Aufgaben:Exercise 2.11: Envelope Demodulation of an SSB Signal - Revision history

Reed at 14:17, 9 April 2022

2022-04-09T14:17:34Z

Reed at 14:16, 9 April 2022

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Reed at 16:16, 31 March 2022

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Reed at 15:57, 31 March 2022

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Reed at 15:55, 31 March 2022

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Guenter at 17:09, 18 February 2022

2022-02-18T17:09:20Z

Guenter: Guenter moved page Aufgaben:Exercise 2.11: Envelope Demodulation of an ESB Signal to Aufgaben:Exercise 2.11: Envelope Demodulation of an SSB Signal

2022-02-18T14:22:58Z

Guenter moved page Exercise 2.11: Envelope Demodulation of an ESB Signal to Exercise 2.11: Envelope Demodulation of an SSB Signal

Guenter at 14:21, 18 February 2022

2022-02-18T14:21:50Z

Guenter at 14:07, 18 February 2022

2022-02-18T14:07:22Z

Guenter at 13:50, 18 February 2022

2022-02-18T13:50:49Z

@@ Line 107: / Line 107: @@
 '''(5)'''&nbsp; A smaller sideband-to-carrier ratio also results in a smaller distortion factor:
 :$$K = \frac{0.058{\,\rm V}}{0.484\,{\rm V}}\hspace{0.15cm}\underline { \approx 12 \%}.$$
-*The simple approximation &nbsp; $K ≈ μ/4$&nbsp; yields here&nbsp; $12.5\%$.
+*The simple approximation &nbsp; $K ≈ μ/4$&nbsp; here yields&nbsp; $12.5\%$.
 *It can be concluded that the above rule of thumb is more accurate for smaller values of &nbsp; $μ$.

@@ Line 78: / Line 78: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The maximum value &nbsp; $a_{\rm max} = 2\ \rm  V$&nbsp; and the minimum value &nbsp; $a_{\rm min} = 0$&nbsp; can be read of the graph or calculated using the equation given:
+'''(1)'''&nbsp; The maximum value &nbsp; $a_{\rm max} = 2\ \rm  V$&nbsp; and the minimum value &nbsp; $a_{\rm min} = 0$&nbsp; can be read off the graph or calculated using the equation given:
 :$$ a_{\rm max}  =  A_{\rm T} \cdot \sqrt{1+ \mu^2 + 2 \mu}= A_{\rm T} \cdot (1+ \mu) = 2\,{\rm V} \hspace{0.05cm},$$
 :$$a_{\rm min}  =  A_{\rm T} \cdot \sqrt{1+ \mu^2 - 2 \mu}= A_{\rm T} \cdot (1- \mu) = 0 \hspace{0.05cm}.$$

@@ Line 117: / Line 117: @@
 #a DSB-AM which has been extremely affected by the channel,
-:simultaneously provides&nbsp; $K_{\rm max} ≈ μ/4$&nbsp;  an upper bound for the DSB-AM.
+:&nbsp; $K_{\rm max} ≈ μ/4$&nbsp; simultaneously gives an upper bound for the DSB-AM.
 *A comparison of the parameters&nbsp; $m = A_{\rm N}/A_{\rm T}$&nbsp; and&nbsp; $μ = A_{\rm N}/(2A_{\rm T})$&nbsp; leads to the result:

@@ Line 54: / Line 54: @@
 $K \ = \ $ { 22.3 3%  } $\ \text{%}$
-{How can you recognize the nonlinear distortions in the present signal &nbsp;$v(t)$?
+{How can you recognize the nonlinear distortions in the signal &nbsp;$v(t)$?
 |type="[]"}
 + The lower cosine half-wave is more peaked than the upper one.

