Difference between revisions of "Aufgaben:Exercise 1.2Z: Linear Distorting System"

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{{quiz-Header|Buchseite=Modulationsverfahren/Qualitätskriterien
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{{quiz-Header|Buchseite=Modulation_Methods/Quality_Criteria
 
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[[File:P_ID957__Mod_Z_1_2.png|right|frame|Zur Herleitung der Verzerrungen <br>bei Rechtecksignalen]]
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[[File:P_ID957__Mod_Z_1_2.png|right|frame|Distortions for square wave signals]]
Modulator, Kanal und Demodulator einer Einrichtung zur Nachrichtenübertragung können durch ein einziges lineares System mit dem Frequenzgang
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Modulator,&nbsp; channel,&nbsp; and demodulator of a communication system can be represented by a single linear system with frequency response
:$$ H(f) = {\rm si }( \pi \cdot f \cdot \Delta t)$$
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:$$ H(f) = {\rm si }( \pi \cdot f \cdot \Delta t)= {\rm sinc }(f \cdot \Delta t)$$
beschrieben werden.&nbsp; Die dazugehörige Impulsantwort ist rechteckförmig, symmetrisch um &nbsp;$t = 0$&nbsp; und weist die Höhe &nbsp;$1/Δt$&nbsp; sowie die (äquivalente) Dauer &nbsp;$Δt$&nbsp; auf:
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The corresponding impulse response is rectangular,&nbsp; symmetrical about &nbsp;$t = 0$&nbsp; and has height &nbsp;$1/Δt$&nbsp; and (equivalent) duration &nbsp;$Δt$&nbsp;:
:$$ h(t) = \left\{ \begin{array}{c} 1/\Delta t \\ 1/(2\Delta t) \\ 0 \\ \end{array} \right. \begin{array}{*{4}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} t\hspace{0.05cm} \right| < \Delta t/2,} \\ {\left| \hspace{0.005cm}t\hspace{0.05cm} \right| = \Delta t/2,} \\ {\left|\hspace{0.005cm} t \hspace{0.05cm} \right| > \Delta t/2.} \\ \end{array}$$
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:$$ h(t) = \left\{ \begin{array}{c} 1/\Delta t \\ 1/(2\Delta t) \\ 0 \\ \end{array} \right. \begin{array}{*{4}c} {\rm{for}} \\ {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} t\hspace{0.05cm} \right| < \Delta t/2,} \\ {\left| \hspace{0.005cm}t\hspace{0.05cm} \right| = \Delta t/2,} \\ {\left|\hspace{0.005cm} t \hspace{0.05cm} \right| > \Delta t/2.} \\ \end{array}$$
  
Es handelt sich um einen Spalttiefpass, der im Kapitel &nbsp;[[Lineare_zeitinvariante_Systeme/Einige_systemtheoretische_Tiefpassfunktionen|Einige systemtheoretische Tiefpassfunktionen]]&nbsp; des Buches „Lineare zeitinvariante Systeme” behandelt wurde.
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This is a rectangular-in-time low-pass filter,&nbsp; as discussed in the chapter &nbsp;[[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Slit_low-pass_filter_.E2.80.93_Rectangular-in-time|Some Low-Pass Functions in Systems Theory]]&nbsp; in the book "Linear and Time Invariant Systems".
  
Am Systemeingang liegt das periodische Rechtecksignal &nbsp;$q(t)$&nbsp; mit der Periodendauer &nbsp;$T_0$&nbsp; an.&nbsp; Die Dauer der einzelnen Rechtecke und die der Lücken sind somit jeweils &nbsp;$T_0/2$.&nbsp; Die Höhe der Rechtecke beträgt &nbsp;$2\ \rm V$.
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The periodic square wave signal &nbsp;$q(t)$&nbsp; of period &nbsp;$T_0$&nbsp; is applied at the system input.&nbsp; Thus, the duration of each rectangle and each gap is &nbsp;$T_0/2$.&nbsp; The height of the rectangles is &nbsp;$2\ \rm V$.
  
Das Signal &nbsp;$v(t)$&nbsp; am Systemausgang wird als Sinkensignal bezeichnet.&nbsp; Dieses ist für zwei verschiedene Parameterwerte der äquivalenten Impulsdauer in der Grafik dargestellt (rote Kurvenverläufe):
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The signal &nbsp;$v(t)$&nbsp; at the system output is called the sink signal.&nbsp; This is represented in the graph for two different parameter values for the equivalent pulse duration (red waveforms):
* Das Signal &nbsp;$v_1(t)$&nbsp; ergibt sich, wenn die äquivalente Impulsdauer von &nbsp;$h(t)$&nbsp; genau &nbsp;$Δt_1$&nbsp; ist.
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* The signal &nbsp;$v_1(t)$&nbsp; results when the equivalent pulse duration of &nbsp;$h(t)$&nbsp; is exactly &nbsp;$Δt_1$&nbsp;.
* Entsprechend ergibt sich das Signal &nbsp;$v_2(t)$&nbsp; mit der äquivalenten Impulsdauer &nbsp;$Δt_2$.
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* Accordingly, the signal &nbsp;$v_2(t)$&nbsp; is obtained with the equivalent pulse duration of &nbsp;$Δt_2$.
  
