Aufgaben:Exercise 2.2Z: Power Consideration - Revision history

Reed: /* Solution */

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Solution

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Reed at 12:42, 24 November 2021

2021-11-24T12:42:37Z

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 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; According to the equations specified on the front page:
+'''(1)'''&nbsp; According to the equations specified on the exercise page:
 :$$P_{\rm 1} = \lim_{T_{\rm M} \rightarrow \infty}\hspace{0.1cm}\frac{1}{T_{\rm M}} \cdot \int_{0}^{ T_{\rm M}} {A_1^2 \cdot \cos^2(\omega_{\rm 1} t + \phi_1) }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
 *For more general calculation,&nbsp; we consider the phase&nbsp; $ϕ_1$,&nbsp; which is actually zero here.&nbsp; Using the equation&nbsp; $\cos^{2}(α) = 0.5 · (1 + \cos(2α))$,&nbsp; we get:

@@ Line 9: / Line 9: @@
 where&nbsp;  $f_2 ≥ f_1$&nbsp; should hold for the frequencies.
-*The graphic on the right  shows the spectrum of the analytical signal &nbsp;$s_+(t)$, which is additively composed of the two components &nbsp;$s_{1+}(t)$&nbsp; and &nbsp;$s_ {2+}(t)$&nbsp;.
+*The graph on the right  shows the spectrum of the analytical signal &nbsp;$s_+(t)$, which is additively composed of the two components &nbsp;$s_{1+}(t)$&nbsp; and &nbsp;$s_ {2+}(t)$&nbsp;.
 *Here,&nbsp; the transmission power &nbsp;$P_{\rm S}$&nbsp; should be understood as the second order moment of the signal &nbsp;$s(t)$,&nbsp; averaged over the largest measurement period possible:

← Older revision		Revision as of 15:30, 24 March 2022
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−	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Zweiseitenband~~-~~Amplitudenmodulation~~	+	{{quiz-Header\|Buchseite=Modulation_Methods/Double-Sideband_Amplitude_Modulation
	}}		}}

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 *The graphic on the right  shows the spectrum of the analytical signal &nbsp;$s_+(t)$, which is additively composed of the two components &nbsp;$s_{1+}(t)$&nbsp; and &nbsp;$s_ {2+}(t)$&nbsp;.
-*Here,&nbsp; the transmission power &nbsp;$P_{\rm S}$&nbsp; should be understood as the root mean square value of the signal &nbsp;$s(t)$,&nbsp; averaged over the largest measurement period possible:
+*Here,&nbsp; the transmission power &nbsp;$P_{\rm S}$&nbsp; should be understood as the second order moment of the signal &nbsp;$s(t)$,&nbsp; averaged over the largest measurement period possible:
 :$$P_{\rm S} = \lim_{T_{\rm M} \rightarrow \infty}\hspace{0.1cm}\frac{1}{T_{\rm M}} \cdot \int_{0}^{ T_{\rm M}} {s^2(t) }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
 *According to this definition:&nbsp; If &nbsp;$s(t)$&nbsp;describes a voltage waveform, &nbsp;$P_{\rm S}$&nbsp; has unit &nbsp; $\rm V^2$&nbsp; and refers to resistance &nbsp;$R = 1 \ \rm Ω$.&nbsp;