← Older revision		Revision as of 15:55, 31 March 2022
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−	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Einseitenbandmodulation~~	+	{{quiz-Header\|Buchseite=Modulation_Methods/Single-Sideband_Modulation
	}}		}}

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 The channel is assumed to be ideal such that the received signal  &nbsp;$r(t)$&nbsp; is identical to the transmitted signal &nbsp;$s(t)$&nbsp;.&nbsp; With the sideband-to-carrier ratio
 :$$ \mu = \frac{A_{\rm N}}{2 \cdot A_{\rm T}}$$
-the equivalent low-pass signal&nbsp; (German:&nbsp; "äquivalentes Tiefpass-Signal" &nbsp; &rArr; &nbsp; subscript:&nbsp; "TP")&nbsp;can be written as:
+the equivalent low-pass signal&nbsp; (German:&nbsp; "äquivalentes Tiefpass-Signal" &nbsp; &rArr; &nbsp; subscript:&nbsp; "TP")&nbsp; can be written as:
 :$$r_{\rm TP}(t) = A_{\rm T} \cdot \left( 1 + \mu \cdot {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\omega_{\rm N}\cdot \hspace{0.03cm}\hspace{0.03cm}t} \right) \hspace{0.05cm}$$

@@ Line 114: / Line 114: @@
 '''(6)'''&nbsp; Thus,&nbsp; the distortion factor is largest when one of the sidebands is entirely cut out.
 *However,&nbsp; since the envelope demodulator has no information to distinguish between
-**a SSB–AM, or
+#a SSB–AM, or
-**a DSB-AM which has been extremely affected by the channel,
+#a DSB-AM which has been extremely affected by the channel,
+:simultaneously provides&nbsp; $K_{\rm max} ≈ μ/4$&nbsp;  an upper bound for the DSB-AM.
 *A comparison of the parameters&nbsp; $m = A_{\rm N}/A_{\rm T}$&nbsp; and&nbsp; $μ = A_{\rm N}/(2A_{\rm T})$&nbsp; leads to the result:

@@ Line 89: / Line 89: @@
 '''(2)'''&nbsp; Ignoring the Fourier coefficients &nbsp; $A_5$,&nbsp; $A_6$,&nbsp; etc., we obtain:
 :$$K = \frac{\sqrt{A_2^2 + A_3^2+ A_4^2 }}{A_1}= \frac{\sqrt{0.170^2 + 0.073^2 + 0.040^2 }{\,\rm V}}{0.849\,{\rm V}}\hspace{0.15cm}\underline { \approx 22.3 \%}.$$
-*Here, the approximation&nbsp; $K ≈ μ/4$&nbsp; yields the value $25\%$.
+*Here,&nbsp; the approximation&nbsp; $K ≈ μ/4$&nbsp; yields the value $25\%$.
-'''(3)'''&nbsp; Only the <u>first answer</u> is correct.
+'''(3)'''&nbsp; Only the&nbsp; <u>first answer</u>&nbsp; is correct.
-*Due to the high-pass within the envelope demodulator, the DC signal component would also be equal to zero if no distortions were present.
+*Due to the high-pass within the envelope demodulator,&nbsp; the DC signal component would also be equal to zero if no distortions were present.
@@ Line 107: / Line 107: @@
 '''(5)'''&nbsp; A smaller sideband-to-carrier ratio also results in a smaller distortion factor:
 :$$K = \frac{0.058{\,\rm V}}{0.484\,{\rm V}}\hspace{0.15cm}\underline { \approx 12 \%}.$$
-*The simple approximation &nbsp; $K ≈ μ/4$&nbsp; here yields &nbsp; $12.5\%$.
+*The simple approximation &nbsp; $K ≈ μ/4$&nbsp; yields here&nbsp; $12.5\%$.
-*It can be concluded that the above rule of thumb is more accurate for smaller values of &nbsp; $μ$&nbsp;.
+*It can be concluded that the above rule of thumb is more accurate for smaller values of &nbsp; $μ$.
-'''(6)'''&nbsp; Thus, the distortion factor is largest when one of the sidebands is entirely cut out.
+'''(6)'''&nbsp; Thus,&nbsp; the distortion factor is largest when one of the sidebands is entirely cut out.
-*However, since the envelope demodulator has no information to distinguish between
+*However,&nbsp; since the envelope demodulator has no information to distinguish between
 **a SSB–AM, or
 **a DSB-AM which has been extremely affected by the channel,
 *A comparison of the parameters &nbsp; $m = A_{\rm N}/A_{\rm T}$&nbsp; and&nbsp; $μ = A_{\rm N}/(2A_{\rm T})$&nbsp; leads to the result:
 :$$K_{\rm max} = \frac{\mu}{4} = \frac{m}{8} \hspace{0.15cm}\underline {=6.25 \%}.$$
 {{ML-Fuß}}