  
Die Veränderung vom Rechtecksignal &nbsp;$q(t)$&nbsp; zum dreieck- bzw. trapezförmigen Sinkensignal &nbsp;$v(t)$&nbsp; ist auf lineare Verzerrungen zurückzuführen und wird durch das Fehlersignal &nbsp;$ε(t) = v(t) - q(t)$&nbsp; erfasst.
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The change from the square wave signal &nbsp;$q(t)$&nbsp; to the triangular or trapezoidal sink signal &nbsp;$v(t)$&nbsp; is due to linear distortions and is captured by the error signal &nbsp;
 +
:$$ε(t) = v(t) - q(t).$$  
  
Mit den Leistungen &nbsp;$P_q$&nbsp; und &nbsp;$P_ε$&nbsp; der Signale &nbsp;$q(t)$&nbsp; und &nbsp;$ε(t)$&nbsp; kann das Sinken–SNR berechnet werden:
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Using the signal powers  &nbsp;$P_q$&nbsp; and &nbsp;$P_ε$&nbsp; of &nbsp;$q(t)$&nbsp; and &nbsp;$ε(t)$,&nbsp; respectively, the sink SNR can be calculated:
  
 
:$$\rho_{v} =P_{q}/{P_{\varepsilon }} \hspace{0.05cm}.$$
 
:$$\rho_{v} =P_{q}/{P_{\varepsilon }} \hspace{0.05cm}.$$
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Hints:  
 
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*This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Quality_Criteria|Quality Criteria]].&nbsp; Particular reference is made to the page &nbsp;[[Modulation_Methods/Quality_Criteria#Signal.E2.80.93to.E2.80.93noise_.28power.29_ratio|Signal–to–noise (power) ratio]] and to the chapter &nbsp;[[Linear_and_Time_Invariant_Systems/Linear_Distortions|Linear Distortions]]&nbsp; in the book&nbsp; "Linear and Time Invarian Systems".
 
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*The powers &nbsp;$P_q$&nbsp; and &nbsp;$P_ε$&nbsp; are the root mean square values of the signals &nbsp;$q(t)$&nbsp; and &nbsp;$ε(t)$&nbsp; and can be determined for periodic signals with period duration &nbsp;$T_0$&nbsp; as follows:
 
 
''Hinweise:''
 
*Die Aufgabe gehört zum  Kapitel&nbsp; [[Modulationsverfahren/Qualitätskriterien|Qualitätskriterien]].&nbsp; Bezug genommen wird insbesondere auf die Seite &nbsp;[[Modulationsverfahren/Qualitätskriterien#Signal.E2.80.93zu.E2.80.93St.C3.B6r.E2.80.93Leistungsverh.C3.A4ltnis|Signal-zu-Stör-Leistungsverhältnis]] und auf das Kapitel &nbsp;[[Lineare_zeitinvariante_Systeme/Lineare_Verzerrungen|Lineare Verzerrungen]] im Buch &bdquo;Lineare zeitinvariante Systeme&rdquo;.
 
*Die Leistungen &nbsp;$P_q$&nbsp; und &nbsp;$P_ε$&nbsp; sind die quadratischen Mittelwerte der Signale &nbsp;$q(t)$&nbsp; und &nbsp;$ε(t)$&nbsp; und können bei periodischen Signalen mit der Periodendauer &nbsp;$T_0$&nbsp; wie folgt ermittelt werden:
 
 
:$$P_{q} = \overline{q(t)^2} = \frac{1}{T_{\rm 0}} \cdot \int_{0}^{ T_{\rm 0}} {q(t)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}, \hspace{0.5cm} P_{\varepsilon} = \overline{\varepsilon(t)^2} = \frac{1}{T_{\rm 0}} \cdot \int_{0}^{ T_{\rm 0}} {\varepsilon(t)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
 
:$$P_{q} = \overline{q(t)^2} = \frac{1}{T_{\rm 0}} \cdot \int_{0}^{ T_{\rm 0}} {q(t)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}, \hspace{0.5cm} P_{\varepsilon} = \overline{\varepsilon(t)^2} = \frac{1}{T_{\rm 0}} \cdot \int_{0}^{ T_{\rm 0}} {\varepsilon(t)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
*Die Angabe von Leistungen in &nbsp;$\rm V^2$&nbsp; bedeutet, dass die Signale auf den Widerstand &nbsp;$R = 1\ \rm \Omega$&nbsp; bezogen werden.
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*Specifying the powers in &nbsp;$\rm V^2$,&nbsp; means that the signals refer to a resistance of &nbsp;$R = 1\ \rm \Omega$&nbsp;.
   
