Aufgaben:Exercise 2.12: Non-coherent Demodulation - Revision history

Hwang at 12:46, 17 November 2022

2022-11-17T12:46:50Z

Reed at 14:26, 9 April 2022

2022-04-09T14:26:49Z

Guenter at 15:38, 22 February 2022

2022-02-22T15:38:45Z

Guenter at 16:16, 18 February 2022

2022-02-18T16:16:21Z

Guenter at 16:07, 18 February 2022

2022-02-18T16:07:46Z

Guenter at 15:51, 18 February 2022

2022-02-18T15:51:58Z

Reed at 19:54, 22 December 2021

2021-12-22T19:54:07Z

Reed: /* Solution */

2021-12-22T19:52:57Z

Solution

Reed at 19:49, 22 December 2021

2021-12-22T19:49:29Z

Reed at 19:49, 22 December 2021

2021-12-22T19:49:16Z

@@ Line 33: / Line 33: @@
 Hints:
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Further_AM_Variants|Further AM Variants]].
-*Particular reference is made to the page&nbsp;  [[Modulation_Methods/Further_AM_Variants#Incoherent_.28non-coherent.29_Demodulation|Incoherent (non-coherent) Demodulation]].
+*Particular reference is made to the section&nbsp;  [[Modulation_Methods/Further_AM_Variants#Incoherent_.28non-coherent.29_Demodulation|Incoherent (non-coherent) Demodulation]].
 *The following trigonometric transformations are given:

← Older revision		Revision as of 14:26, 9 April 2022
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−	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Weitere AM–Varianten~~	+	{{quiz-Header\|Buchseite=Modulation Methods/Further AM Variants
	}}		}}

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 }}
-[[File:P_ID1088__Mod_A_2_12.png|right|frame|ASK Demodulation <br>(non-coherent) ]]
+[[File:EN_Mod_A_2_12.png|right|frame|ASK Demodulation <br>(non-coherent) ]]
 Consider an amplitude modulated signal:
 :$$ s(t) = q(t)  \cdot \cos(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$

@@ Line 75: / Line 75: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Applying the trigonometric transformations given on the exercise page and taking into account the two lowpass filters (the components around twice the carrier frequency are removed), we obtain:
+'''(1)'''&nbsp;  Applying the trigonometric transformations given on the exercise page and taking into account the two low-pass filters&nbsp; (the components around twice the carrier frequency are removed),&nbsp; we obtain:
 :$$b_1(t)  =  q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot 2 \cdot \cos(\omega_{\rm T} \cdot t) = q(t) \cdot \cos(\Delta \phi_{\rm T})\hspace{0.05cm},$$
 :$$ b_2(t)  =  q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot (-2) \cdot \sin(\omega_{\rm T} \cdot t) = q(t) \cdot \sin(\Delta \phi_{\rm T})\hspace{0.05cm}.$$
-*Thus, <u>the first and fourth answers</u> are correct.
+*Thus,&nbsp; <u>the first and fourth answers</u>&nbsp; are correct.
@@ Line 94: / Line 94: @@
-'''(4)'''&nbsp;  The result&nbsp; $b(t) = q^2(t)$ – see subtask '''(2)'''&nbsp; – here leads to the result:
+'''(4)'''&nbsp;  The result&nbsp; $b(t) = q^2(t)$ – see subtask '''(2)'''&nbsp; – leads here to the result:
 :$$b_{\rm min}\hspace{0.15cm}\underline{ = 9},$$
 :$$b_{\rm max}\hspace{0.15cm}\underline{ =9}.$$
-This shows that the demodulator considered here only functions, if at all times &nbsp; $q(t) ≥ 0$&nbsp; or&nbsp;, $q(t) ≤ 0$&nbsp; holds, and this is known at the receiver.
 {{ML-Fuß}}

