Aufgaben:Exercise 1.2Z: Linear Distorting System - Revision history

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@@ Line 76: / Line 76: @@
 ===Solution===
 {{ML-Kopf}}
-'''(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; only yields a triangular signal &nbsp; $v(t)$,&nbsp; only if the rectangles involved have equal width.&nbsp; It follows:
+'''(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:
 :$$\Delta t_1 = T_0 /2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \Delta t_1 / T_0\hspace{0.15cm}\underline {= 0.5} \hspace{0.05cm}.$$

@@ Line 9: / Line 9: @@
 :$$ h(t) = \left\{ \begin{array}{c} 1/\Delta t \\ 1/(2\Delta t) \\ 0 \\ \end{array} \right. \begin{array}{*{4}c} {\rm{for}} \\ {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} t\hspace{0.05cm} \right| < \Delta t/2,} \\ {\left| \hspace{0.005cm}t\hspace{0.05cm} \right| = \Delta t/2,} \\ {\left|\hspace{0.005cm} t \hspace{0.05cm} \right| > \Delta t/2.} \\ \end{array}$$
-This is a rectangular-in-time low-pass filter,&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”.
+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".
 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;$2\ \rm V$.

@@ Line 29: / Line 29: @@
 Hints:
-*This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Qualitätskriterien|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".
+*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".
 *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:
 :$$P_{q} = \overline{q(t)^2} = \frac{1}{T_{\rm 0}} \cdot \int_{0}^{ T_{\rm 0}} {q(t)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}, \hspace{0.5cm} P_{\varepsilon} = \overline{\varepsilon(t)^2} = \frac{1}{T_{\rm 0}} \cdot \int_{0}^{ T_{\rm 0}} {\varepsilon(t)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$

@@ Line 76: / Line 76: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)''' &nbsp; In general, Convolution of the periodic rectangular signal &nbsp; $q(t)$&nbsp; with the similarly rectangular impulse response&nbsp; $h(t)$&nbsp; only yields a triangular signal &nbsp; $v(t)$, only if the rectangles involved have equal width. It follows:
+'''(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; only yields a triangular signal &nbsp; $v(t)$,&nbsp; only if the rectangles involved have equal width.&nbsp; It follows:
 :$$\Delta t_1 = T_0 /2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \Delta t_1 / T_0\hspace{0.15cm}\underline {= 0.5} \hspace{0.05cm}.$$
+[[File:P_ID958__Mod_Z_1_2_b.png|right|frame|Error signals for the two receive filters of different widths]]
 '''(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;:
 :$${\varepsilon}_\text{ 1, max} \hspace{0.15cm}\underline {= {1}\;{\rm V}} \hspace{0.05cm}.$$
+'''(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:
 :$$P_{\varepsilon{\rm 1}} = \frac{1}{T_{\rm 0}/4} \hspace{-0.05cm}\cdot \hspace{-0.05cm}\int_{0}^{ T_{\rm 0}/4} {\varepsilon_1(t)^2 }\hspace{0.1cm}{\rm d}t = \frac{1 \,{\rm V}^2}{T_{\rm 0}/4} \hspace{-0.05cm}\cdot \hspace{-0.05cm} \int_{0}^{ T_{\rm 0}/4} {\left( 1 - \frac{t}{T_{\rm 0}/4}\right)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
 *Substituting &nbsp; $x = 4 · t/T_0$&nbsp;, this can also be written as:
 :$$P_{\varepsilon{\rm 1}} = 1 \,{\rm V}^2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} \int_{0}^{ 1} \hspace{-0.2cm}{\left( 1 - 2x + x^2\right)}\hspace{0.1cm}{\rm d}x \hspace{0.05cm}= 1 \,{\rm V}^2 \hspace{-0.05cm}\cdot \hspace{-0.05cm} \left( 1 - 1 + \frac{1}{3}\right)\hspace{0.15cm}\underline {= 0.333} \,{\rm V}^2\hspace{0.05cm}.$$
 '''(4)''' &nbsp; Averaging over one period of the squared source signal yields:
@@ Line 99: / Line 93: @@
 :$$\rho_{v{\rm 1}} = \frac{P_{q}}{P_{\varepsilon {\rm 1}}} = \frac{2 \,{\rm V}^2}{0.333 \,{\rm V}^2}\hspace{0.15cm}\underline {= 6} \hspace{0.05cm}.$$
-'''(5)''' &nbsp; According to the sketch presented above, a rectangle of duration &nbsp; $0.5 \cdot T_0$&nbsp; now becomes a trapezoid of absolute duration&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;
-*Thus, according to the laws of convolution, it is obvious that the equivalent pulse duration must be&nbsp; $Δt_2/T_0\hspace{0.15cm}\underline { = 0.25}$&nbsp;.
+*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:
 :$$P_{\varepsilon{\rm 2}} ={P_{\varepsilon{\rm 1}}}/{2} \hspace{0.15cm}\underline {= 0.167} \,{\rm V}^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v{\rm 2}} = {P_{q}}/{P_{\varepsilon {\rm 2}}}\hspace{0.15cm}\underline {= 12} \hspace{0.05cm}.$$
 '''(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; 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)'''&nbsp; .
+*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 subtask &nbsp; '''(6)'''&nbsp; it was shown, that for &nbsp; $Δt = T_0/4$&nbsp; the distortion power&nbsp; $P_{ε2}$&nbsp; is only half.
+*It was clearly illustrated that a linear relationship holds.&nbsp; For&nbsp; $Δt ≤ T_0/2$&nbsp; 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&nbsp; $Δt = T_0/20$&nbsp; results in:
+*Thus,&nbsp; the special case&nbsp; $Δt = T_0/20$&nbsp; results in:
 :$$P_{\varepsilon{\rm 3}} = \frac{2 \,{\rm V}^2}{60} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \rho_{v{\rm 3}} = \frac{P_{q}}{P_{\varepsilon {\rm 3}}}\hspace{0.15cm}\underline {= 60} \hspace{0.05cm}.$$