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*For abbreviation we define the following functions:
 +
#&nbsp;$\text{sinc&ndash;function}$&nbsp; (predominantly used in Anglo-American literature) &nbsp; &nbsp; &rArr; &nbsp; &nbsp; ${\rm sinc}( x ) =  {\sin  (\pi  x) }/(\pi  x ),$
 +
#&nbsp;$\text{si&ndash;function}$&nbsp; or $\text{splitting function}$ &nbsp;(predominantly used in German literature) &nbsp; &nbsp; &rArr; &nbsp; &nbsp; ${\rm si}\left( x \right) = \sin \left( x \right)/x = {\rm sinc}(x/\pi ).$  
  
  
  
===Fragebogen===
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===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Wie groß ist die äquivalente Impulsdauer &nbsp;$Δt_1$&nbsp; innerhalb des Signals &nbsp;$v_1(t)$, bezogen auf die Periode&nbsp; $T_0$?
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{How long is the equivalent pulse duration &nbsp;$Δt_1$&nbsp; within the signal &nbsp;$v_1(t)$,&nbsp; relative to the period &nbsp; $T_0$?
 
|type="{}"}
 
|type="{}"}
 
$Δt_1/T_0 \ = \ $ { 0.5 3% }
 
$Δt_1/T_0 \ = \ $ { 0.5 3% }
  
  
{Wie groß ist der Maximalwert des Fehlersignals &nbsp;$ε_1(t) = v_1(t) - q(t)$?
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{What is the maximum value of the error signal &nbsp;$ε_1(t) = v_1(t) - q(t)$?
 
|type="{}"}
 
|type="{}"}
 
$ε_\text{1, max} \ = \ $ { 1 3% } $\ \rm V$
 
$ε_\text{1, max} \ = \ $ { 1 3% } $\ \rm V$
  
{Wie groß ist die „Leistung” &nbsp;$P_{ε1}$&nbsp; des Fehlersignals, also die mittlere quadratische Abweichung zwischen &nbsp;$v_1(t)$&nbsp; und &nbsp;$q(t)$?
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{What is the "power" &nbsp;$P_{ε1}$&nbsp; of the error signal,&nbsp; i.e.,&nbsp; the mean square deviation between &nbsp;$v_1(t)$&nbsp; and &nbsp;$q(t)$?
 
|type="{}"}
 
|type="{}"}
 
$P_{ε1} \ = \ $ { 0.333 3% } $\ \rm V^2$
 
$P_{ε1} \ = \ $ { 0.333 3% } $\ \rm V^2$
  
{Berechnen Sie die Nutzleistung &nbsp;$P_q$&nbsp; und das Sinken–SNR &nbsp;$ρ_{v1}$.
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{Calculate the useful output &nbsp;$P_q$&nbsp; and the sink SNR &nbsp;$ρ_{v1}$.
 
|type="{}"}
 
|type="{}"}
 
$P_q\ = \ $ { 2 3% } $\ \rm V^2$
 
$P_q\ = \ $ { 2 3% } $\ \rm V^2$
 
$ρ_{v1} \ = \ $ { 6 3% }
 
$ρ_{v1} \ = \ $ { 6 3% }
  
{Wie groß ist die äquivalente Impulsdauer &nbsp;$Δt_2$&nbsp; innerhalb des Signals &nbsp;$v_2(t)$, bezogen auf die Periode &nbsp;$T_0$?
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{How long is the equivalent pulse duration &nbsp;$Δt_2$&nbsp; within the signal &nbsp;$v_2(t)$,&nbsp; relative to the period &nbsp;$T_0$?
 
|type="{}"}
 
|type="{}"}
 
$Δt_2/T_0 \ = \ $  { 0.25 3% }  
 
$Δt_2/T_0 \ = \ $  { 0.25 3% }  
  
{Ermitteln Sie das Fehlersignal &nbsp;$ε_2(t) = v_2(t) - q(t)$, die Verzerrungsleistung &nbsp;$P_{ε2}$&nbsp; und das Sinken–SNR &nbsp;$ρ_{v2}$.
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{Determine the error signal &nbsp;$ε_2(t) = v_2(t) - q(t)$,&nbsp; the distortion power &nbsp;$P_{ε2}$&nbsp; and the sink SNR &nbsp;$ρ_{v2}$.
 
