Aufgaben:Exercise 4.4: Coaxial Cable - Frequency Response - Revision history

Guenter at 16:21, 12 November 2021

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Guenter at 16:20, 12 November 2021

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Guenter at 16:10, 12 November 2021

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Guenter at 16:06, 12 November 2021

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Guenter at 16:02, 12 November 2021

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Hwang at 14:24, 8 November 2021

2021-11-08T14:24:53Z

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Guenter at 14:51, 29 October 2021

2021-10-29T14:51:13Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.4: Koaxialkabel – Frequenzgang to Aufgaben:Exercise 4.4: Coaxial Cable - Frequency Response

2021-10-29T14:47:34Z

Guenter moved page Aufgabe 4.4: Koaxialkabel – Frequenzgang to Exercise 4.4: Coaxial Cable - Frequency Response

Javier: Text replacement - "„" to """

2021-05-28T14:39:15Z

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 }}
-[[File:LZI_A_4_4_vers3.png|right|frame|Various coaxial cables]]
+[[File:LZI_A_4_4_vers3.png|right|frame|Various coaxial cable types]]
 A so-called normal coaxial cable of length &nbsp;$l$&nbsp; with
 *core diameter&nbsp; $\text{2.6 mm}$,&nbsp; and

@@ Line 103: / Line 103: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>Solutions 1 and 4</u> are correct:
+'''(1)'''&nbsp; <u>Solutions 1 and 4</u>&nbsp; are correct:
 *The&nbsp; $\alpha_0$&ndash;term causes only a frequency-independent attenuation.
-*The&nbsp; $\beta_1$&ndash;term (linear phase) results in a frequency-independent delay.
+*The&nbsp; $\beta_1$&ndash;term&nbsp; (linear phase)&nbsp; results in a frequency-independent delay.
-*All other terms contribute to the (linear) distortions.
+*All other terms contribute to the&nbsp; (linear)&nbsp; distortions.
@@ Line 120: / Line 118: @@
    \hspace{0.05cm}.$$
+'''(3)'''&nbsp; The following applies to the attenuation curve&nbsp; when all terms are taken into account:
 :$${a}_{\rm K}(f)  =  [\alpha_0 + \alpha_1  \cdot f + \alpha_2  \cdot
    \sqrt{f}\hspace{0.05cm}] \cdot l
@@ Line 130: / Line 125: @@
-'''(4)'''&nbsp; According to the calculation in subtask&nbsp; '''(3)'''&nbsp;, the attenuation value&nbsp; ${a}_{\rm K}(f = 70\ \rm MHz)\hspace{0.15cm}\underline{=4.555 \ \rm Np}$ is obtained here.
+'''(5)'''&nbsp; For any positive quantity&nbsp; $x$&nbsp; the following holds:
 :$$x_{\rm Np} = {\rm ln} \hspace{0.10cm} x =  \frac{{\rm lg} \hspace{0.10cm} x}{{\rm lg} \hspace{0.10cm} {\rm e}}
    =   \frac{1}{{20 \cdot \rm lg} \hspace{0.10cm} {\rm e}} \cdot
@@ Line 145: / Line 136: @@
+'''(6)'''&nbsp; <u>Solutions 1, 4 and 5</u> are correct.&nbsp; Explanation:
-'''(6)'''&nbsp; <u>Solutions 1, 4 and 5</u> are correct. Explanation:
+*With the restriction to the attenuation term with&nbsp; $\alpha_2$,&nbsp; the following applies to the frequency response:
 :$$H_{\rm K}(f)   =
    {\rm e}^{- \alpha_2  \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
@@ Line 154: / Line 144: @@
    {\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2  \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
    \sqrt{f}}   \hspace{0.05cm}.$$
-*If the&nbsp; $\beta_1$&ndash;phase term is omitted, nothing changes with respect to the distortions. &nbsp; Only the phase and group delay would be (both equal) smaller by the value&nbsp; $\tau_1 = (\beta_1 \cdot l)/(2\pi)$&nbsp;.
+*If the&nbsp; $\beta_1$&ndash;phase term is omitted,&nbsp; nothing changes with respect to the distortions. &nbsp; Only the phase and group delay would be&nbsp; (both equal)&nbsp; smaller by the value&nbsp; $\tau_1 = (\beta_1 \cdot l)/(2\pi)$.
-*If, on the other hand, we omit the&nbsp; $\beta_2$&ndash;term, we obtain completely different ratios:
+*If,&nbsp; on the other hand,&nbsp; we omit the&nbsp; $\beta_2$&ndash;term,&nbsp; we obtain completely different conditions:
-::'''(a)''' The frequency response&nbsp; $H_{\rm K}(f)$&nbsp; no longer fulfills the requirement of a causal system; in such a case,&nbsp; $H_{\rm K}(f)$&nbsp; would have to be in minimum phase.
+::'''(a)''' The frequency response&nbsp; $H_{\rm K}(f)$&nbsp; no longer fulfills the requirement of a causal system;&nbsp; in such a case,&nbsp; $H_{\rm K}(f)$&nbsp; would have to be in minimum phase.
-::'''(b)''' The impulse response&nbsp;  $h_{\rm K}(t)$&nbsp; is symmetrical at&nbsp; $t = 0$ with real frequency response, which does not correspond to the conditions.
+::'''(b)''' The impulse response&nbsp;  $h_{\rm K}(t)$&nbsp; is symmetrical at&nbsp; $t = 0$&nbsp; with real frequency response,&nbsp; which does not correspond to the conditions.
-*Therefore, as an approximation for the coaxial cable frequency response, the following is allowed:
+*Therefore,&nbsp; as an approximation for the coaxial cable frequency response,&nbsp; the following is allowed:
 :$${a}_{\rm K}(f) = \alpha_2  \cdot l \cdot
    \sqrt{f},$$
@@ Line 168: / Line 158: @@
 *For a digital system with bit rate&nbsp; $R = 140 \ \rm Mbit/s$ &nbsp; &#8658; &nbsp; $R/2 = 70 \ \rm Mbit/s$&nbsp; and cable length&nbsp; $l = 2 \ \rm km$&nbsp;, &nbsp; $a_\star \approx 40 \ \rm dB$&nbsp; holds (see solution to the last sub-task).
 *A system with four times the bit rate&nbsp; $R/2 = 280 \ \rm Mbit/s$&nbsp; and half the length&nbsp; $(l = 1 \ \rm km)$&nbsp; results in the same characteristic cable attenuation.
-*In contrast, the following holds for a system with&nbsp; $R/2 = 35 \ \rm Mbit/s$&nbsp; and&nbsp; $l = 2 \ \rm km$:
+*In contrast,&nbsp; the following holds for a system with&nbsp; $R/2 = 35 \ \rm Mbit/s$&nbsp; and&nbsp; $l = 2 \ \rm km$:
 :$${a}_\star = 0.2722 \hspace{0.15cm}\frac {\rm Np}{\rm km \cdot \sqrt{MHz}} \cdot {\rm 2\,km}\cdot\sqrt{\rm 35\,MHz}
   \cdot 8.6859 \,\frac {\rm dB}{\rm Np} \approx 28\,{\rm dB}

@@ Line 50: / Line 50: @@
 Notes:
 *The exercise belongs to the chapter&nbsp;   [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Koaxialkabeln|Properties of Coaxial Cables]].
@@ Line 78: / Line 74: @@
-{What is the attenuation (in Np) at frequency &nbsp;$f = 70 \ \rm MHz$ when the cable length is &nbsp;$\underline{l = 2 \ \rm km}$&nbsp;?
+{What is the attenuation&nbsp; (in Np)&nbsp; at frequency &nbsp;$f = 70 \ \rm MHz$&nbsp; when the cable length is &nbsp;$\underline{l = 2 \ \rm km}$?
 |type="{}"}
 $a_{\rm K}(f = 70\ \rm MHz) \ = \ $ { 4.619 3% } $\ \rm Np$
-{Assuming all other things are equal, what is the attenuation when only the &nbsp;$\alpha_2$&ndash;term is considered?
+{Assuming all other things are equal,&nbsp; what is the attenuation when only the &nbsp;$\alpha_2$&ndash;term is considered?
 |type="{}"}
 $a_{\rm K}(f = 70\ \rm MHz) \ = \ $ { 4.555 3% } $\ \rm Np$
-{What is the formula for the conversion between &nbsp;$\rm Np$ and &nbsp;$\rm dB$? &nbsp;What is the&nbsp;$\rm dB$ value that results for the attenuation calculated in&nbsp; '''(4)'''&nbsp;?
+{What is the formula for the conversion between &nbsp;$\rm Np$&nbsp; and &nbsp;$\rm dB$? &nbsp;What is the&nbsp;$\rm dB$ value that results for the attenuation calculated in&nbsp; '''(4)'''?
 |type="{}"}
 $a_{\rm K}(f = 70\ \rm MHz) \ = \ $ { 39.56 3% } $\ \rm dB$
@@ Line 95: / Line 91: @@
 {Which statements are true if we restrict ourselves to the &nbsp;$\alpha_2$&ndash;value with respect to the attenuation function?
 |type="[]"}
-+ One can also do without the phase term with &nbsp;$\beta_1$&nbsp;.
++ One can also do without the phase term with &nbsp;$\beta_1$.
-- One can also do without the phase term with &nbsp;$\beta_2$&nbsp;.
+- One can also do without the phase term with &nbsp;$\beta_2$.
 - $a_\star \approx 40 \ \rm dB$&nbsp; holds for a system with &nbsp;$R = 70 \ \rm Mbit/s$&nbsp; and &nbsp;$l = 2 \ \rm  km$.
 + $a_\star \approx 40 \ \rm dB$&nbsp; holds for a system with &nbsp;$R = 140 \ \rm Mbit/s$&nbsp; and &nbsp;$l = 2 \ \rm  km$.

← Older revision		Revision as of 16:06, 12 November 2021
Line 57:		Line 57:
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Koaxialkabeln\|Properties of Coaxial Cables]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Koaxialkabeln\|Properties of Coaxial Cables]].

−	*You can use the interactive applet  [[Applets:~~Dämpfung_von_Kupferkabeln~~\|~~Dämpfung von Kupferkabeln~~]]  to check your results.	+	*You can use the interactive  "HTML 5/JS" applet  [[Applets:Attenuation_of_Copper_Cables\|Applets:Attenuation of Copper Cables]]  to check your results.

@@ Line 5: / Line 5: @@
 [[File:LZI_A_4_4_vers3.png|right|frame|Various coaxial cables]]
 A so-called normal coaxial cable of length &nbsp;$l$&nbsp; with
-*the core diameter&nbsp; $\text{2.6 mm}$,&nbsp; and
+*core diameter&nbsp; $\text{2.6 mm}$,&nbsp; and
-*the outer diameter&nbsp; $\text{9.5 mm}$
+*outer diameter&nbsp; $\text{9.5 mm}$
@@ Line 17: / Line 17: @@
    {\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2  \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
    \sqrt{f}}   \hspace{0.05cm}.$$
-The attenuation parameters &nbsp;$\alpha_0$, &nbsp;$\alpha_1$ and &nbsp;$\alpha_2$ are to be used in "Neper per kilometer" (Np/km)  and the phase parameters &nbsp;$\beta_1$ and &nbsp;$\beta_2$ in "Radian per kilometer" (rad/km). The following numerical values apply:
+The attenuation parameters &nbsp;$\alpha_0$, &nbsp;$\alpha_1$ and &nbsp;$\alpha_2$ are to be used in&nbsp; "Neper per kilometer"&nbsp; (Np/km)&nbsp;  and the phase parameters &nbsp;$\beta_1$ and &nbsp;$\beta_2$ in&nbsp; "Radian per kilometer"&nbsp; (rad/km).&nbsp; The following numerical values apply:
 :$$\alpha_0 = 0.00162 \hspace{0.15cm}{\rm Np}/{\rm km} \hspace{0.05cm},$$
 :$$\alpha_1 = 0.000435 \hspace{0.15cm} {\rm Np}/{{\rm km} \cdot {\rm MHz}} \hspace{0.05cm},$$
 :$$\alpha_2 = 0.2722 \hspace{0.15cm}{\rm Np}/{{\rm km} \cdot \sqrt{\rm MHz}} \hspace{0.05cm}.$$
-Often one uses for the system-theoretical description of a linear time-invariant system
+For the system-theoretical description of a coaxial cable&nbsp; (German:&nbsp; "Koaxialkabel" &nbsp; &rArr; &nbsp; subscipt&nbsp; "K"),&nbsp; one uses
-* the attenuation function (in Np or dB):
+* the attenuation function&nbsp; (in Np or dB):
 :$${ a}_{\rm K}(f) = - {\rm ln} \hspace{0.10cm}|H_{\rm K}(f)|= - 20 \cdot {\rm lg} \hspace{0.10cm}|H_{\rm K}(f)|
      \hspace{0.05cm},$$
-* the phase function (in rad or degree):
+* the phase function&nbsp; (in rad or degree):
 :$$b_{\rm K}(f) = - {\rm arc} \hspace{0.10cm}H_{\rm K}(f)
      \hspace{0.05cm}.$$
-In practice, one often uses the approximation
+In practice,&nbsp; one often uses the approximation
 :$$H_{\rm K}(f) =
    {\rm e}^{- \alpha_2  \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
@@ Line 43: / Line 43: @@
 This is allowed because &nbsp;$\alpha_2$&nbsp; and &nbsp;$\beta_2$&nbsp; have exactly the same numerical value and differ only by different pseudo units.
-With the definition of the characteristic cable attenuation (in Neper or decibel)
+With the definition of the characteristic cable attenuation&nbsp; (in Neper or decibel)
 :$${a}_{\rm \star(Np)} = {a}_{\rm K}(f = {R}/{2}) = 0.1151 \cdot {a}_{\rm \star(dB)}$$
 digital systems with different bit rate &nbsp;$R$&nbsp; and cable length &nbsp;$l$&nbsp; can be treated uniformly.

← Older revision		Revision as of 14:51, 29 October 2021
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−	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.2 ~~Koaxialkabel~~^]]	+	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.2 Coaxial Cable^]]