Difference between revisions of "Linear and Time Invariant Systems/Properties of Coaxial Cables"

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{{Header
 
{{Header
|Untermenü=Eigenschaften elektrischer Leitungen
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|Untermenü=Properties of Electrical Cables
|Vorherige Seite=Einige Ergebnisse der Leitungstheorie
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|Vorherige Seite=Some Results from Line Transmission Theory
|Nächste Seite=Kupfer–Doppelader
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|Nächste Seite=Properties of Balanced Copper Pairs
 
}}
 
}}
==Übertragungsmaß von Koaxialkabeln==
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==Complex propagation function of coaxial cables==
Koaxialkabel bestehen aus einem Innenleiter und – durch ein Dielektrikum getrennt – einem Außenleiter. Es wurden zwei unterschiedliche Kabeltypen standardisiert, wobei zur Kennzeichnung die Durchmesser von Innen– und Außenleiter herangezogen werden:  
+
<br>
*das Normalkoaxialkabel, dessen Innenleiter einen Durchmesser von 2.6 mm besitzt und der Außendurchmesser 9.5 mm beträgt,  
+
Coaxial cables consist of an inner conductor and &ndash; separated by a dielectric &ndash; an outer conductor.&nbsp; Two different types of cable have been standardized,&nbsp; with the diameters of the inner and outer conductors mentioned for identification purposes:  
*das Kleinkoaxialkabel mit den Abmessungen 1.2 mm und 4.4 mm.
+
*the&nbsp; &raquo;standard coaxial cable&laquo;&nbsp; whose inner conductor has a diameter of&nbsp; $\text{2.6 mm}$&nbsp; and whose outer diameter is&nbsp; $\text{9.5 mm}$,
  
 +
*the&nbsp; &raquo;small coaxial cable&laquo;&nbsp; with diameters&nbsp; $\text{1.2 mm}$&nbsp; and&nbsp; $\text{4.4 mm}$.
  
Der Kabelfrequenzgang $H_{\rm K}(f)$ ergibt sich aus der Kabellänge $l$ und dem Übertragungsmaß
 
$$\gamma(f)  = \alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot \sqrt {f}+ {\rm j}\cdot (\beta_1 \cdot f + \beta_2 \cdot \sqrt {f})\hspace{0.05cm}$$
 
$$\Rightarrow \hspace{0.3cm}H_{\rm K}(f)  = {\rm e}^{-\gamma(f)\hspace{0.05cm} \cdot \hspace{0.05cm} l} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}|H_{\rm K}(f)|  = {\rm e}^{-\alpha(f)\hspace{0.05cm} \cdot \hspace{0.05cm} l}\hspace{0.05cm}.$$
 
  
 +
The cable frequency response &nbsp;$H_{\rm K}(f)$&nbsp; results from the cable length &nbsp;$l$&nbsp; and the complex propagation function $($per unit length$)$
 +
:$$\gamma(f)  = \alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot \sqrt {f}+ {\rm j}\cdot (\beta_1 \cdot f + \beta_2 \cdot \sqrt {f})\hspace{0.05cm}\hspace{0.3cm}
 +
\Rightarrow \hspace{0.3cm}H_{\rm K}(f)  = {\rm e}^{-\gamma(f)\hspace{0.05cm} \cdot \hspace{0.05cm} l} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}|H_{\rm K}(f)|  = {\rm e}^{-\alpha(f)\hspace{0.05cm} \cdot \hspace{0.05cm} l}\hspace{0.05cm}.$$
  
Die kabelspezifischen Konstanten für das Normalkoaxialkabel 2.6/9.5 mm sind:
+
The cable specific constants for the&nbsp; &raquo;'''standard coaxial cable'''&laquo;&nbsp; $\text{(2.6/9.5 mm)}$&nbsp; are:
$$\begin{align*}\alpha_0  & = 0.00162\, \frac{ {\rm Np} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_1 = 0.000435\, \frac{ {\rm Np} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_2 = 0.2722\, \frac{ {\rm Np} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}, \\ \beta_1 & = 21.78\, \frac{ {\rm rad} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \beta_2 = 0.2722\, \frac{ {\rm rad} }{ {\rm km \cdot \sqrt{MHz} } } \hspace{0.05cm}.\end{align*}$$
+
:$$\begin{align*}\alpha_0  & = 0.00162\, \frac{ {\rm Np} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_1 = 0.000435\, \frac{ {\rm Np} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_2 = 0.2722\, \frac{ {\rm Np} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}, \\ \beta_1 & = 21.78\, \frac{ {\rm rad} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \beta_2 = 0.2722\, \frac{ {\rm rad} }{ {\rm km \cdot \sqrt{MHz} } } \hspace{0.05cm}.\end{align*}$$
  
Entsprechend lauten die kilometrischen Dämpfungs– und Phasenkonstanten für das Kleinkoaxialkabel:
+
Accordingly,&nbsp; the kilometric attenuation and phase constants for the&nbsp; &raquo;'''small coaxial cable'''&laquo;&nbsp; $\text{(1.2/4.4 mm)}$:
$$\begin{align*}\alpha_0  & = 0.00783\, \frac{ {\rm Np} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm}
+
:$$\begin{align*}\alpha_0  & = 0.00783\, \frac{ {\rm Np} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm}
 
  \alpha_1 = 0.000443\, \frac{ {\rm Np} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm}  \alpha_2 = 0.5984\, \frac{ {\rm Np} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}, \\ \beta_1 & = 22.18\, \frac{ {\rm rad} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \beta_2 = 0.5984\, \frac{ {\rm rad} }{ {\rm km \cdot \sqrt{MHz} } } \hspace{0.05cm}.\end{align*}$$
 