|type="{}"}
 
|type="{}"}
 
$P_{ε2} \ = \ $ { 0.167 3% } $\ \rm V^2$
 
$P_{ε2} \ = \ $ { 0.167 3% } $\ \rm V^2$
 
$ρ_{v2} \ = \ $ { 12 3% }  
 
$ρ_{v2} \ = \ $ { 12 3% }  
  
{Verallgemeinern Sie Ihre Ergebnisse für eine beliebige äquivalente Impulsdauer &nbsp;$Δt$.&nbsp; Welches Sinken–SNR &nbsp;$ρ_{v3}$&nbsp; ergibt sich für &nbsp;$Δt_3 = T_0/20$?
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{Generalize your results for an arbitrary equivalent pulse duration &nbsp;$Δt$.&nbsp; What sink SNR &nbsp;$ρ_{v3}$&nbsp; results from &nbsp;$Δt_3 = T_0/20$?
 
|type="{}"}
 
|type="{}"}
 
$ρ_{v3} \ = \ ${ 60 3% }  
 
$ρ_{v3} \ = \ ${ 60 3% }  
 
</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)''' &nbsp; Allgemein gilt&nbsp; $v(t) = q(t) ∗ h(t)$.&nbsp; Die Faltung des periodischen Rechtecksignals&nbsp; $q(t)$&nbsp; mit der ebenfalls rechteckigen Impulsantwort&nbsp; $h(t)$&nbsp; liefert nur dann ein Dreiecksignal&nbsp; $v(t)$, wenn die miteinander gefalteten Rechtecke gleiche Breite haben. Daraus folgt:
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'''(1)''' &nbsp; In general,&nbsp; convolution of the periodic rectangular signal &nbsp; $q(t)$&nbsp; with the similarly rectangular impulse response&nbsp; $h(t)$&nbsp; yields a triangular signal &nbsp; $v(t)$,&nbsp; only if the rectangles involved have equal width.&nbsp; It follows:
 
:$$\Delta t_1 = T_0 /2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \Delta t_1 / T_0\hspace{0.15cm}\underline {= 0.5} \hspace{0.05cm}.$$
 
:$$\Delta t_1 = T_0 /2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \Delta t_1 / T_0\hspace{0.15cm}\underline {= 0.5} \hspace{0.05cm}.$$
  
 
+
[[File:P_ID958__Mod_Z_1_2_b.png|right|frame|Error signals for the two receive filters of different widths]]
 
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'''(2)''' &nbsp; The error signal &nbsp; $ε_1(t)$&nbsp; is shown in the adjacent graph.&nbsp; It can be seen that&nbsp; $ε_1(t)$&nbsp;can assume all values in the range &nbsp; $±1 \ \rm V$&nbsp;:
[[File:P_ID958__Mod_Z_1_2_b.png|right|frame|Fehlersignale bei den beiden betrachteten Empfangsfiltern unterschiedlicher Breite]]
 
'''(2)''' &nbsp; Das Fehlersignal&nbsp; $ε_1(t)$&nbsp; ist in nebenstehender Grafik oben dargestellt.&nbsp; Man erkennt, dass&nbsp; $ε_1(t)$&nbsp; alle Werte zwischen&nbsp; $±1 \ \rm V$&nbsp; annehmen kann:
 
 
:$${\varepsilon}_\text{ 1, max} \hspace{0.15cm}\underline {= {1}\;{\rm V}} \hspace{0.05cm}.$$
 
:$${\varepsilon}_\text{ 1, max} \hspace{0.15cm}\underline {= {1}\;{\rm V}} \hspace{0.05cm}.$$
  
 
+
'''(3)''' &nbsp; It is sufficient to average over the time range from&nbsp; $t = 0$&nbsp; to&nbsp; $t =T_0/4$,&nbsp; since all other subintervals contribute identically:
 
 
'''(3)''' &nbsp; Es genügt die Mittelung über den Zeitbereich von&nbsp; $t = 0$&nbsp; bis&nbsp; $t =T_0/4$, da alle anderen Teilintervalle genau gleiche Beiträge liefern:
 
 
:$$P_{\varepsilon{\rm 1}} = \frac{1}{T_{\rm 0}/4} \hspace{-0.05cm}\cdot \hspace{-0.05cm}\int_{0}^{ T_{\rm 0}/4} {\varepsilon_1(t)^2 }\hspace{0.1cm}{\rm d}t = \frac{1 \,{\rm V}^2}{T_{\rm 0}/4} \hspace{-0.05cm}\cdot \hspace{-0.05cm} \int_{0}^{ T_{\rm 0}/4} {\left( 1 - \frac{t}{T_{\rm 0}/4}\right)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
 