@@ Line 6: / Line 6: @@
 Consider an amplitude modulated signal:
 :$$ s(t) = q(t)  \cdot \cos(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$
-Reaching the receiver based on the channel propagation time, the signal is
+Reaching the receiver based on the channel propagation time,&nbsp;  the signal is
 :$$ r(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \hspace{0.05cm}.$$
-The arrangement shown here allows perfect demodulation – that is, &nbsp;$v(t) = q(t)$ – without knowledge of the phase &nbsp;$Δϕ_T$, but only if the source signal &nbsp;$q(t)$&nbsp; satisfies certain conditions.
+The arrangement shown here allows perfect demodulation – that is: &nbsp;$v(t) = q(t)$ – without knowledge of the phase &nbsp;$Δϕ_T$,&nbsp; but only if the source signal &nbsp;$q(t)$&nbsp; satisfies certain conditions.
 The two receiver-side carrier signals are:
@@ Line 14: / Line 14: @@
 :$$ z_{\rm 2, \hspace{0.08cm}E}(t)  =  -2 \cdot \sin(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$
-$\rm TP_1$&nbsp; and &nbsp;$\rm TP_2$&nbsp; denote two ideal (rectangular) lowpass filters, each with cutoff frequency equal to the carrier frequency &nbsp;$f_{\rm T}$&nbsp;.
+$\rm LP_1$&nbsp; and &nbsp;$\rm LP_2$&nbsp; denote two ideal&nbsp; (rectangular)&nbsp; low-pass filters,&nbsp; each with cutoff frequency equal to the carrier frequency &nbsp;$f_{\rm T}$.
-As (digital) source signals we consider:
+We consider as (digital) source signals:
-* the unipolar square wave &nbsp;$q_1(t)$&nbsp; with dimensionless amplitude values &nbsp;$0$&nbsp; and &nbsp;$3$,
+# the unipolar square wave  signal &nbsp;$q_1(t)$&nbsp; with dimensionless amplitude values &nbsp;$0$&nbsp; and &nbsp;$3$,
-* the bipolar square wave signal &nbsp;$q_2(t)$&nbsp; with the dimensionless amplitude values &nbsp;$±3$.
+# the bipolar square wave signal &nbsp;$q_2(t)$&nbsp; with the dimensionless amplitude values &nbsp;$±3$.
-With respect to &nbsp;$s(t)$&nbsp;, these two signals result in an &nbsp;[[Modulation_Methods/Linear_Digital_Modulation#ASK_.E2.80.93_Amplitude_Shift_Keying|ASK signal]]&nbsp; and a &nbsp;[[Modulation_Methods/Linear_Digital_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying|BPSK signal]], respectively.
+With respect to &nbsp;$s(t)$,&nbsp; these two signals result in
 The nonlinear function &nbsp;$v = g(b)$&nbsp; is to be determined in this exercise.
@@ Line 28: / Line 31: @@
 Hints:
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Further_AM_Variants|Further AM Variants]].
 *Particular reference is made to the page&nbsp;  [[Modulation_Methods/Further_AM_Variants#Incoherent_.28non-coherent.29_Demodulation|Incoherent (non-coherent) Demodulation]].
@@ Line 46: / Line 45: @@
 <quiz display=simple>
-{What are the signals &nbsp;$b_1(t)$&nbsp; and &nbsp;$b_2(t)$&nbsp; in both branches – after multiplier and lowpass respectively?&nbsp; Which statements apply?
+{What are the signals &nbsp;$b_1(t)$&nbsp; and &nbsp;$b_2(t)$&nbsp; in both branches – after multiplier and low-pass respectively?&nbsp; Which statements apply?
 |type="[]"}
 + $b_1(t) = q(t) · \cos(Δϕ_{\rm T})$.
@@ Line 54: / Line 53: @@
 -  $b_1(t) = b_2(t) = q(t)$.
-{What values of &nbsp;$b_{\rm min}$&nbsp; and &nbsp;$b_{\rm max}$&nbsp; does the signal &nbsp;$b(t)$&nbsp; take on, when the unipolar source signal &nbsp;$q_1(t)$&nbsp; is applied to the input?
+{What values of &nbsp;$b_{\rm min}$&nbsp; and &nbsp;$b_{\rm max}$&nbsp; does the signal &nbsp;$b(t)$&nbsp; take on,&nbsp; when the unipolar source signal &nbsp;$q_1(t)$&nbsp; is applied to the input?
 |type="{}"}
 $b_{\rm min} \ = \ $  { 0. }
 $b_{\rm max} \ = \ $ { 9 3% }
-{How should the characteristic curve &nbsp;$v = g(b)$&nbsp; be chosen, so that &nbsp;$v(t) = q(t)$&nbsp; holds?
+{How should the characteristic curve &nbsp;$v = g(b)$&nbsp; be chosen,&nbsp; so that &nbsp;$v(t) = q(t)$&nbsp; holds?
 |type="()"}
 - $v=g(b) = b^2$.
@@ Line 66: / Line 65: @@
-{What values of &nbsp;$b_{\rm min}$&nbsp; and &nbsp;$b_{\rm max}$&nbsp; does the signal &nbsp;$b(t)$&nbsp; take on, when the bipolar source signal &nbsp;$q_2(t)$&nbsp; is applied to the input?
+{What values of &nbsp;$b_{\rm min}$&nbsp; and &nbsp;$b_{\rm max}$&nbsp; does the signal &nbsp;$b(t)$&nbsp; take on,&nbsp; when the bipolar source signal &nbsp;$q_2(t)$&nbsp; is applied to the input?
 |type="{}"}
 $b_{\rm min} \ = \ $ { 9 3% }