  \alpha_1 = 0.000443\, \frac{ {\rm Np} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm}  \alpha_2 = 0.5984\, \frac{ {\rm Np} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}, \\ \beta_1 & = 22.18\, \frac{ {\rm rad} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \beta_2 = 0.5984\, \frac{ {\rm rad} }{ {\rm km \cdot \sqrt{MHz} } } \hspace{0.05cm}.\end{align*}$$
  
 +
These values can be calculated from the geometric dimensions of the cables and have been confirmed by measurements at the Fernmeldetechnisches Zentralamt in Darmstadt - see [Wel77]<ref>Wellhausen, H. W.:&nbsp; Dämpfung, Phase und Laufzeiten bei Weitverkehrs–Koaxialpaaren.&nbsp; Frequenz 31, S. 23-28, 1977.</ref>.&nbsp;  They apply to a temperature of&nbsp; $20^\circ\ \text{C (293 K)}$&nbsp; and frequencies greater than&nbsp; $\text{200 kHz}$.&nbsp; 
  
Diese Werte können aus den geometrischen Abmessungen der Kabel berechnet werden und wurden durch Messungen am Fernmeldetechnischen Zentralamt in Darmstadt bestätigt – siehe [Wel77]<ref>Wellhausen, H. W.: ''Dämpfung, Phase und Laufzeiten bei Weitverkehrs–Koaxialpaaren.''. Frequenz 31, S. 23-28, 1977.</ref>.  Sie gelten für eine Temperatur von 20°C (293 K) und Frequenzen größer als 200 kHz. Es besteht folgender Zusammenhang zu den $\href{http://en.lntwww.de/cgi-bin/extern/uni.pl?uno=hyperlink&due=block&b_id=2433&hyperlink_typ=block_verweis&hyperlink_fenstergroesse=blockverweis_gross&session_id=7761LSYGTN1463149483}{Leitungsbelägen}$:
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There is the following connection to the&nbsp; [[Linear_and_Time_Invariant_Systems/Some_Results_from_Line_Transmission_Theory#Equivalent_circuit_diagram_of_a_short_transmission_line_section|&raquo;primary line parameters&laquo;]]:
*Die vom frequenzunabhängigem Anteil $R’$ herrührenden Ohmschen Verluste werden durch $α_0$ modelliert und verursachen eine (bei Koaxialkabeln geringe) frequenzunabhängige Dämpfung.  
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#The ohmic losses originating from the frequency-independent component &nbsp;$R\hspace{0.05cm}'$&nbsp; are modeled by the parameter &nbsp;$α_0$&nbsp; and cause a&nbsp; $($for coaxial cables small$)$&nbsp; frequency-independent attenuation.  
*Der Anteil $α_1 · f$ des Dämpfungsmaßes ist auf die Ableitungsverluste $(G’)$ zurückzuführen und der frequenzproportionale Term $β_1 · f$ bewirkt nur eine Phasenlaufzeit, aber keine Verzerrungen.  
+
#The component &nbsp;$α_1 · f$&nbsp; of the&nbsp; &raquo;attenuation function (per unit length)&laquo;&nbsp; is due to the derivation losses &nbsp;$(G\hspace{0.08cm}’)$&nbsp; and the frequency-proportional term &nbsp;$β_1 · f$&nbsp; causes only delay but no distortion.
*Die Anteile $α_2$ und $β_2$ gehen auf den Skineffekt zurück, der bewirkt, dass bei höherfrequentem Wechselstrom die Stromdichte im Leiterinneren niedriger ist als an der Oberfläche. Dadurch steigt der Widerstandsbelag $R’$ einer elektrischen Leitung mit der Wurzel aus der Frequenz an.  
+
#The components &nbsp;$α_2$&nbsp; and &nbsp;$β_2$&nbsp; are due to the&nbsp; [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference#Frequency_response_of_a_coaxial_cable|&raquo;'''skin effect'''&laquo;]],&nbsp; which causes the current density inside the conductor to be lower than at the surface in the case of higher-frequency alternating current.&nbsp; As a result,&nbsp; the&nbsp; &raquo;serial resistance (per unit length)&laquo;&nbsp; &nbsp;$R\hspace{0.05cm}’$&nbsp; of an electric line increases with the square root of the frequency.
  
==Charakteristische Kabeldämpfung (1)==
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==Characteristic cable attenuation==
Wir betrachten zunächst die linke Grafik; das rechte Diagramm wird weiter unten beschrieben. Links dargestellt ist das Dämpfungsmaß der zwei Koaxialkabeltypen im Frequenzbereich bis 500 MHz:
+
<br>
$${\rm a}_{\rm K}(f)  =\left [ \alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot \sqrt {f} \hspace{0.05cm} \right ] \cdot l \hspace{0.05cm}.$$
+
The graph shows the frequency-dependent attenuation curve for the normal coaxial cable and the small coaxial cable.&nbsp; Shown on the left is the cable attenuation per unit length of the two coaxial cable types in the frequency range up to&nbsp; $\text{500 MHz}$:
Die Ordinatenbeschriftung ist hierbei in Np/km angegeben. Oft erfolgt sie auch in dB/km, wobei die Umrechnung 1 dB = ln(10)/20 = 0.11513 Np gilt.
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[[File:EN_LZI_4_2_S2_neu.png |right|frame| Attenuation function and characteristic attenuation of coaxial cables]]
 +
:$${\alpha}_{\rm K}(f)  \hspace{-0.05cm} = \alpha_0 \hspace{-0.05cm}+ \hspace{-0.05cm} \alpha_1 \cdot f \hspace{-0.05cm}+ \hspace{-0.05cm} \alpha_2 \hspace{-0.05cm}\cdot \hspace{-0.05cm}\sqrt {f} \hspace{0.01cm}  \hspace{0.01cm}.$$
 +
:$$\Rightarrow \hspace{0.3cm}{\rm a}_{\rm K}(f) =\alpha_{\rm K}(f) \cdot l $$
  
[[File:P_ID1802__LZI_4_2_S2_v2.png | Dämpfungsmaß und charakteristische Dämpfung von Koaxialkabeln]]
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'''Notes on the representation chosen here'''
  
Man erkennt aus dieser Darstellung, dass der Fehler bei Vernachlässigung des frequenzunabhängigen Anteils $α_0$ und des frequenzproportionalen Terms $(α_1f)$ noch tolerabel ist. Im Folgenden gehen wir deshalb von der folgenden vereinfachten Dämpfungsfunktion aus:
+
#To make the difference between the attenuation function per unit length&nbsp; &raquo;alpha&laquo;&nbsp; and the function&nbsp; &raquo;a&laquo;&nbsp; $($after multiplication by length$)$&nbsp; more recognizable,&nbsp; the attenuation function is written here as&nbsp; ${\rm a}_{\rm K}(f)$&nbsp; and not&nbsp; $($''italics''$)$&nbsp; as&nbsp; ${a}_{\rm K}(f)$.
$${\rm a}_{\rm K}(f)  = \alpha_2 \cdot \sqrt {f} \cdot l =  {\rm a}_{\rm \star}\cdot \sqrt
+
#The ordinate labeling is given here in&nbsp; &raquo;Np/km&laquo;.&nbsp; Often it is also done in&nbsp; &raquo;dB/km&laquo;,&nbsp; with the following conversion:
 +
::$$\ln(10)/20 = 0.11513\text{...} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 1 \ \rm dB = 0.11513\text{...  Np.} $$
 +
 
 +
'''Interpretation of the left graph'''
 +
 
 +
*It can be seen from the curves shown that the error is still tolerable when neglecting the frequency-independent component &nbsp;$α_0$&nbsp; and the frequency-proportional term &nbsp;$(α_1\cdot f)$.
 +
 +
*Sometimes,&nbsp; we assume the&nbsp; &raquo;'''simplified attenuation function'''&laquo;:
 +
:$${\rm a}_{\rm K}(f)  = \alpha_2 \cdot \sqrt {f} \cdot l =  {\rm a}_{\rm \star}\cdot \sqrt
 
  { {2f}/{R}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}|H_{\rm K}(f)|  = {\rm e}^{- {\rm a}_{\rm K}(f)}\hspace{0.05cm}, \hspace{0.2cm}  {\rm a}_{\rm K}(f)\hspace{0.15cm}{\rm in }\hspace{0.15cm}{\rm Np}\hspace{0.05cm}.$$
 
  { {2f}/{R}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}|H_{\rm K}(f)|  = {\rm e}^{- {\rm a}_{\rm K}(f)}\hspace{0.05cm}, \hspace{0.2cm}  {\rm a}_{\rm K}(f)\hspace{0.15cm}{\rm in }\hspace{0.15cm}{\rm Np}\hspace{0.05cm}.$$
  
Beachten Sie, dass das Dämpfungsmaß mit „alpha” bezeichnet wird und die Dämpfungsfunktion (nach Multiplikation mit der Länge) mit „a” , was beim verwendeten Zeichensatz schwer zu erkennen ist.  
+
{{BlaueBox|TEXT= 
 +
$\text{Definition:}$&nbsp;
 +
We denote  as&nbsp; &raquo;'''characteristic cable attenuation'''&laquo;&nbsp; $\rm a_∗$&nbsp;
 +
*the attenuation of a coaxial cable at half the bit rate
 +
 
 +
*due to the&nbsp; $α_2$ term alone &nbsp; &rArr; &nbsp; &raquo;skin effect&laquo;,&nbsp; thus neglecting the&nbsp; $α_0$&nbsp; and the&nbsp; $α_1$ term:
 +
:$${\rm a}_{\rm \star} = {\rm a}_{\rm K}(f = {R}/{2}) = \alpha_2 \cdot \sqrt {{R}/{2}} \cdot l\hspace{0.05cm}.$$
 +
This value is particularly suitable for comparing different conducted transmission systems with different 
 +
#coaxial cable types&nbsp; $($normal or small coaxial cable$)$,&nbsp; each identified by the parameter&nbsp; $\alpha_2$,
 +
#bit rates&nbsp; $(R)$,&nbsp; and 
 +
#cable lengths&nbsp; $(l)$.}}
 +
 
  
Die charakteristische Kabeldämpfung $a_∗$ eignet sich insbesondere für den Vergleich verschiedener leitungsgebundener Übertragungssysteme mit unterschiedlichen Bitraten $(R)$, Kabeltypen (zum Beispiel Normal– oder Kleinkoaxialkabel) und Kabellängen $l$. Bei all diesen Übertragungssystemen beschreibt $a_∗$ die Dämpfung bei der halben Bitrate unter Vernachlässigung des $α_0$ – und des $α_1$ –Terms:
+
'''Interpretation of the right graph'''
$${\rm a}_{\rm \star} = {\rm a}_{\rm K}(f = {R}/{2})  = \alpha_2 \cdot \sqrt {{R}/{2}} \cdot l\hspace{0.05cm}.$$
 
Die Beschreibung des rechten Diagramms folgt anschließend.
 
  
==Charakteristische Kabeldämpfung (2)==
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The right&ndash;hand  diagram above shows the characteristic cable attenuation&nbsp; $\rm a_∗$&nbsp; in&nbsp; &raquo;Neper&laquo;&nbsp; $\rm (Np)$&nbsp; as a function of the bit rate&nbsp; $R$&nbsp; and the cable length&nbsp; $l$&nbsp; for
Das rechte Diagramm zeigt die charakteristische Kabeldämpfung $a_∗$ in Neper (Np) in Abhängigkeit der Bitrate $R$ und der Kabellänge $l$  
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*the normal coaxial cable&nbsp; $\text{(2.6/9.5 mm)}$ &nbsp; &rArr; &nbsp; left ordinate labeling,&nbsp; and
*beim Normalkoaxialkabel (linke Ordinatenbeschriftung) und
 
*beim Kleinkoaxialkabel (rechte Ordinatenbeschriftung).
 
  
 +
*for the small coaxial cable&nbsp; $\text{(1.2/4.4 mm)}$ &nbsp; &rArr; &nbsp; right ordinate labeling.
  
[[File:P_ID2845__LZI_4_2_S2_v2.png | Dämpfungsmaß und charakteristische Dämpfung von Koaxialkabeln]]
 
  
 +
This diagram shows the PCM systems of hierarchy levels &nbsp; $3$&nbsp; to&nbsp; $5$&nbsp; proposed by the&nbsp; [https://en.wikipedia.org/wiki/ITU-T $\text{ITU-T}$]&nbsp; $($&raquo;ITU Telecommunication Standardization Sector&laquo;$)$&nbsp; in the 1970s.&nbsp; One recognizes:
 +
#For all these systems for PCM speech transmission,&nbsp; the characteristic cable attenuation assumes values between&nbsp; $7 \ \rm Np \ \ (≈ 61 \ dB)$&nbsp; and&nbsp; $10.6 \ \rm Np \ \ (≈ 92 \ dB)$&nbsp;.
 +
#The system &nbsp;$\text{PCM 480}$&nbsp; – designed for 480 simultaneous telephone calls - with bit rate &nbsp;$R ≈ 35 \ \rm Mbit/s$&nbsp; was specified for both the normal coaxial cable&nbsp; $($with&nbsp; $l = 9.3 \ \rm km)$&nbsp; and the small coaxial cable&nbsp; $($with&nbsp; $l = 4 \ \rm km)$.&nbsp; The &nbsp;$\rm a_∗$&ndash;values &nbsp;$10.4\ \rm  Np$&nbsp; resp.  &nbsp;$9.9\ \rm  Np$&nbsp; are in the same order of magnitude.
 +
#The system &nbsp;$\text{PCM 1920}$&nbsp; of the fourth hierarchy level&nbsp; $($specified for the normal coaxial cable$)$&nbsp; with &nbsp;$R ≈ 140 \ \rm Mbit/s$&nbsp; and  &nbsp;$l = 4.65 \ \rm km$&nbsp;  is parameterized by &nbsp;$\rm a_∗ = 10.6 \ \rm Np$&nbsp; or &nbsp;$10.6 \ {\rm Np}· 8.688 \ \rm dB/Np ≈ 92\ \rm dB$&nbsp;.
 +
#Although the system &nbsp;$\text{PCM 7680}$&nbsp; has a four times greater bit rate  &nbsp;$R ≈ 560 \ \rm Mbit/s$&nbsp;,&nbsp; the characteristic cable attenuation of &nbsp; $\rm a_∗ ≈ 61 \ dB$&nbsp; is due to the better medium&nbsp; &raquo;normal coaxial cable&laquo; and the shorter cable sections&nbsp;$(l = 1. 55 \ \rm km)$&nbsp; by a factor of&nbsp; $3$&nbsp; significantly lower  .
 +
#These numerical values also show that for coaxial cable systems,&nbsp; the cable length &nbsp;$l$&nbsp; is more critical than the bit rate &nbsp;$R$.&nbsp; If one wants to double the cable length,&nbsp; one has to reduce the bit rate by a factor &nbsp;$4$.
  
In der Grafik eingezeichnet sind die vom CCITT in den 1970–Jahren vorgeschlagenen PCM–Systeme der Hierarchiestufen 3 bis 5. Man erkennt:
 
*Bei all diesen Systemen zur PCM–Sprachübertragung nimmt die charakteristische Kabeldämpfung Werte zwischen 7 Np (≈ 61 dB) und 10.6 Np (≈ 92 dB) an.
 
*Das System PCM 480 – ausgelegt für 480 gleichzeitige Telefonate – mit der Bitrate $R$ ≈ 35 Mbit/s wurde sowohl für das Normalkoaxialkabel (mit der Leitungslänge $l =$ 9.3 km) als auch für das Kleinkoaxialkabel (mit 4 km Länge) spezifiziert. Die $a_∗$–Werte 10.4 Np bzw. 9.9 Np liegen in der gleichen Größenordnung.
 
*Das Übertragungssystem PCM 1920 der vierten Hierarchiestufe $(R$ ≈ 140 Mbit/s, $l =$ 4.65 km, Normalkoaxialkabel) wird durch $a_∗$ = 10.6 Np bzw. 10.6 · 8.688 dB/Np ≈ 92 dB parametrisiert.
 
*Obwohl das System PCM 7680 demgegenüber zwar die vierfache Kapazität $(R$ ≈ 560 Mbit/s) aufweist, ist die charakteristische Kabeldämpfung mit $a_∗$ ≈ 61 dB aufgrund der um den Faktor 3 kürzeren Kabelabschnitte $(l$ = 1.55 km, Normalkoaxialkabel) deutlich geringer.
 
*Aus diesen Zahlenwerten geht auch hervor, dass bei Koaxialkabelsystemen die Kabellänge $l$ kritischer ist als die Bitrate $R$. Will man die Kabellänge verdoppeln, muss man die Bitrate um den Faktor 4 herabsetzen.
 
  
 +
You can view the topic described here with the interactive HTLM 5/JavaScript applet &nbsp;[[Applets:Attenuation_of_Copper_Cables|&raquo;Attenuation of Copper Cables&laquo;]]&nbsp;.
  
Die hier beschriebene Thematik können Sie sich auch mit folgendem Interaktionsmodul verdeutlichen:
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==Impulse response of a coaxial cable==
$\href{http://en.lntwww.de/cgi-bin/extern/uni.pl?uno=hyperlink&due=block&b_id=2133&hyperlink_typ=block_verweis&hyperlink_fenstergroesse=blockverweis_gross&session_id=7761LSYGTN1463149483}{Dämpfung von Kupferkabeln}$
+
<br>
 +
To calculate the impulse response,&nbsp; the first two of the in total five&nbsp;  components of the complex propagation function&nbsp; $($per unit length$)$&nbsp; can be neglected&nbsp; $($the reasoning can be found in the previous section$)$.&nbsp; So we start from the following equation:
 +
:$$\gamma(f)  = \alpha_0 + \alpha_1 \cdot f +  {\rm j} \cdot \beta_1 \cdot f  +\alpha_2 \cdot \sqrt {f}+ {\rm j}\cdot \beta_2 \cdot \sqrt {f} \approx    {\rm j} \cdot \beta_1 \cdot f  +\alpha_2 \cdot \sqrt {f}+ {\rm j}\cdot \beta_2 \cdot \sqrt {f} \hspace{0.05cm}.$$
  
==Impulsantworten von Koaxialkabeln (1)==
+
Considering
Zur Berechnung der Impulsantwort können von den fünf Anteilen des Übertragungsmaßes die beiden ersten Dämpfungsanteile vernachlässigt werden (Begründung siehe vorheriger Abschnitt):
+
#the cable length &nbsp;$l$,
$$\begin{align*}\gamma(f) & = \alpha_0 + \alpha_1 \cdot f +  {\rm j} \cdot \beta_1 \cdot f  +\alpha_2 \cdot \sqrt {f}+ {\rm j}\cdot \beta_2 \cdot \sqrt {f} \approx \\ & \approx &   {\rm j} \cdot \beta_1 \cdot f  +\alpha_2 \cdot \sqrt {f}+ {\rm j}\cdot \beta_2 \cdot \sqrt {f} \hspace{0.05cm}.\end{align*}$$
+
#the characteristic cable attenuation &nbsp;$\rm a_∗$,&nbsp; and
 +
#that &nbsp;$α_2$&nbsp; $($in&nbsp;  &raquo;Np&laquo;$)$&nbsp; and &nbsp;$β_2$&nbsp; (in&nbsp; &raquo;rad&laquo;$)$&nbsp; are numerically equal,
  
  
Unter Berücksichtigung
+
thus applies to the&nbsp; &raquo;frequency response of the coaxial cable&laquo;:
*der Kabellänge $l$,
+
:$$H_{\rm K}(f)    = {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1 f} \cdot {\rm e}^{-{\rm a}_{\rm \star}\hspace{0.05cm}\cdot \hspace{0.05cm} \sqrt{2f/R} }\cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}{\rm a}_{\rm \star}\hspace{0.05cm}\cdot \hspace{0.02cm} \sqrt{2f/R}}= {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1 f} \cdot {\rm e}^{-2{\rm a}_{\rm \star}\hspace{0.03cm}\cdot \hspace{0.03cm} \sqrt{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}f/R}} \hspace{0.05cm}.$$
*der charakteristischen Kabeldämpfung $a_∗$ und
+
The following abbreviations are used here:
*der Tatsache, dass $α_2$ (in Np) und $β_2$ (in rad) zahlenmäßig gleich sind,
+
:$$b_1\hspace{0.1cm}{(\rm in }\hspace{0.15cm}{\rm rad)}= \beta_1 \cdot l \hspace{0.05cm}, \hspace{0.8cm} {\rm a}_{\rm \star}\hspace{0.1cm}{(\rm in }\hspace{0.15cm}{\rm Np)}= \alpha_2 \cdot \sqrt {R/2} \cdot l \hspace{0.05cm}.$$
  
 +
The time domain representation is obtained by applying the&nbsp; [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_second_Fourier_integral|&raquo;inverse Fourier transform&laquo;]]&nbsp; and the&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation|&raquo;convolution theorem&laquo;]]:
 +
:$$h_{\rm K}(t)  = \mathcal{F}^{-1} \left \{ {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1
 +
f}\right \} \star\mathcal{F}^{-1} \left \{ {\rm e}^{-2{\rm a}_{\rm \star}\hspace{0.03cm}\cdot \hspace{0.03cm} \sqrt{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}f/R} }\right \} \hspace{0.05cm}.$$
  
gilt somit für den Frequenzgang des Koaxialkabels:
+
It must be taken into account here:
$$\begin{align*}H_{\rm K}(f)   & = {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1 f} \cdot {\rm e}^{-{\rm a}_{\rm \star}\hspace{0.05cm}\cdot \hspace{0.05cm} \sqrt{2f/R} }\cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}{\rm a}_{\rm \star}\hspace{0.05cm}\cdot \hspace{0.02cm} \sqrt{2f/R}}=\\ & = {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1 f} \cdot {\rm e}^{-2{\rm a}_{\rm \star}\hspace{0.03cm}\cdot \hspace{0.03cm} \sqrt{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}f/R}} \hspace{0.05cm},\end{align*}$$
+
*The first term yields the Dirac delta function &nbsp;$δ(t - τ_{\rm P})$&nbsp; shifted by the phase delay &nbsp;$τ_{\rm P} = b_1/2π$&nbsp;.
$${\rm wobei} \hspace{0.15cm}b_1\hspace{0.1cm}{(\rm in }\hspace{0.15cm}{\rm rad)}= \beta_1 \cdot l \hspace{0.05cm}, \hspace{0.2cm} {\rm a}_{\rm \star}\hspace{0.1cm}{(\rm in }\hspace{0.15cm}{\rm Np)}= \alpha_2 \cdot \sqrt {R/2} \cdot l \hspace{0.05cm}.$$
+
 
 +
*The second term can be given analytically closed.&nbsp; We write &nbsp;$h_{\rm K}(t + τ_P)$,&nbsp; so that the phase delay &nbsp;$τ_{\rm P}$&nbsp; need not be considered further.
 +
:$$h_{\rm K}(t + \tau_{\rm P})  = \frac {{\rm a}_{\rm \star}}{\pi \cdot \sqrt{2 \cdot R \cdot t^3}}\cdot {\rm exp} \left [ -\frac {{\rm a}_{\rm \star}^2}{ {2\pi \cdot R\cdot t}} \right ]\hspace{0.05cm},\hspace{0.2cm}{\rm a}_{\rm \star}\hspace{0.15cm}{\rm in\hspace{0.15cm} Np}\hspace{0.05cm}.$$
 +
*Since the bit rate &nbsp;$R$&nbsp; also  has already been considered in the&nbsp;$\rm a_∗$&nbsp; definition;&nbsp;  this equation can be easily represented with the normalized time &nbsp;$t\hspace{0.05cm}' = t/T$&nbsp;:
 +
:$$h_{\rm K}(t\hspace{0.05cm}' + \tau_{\rm P}\hspace{0.05cm} ')  = \frac {1}{T} \cdot \frac {{\rm a}_{\rm \star}}{\pi \cdot \sqrt{\cdot t\hspace{0.05cm}'\hspace{0.05cm}^3}}\cdot {\rm exp} \left [ -\frac {{\rm a}_{\rm \star}^2}{ {2\pi \cdot t\hspace{0.05cm}'}} \right ]\hspace{0.05cm},\hspace{0.2cm}{\rm a}_{\rm \star}\hspace{0.15cm}{\rm in\hspace{0.15cm} Np}\hspace{0.05cm}.$$
 +
:Here&nbsp; $T = 1/R$&nbsp; denotes&nbsp; &raquo;the symbol duration of a binary system&laquo;&nbsp; and it holds &nbsp;$τ_{\rm P} \hspace{0.05cm}' = τ_{\rm P}/T$.
  
Zur Zeitbereichsdarstellung kommt man durch Anwendung der $\href{http://en.lntwww.de/cgi-bin/extern/uni.pl?uno=hyperlink&due=block&b_id=600&hyperlink_typ=block_verweis&hyperlink_fenstergroesse=blockverweis_gross&session_id=7761LSYGTN1463149483}{Fourierrücktransformation}$ und des Faltungssatzes:
+
{{GraueBox|TEXT= 
$$h_{\rm K}(t)  = {\rm F}^{-1} \left \{ {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1
+
$\text{Example 1:}$&nbsp;
f}\right \} \star {\rm F}^{-1} \left \{ {\rm e}^{-2{\rm a}_{\rm \star}\hspace{0.03cm}\cdot \hspace{0.03cm} \sqrt{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}f/R} }\right \} \hspace{0.05cm}.$$
+
The results of this section are illustrated by the following graph as an example.
 +
[[File:EN_LZI_4_2_S3_neu.png|right|frame| Impulse response of a coaxial cable with &nbsp;$\rm a_∗ = 60 \ dB$]]
 +
#The normalized impulse response &nbsp;$T · h_{\rm K}(t)$&nbsp; of a coaxial cable with &nbsp;$\rm a_∗ = 60 \ dB  \ \ (6.9\ Np)$ is shown.
 +
#The attenuation coefficients &nbsp;$α_0$&nbsp; and &nbsp;$α_1$&nbsp; can thus be neglected,&nbsp; as shown in the last section.
 +
#For the left graph, the parameter &nbsp;$β_1 = 0$&nbsp; was also set.
  
Hierbei ist zu berücksichtigen:
 
*Der erste Term liefert die um die Phasenlaufzeit $τ_{\rm P} = b_1/2π$ verschobene Diracfunktion $δ(t – τ_{\rm P})$.
 
*Der zweite Term lässt sich analytisch geschlossen angeben. Wir schreiben hierfür $h_{\rm K}(t + τ_P)$. Im Gegensatz zu $h_{\rm K}(t)$ ist hier die Phasenlaufzeit $τ_{\rm P}$ nicht berücksichtigt.
 
$$h_{\rm K}(t + \tau_{\rm P})  = \frac {{\rm a}_{\rm \star}}{\pi \cdot \sqrt{2 \hspace{0.05cm}R  \hspace{0.05cm}t^3}}\cdot {\rm exp} \left [ -\frac {{\rm a}_{\rm \star}^2}{ {2\pi \hspace{0.05cm}R\hspace{0.05cm} t}} \right ]\hspace{0.05cm},\hspace{0.2cm}{\rm a}_{\rm \star}\hspace{0.15cm}{\rm in\hspace{0.15cm} Np}\hspace{0.05cm}.$$
 
*Da die Bitrate $R$ bereits bei der Definition der charakteristischen Kabeldämpfung $a_∗$ berücksichtigt wurde, lässt sich diese Gleichung mit der normierten Zeit $t' = t/T$ einfach darstellen, wobei $T = 1/R$ die Symboldauer eines Binärsystems angibt. Desweiteren gilt $τ_{\rm P} ' = τ_{\rm P}/T:$
 
$$h_{\rm K}(t' + \tau_{\rm P} ')  = \frac {1}{T} \cdot \frac {{\rm a}_{\rm \star}}{\pi \cdot \sqrt{2  \hspace{0.05cm}t'^3}}\cdot {\rm exp} \left [ -\frac {{\rm a}_{\rm \star}^2}{ {2\pi \hspace{0.05cm}t'}} \right ]\hspace{0.05cm},\hspace{0.2cm}{\rm a}_{\rm \star}\hspace{0.15cm}{\rm in\hspace{0.15cm} Np}\hspace{0.05cm}.$$
 
  
==Impulsantworten von Koaxialkabeln (2)==
+
Because of the parameterization by the coefficient &nbsp;$\rm a_∗$,&nbsp; and the time normalization to the symbol duration &nbsp;$T$&nbsp; the left curve is equally valid
Dargestellt wird die normierte Impulsantwort $T · h_{\rm K}(t)$ eines Koaxialkabels mit $a_∗ =$ 60 dB (6.9 Np)  $⇒  α_0$ und $α_1$ werden hier vernachlässigt. Für die linke Grafik wurde zudem der Parameter $β_1$ zu 0 gesetzt.
+
#for systems with small or normal coaxial cable,&nbsp;
 +
#different lengths,
 +
#different bit rates.&nbsp;
  
  
[[File:P_ID1803__LZI_4_2_S3_neu.png | Impulsantwort eines Koaxialkabels mit a∗ = 60 dB]]
+
For example for 
 +
*normal coaxial cable&nbsp; $\text{2.6/9.5 mm}$,&nbsp; bit rate &nbsp;$R = 140 \ \rm Mbit/s$,&nbsp; cable length &nbsp;$l = 3 \ \rm km$  &nbsp; ⇒ &nbsp;  $\text{system A}$,
 +
*small coaxial cable&nbsp; $\text{1.2/4.4 mm}$,&nbsp; bit rate &nbsp;$R = 35 \ \rm Mbit/s$, <br>cable length &nbsp;$l = 2.8 \ \rm km$  &nbsp; ⇒ &nbsp;  $\text{system B}$.  
 +
<br clear=all>
 +
It can be seen in the left&ndash;hand diagram that even at this moderate cable attenuation &nbsp;$\rm a_∗ = 60 \ \rm dB$&nbsp; the impulse response already extends over more than &nbsp; $200$&nbsp; symbol durations due to the skin effect &nbsp;$(α_2 = β_2 ≠ 0)$.&nbsp; Since the integral over &nbsp;$h_{\rm K}(t)$&nbsp; is equal to &nbsp;$H_{\rm K}(f = 0) = 1$,&nbsp; the maximum value becomes very small:
 +
:$${\rm Max}\big [h_{\rm K}(t)\big ] \approx 0.03.$$
  
 +
In the diagram on the right,&nbsp; the effects of the phase parameter &nbsp;$β_1$&nbsp; can be seen.&nbsp; Note the different time scales of the left and the right diagram: 
 +
*For&nbsp; $\text{system A}$ &nbsp;$(β_1 = 21.78 \ \rm rad/(km · MHz)$, &nbsp;$T = 7.14\ \rm  ns)$&nbsp; &nbsp;$β_1$&nbsp; leads to a phase delay of
 +
:$$\tau_{\rm A}= \frac {\beta_1 \cdot l}{2\pi} =\frac {21.78\, { {\rm rad} }/{ {(\rm km \cdot MHz)} }\cdot 3\,{\rm km} }{2\pi} = 10.4\,{\rm \mu s}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\tau_{\rm A}\hspace{0.05cm}' = {\tau_{\rm A} }/{T} \approx 1457\hspace{0.05cm}.$$
 +
*On the other hand,&nbsp; the following can be obtained for&nbsp; $\text{system B}$ &nbsp;$(β_1 = 22.18 \ \rm  rad/(km · MHz)$, &nbsp;$T = 30 \ \rm  ns)$:
 +
:$$\tau_{\rm B}= \frac {\beta_1 \cdot l}{2\pi} =\frac {22.18\, { {\rm rad} }/{ {(\rm km \cdot MHz)} }\cdot 2.8\,{\rm km} }{2\pi} = 9.9\,{\rm &micro; s}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\tau_{\rm B}\hspace{0.05cm}' ={\tau_{\rm B} }/{T} \approx 330\hspace{0.05cm}.$$
  
Wegen der Parametrisierung mittels der charakteristischen Kabeldämpfung $a_∗$ und der Normierung der Zeit auf die Symboldauer $T$ gilt diese linke Kurve für Systeme mit Klein– bzw. Normalkoaxialkabel, für unterschiedliche Längen und verschiedene Bitraten gleichermaßen, zum Beispiel für ein
+
Although &nbsp;$τ_{\rm A} ≈ τ_{\rm B}$&nbsp; holds,&nbsp; completely different ratios result because of the time normalization to &nbsp;$T = 1/R$&nbsp;. }}
*Normalkoaxialkabel 2.6/9.5 mm, Bitrate $R =$ 140 Mbit/s, Kabellänge $l =$ 3 km  ⇒  '''System A''',
 
*Kleinkoaxialkabel 1.2/4.4 mm, Bitrate $R =$ 35 Mbit/s, Kabellänge $l =$ 2.8 km  ⇒  '''System B'''.  
 
  
  
Man erkennt, dass sich selbst bei dieser moderaten Kabeldämpfung $a_∗ =$ 60 dB die Impulsantwort aufgrund des Skineffektes $(α_2 = β_2 ≠$ 0) schon über mehr als 200 Symboldauern erstreckt. Da das Integral über $h_{\rm K}(t)$ gleich $H_{\rm K}(f = 0) =$ 1 ist, wird der Maximalwert von $h_{\rm K}(t)$ sehr klein (≈ 0.03).  
+
{{BlaueBox|TEXT=
 +
$\text{Conclusion:}$&nbsp; When simulating and optimizing transmission systems,&nbsp; <br>'''one usually omits the linear phase term''' &nbsp;$b_1 = β_1 · f$,&nbsp; since this results exclusively in a&nbsp; $($often not disturbing$)$&nbsp; phase delay,&nbsp; '''but no signal distortions'''.}}
  