:$$P_{\varepsilon{\rm 1}} = \frac{1}{T_{\rm 0}/4} \hspace{-0.05cm}\cdot \hspace{-0.05cm}\int_{0}^{ T_{\rm 0}/4} {\varepsilon_1(t)^2 }\hspace{0.1cm}{\rm d}t = \frac{1 \,{\rm V}^2}{T_{\rm 0}/4} \hspace{-0.05cm}\cdot \hspace{-0.05cm} \int_{0}^{ T_{\rm 0}/4} {\left( 1 - \frac{t}{T_{\rm 0}/4}\right)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
*Mit der Substitution&nbsp; $x = 4 · t/T_0$&nbsp; kann hierfür auch geschrieben werden:
+
*Substituting &nbsp; $x = 4 · t/T_0$&nbsp;, this can also be written as:
 
:$$P_{\varepsilon{\rm 1}} = 1 \,{\rm V}^2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} \int_{0}^{ 1} \hspace{-0.2cm}{\left( 1 - 2x + x^2\right)}\hspace{0.1cm}{\rm d}x \hspace{0.05cm}= 1 \,{\rm V}^2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} \left( 1 - 1 + \frac{1}{3}\right)\hspace{0.15cm}\underline {= 0.333} \,{\rm V}^2\hspace{0.05cm}.$$
 
:$$P_{\varepsilon{\rm 1}} = 1 \,{\rm V}^2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} \int_{0}^{ 1} \hspace{-0.2cm}{\left( 1 - 2x + x^2\right)}\hspace{0.1cm}{\rm d}x \hspace{0.05cm}= 1 \,{\rm V}^2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} \left( 1 - 1 + \frac{1}{3}\right)\hspace{0.15cm}\underline {= 0.333} \,{\rm V}^2\hspace{0.05cm}.$$
  
 
+
'''(4)''' &nbsp; Averaging over one period of the squared source signal yields:
 
 
'''(4)''' &nbsp; Die Mittelung über eine Periode des quadrierten Quellensignals liefert:
 
 
:$$P_{q} = \frac{1}{T_0} \cdot \left[(2\,{\rm V})^2 \cdot \frac{T_0}{2}+(0\,{\rm V})^2 \cdot \frac{T_0}{2} \right]\hspace{0.15cm}\underline {= 2\,{\rm V^2}}\hspace{0.05cm}.$$
 
:$$P_{q} = \frac{1}{T_0} \cdot \left[(2\,{\rm V})^2 \cdot \frac{T_0}{2}+(0\,{\rm V})^2 \cdot \frac{T_0}{2} \right]\hspace{0.15cm}\underline {= 2\,{\rm V^2}}\hspace{0.05cm}.$$
*Das Sinken–SNR beträgt somit
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*The sink SNR is therefore
 
:$$\rho_{v{\rm 1}} = \frac{P_{q}}{P_{\varepsilon {\rm 1}}} = \frac{2 \,{\rm V}^2}{0.333 \,{\rm V}^2}\hspace{0.15cm}\underline {= 6} \hspace{0.05cm}.$$
 
:$$\rho_{v{\rm 1}} = \frac{P_{q}}{P_{\varepsilon {\rm 1}}} = \frac{2 \,{\rm V}^2}{0.333 \,{\rm V}^2}\hspace{0.15cm}\underline {= 6} \hspace{0.05cm}.$$
  
 +
'''(5)''' &nbsp; According to the sketch presented above,&nbsp; a rectangle of duration &nbsp; $0.5 \cdot T_0$&nbsp; now becomes a trapezoid of absolute duration&nbsp; $0.75 · T_0$.
 +
*Thus,&nbsp; according to the laws of convolution,&nbsp; it is obvious that the equivalent pulse duration must be&nbsp; $Δt_2/T_0\hspace{0.15cm}\underline { = 0.25}$&nbsp;.
  
  
'''(5)''' &nbsp; Entsprechend der Skizze auf dem Angabenblatt wird nun aus einem Rechteck der Dauer&nbsp; $0.5 \cdot T_0$&nbsp; ein Trapez der absoluten Dauer&nbsp; $0.75 · T_0$.
+
'''(6)''' &nbsp; The lower plot in the above graph shows that &nbsp; $ε_2(t)$&nbsp; is composed of four triangles within a period of &nbsp; $T_0$&nbsp; just like&nbsp; $ε_1(t)$,&nbsp; though they are only half as wide.&nbsp;  
*Damit ist nach den Gesetzen der Faltung offensichtlich, dass die äquivalente Impulsdauer&nbsp; $Δt_2/T_0\hspace{0.15cm}\underline { = 0.25}$&nbsp; sein muss.
+
*Thus,&nbsp; in half the time, &nbsp; $ε_2(t) = 0$.
 
+
*Because of&nbsp; $ε_\text{2, max} = ε_\text{1, max} = 1 \ \rm V$,&nbsp; one obtains:
 
 
 
 
'''(6)''' &nbsp; Die untere Skizze in obiger Grafik zeigt, dass sich&nbsp; $ε_2(t)$&nbsp; ebenso wie&nbsp; $ε_1(t)$&nbsp; innerhalb einer Periodendauer&nbsp; $T_0$&nbsp; aus vier Dreiecken zusammensetzt, doch sind diese nur halb so breit.&nbsp;
 
*In der Hälfte der Zeit ist nämlich&nbsp; $ε_2(t) = 0$.
 
 
 
*Wegen&nbsp; $ε_\text{2, max} = ε_\text{1, max} = 1 \ \rm V$&nbsp; erhält man:
 
 
:$$P_{\varepsilon{\rm 2}} ={P_{\varepsilon{\rm 1}}}/{2} \hspace{0.15cm}\underline {= 0.167} \,{\rm V}^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v{\rm 2}} = {P_{q}}/{P_{\varepsilon {\rm 2}}}\hspace{0.15cm}\underline {= 12} \hspace{0.05cm}.$$
 
:$$P_{\varepsilon{\rm 2}} ={P_{\varepsilon{\rm 1}}}/{2} \hspace{0.15cm}\underline {= 0.167} \,{\rm V}^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v{\rm 2}} = {P_{q}}/{P_{\varepsilon {\rm 2}}}\hspace{0.15cm}\underline {= 12} \hspace{0.05cm}.$$
  
  
 
+
'''(7)''' &nbsp; For&nbsp; $Δt = T_0/2$&nbsp; the distortion power&nbsp; $P_{ε1} = 1/3 \ \rm  V^{ 2 }$&nbsp;was calculated in subtask&nbsp; '''(3)'''.  
'''(7)''' &nbsp; Für&nbsp; $Δt = T_0/2$&nbsp; wurde in der Teilaufgabe&nbsp; '''(3)'''&nbsp; die Verzerrungsleistung&nbsp; $P_{ε1} = 1/3 \ \rm  V^{ 2 }$&nbsp; berechnet.  
+
*In subtask&nbsp; '''(6)'''&nbsp; it was shown,&nbsp; that for &nbsp; $Δt = T_0/4$&nbsp; the distortion power&nbsp; $P_{ε2}$&nbsp; is only half.
*In der Teilaufgabe&nbsp; '''(6)'''&nbsp; wurde gezeigt, dass bei&nbsp; $Δt = T_0/4$&nbsp; die Verzerrungsleistung&nbsp; $P_{ε2}$&nbsp; nur halb so groß ist.
+
*It was clearly illustrated that a linear relationship holds.&nbsp; For&nbsp; $Δt ≤ T_0/2$&nbsp; we get the following empirical equations:
 
 
*Anschaulich wurde erläutert, dass ein linearer Zusammenhang besteht.&nbsp; Daraus folgen für&nbsp; $Δt ≤ T_0/2$&nbsp; die empirischen Gleichungen:
 
 
:$$P_{\varepsilon} = \frac{2 \,{\rm V}^2}{3} \cdot \frac{\Delta t}{T_0} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v} = \frac{P_{q}}{P_{\varepsilon }}= \frac{3}{\Delta t/T_0} \hspace{0.05cm}.$$
 
:$$P_{\varepsilon} = \frac{2 \,{\rm V}^2}{3} \cdot \frac{\Delta t}{T_0} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v} = \frac{P_{q}}{P_{\varepsilon }}= \frac{3}{\Delta t/T_0} \hspace{0.05cm}.$$
*Der Sonderfall&nbsp; $Δt = T_0/20$&nbsp; führt somit zu den Resultaten:
+
*Thus,&nbsp; the special case&nbsp; $Δt = T_0/20$&nbsp; results in:
 
:$$P_{\varepsilon{\rm 3}} = \frac{2 \,{\rm V}^2}{60} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v{\rm 3}} = \frac{P_{q}}{P_{\varepsilon {\rm 3}}}\hspace{0.15cm}\underline {= 60} \hspace{0.05cm}.$$
 
:$$P_{\varepsilon{\rm 3}} = \frac{2 \,{\rm V}^2}{60} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v{\rm 3}} = \frac{P_{q}}{P_{\varepsilon {\rm 3}}}\hspace{0.15cm}\underline {= 60} \hspace{0.05cm}.$$
  
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[[Category:Aufgaben zu Modulationsverfahren|^1.2 Qualitätskriterien^]]
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[[Category:Modulation Methods: Exercises|^1.2 Quality Criteria^]]

Latest revision as of 17:50, 23 March 2022

Distortions for square wave signals

Modulator,  channel,  and demodulator of a communication system can be represented by a single linear system with frequency response

$$ H(f) = {\rm si }( \pi \cdot f \cdot \Delta t)= {\rm sinc }(f \cdot \Delta t)$$