@@ Line 58: / Line 58: @@
 '''(1)'''&nbsp; According to the equations specified on the front page:
 :$$P_{\rm 1} = \lim_{T_{\rm M} \rightarrow \infty}\hspace{0.1cm}\frac{1}{T_{\rm M}} \cdot \int_{0}^{ T_{\rm M}} {A_1^2 \cdot \cos^2(\omega_{\rm 1} t + \phi_1) }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
-*For more general calculation,&nbsp; we consider the phase&nbsp; $ϕ_1$,&nbsp; which is actually zero here.&nbsp; Using the equation $\cos^{2}(α) = 0.5 · (1 + \cos(2α))$,&nbsp; we get:
+*For more general calculation,&nbsp; we consider the phase&nbsp; $ϕ_1$,&nbsp; which is actually zero here.&nbsp; Using the equation&nbsp; $\cos^{2}(α) = 0.5 · (1 + \cos(2α))$,&nbsp; we get:
 :$$ P_{\rm 1} = \lim_{T_{\rm M} \rightarrow \infty}\hspace{0.1cm}\frac{1}{T_{\rm M}} \cdot \int_{0}^{ T_{\rm M}} {\frac{A_1^2}{2}}\hspace{0.1cm}{\rm d}t + \lim_{T_{\rm M} \rightarrow \infty}\hspace{0.1cm}\frac{1}{T_{\rm M}} \cdot \int_{0}^{ T_{\rm M}} {\frac{A_1^2}{2}\cdot \cos(2\omega_{\rm 1} t + 2\phi_1)}\hspace{0.1cm}{\rm d}t\hspace{0.05cm}.$$
 *Regardless of the phase&nbsp; $ϕ_1$,&nbsp; the second term does not contribute to the division by&nbsp; $T_{\rm M}$&nbsp; and subsequent boundary transition due to integration over the cosine function.&nbsp; Thus,&nbsp; we get:

@@ Line 56: / Line 56: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; According to the equations specified on the page:
+'''(1)'''&nbsp; According to the equations specified on the front page:
 :$$P_{\rm 1} = \lim_{T_{\rm M} \rightarrow \infty}\hspace{0.1cm}\frac{1}{T_{\rm M}} \cdot \int_{0}^{ T_{\rm M}} {A_1^2 \cdot \cos^2(\omega_{\rm 1} t + \phi_1) }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
-*For more general calculation, we consider the phase&nbsp; $ϕ_1$&nbsp;, which is actually zero here.&nbsp; Using the equation $\cos^{2}(α) = 0.5 · (1 + \cos(2α))$&nbsp;, we get:
+*For more general calculation,&nbsp; we consider the phase&nbsp; $ϕ_1$,&nbsp; which is actually zero here.&nbsp; Using the equation $\cos^{2}(α) = 0.5 · (1 + \cos(2α))$,&nbsp; we get:
 :$$ P_{\rm 1} = \lim_{T_{\rm M} \rightarrow \infty}\hspace{0.1cm}\frac{1}{T_{\rm M}} \cdot \int_{0}^{ T_{\rm M}} {\frac{A_1^2}{2}}\hspace{0.1cm}{\rm d}t + \lim_{T_{\rm M} \rightarrow \infty}\hspace{0.1cm}\frac{1}{T_{\rm M}} \cdot \int_{0}^{ T_{\rm M}} {\frac{A_1^2}{2}\cdot \cos(2\omega_{\rm 1} t + 2\phi_1)}\hspace{0.1cm}{\rm d}t\hspace{0.05cm}.$$
-*Regardless of the phase&nbsp; $ϕ_1$&nbsp;, the second term does not contribute to the division by&nbsp; $T_{\rm M}$&nbsp; and subsequent boundary transition due to integration over the cosine function. Thus, we get:
+*Regardless of the phase&nbsp; $ϕ_1$,&nbsp; the second term does not contribute to the division by&nbsp; $T_{\rm M}$&nbsp; and subsequent boundary transition due to integration over the cosine function.&nbsp; Thus,&nbsp; we get:
 :$$P_{\rm 1} = \frac{A_1^2}{2} = \frac{(2\,{\rm V})^2}{2} \hspace{0.15cm}\underline {= 2\,{\rm V}^2}\hspace{0.05cm}.$$
-'''(2)'''&nbsp; With&nbsp; $R = 50\ \rm  Ω$&nbsp;, we get the "unnormalized" power:
+'''(2)'''&nbsp; With&nbsp; $R = 50\ \rm  Ω$,&nbsp; we get the&nbsp; "unnormalized"&nbsp; power:
 :$$P_{\rm 1} = \frac{2\,{\rm V}^2}{50\,{\rm \Omega}} \hspace{0.15cm}\underline {= 40\,{\rm mW}}\hspace{0.05cm}.$$
-'''(3)'''&nbsp; It has already been shown in the solution to subtask &nbsp; '''(1)'''&nbsp; that the phase has no influence on the power.  It follows that:
+'''(3)'''&nbsp; It has already been shown in the solution to subtask&nbsp; '''(1)'''&nbsp; that the phase has no influence on the power.&nbsp;  It follows that:
 :$$P_{\rm 2} = \frac{A_2^2}{2} \hspace{0.15cm}\underline {= 0.5\,{\rm V}^2}\hspace{0.05cm}.$$
-'''(4)'''&nbsp;To calculate power, we must average over $s^{2}(t)$, where:
+'''(4)'''&nbsp;To calculate this power,&nbsp; we have to average over $s^{2}(t)$,&nbsp; where:
 :$$s^2(t) = s_1^2(t) + s_2^2(t) + 2 \cdot s_1(t) \cdot s_2(t).$$
-*Due to the division by the measurement duration&nbsp; $T_{\rm M}$&nbsp; and the required boundary transition, the last term does not contribute regardless of the phase &nbsp; $ϕ$&nbsp;.&nbsp; Thus:
+*Due to the division by the measurement duration&nbsp; $T_{\rm M}$&nbsp; and the required boundary transition,&nbsp; the last term does not contribute regardless of the phase &nbsp; $ϕ$&nbsp;.&nbsp; Thus:
 :$$P_{\rm S} = P_{\rm 1} + P_{\rm 2} \hspace{0.15cm}\underline {= 2.5\,{\rm V}^2}\hspace{0.05cm}.$$
-'''(5)'''&nbsp; When&nbsp; $f_2 = f_1$&nbsp;, the spectrum of the analytical signal is:
+'''(5)'''&nbsp; When&nbsp; $f_2 = f_1$,&nbsp; the spectrum of the analytical signal is:
 :$$S_+(f) = (A_{\rm 1} + A_{\rm 2} \cdot {\rm e}^{{\rm j}\hspace{0.03cm} \cdot \hspace{0.03cm} \phi})\cdot \delta (f - f_1) \hspace{0.05cm}.$$
 *This results in the signal:
@@ Line 86: / Line 86: @@
 :$$A_3  =  \sqrt{ \left(A_1 + A_2 \cdot \cos(\phi)\right)^2 + A_2^2 \cdot \sin^2(\phi)} =
 \sqrt{ A_1^2 + A_2^2 + 2 \cdot A_1 \cdot A_2 \cdot \cos(\phi)}\hspace{0.05cm}.$$
-*For&nbsp; $ϕ = 0$&nbsp;, the sum of the amplitudes is scalar:
+*For&nbsp; $ϕ = 0$,&nbsp; the sum of the amplitudes is scalar:
 :$$A_3 = \sqrt{ A_1^2 + A_2^2 + 2 \cdot A_1 \cdot A_2 } = A_1 + A_2 = 3\,{\rm V}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_{\rm S} \hspace{0.15cm}\underline {= 4.5\,{\rm V}^2}\hspace{0.05cm}.$$
-*On the other hand, the amplitudes for&nbsp; $ϕ = 90^\circ$&nbsp; are added as vectors&nbsp; &rArr; &nbsp; same result as in subtask&nbsp; '''(4)''':
+*On the other hand,&nbsp; the amplitudes for&nbsp; $ϕ = 90^\circ$&nbsp; are added as vectors&nbsp; &rArr; &nbsp; same result as in subtask&nbsp; '''(4)''':
 :$$ A_3 = \sqrt{ A_1^2 + A_2^2 } = \sqrt{5}\,{\rm V}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_{\rm S} = \frac{5\,{\rm V}^2}{2}\hspace{0.15cm}\underline {= 2.5\,{\rm V}^2}\hspace{0.05cm}.$$
-*For&nbsp; $ϕ = 180^\circ$&nbsp;, the cosine oscillations overlap destructively:
+*For&nbsp; $ϕ = 180^\circ$,&nbsp; the cosine oscillations overlap destructively:
 :$$A_3 = A_1 - A_2 = 1\,{\rm V}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} P_{\rm S} \hspace{0.15cm}\underline {= 0.5\,{\rm V}^2}\hspace{0.05cm}.$$
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID991__Mod_Z_2_2.png|right|frame|Analytical signal - line spectrum]]
+[[File:P_ID991__Mod_Z_2_2.png|right|frame|Analytical signal - Line spectrum]]
 Let us consider two harmonic oscillations
 :$$ s_1(t)  = A_1 \cdot \cos(\omega_{\rm 1} \cdot t ) \hspace{0.05cm},$$
@@ Line 35: / Line 35: @@
 $P_1 \ = \ $ { 2 3% } $\ \rm V^{ 2 }$
-{Let&nbsp;$R = 50 \ \rm Ω$.&nbsp; What is the physical power of the signal &nbsp;$s_1(t)$?
+{Let&nbsp; $R = 50 \ \rm Ω$.&nbsp; What is the physical power of the signal &nbsp;$s_1(t)$?
 |type="{}"}
 $P_1 \ = \ $ { 40 3% } $\ \text{mW}$

@@ Line 9: / Line 9: @@
 where&nbsp;  $f_2 ≥ f_1$&nbsp; should hold for the frequencies.
-The graph shows the spectrum of the analytical signal &nbsp;$s_+(t)$, which is additively composed of the two components &nbsp;$s_{1+}(t)$&nbsp; and &nbsp;$s_ {2+}(t)$&nbsp;.
+*The graphic on the right  shows the spectrum of the analytical signal &nbsp;$s_+(t)$, which is additively composed of the two components &nbsp;$s_{1+}(t)$&nbsp; and &nbsp;$s_ {2+}(t)$&nbsp;.
-Here, the transmission power &nbsp;$P_{\rm S}$&nbsp; should be understood as the root mean square value of the signal &nbsp;$s(t)$, averaged over the largest measurement period possible:
+*Here,&nbsp; the transmission power &nbsp;$P_{\rm S}$&nbsp; should be understood as the root mean square value of the signal &nbsp;$s(t)$,&nbsp; averaged over the largest measurement period possible:
 :$$P_{\rm S} = \lim_{T_{\rm M} \rightarrow \infty}\hspace{0.1cm}\frac{1}{T_{\rm M}} \cdot \int_{0}^{ T_{\rm M}} {s^2(t) }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
-According to this definition, if &nbsp;$s(t)$&nbsp;describes a voltage waveform, &nbsp;$P_{\rm S}$&nbsp; has unit &nbsp; $\rm V^2$&nbsp; and refers to resistance &nbsp;$R = 1 \ \rm Ω$.&nbsp; Dividing by&nbsp;$R$&nbsp; gives the physical power in &nbsp; $\rm W$.
+*According to this definition:&nbsp; If &nbsp;$s(t)$&nbsp;describes a voltage waveform, &nbsp;$P_{\rm S}$&nbsp; has unit &nbsp; $\rm V^2$&nbsp; and refers to resistance &nbsp;$R = 1 \ \rm Ω$.&nbsp;
@@ Line 19: / Line 20: @@
 Hints:
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Double-Sideband_Amplitude_Modulation|Double-Sideband Amplitude Modulation]].
 *Reference is also made to the chapter&nbsp;  [[Modulation_Methods/Quality_Criteria|Quality Criteria]].
-*Use the numerical values &nbsp;$A_1 = 2\ \rm  V$, &nbsp;$A_2 = 1 \ \rm V$&nbsp; and &nbsp;$R = 50 \ \rm Ω$.
+*Use the numerical values &nbsp;$A_1 = 2\ \rm  V$, &nbsp;$A_2 = 1 \ \rm V$,&nbsp; and &nbsp;$R = 50 \ \rm Ω$.