Difference between revisions of "Aufgaben:Exercise 3.3Z: Characteristics Determination"

Latest revision as of 16:03, 9 April 2022

Spectrum of the analytical signal

Let us consider the phase modulation of the harmonic oscillation

$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t + \phi_{\rm N}) \hspace{0.05cm},$$

which, given a normalized carrier amplitude $(A_{\rm T} = 1)$ , leads to the following transmitted signal:

$$ s(t) = \cos \hspace{-0.1cm}\big[\omega_{\rm T} \cdot t + K_{\rm PM} \cdot q(t) \big]\hspace{0.05cm}.$$

The spectrum of the corresponding analytical signal $s_{\rm TP}(t)$ is generally:

$$S_{\rm TP}(f) = \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}(\phi_{\rm N}\hspace{0.05cm}+\hspace{0.05cm} 90^\circ) }\cdot \hspace{0.05cm} \delta (f - n \cdot f_{\rm N})\hspace{0.05cm}$$

Here, $η = K_{\rm PM} · A_{\rm N}$ is called the modulation index.

In the graph, the real and imaginary parts of the spectrum $S_+(f)$ of the analytical signal $s_+(t)$ are shown separately. This should be used to determine the characteristics $f_{\rm T}$, $f_{\rm N}$, $ϕ_{\rm N}$ and $η$ .

Hints:

This exercise belongs to the chapter Phase Modulation.
Particular reference is made to the page Equivalent low-pass signal in phase modulation.

For the calculation of the modulation index, you can take advantage of the following property of the Bessel function:

$${\rm J}_n (\eta) = \frac{2 \cdot (n-1)}{\eta} \cdot {\rm J}_{n-1} (\eta) - {\rm J}_{n-2} (\eta) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm J}_{2} (\eta)= {2}/{\eta} \cdot {\rm J}_{1} (\eta) - {\rm J}_{0} (\eta) \hspace{0.05cm}.$$

Questions

$f_{\rm T} \ = \ $

$\ \rm kHz$

$f_{\rm N} \ = \ $

$\ \rm kHz$

$|S_{\rm TP}(f = 3 \ \rm kHz)| \ = \ $

${\rm arc} \ S_{\rm TP}(f = 3\ \rm kHz) \ = \ $

$\ \rm Grad$

$|S_{\rm TP}(f = 6 \ \rm kHz)| \ = \ $

${\rm arc} \ S_{\rm TP}(f = 6\ \rm kHz) \ = \ $

$\ \rm Grad$

$ϕ_{\rm N} \ = \ $

$\ \rm Grad$

$η \ = \ $

Solution

(1) Regarding $|S_+(f)|$ there is a symmetry with respect to the carrier frequency $f_{\rm T}\hspace{0.15cm}\underline { = 40 \ \rm kHz}$. The distance between the spectral lines is $f_{\rm N}\hspace{0.15cm}\underline { = 3 \ \rm kHz}$.

(2) Considering $S_{\rm TP}(f = 3{\ \rm kHz}) = S_+(f = 43 \ \rm kHz)$ , it holds that:

$$|S_{\rm TP}(f = 3\,{\rm kHz})| = \sqrt{0.279^2 + 0.483^2} \hspace{0.15cm}\underline {= 0.558}\hspace{0.05cm},$$

$$ {\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 3\,{\rm kHz}) = \arctan \frac{0.483}{0.279} = \arctan 1.732\hspace{0.15cm}\underline { = 60^\circ} \hspace{0.05cm}.$$

(3) Analogously to in question (2) , at frequency $f = 6 \ \rm kHz$ we obtain:

$$|S_{\rm TP}(f = 6\,{\rm kHz})| = \sqrt{(-0.116)^2 + 0.201^2} \hspace{0.15cm}\underline {= 0.232}\hspace{0.05cm},$$

$${\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 6\,{\rm kHz}) = \arctan \frac{-0.116}{0.201} = 180^\circ - \arctan 1.732 \hspace{0.15cm}\underline {= 120^\circ} \hspace{0.05cm}.$$

(4) When $n = 1$ ⇒ $f = 3 \ \rm kHz$ as in question (2), the phase is:

$$ \phi_{\rm N} + 90^\circ = 60^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} = -30^\circ\hspace{0.05cm}.$$

Checking this result when $n = 2$ ⇒ $f = 6 \ \rm kHz$ as in question (3) yields the same value:

$$ 2\cdot (\phi_{\rm N} + 90^\circ) = 120^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} \hspace{0.15cm}\underline {= -30^\circ}\hspace{0.05cm}.$$

(5) The given equation can be rewritten as follows:

$$\eta = \frac{2 \cdot {\rm J}_{1}{(\eta)}}{{\rm J}_{0}(\eta) + {\rm J}_{2}(\eta)} \hspace{0.05cm}.$$

With ${\rm J}_0(η) = 0.512$, ${\rm J}_1(η) = 0.558$ and ${\rm J}_2(η) = 0.232$ we thus get:

$$ \eta = \frac{2 \cdot 0.558}{0.512 + 0.232}\hspace{0.15cm}\underline { = 1.5}\hspace{0.05cm}.$$

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Modulationsverfahren/Phasenmodulation (PM)
+{{quiz-Header|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
 }}
-[[File:P_ID1085__Mod_Z_3_3.png|right|frame|Spektrum des analytischen Signals]]
+[[File:P_ID1085__Mod_Z_3_3.png|right|frame|Spectrum of the analytical signal]]
-Wir betrachten die Phasenmodulation der harmonischen Schwingung
+Let us consider the phase modulation of the harmonic oscillation
 :$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t + \phi_{\rm N}) \hspace{0.05cm},$$
-die bei Voraussetzung einer normierten Trägeramplitude ($A_{\rm T} = 1$) zu folgendem Sendesignal führt:
+which, given a normalized carrier amplitude &nbsp;$(A_{\rm T} = 1)$&nbsp;, leads to the following transmitted signal:
-:$$ s(t) = \cos \left(\omega_{\rm T} \cdot t + K_{\rm PM} \cdot q(t) \right)\hspace{0.05cm}.$$
+:$$ s(t) = \cos \hspace{-0.1cm}\big[\omega_{\rm T} \cdot t + K_{\rm PM} \cdot q(t) \big]\hspace{0.05cm}.$$
-Das Spektrum des dazugehörigen analytischen Signals $s_{\rm TP}(t)$ lautet allgemein:
+The spectrum of the corresponding analytical signal&nbsp;$s_{\rm TP}(t)$&nbsp; is generally:
 :$$S_{\rm TP}(f) = \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}(\phi_{\rm N}\hspace{0.05cm}+\hspace{0.05cm} 90^\circ) }\cdot \hspace{0.05cm} \delta (f - n \cdot f_{\rm N})\hspace{0.05cm}$$
-Hierbei bezeichnet man $η = K_{\rm PM} · A_{\rm N}$ als den Modulationsindex.
+Here, &nbsp;$η = K_{\rm PM} · A_{\rm N}$&nbsp; is called the modulation index.
-In der Grafik ist das Spektrum $S_+(f)$ des analytischen Signals $s_+(t)$ getrennt nach Real- und Imaginärteil dargestellt. Aus diesem sollen die Kenngrößen $f_{\rm T}$, $f_{\rm N}$, $ϕ_{\rm N}$ und $η$ ermittelt werden.
+In the graph, the real and imaginary parts of the spectrum &nbsp;$S_+(f)$&nbsp; of the analytical signal &nbsp;$s_+(t)$&nbsp;are shown separately. This should be used to determine the characteristics &nbsp;$f_{\rm T}$, &nbsp;$f_{\rm N}$, &nbsp;$ϕ_{\rm N}$&nbsp; and &nbsp;$η$&nbsp;.
-''Hinweise:''
-*Die Aufgabe gehört zum  Kapitel [[Modulationsverfahren/Phasenmodulation_(PM)|Phasenmodulation]].
-*Bezug genommen wird insbesondere auf die Seite  [[Modulationsverfahren/Phasenmodulation_(PM)#.C3.84quivalentes_TP.E2.80.93Signal_bei_Phasenmodulation|Äquivalentes Tiefpass-Signal bei Phasenmodulation]].
-*Sollte die Eingabe des Zahlenwertes &bdquo;0&rdquo; erforderlich sein, so geben Sie bitte &bdquo;0.&rdquo; ein.
-*Zur Berechnung des Modulationsindex können Sie folgende Eigenschaft der Besselfunktion ausnutzen:
+''Hints:''
+*This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Phase_Modulation_(PM)|Phase Modulation]].
+*Particular reference is made to the page&nbsp;  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation|Equivalent low-pass signal in phase modulation]].
+*For the calculation of the modulation index, you can take advantage of the following property of the Bessel function:
 :$${\rm J}_n (\eta) = \frac{2 \cdot (n-1)}{\eta} \cdot {\rm J}_{n-1} (\eta) - {\rm J}_{n-2} (\eta) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm J}_{2} (\eta)= {2}/{\eta} \cdot {\rm J}_{1} (\eta) - {\rm J}_{0} (\eta) \hspace{0.05cm}.$$
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Wie groß sind die Frequenzen $f_T$ und $f_N$?
+{What are the frequencies &nbsp;$f_{\rm T}$&nbsp; and &nbsp;$f_{\rm N}$?
 |type="{}"}
 $f_{\rm T} \ = \ $  { 40 3% } $\ \rm kHz$
 $f_{\rm N} \ = \ $ { 3 3% } $\ \rm kHz$
-{Berechnen Sie den Betrag und die Phase von $S_{\rm TP}(f = 3 \ \rm kHz)$.
+{Calculate the magnitude and the phase of &nbsp;$S_{\rm TP}(f = 3 \ \rm kHz)$.
 |type="{}"}
 $|S_{\rm TP}(f = 3 \ \rm kHz)| \ = \ $  { 0.558 3% }
-$arc \ S_{\rm TP}(f = 3\ \rm  kHz) \ = \ $ { 60 3% } $\ \rm Grad$
+${\rm arc} \ S_{\rm TP}(f = 3\ \rm  kHz) \ = \ $ { 60 3% } $\ \rm Grad$
-{Berechnen Sie den Betrag und die Phase von $S_{\rm TP}(f = 6 \ \rm kHz)$.
+{Calculate the magnitude and the phase of &nbsp;$S_{\rm TP}(f = 6 \ \rm kHz)$.
 |type="{}"}
 $|S_{\rm TP}(f = 6 \ \rm kHz)| \ = \ $ { 0.232 3% }
-$arc \ S_{\rm TP}(f = 6\ \rm  kHz) \ = \ $ { 120 3% } $\ \rm Grad$
+${\rm arc} \ S_{\rm TP}(f = 6\ \rm  kHz) \ = \ $ { 120 3% } $\ \rm Grad$
-{Wie groß ist die Phase des Quellensignals $q(t)$?
+{What is the phase of the source signal &nbsp;$q(t)$?
 |type="{}"}
 $ϕ_{\rm N} \ = \ $ { -30.9--29.1 } $\ \rm Grad$
-{Wie groß ist der Modulationsindex $η$?
+{What is the magnitude of the modulation index &nbsp;$η$&nbsp;?
 |type="{}"}
 $η \ = \ $ { 1.5 3% }
@@ Line 53: / Line 60: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''1.''' Bezüglich $|S_+(f)|$ gibt es eine Symmetrie zur Trägerfrequenz $f_T = 40 kHz$. Der Abstand zwischen den Spektrallinien beträgt $f_N = 3 kHz$.
+'''(1)'''&nbsp; Regarding &nbsp; $|S_+(f)|$&nbsp; there is a symmetry with respect to the carrier frequency&nbsp; $f_{\rm T}\hspace{0.15cm}\underline { = 40 \ \rm kHz}$.&nbsp;  The distance between the spectral lines is&nbsp; $f_{\rm N}\hspace{0.15cm}\underline { = 3 \ \rm kHz}$.
+'''(2)'''&nbsp; Considering&nbsp; $S_{\rm TP}(f = 3{\ \rm kHz}) = S_+(f = 43 \ \rm kHz)$&nbsp;, it holds that:
+:$$|S_{\rm TP}(f = 3\,{\rm kHz})| = \sqrt{0.279^2 + 0.483^2} \hspace{0.15cm}\underline {= 0.558}\hspace{0.05cm},$$
+:$$ {\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 3\,{\rm kHz}) = \arctan \frac{0.483}{0.279} = \arctan 1.732\hspace{0.15cm}\underline { = 60^\circ} \hspace{0.05cm}.$$
+'''(3)'''&nbsp; Analogously to in question&nbsp; '''(2)'''&nbsp;, at frequency &nbsp; $f = 6 \ \rm kHz$ we obtain:
+:$$|S_{\rm TP}(f = 6\,{\rm kHz})| = \sqrt{(-0.116)^2 + 0.201^2} \hspace{0.15cm}\underline {= 0.232}\hspace{0.05cm},$$
+:$${\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 6\,{\rm kHz}) = \arctan \frac{-0.116}{0.201} = 180^\circ - \arctan 1.732 \hspace{0.15cm}\underline {= 120^\circ} \hspace{0.05cm}.$$
-'''2.''' Unter Berücksichtigung von $S_{TP}(f = 3 kHz) = S_+(f = 43 kHz)$ gilt:
+'''(4)'''&nbsp; When &nbsp; $n = 1$ &nbsp; &rArr; &nbsp; $f = 3 \ \rm kHz$&nbsp; as in question &nbsp; '''(2)''', the phase is:
-$$|S_{\rm TP}(f = 3\,{\rm kHz})| = \sqrt{0.279^2 + 0.483^2} \hspace{0.15cm}\underline {= 0.558}\hspace{0.05cm},$$
+:$$ \phi_{\rm N} + 90^\circ = 60^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} = -30^\circ\hspace{0.05cm}.$$
-$$ {\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 3\,{\rm kHz}) = \arctan \frac{0.483}{0.279} = \arctan 1.732\hspace{0.15cm}\underline { = 60^\circ} \hspace{0.05cm}.$$
+*Checking this result when &nbsp; $n = 2$ &nbsp; &rArr; &nbsp; $f = 6 \ \rm kHz$&nbsp; as in question&nbsp; '''(3)'''&nbsp; yields the same value:
+:$$ 2\cdot (\phi_{\rm N} + 90^\circ) = 120^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} \hspace{0.15cm}\underline {= -30^\circ}\hspace{0.05cm}.$$
-'''3.''' n analoger Weise zur Teilaufgabe b) erhält man für $6 kHz$:
-$$|S_{\rm TP}(f = 6\,{\rm kHz})| = \sqrt{(-0.116)^2 + 0.201^2} \hspace{0.15cm}\underline {= 0.232}\hspace{0.05cm},$$
-$${\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 6\,{\rm kHz}) = \arctan \frac{-0.116}{0.201} = 180^\circ - \arctan 1.732 \hspace{0.15cm}\underline {= 120^\circ} \hspace{0.05cm}.$$
-'''4.''' Die Phase lautet für $n = 1$ (siehe Teilaufgabe b):
-$$ \phi_{\rm N} + 90^\circ = 60^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} = -30^\circ\hspace{0.05cm}.$$
-Die Überprüfung dieses Ergebnisses mit $n = 2$ liefert den gleichen Wert:
-$$ 2\cdot (\phi_{\rm N} + 90^\circ) = 120^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} \hspace{0.15cm}\underline {= -30^\circ}\hspace{0.05cm}.$$
-'''5.'''  Die angegebene Gleichung kann wie folgt umgeformt werden:
+'''(5)'''&nbsp; The given equation can be rewritten as follows:
-$$\eta = \frac{2 \cdot {\rm J}_{1}{(\eta)}}{{\rm J}_{0}(\eta) + {\rm J}_{2}(\eta)} \hspace{0.05cm}.$$
+:$$\eta = \frac{2 \cdot {\rm J}_{1}{(\eta)}}{{\rm J}_{0}(\eta) + {\rm J}_{2}(\eta)} \hspace{0.05cm}.$$
-Mit $J_0(η) = 0.512$, $J_1(η) = 0.558$ und $J_2(η) = 0.232$ erhält man somit:
+*With&nbsp; ${\rm J}_0(η) = 0.512$,&nbsp; ${\rm J}_1(η) = 0.558$&nbsp; and&nbsp; ${\rm J}_2(η) = 0.232$&nbsp; we thus get:
-$$ \eta = \frac{2 \cdot 0.558}{0.512 + 0.232}\hspace{0.15cm}\underline { = 1.5}\hspace{0.05cm}.$$
+:$$ \eta = \frac{2 \cdot 0.558}{0.512 + 0.232}\hspace{0.15cm}\underline { = 1.5}\hspace{0.05cm}.$$
@@ Line 80: / Line 95: @@
-[[Category:Aufgaben zu Modulationsverfahren|^3.1 Phasenmodulation (PM)^]]
+[[Category:Modulation Methods: Exercises|^3.1 Phase Modulation^]]