The corresponding impulse response is rectangular,  symmetrical about  $t = 0$  and has height  $1/Δt$  and (equivalent) duration  $Δt$ :

$$ h(t) = \left\{ \begin{array}{c} 1/\Delta t \\ 1/(2\Delta t) \\ 0 \\ \end{array} \right. \begin{array}{*{4}c} {\rm{for}} \\ {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} t\hspace{0.05cm} \right| < \Delta t/2,} \\ {\left| \hspace{0.005cm}t\hspace{0.05cm} \right| = \Delta t/2,} \\ {\left|\hspace{0.005cm} t \hspace{0.05cm} \right| > \Delta t/2.} \\ \end{array}$$

This is a rectangular-in-time low-pass filter,  as discussed in the chapter  Some Low-Pass Functions in Systems Theory  in the book "Linear and Time Invariant Systems".

The periodic square wave signal  $q(t)$  of period  $T_0$  is applied at the system input.  Thus, the duration of each rectangle and each gap is  $T_0/2$.  The height of the rectangles is  $2\ \rm V$.

The signal  $v(t)$  at the system output is called the sink signal.  This is represented in the graph for two different parameter values for the equivalent pulse duration (red waveforms):

  • The signal  $v_1(t)$  results when the equivalent pulse duration of  $h(t)$  is exactly  $Δt_1$ .
  • Accordingly, the signal  $v_2(t)$  is obtained with the equivalent pulse duration of  $Δt_2$.


The change from the square wave signal  $q(t)$  to the triangular or trapezoidal sink signal  $v(t)$  is due to linear distortions and is captured by the error signal  

$$ε(t) = v(t) - q(t).$$

Using the signal powers  $P_q$  and  $P_ε$  of  $q(t)$  and  $ε(t)$,  respectively, the sink SNR can be calculated:

$$\rho_{v} =P_{q}/{P_{\varepsilon }} \hspace{0.05cm}.$$



Hints:

  • This exercise belongs to the chapter  Quality Criteria.  Particular reference is made to the page  Signal–to–noise (power) ratio and to the chapter  Linear Distortions  in the book  "Linear and Time Invarian Systems".
  • The powers  $P_q$  and  $P_ε$  are the root mean square values of the signals  $q(t)$  and  $ε(t)$  and can be determined for periodic signals with period duration  $T_0$  as follows:
$$P_{q} = \overline{q(t)^2} = \frac{1}{T_{\rm 0}} \cdot \int_{0}^{ T_{\rm 0}} {q(t)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}, \hspace{0.5cm} P_{\varepsilon} = \overline{\varepsilon(t)^2} = \frac{1}{T_{\rm 0}} \cdot \int_{0}^{ T_{\rm 0}} {\varepsilon(t)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
  • Specifying the powers in  $\rm V^2$,  means that the signals refer to a resistance of  $R = 1\ \rm \Omega$ .
  • For abbreviation we define the following functions:
  1.  $\text{sinc–function}$  (predominantly used in Anglo-American literature)     ⇒     ${\rm sinc}( x ) = {\sin (\pi x) }/(\pi x ),$
  2.  $\text{si–function}$  or $\text{splitting function}$  (predominantly used in German literature)     ⇒     ${\rm si}\left( x \right) = \sin \left( x \right)/x = {\rm sinc}(x/\pi ).$


Questions

1

How long is the equivalent pulse duration  $Δt_1$  within the signal  $v_1(t)$,  relative to the period   $T_0$?

$Δt_1/T_0 \ = \ $

2

What is the maximum value of the error signal  $ε_1(t) = v_1(t) - q(t)$?

$ε_\text{1, max} \ = \ $

$\ \rm V$

3

What is the "power"  $P_{ε1}$  of the error signal,  i.e.,  the mean square deviation between  $v_1(t)$  and  $q(t)$?

$P_{ε1} \ = \ $

$\ \rm V^2$

4

Calculate the useful output  $P_q$  and the sink SNR  $ρ_{v1}$.

$P_q\ = \ $

$\ \rm V^2$
$ρ_{v1} \ = \ $

5

How long is the equivalent pulse duration  $Δt_2$  within the signal  $v_2(t)$,  relative to the period  $T_0$?

$Δt_2/T_0 \ = \ $

6

Determine the error signal  $ε_2(t) = v_2(t) - q(t)$,  the distortion power  $P_{ε2}$  and the sink SNR  $ρ_{v2}$.

$P_{ε2} \ = \ $

$\ \rm V^2$
$ρ_{v2} \ = \ $

7

Generalize your results for an arbitrary equivalent pulse duration  $Δt$.  What sink SNR  $ρ_{v3}$  results from  $Δt_3 = T_0/20$?

