does not occur directly.
-*However, the graphs Grafiken do indicate that &nbsp;  $f_{\rm T} · T = 6$&nbsp; is also used here.
+*However, the graphs do indicate that &nbsp;  $f_{\rm T} · T = 6$&nbsp; is also used here.

@@ Line 65: / Line 65: @@
 '''(1)'''&nbsp; <u>Answer 2</u> is correct:
 *For a rectangular (digital) source signal, phase modulation (PM) can be recognised by the typical phase jumps – see the signal waveform&nbsp; $s_2(t)$.
-*Frequency modulation (FM), on the other hand, has diverse instantaneous frequencies at different times as in&nbsp; $s_1(t)$.
+*Frequency modulation (FM), on the other hand, has diverse instantaneous frequencies at different times, as in&nbsp; $s_1(t)$.

@@ Line 34: / Line 34: @@
 {Which of the signals is due to phase modulation and which is due to frquency modulation?
 |type="()"}
-- $s_1(t)$&nbsp; describes a phase modulation.
+- $s_1(t)$&nbsp; represents a phase modulation.
-+ $s_1(t)$&nbsp; describes a frequency modulation.
++ $s_1(t)$&nbsp; represents a frequency modulation.
 {What is the carrier phase &nbsp;$ϕ_{\rm T}$ that could be measured without a message signal &nbsp; &rArr;  &nbsp; $q(t) \equiv 0$&nbsp;?

@@ Line 64: / Line 64: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; <u>Answer 2</u> is correct:
-*For a rectangular (digital) source signal, phase modulation (PM) can be recognised bz the typical phase jumps – see the signal waveform&nbsp; $s_2(t)$.
+*For a rectangular (digital) source signal, phase modulation (PM) can be recognised by the typical phase jumps – see the signal waveform&nbsp; $s_2(t)$.
 *Frequency modulation (FM), on the other hand, has a diverse instantaneous frequencies at different times as in&nbsp; $s_1(t)$.

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 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Richtig ist die <u>Antwort 2</u>:
+'''(1)'''&nbsp; <u>Answer 2</u> is correct:
-*Bei einem rechteckförmigen (digitalen) Quellensignal erkennt man die Phasenmodulation (PM) an den typischen Phasensprüngen – siehe Signalverlauf&nbsp; $s_2(t)$.
+*For a rectangular (digital) source signal, phase modulation (PM) can be recognised bz the typical phase jumps – see the signal waveform&nbsp; $s_2(t)$.
-*Die Frequenzmodulation (FM) hat dagegen zu den verschiedenen Zeiten unterschiedliche Augenblicksfrequenzen wie bei&nbsp; $s_1(t)$.
+*Frequency modulation (FM), on the other hand, has a diverse instantaneous frequencies at different times as in&nbsp; $s_1(t)$.
-'''(2)'''&nbsp;  Mit&nbsp; $q(t) = 0$&nbsp; erhält man entsprechend den gegebenen Gleichungen sowohl für PM als auch für FM
+'''(2)'''&nbsp;  When&nbsp; $q(t) = 0$&nbsp;, the equations provided for both PM and FM give
 :$$s(t) = A_{\rm T} \cdot \cos (\omega_{\rm T} \cdot t ) \hspace{0.3cm}\Rightarrow\hspace{0.3cm} \phi_{\rm T} \hspace{0.15cm}\underline {= 0}\hspace{0.05cm}.$$
-'''(3)'''&nbsp;  Die Trägerfrequenz&nbsp; $f_{\rm T}$&nbsp; kann direkt nur aus dem PM–Signal&nbsp; $s_2(t)$&nbsp; ermittelt werden.
+'''(3)'''&nbsp;  The carrier frequency&nbsp; $f_{\rm T}$&nbsp; can be directly determined only from the PM signal &nbsp; $s_2(t)$&nbsp;.
-*Durch Abzählen der Schwingungen von&nbsp; $s_2(t)$&nbsp; im Zeitintervall&nbsp; $T$&nbsp; erkennt man, dass&nbsp; $f_{\rm T} · T\hspace{0.15cm}\underline{ = 6}$&nbsp; verwendet wurde.
+*By counting the oscillations of&nbsp; $s_2(t)$&nbsp; in the time interval&nbsp; $T$&nbsp;, it can be seen that&nbsp; $f_{\rm T} · T\hspace{0.15cm}\underline{ = 6}$&nbsp; was used.
-*Bei der Frequenzmodulation eines bipolaren Quellensignals tritt&nbsp; $f_{\rm T}$&nbsp; nicht direkt auf.
+*When frequency modulating a bipolar source signal, &nbsp; $f_{\rm T}$&nbsp; does not occur directly.
-*Die Grafiken lassen allerdings darauf schließen, dass hier ebenfalls&nbsp;  $f_{\rm T} · T = 6$&nbsp; zugrunde liegt.
+*however, the graphs Grafiken do indicate that &nbsp;  $f_{\rm T} · T = 6$&nbsp; is also used here.
-'''(4)'''&nbsp;  Der Amplitudenwert&nbsp; $A = 2 \ \rm V$&nbsp; führt zur Phase&nbsp; $90^\circ$&nbsp; bzw.&nbsp; $π/2$&nbsp; (Minus–Sinusverlauf).&nbsp; Daraus folgt:
+'''(4)'''&nbsp;  The amplitude value&nbsp; $A = 2 \ \rm V$&nbsp; results in the phase&nbsp; $90^\circ$&nbsp; or&nbsp; $π/2$&nbsp; (minus sine wave).&nbsp; This gives:
 :$$K_{\rm PM} = \frac {\pi /2}{2\,{\rm V}} \hspace{0.15cm}\underline {= 0.785\,{\rm V}^{-1}} \hspace{0.05cm}.$$
-'''(5)'''&nbsp;  Die Grafik für&nbsp; $s_1(t)$&nbsp; zeigt, dass innerhalb eines Zeitintervalls&nbsp; $T$&nbsp; entweder vier oder acht Schwingungen auftreten: &nbsp; $4 \le f_{\rm A}(t) \cdot T \le 8\hspace{0.05cm}.$
+'''(5)'''&nbsp;  The graph for&nbsp; $s_1(t)$&nbsp; shows that either four or eight oscillations arise within a time interval&nbsp; $T$&nbsp;: &nbsp; $4 \le f_{\rm A}(t) \cdot T \le 8\hspace{0.05cm}.$
-*Unter Berücksichtigung der (normiertern) Trägerfrequenz&nbsp; $f_{\rm T} · T = 6$&nbsp; ergibt sich für den (normierten) Frequenzhub:
+*Considering the (normalized) carrier frequency&nbsp; $f_{\rm T} · T = 6$&nbsp;, the (normalized) frequency deviation is:
 :$$\Delta f_{\rm A} \cdot T \hspace{0.15cm}\underline {=2}\hspace{0.05cm}.$$
-'''(6)'''&nbsp;  Der Frequenzhub kann auch wie folgt dargestellt werden:
+'''(6)'''&nbsp;  Der Frequency deviation can also be represented as follows:
 :$$\Delta f_{\rm A} = \frac {K_{\rm FM}}{2\pi}\cdot A \hspace{0.05cm}.$$
 *Mit&nbsp; $Δf_{\rm A} · {\rm A}  = 2$&nbsp; erhält man somit: