Aufgaben:Exercise 4.7: Copper Twin Wire 0.5 mm - Revision history

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Guenter at 15:05, 29 October 2021

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Guenter at 15:02, 29 October 2021

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Guenter: Guenter moved page Aufgaben:Aufgabe 4.7: Kupfer-Doppelader 0.5 mm to Aufgaben:Exercise 4.7: Copper Twin Wire 0.5 mm

2021-10-29T14:57:02Z

Guenter moved page Aufgabe 4.7: Kupfer-Doppelader 0.5 mm to Exercise 4.7: Copper Twin Wire 0.5 mm

@@ Line 121: / Line 121: @@
 *The partial impulse response&nbsp;  $h_1(t ) \star h_2(t )$&nbsp; takes into account the influence of&nbsp; $\alpha_1$,&nbsp; $\alpha_2$&nbsp; and&nbsp; $\beta_2$&nbsp; and thus all terms leading to distortions.
 *In contrast, &nbsp; $\alpha_0$&nbsp; leads only to a frequency-independent attenuation and&nbsp; $\beta_1$&nbsp; only to a constant running time for all frequencies.
-*Solution 2 does not apply: &nbsp;  First&nbsp; (for small&nbsp; $t$&ndash;values)&nbsp; &nbsp; $h_1(t ) \star h_2(t )$&nbsp; is smaller than&nbsp; $h_2(t )$.&nbsp; Then, for large&nbsp; $t$&ndash;values, the blue curve lies above the red one.
+*Solution 2 does not apply: &nbsp;  First&nbsp; (for small&nbsp; $t$&ndash;values)&nbsp; &nbsp; $h_1(t ) \star h_2(t )$&nbsp; is smaller than&nbsp; $h_2(t )$.&nbsp;
 *This means: &nbsp; $\alpha_1$&nbsp; and thus also&nbsp; $h_1(t )$&nbsp; actually cause additional distortions,&nbsp; even when they are not very significant.
 {{ML-Fuß}}

@@ Line 64: / Line 64: @@
 {Calculate the phase delay &nbsp;$\tau_P$,&nbsp; related to the symbol duration &nbsp;$T$.
 |type="{}"}
-$\tau_{{\rm P}/T} \ = \ $ { 73 3% }
+$\tau_{\rm P}/T \ = \ $ { 73 3% }

@@ Line 90: / Line 90: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; With&nbsp; $a_0 = \alpha_0 \cdot l \approx 0.76 \ \rm Np$,&nbsp; for the constant&nbsp; $K$, which indicates the influence of the coefficient&nbsp; $ \alpha_0$&nbsp; on the impulse response, we obtain:
+'''(1)'''&nbsp; With&nbsp; ${\rm a}_0 = \alpha_0 \cdot l \approx 0.76 \ \rm Np$,&nbsp; for the constant&nbsp; $K$,&nbsp; which indicates the influence of the coefficient&nbsp; $ \alpha_0$&nbsp; on the impulse response,&nbsp; we obtain:
-:$$K = {\rm e}^{-{a}_0 }= {\rm e}^{-0.76} \hspace{0.15cm}\underline{= 0.468} \hspace{0.05cm}.$$
+:$$K = {\rm e}^{-{\rm a}_0 }= {\rm e}^{-0.76} \hspace{0.15cm}\underline{= 0.468} \hspace{0.05cm}.$$
-'''(2)'''&nbsp; For the phase delay, using the given equation:
+'''(2)'''&nbsp; For the phase delay,&nbsp; using the given equation:
 :$$\tau_{\rm P} = \frac{\beta_1 \cdot l}{2 \pi}=  \frac{30.6 \cdot 1.5}{2 \pi}\, {\rm &micro; s}\approx 7.31\, {\rm &micro;
   s}\hspace{0.05cm},$$
@@ Line 103: / Line 103: @@
-'''(3)'''&nbsp; The impulse response of a coaxial cable is approximately equal to&nbsp; $h_2(t)$, if the cable has the following characteristic cable attenuation:
+'''(3)'''&nbsp; The impulse response of a coaxial cable is approximately equal to&nbsp; $h_2(t)$,&nbsp; if the cable has the following characteristic cable attenuation:
-:$${a}_\star ={a}_2  =  \alpha_2 \cdot l \cdot \sqrt{{R}/{2}} =
+:$${\rm a}_\star ={\rm a}_2  =  \alpha_2 \cdot l \cdot \sqrt{{R}/{2}} =
 .1467\,\, \frac{\rm Np}{\rm km \cdot \sqrt{MHz}} \cdot 1.5\,{\rm km} \cdot \sqrt{\frac{10\,{\rm MHz}}{2}}
    =  2.93\,{\rm Np} = 2.93\,{\rm Np} \cdot8.686\,\frac {\rm dB}{\rm Np} \hspace{0.15cm}\underline{ =25.5\,{\rm dB}}\hspace{0.05cm}.$$
@@ Line 110: / Line 110: @@
-'''(4)'''&nbsp; <u>Statements 1 and 2</u> are correct:
+'''(4)'''&nbsp; <u>Statements 1 and 2</u>&nbsp; are correct:
-*The Fourier transform&nbsp; $H_1(f) = {\rm e}^{-A \hspace{0.05cm} \cdot \hspace{0.05cm} |f|}$ with $A = 2 \hspace{0.05cm}\cdot \hspace{0.05cm} {a}_1/R$&nbsp; is real and even, so&nbsp; $h_1(t)$&nbsp; is also real and even.
+*The Fourier transform&nbsp; $H_1(f) = {\rm e}^{-A \hspace{0.05cm} \cdot \hspace{0.05cm} |f|}$&nbsp; with&nbsp; $A = 2 \hspace{0.05cm}\cdot \hspace{0.05cm} {a}_1/R$&nbsp; is real and even,&nbsp; so&nbsp; $h_1(t)$&nbsp; is also real and even.
 *Due to the low&ndash;pass characteristic of&nbsp; $H_1(f)$,&nbsp; the maximum is at&nbsp; $t = 0$.
-*The last statement, on the other hand, is incorrect: &nbsp; the integral over&nbsp; $h_1(t)$&nbsp; in the entire time domain&nbsp; $ \pm \infty$&nbsp; is equal to&nbsp; $H_1(f=0) = 1$.
+*The last statement,&nbsp; on the other hand,&nbsp; is incorrect: &nbsp; The integral over&nbsp; $h_1(t)$&nbsp; in the entire time domain&nbsp; $ \pm \infty$&nbsp; is equal to&nbsp; $H_1(f=0) = 1$.
-'''(5)'''&nbsp; Only <u>solution 1</u> is correct:
+'''(5)'''&nbsp; Only <u>solution 1</u>&nbsp; is correct:
 *The partial impulse response&nbsp;  $h_1(t ) \star h_2(t )$&nbsp; takes into account the influence of&nbsp; $\alpha_1$,&nbsp; $\alpha_2$&nbsp; and&nbsp; $\beta_2$&nbsp; and thus all terms leading to distortions.
 *In contrast, &nbsp; $\alpha_0$&nbsp; leads only to a frequency-independent attenuation and&nbsp; $\beta_1$&nbsp; only to a constant running time for all frequencies.
-*Solution 2, on the other hand, does not apply: &nbsp;  first&nbsp; (for small&nbsp; $t$&ndash;values)&nbsp; &nbsp; $h_1(t ) \star h_2(t )$&nbsp; is smaller than&nbsp; $h_2(t )$. Then, for large&nbsp; $t$&ndash;values, the blue curve lies above the red one.
+*Solution 2 does not apply: &nbsp;  First&nbsp; (for small&nbsp; $t$&ndash;values)&nbsp; &nbsp; $h_1(t ) \star h_2(t )$&nbsp; is smaller than&nbsp; $h_2(t )$.&nbsp; Then, for large&nbsp; $t$&ndash;values, the blue curve lies above the red one.
-*This means: &nbsp; $\alpha_1$&nbsp; and thus also&nbsp; $h_1(t )$&nbsp; actually cause additional distortions, even when they are not very significant.
+*This means: &nbsp; $\alpha_1$&nbsp; and thus also&nbsp; $h_1(t )$&nbsp; actually cause additional distortions,&nbsp; even when they are not very significant.
 {{ML-Fuß}}

@@ Line 14: / Line 14: @@
   \hspace{0.05cm}.$$
 *The following quantities are used,&nbsp; which can be derived from the attenuation and phase function per unit length:
-:$${\rm a}_0  =  \alpha_0 \cdot l\hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm}\alpha_0 = 0.5066\,\, \frac{\rm Np}{\rm km}\hspace{0.05cm},$$
+:$${\rm a}_0  =  \alpha_0 \cdot l\hspace{0.05cm},\hspace{0.2cm}{\rm with} \hspace{0.15cm}\alpha_0 = 0.5066\,\, \frac{\rm Np}{\rm km}\hspace{0.05cm},$$
-:$$  \tau_{\rm P}  =  \frac{\beta_1 \cdot l}{2 \pi} \hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm}\beta_1 = 30.6\,\,
+:$$  \tau_{\rm P}  =  \frac{\beta_1 \cdot l}{2 \pi} \hspace{0.05cm},\hspace{0.2cm}{\rm with} \hspace{0.15cm}\beta_1 = 30.6\,\,
    \frac{\rm rad}{\rm km \cdot MHz}\hspace{0.05cm},$$
-:$$  {\rm a}_1  =  \alpha_1 \cdot l \cdot {{R}/{2}}\hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm}
+:$$  {\rm a}_1  =  \alpha_1 \cdot l \cdot {{R}/{2}}\hspace{0.05cm},\hspace{0.2cm}{\rm with} \hspace{0.15cm}
    \alpha_1 = 0.136\,\, \frac{\rm Np}{\rm km \cdot MHz}\hspace{0.05cm},$$
-:$$  {\rm a}_2  =  \alpha_2 \cdot l \cdot \sqrt{{R}/{2}}\hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm}
+:$$  {\rm a}_2  =  \alpha_2 \cdot l \cdot \sqrt{{R}/{2}}\hspace{0.05cm},\hspace{0.2cm}{\rm with} \hspace{0.15cm}
    \alpha_2 = 1.1467\,\, \frac{\rm Np}{\rm km \cdot \sqrt{MHz}}\hspace{0.05cm},$$
-:$$  {b}_2  =  \beta_2 \cdot l \cdot \sqrt{{R}/{2}}\hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm}
+:$$  {b}_2  =  \beta_2 \cdot l \cdot \sqrt{{R}/{2}}\hspace{0.05cm},\hspace{0.2cm}{\rm with} \hspace{0.15cm}
    \beta_2 = 1.1467\,\, \frac{\rm rad}{\rm km \cdot
    \sqrt{MHz}}\hspace{0.05cm}.$$

@@ Line 64: / Line 64: @@
 {Calculate the phase delay &nbsp;$\tau_P$,&nbsp; related to the symbol duration &nbsp;$T$.
 |type="{}"}
-$\tau_{{ßrm P}/T} \ = \ $ { 73 3% }
+$\tau_{{\rm P}/T} \ = \ $ { 73 3% }

@@ Line 6: / Line 6: @@
 The time response of a copper twin wire with diameter &nbsp;$d = 0.5 \ \rm mm$&nbsp; is to be analyzed.
-The frequency response with the line length &nbsp;$l = 1.5 \ \rm km$&nbsp; and the bit rate &nbsp;$R = 10  \rm Mbit/s$:
+*The frequency response with the line length &nbsp;$l = 1.5 \ \rm km$&nbsp; and the bit rate &nbsp;$R = 10  \ \rm Mbit/s$:
 :$$H_{\rm K}(f)  =  {\rm e}^{-{\rm a}_0 } \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} 2 \pi \cdot f \hspace{0.05cm} \cdot  \hspace{0.01cm}\tau_{\rm P}}
    \cdot {\rm
@@ Line 13: / Line 13: @@
    e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} b_2 \hspace{0.05cm}\cdot \hspace{0.05cm}\sqrt{2f/R}}
   \hspace{0.05cm}.$$
-The following quantities are used, which can be derived from the damping and phase function per unit length:
+*The following quantities are used,&nbsp; which can be derived from the attenuation and phase function per unit length:
-:$${a}_0  =  \alpha_0 \cdot l\hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm}\alpha_0 = 0.5066\,\, \frac{\rm Np}{\rm km}\hspace{0.05cm},$$
+:$${\rm a}_0  =  \alpha_0 \cdot l\hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm}\alpha_0 = 0.5066\,\, \frac{\rm Np}{\rm km}\hspace{0.05cm},$$
 :$$  \tau_{\rm P}  =  \frac{\beta_1 \cdot l}{2 \pi} \hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm}\beta_1 = 30.6\,\,
    \frac{\rm rad}{\rm km \cdot MHz}\hspace{0.05cm},$$
-:$$  {a}_1  =  \alpha_1 \cdot l \cdot {{R}/{2}}\hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm}
+:$$  {\rm a}_1  =  \alpha_1 \cdot l \cdot {{R}/{2}}\hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm}
    \alpha_1 = 0.136\,\, \frac{\rm Np}{\rm km \cdot MHz}\hspace{0.05cm},$$
-:$$  {a}_2  =  \alpha_2 \cdot l \cdot \sqrt{{R}/{2}}\hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm}
+:$$  {\rm a}_2  =  \alpha_2 \cdot l \cdot \sqrt{{R}/{2}}\hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm}
    \alpha_2 = 1.1467\,\, \frac{\rm Np}{\rm km \cdot \sqrt{MHz}}\hspace{0.05cm},$$
 :$$  {b}_2  =  \beta_2 \cdot l \cdot \sqrt{{R}/{2}}\hspace{0.05cm},\hspace{0.2cm}{\rm mit} \hspace{0.15cm}
@@ Line 25: / Line 25: @@
    \sqrt{MHz}}\hspace{0.05cm}.$$
-The impulse response can thus be expressed in the form
+*The impulse response can thus be expressed in the form
 :$$h_{\rm K}(t )  = K \cdot \big [ \delta(t - \tau_{\rm P})\star h_{1}(t) \star h_{2}(t) \big ]$$
 where
-* the partial impulse response &nbsp;$h_1(t)$&nbsp; is due to the third term in the above equation, and
+* the partial impulse response &nbsp;$h_1(t)$&nbsp; is due to the third term in the frequency domain equation,&nbsp; and
 * $h_2(t)$&nbsp; indicates the joint time-domain representation of the last two terms.
-The graph shows as red curve the part &nbsp;$h_2(t)$&nbsp; of the impulse response and the convolution product &nbsp;$h_1(t) \star h_2(t)$&nbsp; &rArr; &nbsp; blue curve). <br>Here &nbsp;$h_2(t)$&nbsp; is equal to [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Koaxialkabeln#Impulsantwort_eines_Koaxialkabels|coaxial cable impulse response]] with the characteristic cable attenuation &nbsp;${a}_\star = {a}_2$.
+The graph shows as red curve the part &nbsp;$h_2(t)$&nbsp; of the impulse response and the convolution product &nbsp;$h_1(t) \star h_2(t)$&nbsp; &rArr; &nbsp; blue curve). <br>Here &nbsp;$h_2(t)$&nbsp; is equal to the&nbsp; [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Koaxialkabeln#Impulsantwort_eines_Koaxialkabels|coaxial cable impulse response]]&nbsp; with the characteristic cable attenuation &nbsp;${\rm a}_\star = {\rm a}_2$.
@@ Line 45: / Line 45: @@
 *The exercise belongs to the chapter&nbsp;   [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Kupfer–Doppeladern|Properties of Balanced Copper Pairs]].
-*The parameters &nbsp;$\alpha_0$, &nbsp;$\alpha_1$ and &nbsp;$\alpha_2$ were converted from the &nbsp;$k$&ndash;parameters as shown in&nbsp; [[Aufgaben:Exercise_4.6:_k-parameters_and_alpha-parameters|Exercise 4.6]]&nbsp; gezeigt.
+*The parameters &nbsp;$\alpha_0$, &nbsp;$\alpha_1$ and &nbsp;$\alpha_2$ were converted from the &nbsp;$k$&ndash;parameters as shown in&nbsp; [[Aufgaben:Exercise_4.6:_k-parameters_and_alpha-parameters|Exercise 4.6]].
 *The phase function parameter &nbsp;$\beta_2$&nbsp; was set here numerically equal to the attenuation function parameter&nbsp; $\alpha_2$.&nbsp;
-*Therefore, the attenuation component &nbsp;${a}_2$&nbsp; and the phase component &nbsp;${b}_2$&nbsp; differ only in units.
+*Therefore,&nbsp; the attenuation component &nbsp;${\rm a}_2$&nbsp; and the phase component &nbsp;${b}_2$&nbsp; differ only in units.
 *On the page&nbsp; [[Linear_and_Time_Invariant_Systems/Properties_of_Balanced_Copper_Pairs#Discussion_of_the_approximate_solution_found|Discussion of the approximate solution found]],&nbsp; it is explained why this measure is necessary.
 *You can use the&nbsp; (German language)&nbsp; interactive SWF applet &nbsp;[[Applets:Zeitverhalten_von_Kupferkabeln|"Zeitverhalten von Kupferkabeln"]] &nbsp; &rArr; &nbsp; "Time behavior of copper cables"&nbsp; to check your results.
@@ Line 62: / Line 62: @@
-{Calculate the phase delay &nbsp;$\tau_P$, related to the symbol duration &nbsp;$T$.
+{Calculate the phase delay &nbsp;$\tau_P$,&nbsp; related to the symbol duration &nbsp;$T$.
 |type="{}"}
-$\tau_P/T \ = \ $ { 73 3% }
+$\tau_{{ßrm P}/T} \ = \ $ { 73 3% }
 {What is the characteristic attenuation &nbsp;$a_\star$&nbsp; of the comparable coaxial cable?
 |type="{}"}
-$a_\star \ = \ $ { 25.5 } $\ \rm dB$
+${\rm a}_\star \ = \ $ { 25.5 } $\ \rm dB$
-{What are the characteristics of the partial impulse response &nbsp;$h_{\rm 1}(t )$&nbsp;?
+{What are the characteristics of the partial impulse response &nbsp;$h_{\rm 1}(t )$?
 |type="[]"}
 +  $h_{\rm 1}(t )$&nbsp; is an even function.

← Older revision		Revision as of 15:05, 29 October 2021
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−	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.3 Copper Twisted Pair^]]	+	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.3 Balanced Copper Twisted Pair^]]

← Older revision		Revision as of 15:02, 29 October 2021
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−	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.3 ~~Kupfer–Doppelader~~^]]	+	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.3 Copper Twisted Pair^]]

← Older revision	Revision as of 14:57, 29 October 2021
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