Aufgaben:Exercise 4.6: k-parameters and alpha-parameters - Revision history

Guenter at 16:11, 23 November 2021

2021-11-23T16:11:41Z

Guenter at 16:03, 23 November 2021

2021-11-23T16:03:15Z

Hwang at 08:56, 22 November 2021

2021-11-22T08:56:16Z

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Guenter: Guenter moved page Aufgaben:Exercise 4.6: k-Parameters and Alpha-Parameters to Aufgaben:Exercise 4.6: k-parameters and alpha-parameters

2021-11-20T13:59:27Z

Guenter moved page Exercise 4.6: k-Parameters and Alpha-Parameters to Exercise 4.6: k-parameters and alpha-parameters

Guenter: Guenter moved page Aufgaben:Exercise 4.6: K-Parameters and Alpha-Parameters to Aufgaben:Exercise 4.6: k-Parameters and Alpha-Parameters

2021-11-19T09:18:33Z

Guenter moved page Exercise 4.6: K-Parameters and Alpha-Parameters to Exercise 4.6: k-Parameters and Alpha-Parameters

Hwang at 15:00, 18 November 2021

2021-11-18T15:00:21Z

Guenter: Guenter moved page Aufgaben:Exercise 4.6: K Parameters and Alpha Parameters to Aufgaben:Exercise 4.6: K-Parameters and Alpha-Parameters

2021-11-07T15:12:34Z

Guenter moved page Exercise 4.6: K Parameters and Alpha Parameters to Exercise 4.6: K-Parameters and Alpha-Parameters

Guenter at 15:05, 29 October 2021

2021-10-29T15:05:03Z

Guenter at 15:01, 29 October 2021

2021-10-29T15:01:30Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.6: k-Parameter und Alpha-Parameter to Aufgaben:Exercise 4.6: K Parameters and Alpha Parameters

2021-10-29T14:53:42Z

Guenter moved page Aufgabe 4.6: k-Parameter und Alpha-Parameter to Exercise 4.6: K Parameters and Alpha Parameters

@@ Line 110: / Line 110: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>Solutions 1 and 6</u> are correct:
+'''(1)'''&nbsp; <u>Solutions 1 and 6</u>&nbsp; are correct:
 *The derivative of the given expected value with respect to&nbsp; $\alpha_1$&nbsp; gives:
 :$$\frac{{\rm d}\,{\rm E}[\varepsilon^2(f)]}{{\rm d}\,{\alpha_1}}  =
@@ Line 130: / Line 130: @@
-'''(2)'''&nbsp; <u>Solutions 2 and 5</u> are correct:
+'''(2)'''&nbsp; <u>Solutions 2 and 5</u>&nbsp; are correct:
 *Using the same procedure as in subtask&nbsp; '''(1)''',&nbsp; we obtain:
 :$$\frac{{\rm d}\,{\rm E}[\varepsilon^2(f)]}{{\rm d}\,{\alpha_2}}  =
@@ Line 148: / Line 148: @@
-'''(3)'''&nbsp; From&nbsp; $C_1 \cdot \alpha_2 + C_2  = D_1 \cdot \alpha_2 + D_2$&nbsp; we obtain a linear equation for&nbsp; $\alpha_2$. With the result from&nbsp; '''(2)'''&nbsp; we can write:
+'''(3)'''&nbsp; <u>Both solutions</u>&nbsp; are correct:
 :$$\alpha_2   =  \frac{D_2 - C_2}{C_1 - D_1} = \frac{- \frac{2.5 \cdot k_2 }{k_3
 +1.5} \cdot \frac{B^{k_3-1}}{f_0^{k_3}} + \frac{3 k_2 }{k_3 +2}
@@ Line 171: / Line 173: @@
 -1}\cdot \frac{k_3 -0.5}{(k_3 + 1.5)(k_3 +
 )}\cdot \frac {k_2}{f_0}\hspace{0.05cm} .$$
-<u>Both solutions</u> are correct.
-*Regardless of the bandwidth, we obtain for&nbsp; $k_3 = 1$:
+*Regardless of the bandwidth,&nbsp; we obtain for&nbsp; $k_3 = 1$:
 :$$\alpha_1    =   (B/f_0)^{k_3
 -1}\cdot \frac{15 \cdot (k_3 -0.5)}{(k_3 + 1.5)(k_3 +
@@ Line 182: / Line 184: @@
 )}\cdot \frac {k_2}{\sqrt{f_0}}\hspace{0.15cm}\underline{= 0} \hspace{0.05cm}
   .$$
-*In contrast, for&nbsp; $k_3 = 0.5$:
+*In contrast,&nbsp; for&nbsp; $k_3 = 0.5$:
 :$$\alpha_1    =   (B/f_0)^{k_3
 -1}\cdot \frac{15 \cdot (k_3 -0.5)}{(k_3 + 1.5)(k_3 +

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_LZI_A_4_6.png|right|frame|Attenuation function per unit length&nbsp; $\text{(0.5 mm}$&nbsp; twin wire$)$ with&nbsp; $k$– and&nbsp; $\alpha$-parameters]]
+[[File:EN_LZI_A_4_6.png|right|frame|Attenuation function per unit length, <br>valid for&nbsp; "copper twin wire"&nbsp; (0.5 mm)]]
-For symmetrical copper twisted pairs, the following empirical formula can be found in&nbsp; [PW95],&nbsp; which is valid for the frequency range &nbsp;$0 \le f \le 30 \ \rm MHz$:
+For symmetrical copper twisted pairs,&nbsp; the following empirical formula can be found in&nbsp; [PW95],&nbsp; which is valid for the frequency range &nbsp;$0 \le f \le 30 \ \rm MHz$:
 :$$\alpha_{\rm I} (f) = k_1 + k_2  \cdot (f/f_0)^{k_3} , \hspace{0.15cm}
   f_0 = 1\,{\rm MHz} .$$
-In contrast, the attenuation function per unit length of a coaxial cable is usually given in the following form:
+In contrast,&nbsp; the attenuation function per unit length of a coaxial cable is usually given in the following form:
 :$$\alpha_{\rm II}(f)  = \alpha_0 + \alpha_1 \cdot f +  \alpha_2 \cdot \sqrt {f}\hspace{0.05cm}.$$
 Especially for the calculation of impulse response and rectangular response it is advantageous also for the copper twisted pairs to choose the second representation form with the cable parameters &nbsp;$\alpha_0$, &nbsp;$\alpha_1$&nbsp; and&nbsp; $\alpha_2$&nbsp; instead of the representation with &nbsp;$k_1$, &nbsp;$k_2$&nbsp; and &nbsp;$k_3$.
-For the conversion, one proceeds as follows:
+For the conversion,&nbsp; one proceeds as follows:
-* From above equations, it is obvious that the coefficient characterizing the DC signal attenuation is &nbsp;$\alpha_0 = k_1$&nbsp;.
+* From above equations,&nbsp; it is obvious that the coefficient characterizing the DC signal attenuation is &nbsp;$\alpha_0 = k_1$.
-* To determine&nbsp; $\alpha_1$&nbsp; and&nbsp; $\alpha_2$,&nbsp; it is assumed that the mean square error should be minimum in the range of a given bandwidth &nbsp;$B$&nbsp;:
+* To determine&nbsp; $\alpha_1$&nbsp; and&nbsp; $\alpha_2$,&nbsp; it is assumed that the mean square error should be minimum in the range of a given bandwidth &nbsp;$B$:
 :$${\rm E}\big[\varepsilon^2(f)\big] =  \int_{0}^{
 B} \left [ \alpha_{\rm II} (f) - \alpha_{\rm I} (f)\right ]^2
@@ Line 51: / Line 51: @@
   \hspace{0.05cm}.$$
-*The red curve shows the function &nbsp;$\alpha(f)$&nbsp; calculated with the parameters. For &nbsp;$f = 30 \ \rm MHz$&nbsp; the attenuation function per unit length is &nbsp;$\alpha(f)= 87.5 \ \rm dB/km$.
+*The red curve shows the function &nbsp;$\alpha(f)$&nbsp; calculated with this parameters.&nbsp; For &nbsp;$f = 30 \ \rm MHz$&nbsp; the attenuation function per unit length is &nbsp;$\alpha(f)= 87.5 \ \rm dB/km$.
 *The blue curve gives the approximation with the &nbsp;$\alpha$&ndash;coefficients.&nbsp; This is almost indistinguishable from the red curve within the drawing accuracy.
@@ Line 58: / Line 58: @@
 Notes:
 *The exercise belongs to the chapter&nbsp;   [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Kupfer–Doppeladern|Properties of Balanced Copper Pairs]].