$ρ_{v3} \ = \ $


Solution

(1)   In general,  convolution of the periodic rectangular signal   $q(t)$  with the similarly rectangular impulse response  $h(t)$  yields a triangular signal   $v(t)$,  only if the rectangles involved have equal width.  It follows:

$$\Delta t_1 = T_0 /2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \Delta t_1 / T_0\hspace{0.15cm}\underline {= 0.5} \hspace{0.05cm}.$$
Error signals for the two receive filters of different widths

(2)   The error signal   $ε_1(t)$  is shown in the adjacent graph.  It can be seen that  $ε_1(t)$ can assume all values in the range   $±1 \ \rm V$ :

$${\varepsilon}_\text{ 1, max} \hspace{0.15cm}\underline {= {1}\;{\rm V}} \hspace{0.05cm}.$$

(3)   It is sufficient to average over the time range from  $t = 0$  to  $t =T_0/4$,  since all other subintervals contribute identically:

$$P_{\varepsilon{\rm 1}} = \frac{1}{T_{\rm 0}/4} \hspace{-0.05cm}\cdot \hspace{-0.05cm}\int_{0}^{ T_{\rm 0}/4} {\varepsilon_1(t)^2 }\hspace{0.1cm}{\rm d}t = \frac{1 \,{\rm V}^2}{T_{\rm 0}/4} \hspace{-0.05cm}\cdot \hspace{-0.05cm} \int_{0}^{ T_{\rm 0}/4} {\left( 1 - \frac{t}{T_{\rm 0}/4}\right)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
  • Substituting   $x = 4 · t/T_0$ , this can also be written as:
$$P_{\varepsilon{\rm 1}} = 1 \,{\rm V}^2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} \int_{0}^{ 1} \hspace{-0.2cm}{\left( 1 - 2x + x^2\right)}\hspace{0.1cm}{\rm d}x \hspace{0.05cm}= 1 \,{\rm V}^2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} \left( 1 - 1 + \frac{1}{3}\right)\hspace{0.15cm}\underline {= 0.333} \,{\rm V}^2\hspace{0.05cm}.$$

(4)   Averaging over one period of the squared source signal yields:

$$P_{q} = \frac{1}{T_0} \cdot \left[(2\,{\rm V})^2 \cdot \frac{T_0}{2}+(0\,{\rm V})^2 \cdot \frac{T_0}{2} \right]\hspace{0.15cm}\underline {= 2\,{\rm V^2}}\hspace{0.05cm}.$$
  • The sink SNR is therefore
$$\rho_{v{\rm 1}} = \frac{P_{q}}{P_{\varepsilon {\rm 1}}} = \frac{2 \,{\rm V}^2}{0.333 \,{\rm V}^2}\hspace{0.15cm}\underline {= 6} \hspace{0.05cm}.$$

(5)   According to the sketch presented above,  a rectangle of duration   $0.5 \cdot T_0$  now becomes a trapezoid of absolute duration  $0.75 · T_0$.

  • Thus,  according to the laws of convolution,  it is obvious that the equivalent pulse duration must be  $Δt_2/T_0\hspace{0.15cm}\underline { = 0.25}$ .


(6)   The lower plot in the above graph shows that   $ε_2(t)$  is composed of four triangles within a period of   $T_0$  just like  $ε_1(t)$,  though they are only half as wide. 

  • Thus,  in half the time,   $ε_2(t) = 0$.
  • Because of  $ε_\text{2, max} = ε_\text{1, max} = 1 \ \rm V$,  one obtains:
$$P_{\varepsilon{\rm 2}} ={P_{\varepsilon{\rm 1}}}/{2} \hspace{0.15cm}\underline {= 0.167} \,{\rm V}^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v{\rm 2}} = {P_{q}}/{P_{\varepsilon {\rm 2}}}\hspace{0.15cm}\underline {= 12} \hspace{0.05cm}.$$


(7)   For  $Δt = T_0/2$  the distortion power  $P_{ε1} = 1/3 \ \rm V^{ 2 }$ was calculated in subtask  (3).

  • In subtask  (6)  it was shown,  that for   $Δt = T_0/4$  the distortion power  $P_{ε2}$  is only half.
  • It was clearly illustrated that a linear relationship holds.  For  $Δt ≤ T_0/2$  we get the following empirical equations:
$$P_{\varepsilon} = \frac{2 \,{\rm V}^2}{3} \cdot \frac{\Delta t}{T_0} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v} = \frac{P_{q}}{P_{\varepsilon }}= \frac{3}{\Delta t/T_0} \hspace{0.05cm}.$$
  • Thus,  the special case  $Δt = T_0/20$  results in:
$$P_{\varepsilon{\rm 3}} = \frac{2 \,{\rm V}^2}{60} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v{\rm 3}} = \frac{P_{q}}{P_{\varepsilon {\rm 3}}}\hspace{0.15cm}\underline {= 60} \hspace{0.05cm}.$$