@@ Line 6: / Line 6: @@
 Let us consider the transmission of the cosine signal
 :$$ q(t) = A_{\rm N} \cdot \cos(\omega_{\rm N} \cdot t)$$
-according to the modulation method "USB–AM with carrier".&nbsp; At the receiver, the high frequency is reset to the LF range with an &nbsp; [[Modulation_Methods/Envelope_Demodulation|envelope demodulator]]&nbsp;.
+according to the modulation method&nbsp; $\rm USB–AM$&nbsp; ("upper-sideband amplitude modulation")&nbsp; with carrier.&nbsp; At the receiver,&nbsp; the high frequency range (HF) is reset to the low frquency range (LF) with an &nbsp; [[Modulation_Methods/Envelope_Demodulation|envelope demodulator]].
-The channel is assumed to be ideal such that the received signal &nbsp;  &nbsp;$r(t)$&nbsp; is identical to the transmit signal &nbsp;$s(t)$&nbsp;.&nbsp; With a sideband-to-carrier ratio
+The channel is assumed to be ideal such that the received signal  &nbsp;$r(t)$&nbsp; is identical to the transmitted signal &nbsp;$s(t)$&nbsp;.&nbsp; With the sideband-to-carrier ratio
 :$$ \mu = \frac{A_{\rm N}}{2 \cdot A_{\rm T}}$$
-the equivalent lowpass signal can be written as:
+the equivalent low-pass signal&nbsp; (German:&nbsp; "äquivalentes Tiefpass-Signal" &nbsp; &rArr; &nbsp; subscript:&nbsp; "TP")&nbsp;can be written as:
 :$$r_{\rm TP}(t) = A_{\rm T} \cdot \left( 1 + \mu \cdot {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\omega_{\rm N}\cdot \hspace{0.03cm}\hspace{0.03cm}t} \right) \hspace{0.05cm}$$
-The envelope – i.e., the magnitude of this complex signal – can be determined by geometric considerations.&nbsp; Independent of the parameter &nbsp;$μ$, one obtains:
+The envelope – i.e.,&nbsp; the magnitude of this complex signal – can be determined by geometric considerations.&nbsp; Independent of the parameter &nbsp;$μ$,&nbsp; one obtains:
 :$$a(t ) = A_{\rm T} \cdot \sqrt{1+ \mu^2 + 2 \mu \cdot \cos(\omega_{\rm N} \cdot t)}\hspace{0.05cm}.$$
-The time-independent envelope &nbsp;$a(t)$&nbsp; for &nbsp;$μ = 1$&nbsp; and &nbsp;$μ = 0.5$&nbsp; is shown in the graph.&nbsp; In each case, the amplitude-matched cosine oscillations, which would be a prerequisite for distortion-free demodulation, are plotted as dashed comparison curves.
+The time-independent envelope &nbsp;$a(t)$&nbsp; for &nbsp;$μ = 1$&nbsp; and &nbsp;$μ = 0.5$&nbsp; is shown in the graph.&nbsp; In each case,&nbsp; the amplitude-matched cosine oscillations,&nbsp; which would be a prerequisite for distortion-free demodulation,&nbsp; are plotted as dashed comparison curves.
-*The periodic signal&nbsp;$a(t)$&nbsp; can be approximated by a&nbsp; [[Signal_Representation/Fourier_Series|Fourier Series]] :
+*The periodic signal&nbsp; $a(t)$&nbsp; can be approximated by a&nbsp; [[Signal_Representation/Fourier_Series|Fourier series]] :
 :$$a(t ) = A_{\rm 0} + A_{\rm 1} \cdot \cos(\omega_{\rm N} \cdot t) + A_{\rm 2} \cdot \cos(2\omega_{\rm N} \cdot t)+ A_{\rm 3} \cdot \cos(3\omega_{\rm N} \cdot t)\hspace{0.05cm}+\text{...}$$
 *The Fourier coefficients were determined using a simulation program. &nbsp; With &nbsp;$μ = 1$&nbsp; the following values were obtained:
 :$$A_{\rm 0} = 1.273\,{\rm V},\hspace{0.3cm} A_{\rm 1} = 0.849\,{\rm V},\hspace{0.3cm}A_{\rm 2} = -0.170\,{\rm V},\hspace{0.3cm} A_{\rm 3} = 0.073\,{\rm V},\hspace{0.3cm}A_{\rm 4} = 0.040\,{\rm V} \hspace{0.05cm}.$$
-*Accordingly, for &nbsp;$μ = 0.5$, the simulation yielded:
+*Accordingly,&nbsp; for &nbsp;$μ = 0.5$,&nbsp; the simulation yielded:
 :$$A_{\rm 0} = 1.064\,{\rm V},\hspace{0.3cm} A_{\rm 1} = 0.484\,{\rm V},\hspace{0.3cm}A_{\rm 2} = 0.058\,{\rm V} \hspace{0.05cm}.$$
 :The values not given here can be ignored when calculating of the distortion factor.
 *The sink signal &nbsp;$v(t)$&nbsp; is obtained from &nbsp;$a(t)$&nbsp; as follows:
 :$$v(t) = 2 \cdot \big [a(t ) - A_{\rm 0} \big ] \hspace{0.05cm}.$$
-:The factor of&nbsp; $2$&nbsp; corrects for the amplitude loss due to SSB-AM, while the subtraction of the DC signal coefficient &nbsp;$A_0$&nbsp; takes into account the influence of the high-pass within the envelope demodulator.
+:The factor of&nbsp; $2$&nbsp; corrects for the amplitude loss due to&nbsp; "single-sideband amplitude modulation",&nbsp; while the subtraction of the DC signal coefficient &nbsp;$A_0$&nbsp; takes into account the influence of the high-pass within the envelope demodulator.
 Hints:
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Single-Sideband_Modulation|Single-Sideband Modulation]].
 *Particular reference is made to the page &nbsp;   [[Modulation_Methods/Single-Sideband_Modulation#Sideband-to-carrier_ratio|Sideband-to-carrier ratio]].
-*Also compare your results to the rule of thumb which states that for the envelope demodulation of an SSB-AM signal with sideband-to-carrier ratio &nbsp;$μ$&nbsp;, the distortion factor is &nbsp;$K ≈ μ/4$&nbsp;.
+*Also compare your results to the rule of thumb which states that <br>"for the envelope demodulation of an SSB-AM signal with sideband-to-carrier ratio &nbsp;$μ$,&nbsp; the distortion factor is &nbsp;$K ≈ μ/4$".
@@ Line 51: / Line 45: @@
 <quiz display=simple>
-{Give the maximum and minimum values of the sink signal &nbsp;$v(t)$&nbsp; when &nbsp;$μ = 1$&nbsp;.
+{Give the maximum and minimum values of the sink signal &nbsp;$v(t)$&nbsp; when &nbsp;$μ = 1$.
 |type="{}"}
 $v_{\rm max} \ = \ $ { 1.454 3% } $\ \rm V$
@@ Line 65: / Line 59: @@
 - The DC component &nbsp;${\rm Ε}\big[v(t)\big ] = 0$.
-{Give the maximum and minimum values of the sink signal &nbsp;$v(t)$&nbsp; when &nbsp;$μ = 0.5$&nbsp;.
+{Give the maximum and minimum values of the sink signal &nbsp;$v(t)$&nbsp; when &nbsp;$μ = 0.5$.
 |type="{}"}
 $v_{\rm max} \ = \ $ { 0.872 3% } $\ \rm V$
@@ Line 74: / Line 68: @@
 $K \ = \ $ { 12 3% } $\ \text{%}$
-{What is the upper bound &nbsp;$K_{\rm max}$&nbsp; of the distortion factor in DSB-AM with &nbsp;$m = 0.5$&nbsp; and envelope demodulation, if one sideband is completely damped by the channel.
+{What is the upper bound &nbsp;$K_{\rm max}$&nbsp; of the distortion factor in&nbsp; "double-sideband amplitude modulation"&nbsp; $\text{(DSB-AM)}$&nbsp; with &nbsp;$m = 0.5$&nbsp; <br>and envelope demodulation,&nbsp; if one sideband is completely damped by the channel.
 |type="{}"}
 $K_{\rm max} \ = \ ${ 6.25 3% } $\ \text{%}$