@@ Line 29: / Line 29: @@
 Hints:
-*This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Qualitätskriterien|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]] in the book&nbsp; "Linear and Time Invarian Systems".
+*This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Qualitätskriterien|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".
 *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:
 :$$P_{q} = \overline{q(t)^2} = \frac{1}{T_{\rm 0}} \cdot \int_{0}^{ T_{\rm 0}} {q(t)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}, \hspace{0.5cm} P_{\varepsilon} = \overline{\varepsilon(t)^2} = \frac{1}{T_{\rm 0}} \cdot \int_{0}^{ T_{\rm 0}} {\varepsilon(t)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$

@@ Line 9: / Line 9: @@
 :$$ h(t) = \left\{ \begin{array}{c} 1/\Delta t \\ 1/(2\Delta t) \\ 0 \\ \end{array} \right. \begin{array}{*{4}c} {\rm{for}} \\ {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} t\hspace{0.05cm} \right| < \Delta t/2,} \\ {\left| \hspace{0.005cm}t\hspace{0.05cm} \right| = \Delta t/2,} \\ {\left|\hspace{0.005cm} t \hspace{0.05cm} \right| > \Delta t/2.} \\ \end{array}$$
-This is a slit low-pass filter, 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”.
+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”.
 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;$2\ \rm V$.
-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 with equivalent pulse duration (red waveforms):
+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):
 * The signal &nbsp;$v_1(t)$&nbsp; results when the equivalent pulse duration of &nbsp;$h(t)$&nbsp; is exactly &nbsp;$Δt_1$&nbsp;.
 * Accordingly, the signal &nbsp;$v_2(t)$&nbsp; is obtained with the equivalent pulse duration of &nbsp;$Δt_2$.
-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;$ε(t) = v(t) - q(t)$&nbsp;.
+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;
-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:
+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:
 :$$\rho_{v} =P_{q}/{P_{\varepsilon }} \hspace{0.05cm}.$$
@@ Line 35: / Line 36: @@
 *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:
 :$$P_{q} = \overline{q(t)^2} = \frac{1}{T_{\rm 0}} \cdot \int_{0}^{ T_{\rm 0}} {q(t)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}, \hspace{0.5cm} P_{\varepsilon} = \overline{\varepsilon(t)^2} = \frac{1}{T_{\rm 0}} \cdot \int_{0}^{ T_{\rm 0}} {\varepsilon(t)^2 }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
-*Specifying the powers in &nbsp;$\rm V^2$&nbsp;, means that the signals refer to a resistance of &nbsp;$R = 1\ \rm \Omega$&nbsp;.
+*Specifying the powers in &nbsp;$\rm V^2$,&nbsp; means that the signals refer to a resistance of &nbsp;$R = 1\ \rm \Omega$&nbsp;.
-*'''si'''
+*For abbreviation we define the following functions:
@@ Line 43: / Line 46: @@
 <quiz display=simple>
-{How long is the equivalent pulse duration &nbsp;$Δt_1$&nbsp; within the signal &nbsp;$v_1(t)$, relative to the period &nbsp; $T_0$?
+{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="{}"}
 $Δt_1/T_0 \ = \ $ { 0.5 3% }
@@ Line 52: / Line 55: @@
 $ε_\text{1, max} \ = \ $ { 1 3% } $\ \rm V$
-{What is the "power" &nbsp;$P_{ε1}$&nbsp; of the error signal, i.e., the mean square deviation between &nbsp;$v_1(t)$&nbsp; and &nbsp;$q(t)$?
+{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="{}"}
 $P_{ε1} \ = \ $ { 0.333 3% } $\ \rm V^2$
@@ Line 61: / Line 64: @@
 $ρ_{v1} \ = \ $ { 6 3% }
-{How long is the equivalent pulse duration &nbsp;$Δt_2$&nbsp; within the signal &nbsp;$v_2(t)$, relative to the period &nbsp;$T_0$?
+{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="{}"}
 $Δt_2/T_0 \ = \ $  { 0.25 3% }
-{Determine the error signal &nbsp;$ε_2(t) = v_2(t) - q(t)$, the distortion power &nbsp;$P_{ε2}$&nbsp; and the sink SNR &nbsp;$ρ_{v2}$.
+{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="{}"}
 $P_{ε2} \ = \ $ { 0.167 3% } $\ \rm V^2$

@@ Line 6: / Line 6: @@
 Modulator,&nbsp; channel,&nbsp; and demodulator of a communication system can be represented by a single linear system with frequency response
 :$$ H(f) = {\rm si }( \pi \cdot f \cdot \Delta t)= {\rm sinc }(f \cdot \Delta t)$$
-The corresponding impulse response is rectangular, symmetrical about &nbsp;$t = 0$&nbsp; and has height &nbsp;$1/Δt$&nbsp; and (equivalent) duration &nbsp;$Δt$&nbsp;:
+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;:
 :$$ h(t) = \left\{ \begin{array}{c} 1/\Delta t \\ 1/(2\Delta t) \\ 0 \\ \end{array} \right. \begin{array}{*{4}c} {\rm{for}} \\ {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} t\hspace{0.05cm} \right| < \Delta t/2,} \\ {\left| \hspace{0.005cm}t\hspace{0.05cm} \right| = \Delta t/2,} \\ {\left|\hspace{0.005cm} t \hspace{0.05cm} \right| > \Delta t/2.} \\ \end{array}$$