@@ Line 76: / Line 76: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Durch Anwendung der auf der Angabenseite gegebenen trigonometrischen Umformungen erhält man unter Berücksichtigung der beiden Tiefpässe (die Anteile um die doppelte Trägerfrequenz werden entfernt):
+'''(1)'''&nbsp;  Applying the trigonometric transformations given on the exercise page and taking into account the two lowpass filters (the components around twice the carrier frequency are removed), we obtain:
 :$$b_1(t)  =  q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot 2 \cdot \cos(\omega_{\rm T} \cdot t) = q(t) \cdot \cos(\Delta \phi_{\rm T})\hspace{0.05cm},$$
 :$$ b_2(t)  =  q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot (-2) \cdot \sin(\omega_{\rm T} \cdot t) = q(t) \cdot \sin(\Delta \phi_{\rm T})\hspace{0.05cm}.$$
-*Richtig sind somit <u>die erste und die vierte Antwort</u>.
+*Thus, <u>the first and fourth answers</u> are correct.
-'''(2)'''&nbsp;  Die Summe der Quadrate der beiden Teilsignale ergibt:
+'''(2)'''&nbsp;  The sum of the squares of the two partial signals gives:
 :$$ b(t) = b_1^2(t) + b_2^2(t)= q^2(t) \cdot \left( \cos^2(\Delta \phi_{\rm T})+ \sin^2(\Delta \phi_{\rm T})\right) = q^2(t)\hspace{0.05cm}.$$
-Die möglichen Amplitudenwerte sind somit: &nbsp;
+The possible amplitude values are thus:
 :$$b_{\rm min}\hspace{0.15cm}\underline{ = 0},$$
 :$$ b_{\rm max}\hspace{0.15cm}\underline{ =9}.$$
@@ Line 91: / Line 91: @@
-'''(3)'''&nbsp;  Richtig ist der <u>zweite Lösungsvorschlag</u>:
+'''(3)'''&nbsp; The <u>second answer</u> is correct:
 :$$v=g(b) = \sqrt{b} \hspace{0.3cm} \Rightarrow  \hspace{0.3cm} v(t) = \sqrt{ q^2(t) } = q(t)\hspace{0.05cm}.$$
-'''(4)'''&nbsp;  Das Ergebnis&nbsp; $b(t) = q^2(t)$ – siehe Teilaufgabe&nbsp; '''(2)'''&nbsp; – führt hier zum Ergebnis:
+'''(4)'''&nbsp;  The result&nbsp; $b(t) = q^2(t)$ – see subtask '''(2)'''&nbsp; – here leads to the result:
 :$$b_{\rm min}\hspace{0.15cm}\underline{ = 9},$$
 :$$b_{\rm max}\hspace{0.15cm}\underline{ =9}.$$
-Dies zeigt, dass der hier betrachtete Demodulator nur dann funktioniert, wenn für alle Zeiten&nbsp; $q(t) ≥ 0$&nbsp; oder&nbsp; $q(t) ≤ 0$&nbsp; gilt und dies dem Empfänger auch bekannt ist.
+This shows, that the demodulator considered here only function when for all times &nbsp; $q(t) ≥ 0$&nbsp; or&nbsp; $q(t) ≤ 0$&nbsp; holds, and this is known at the receiver.
 {{ML-Fuß}}