 *You can use the&nbsp; (German language)&nbsp; interactive SWF applet &nbsp;[[Applets:Dämpfung_von_Kupferkabeln|"Dämpfung von Kupferkabeln"]]&nbsp; &rArr; &nbsp; "Attenuation of copper cables"&nbsp;.
-*[PW95]&nbsp; denotes the following literature reference: &nbsp; Pollakowski, P.; Wellhausen, H.-W.: ''Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz.'' Deutsche Telekom AG, Forschungs- und Technologiezentrum Darmstadt, 1995.
+*[PW95]&nbsp; denotes the following literature reference: &nbsp; Pollakowski, P.; Wellhausen, H.-W.:&nbsp; Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz.&nbsp; Deutsche Telekom AG, Forschungs- und Technologiezentrum Darmstadt, 1995.
@@ Line 71: / Line 68: @@
 <quiz display=simple>
-{Calculate the parameters &nbsp;$C_1$&nbsp; and  &nbsp;$C_2$&nbsp; of the equation &nbsp;$\alpha_1 + C_1 \cdot \alpha_2 + C_2  = 0$ resulting from the derivative &nbsp;${\rm dE\big[\text{...}\big]/d}\alpha_1$.&nbsp; <br>Which results are correct?
+{Calculate the parameters &nbsp;$C_1$&nbsp; and  &nbsp;$C_2$&nbsp; of the equation &nbsp;$\alpha_1 + C_1 \cdot \alpha_2 + C_2  = 0$&nbsp; resulting from the derivative &nbsp;${\rm dE\big[\text{...}\big]/d}\alpha_1$.&nbsp; <br>Which results are correct?
 |type="[]"}
 + $C_1 = 6/5 \cdot B^{-0.5}$,
@@ Line 81: / Line 78: @@
-{Calculate the parameters &nbsp;$D_1$&nbsp; and  &nbsp;$D_2$&nbsp; of the equation &nbsp;$ \alpha_1 + D_1 \cdot \alpha_2 + D_2  = 0$ resulting from the derivative &nbsp;${\rm dE\big[\text{...}\big]/d}\alpha_2$.&nbsp; <br> Which results are correct?
+{Calculate the parameters &nbsp;$D_1$&nbsp; and  &nbsp;$D_2$&nbsp; of the equation &nbsp;$ \alpha_1 + D_1 \cdot \alpha_2 + D_2  = 0$&nbsp; resulting from the derivative &nbsp;${\rm dE\big[\text{...}\big]/d}\alpha_2$.&nbsp; <br> Which results are correct?
 |type="[]"}
 - $D_1 = 6/5 \cdot B^{-0.5}$,
@@ Line 103: / Line 100: @@
-{Using the &nbsp;$\alpha$&ndash;parameters, calculate the attenuation function per unit length for the frequency &nbsp;$f = 30\ \rm MHz$.
+{Using the &nbsp;$\alpha$&ndash;parameters,&nbsp; calculate the attenuation function per unit length for the frequency &nbsp;$f = 30\ \rm MHz$.
 |type="{}"}
 $\alpha_{\rm  II}(f = 30\ \rm MHz) \ = \ $ { 88.1 3% } $\ \rm dB/km$

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Lineare zeitinvariante Systeme/Kupfer–Doppelader
+{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/Properties_of_Balanced_Copper_Pairs
 }}
-[[File:EN_LZI_A_4_6.png|right|frame|Dämpfungsmaß&nbsp; $\text{(0.5 mm}$&nbsp; Doppelader$)$ mit&nbsp; $k$– und&nbsp; $\alpha$-Parameter]]
+[[File:EN_LZI_A_4_6.png|right|frame|Attenuation function per unit length&nbsp; $\text{(0.5 mm}$&nbsp; twin wire$)$ with&nbsp; $k$– and&nbsp; $\alpha$-parameters]]
 Für symmetrische Kupfer&ndash;Doppeladern findet man in&nbsp; [PW95]&nbsp; die folgende empirische Formel, gültig für den Frequenzbereich &nbsp;$0 \le f \le 30 \ \rm MHz$:
 :$$\alpha_{\rm I} (f) = k_1 + k_2  \cdot (f/f_0)^{k_3} , \hspace{0.15cm}

← 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 15:01, 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 09:18, 19 November 2021
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← Older revision	Revision as of 14:53, 29 October 2021
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