 +
==Basic receiver pulse==
  
In der rechten Grafik sind die Auswirkungen des Phasenparameters $β_1$ zu sehen. Beachten Sie bitte auch die unterschiedlichen Zeitmaßstäbe in der linken und der rechten Darstellung:
+
<br>
*Beim System '''A''' $(β_1 =$ 21.78 rad/(km · MHz), $T =$ 7.14 ns) führt $β_1$ zu einer Laufzeit von
+
With the&nbsp; &raquo;basic transmission pulse&laquo; &nbsp;$g_s(t)$ &nbsp; &rArr; &nbsp; &raquo;basic  pulse of the transmitted signal&laquo;&nbsp; $s(t)$&nbsp; and the&nbsp; &raquo;impulse response &nbsp;$h_{\rm K}(t)$&nbsp; of the channel,&nbsp; the result for the&nbsp; &raquo;basic receiver pulse&laquo; &nbsp;$g_r(t)$ &nbsp; &rArr; &nbsp; &raquo;basic  pulse of the received signal&laquo;&nbsp; $r(t)$&nbsp; is:
$$\tau_{\rm A}= \frac {\beta_1 \cdot l}{2\pi} =\frac {21.78\, { {\rm rad} }/{ {(\rm km \cdot MHz)} }\cdot 3\,{\rm km} }{2\pi} = 10.4\,{\rm \mu s}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\tau_{\rm A}' = {\tau_{\rm A}}/{T} \approx 1457\hspace{0.05cm}.$$
+
:$$g_r(t) = g_s(t) \star h_{\rm K}(t)\hspace{0.05cm}.$$
*Dagegen erhält man für das System '''B''' $(β_1 =$ 22.18 rad/(km · MHz), $T =$ 30 ns):
+
If a non-return-to-zero&nbsp; $\rm (NRZ)$&nbsp; rectangular pulse &nbsp;$g_s(t)$&nbsp; with amplitude &nbsp;$s_0$&nbsp; and duration &nbsp;$Δt_s = T$&nbsp; is used at the transmitter,&nbsp; the following results for the basic pulse at the receiver input:
$$\tau_{\rm B}= \frac {\beta_1 \cdot l}{2\pi} =\frac {22.18\, { {\rm rad} }/{ {(\rm km \cdot MHz)} }\cdot 2.8\,{\rm km} }{2\pi} = 9.9\,{\rm \mu s}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\tau_{\rm B}' ={\tau_{\rm B}}/{T} \approx 330\hspace{0.05cm}.$$
+
:$$g_r(t) = 2 s_0 \cdot \left [ {\rm Q} \left (\frac {{\rm a}_{\rm \star}/\sqrt {\pi}}{ \sqrt{ (t/T - 0.5)}}\right  ) -
 +
{\rm Q} \left (\frac {{\rm a}_{\rm \star}/\sqrt {\pi}}{ \sqrt{ (t/T + 0.5)}}\right  ) \right ]\hspace{0.05cm}.$$
 +
Here &nbsp;$\rm a_∗$&nbsp; denotes the&nbsp; &raquo;characteristic cable attenuation&laquo;&nbsp; $($in Neper$)$&nbsp; and  &nbsp;${\rm Q}(x)$&nbsp; the&nbsp; [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability|&raquo;complementary Gaussian error function&laquo;]].
  
 +
{{GraueBox|TEXT= 
 +
$\text{Example 2:}$&nbsp;
 +
The figure shows for the characteristic cable attenuations &nbsp;$\rm a_∗ = 40 \ \rm dB$, ... ,&nbsp; $100 \ \rm dB$&nbsp; $($smaller&nbsp; $\rm a_∗$&ndash;values &nbsp; are not relevant for practice$)$
  
 +
*the normalized coaxial cable impulse response &nbsp;$T · h_{\rm K}(t)$ &nbsp;  &rArr; &nbsp; solid curves,
 +
 +
*the basic receiver pulse&nbsp; $($&raquo;rectangular response&laquo;$)$&nbsp; $g_r(t)$&nbsp; normalized to the transmission amplitude &nbsp;$s_0$ &nbsp;  &rArr; &nbsp; dotted line.
 +
[[File:EN_LZI_4_2_S4.png|right|frame|Impulse response of the coaxial cable and basic receiver pulse]]
  
Obwohl bei den getroffenen Annahmen $τ_{\rm A} ≈ τ_{\rm B}$ gilt, ergeben sich wegen der Zeitnormierung auf $T = 1/R$ völlig unterschiedliche Verhältnisse.
 
  
Bei der Simulation und Optimierung von Nachrichtensystemen verzichtet man meist auf den Phasenterm mit $b_1 = β_1 · l$, da dieser ausschließlich eine Laufzeit, aber keine Signalverzerrung zur Folge hat.
+
One recognizes the following from this diagram:
 +
#With &nbsp;$\rm a_∗ = 40 \ \rm dB$,&nbsp; the normalized basic receiver pulse  &nbsp;$g_r(t)/s_0$&nbsp; is at the peak slightly&nbsp; $($about a factor of $0.95)$&nbsp; smaller than the normalized impulse response &nbsp;$T · h_{\rm K}(t)$.&nbsp; Here is a small difference between impulse response and basic receiver pulse .
 +
#In contrast,&nbsp; for  &nbsp;$a_∗ ≥ 60 \ \rm dB$,&nbsp; the basic receiver pulse and the impulse response are indistinguishable within the drawing accuracy.
 +
#For a return-to-zero&nbsp; $\rm (RZ)$&nbsp; pulse,&nbsp; the above equation for &nbsp;$g_r(t)$&nbsp;  would still need to be multiplied by the factor &nbsp;$Δt_s/T$.&nbsp; In this case &nbsp;$g_r(t)/s_0$&nbsp; is smaller than &nbsp;$T · h_{\rm K}(t)$&nbsp; by at least this factor.
 +
#The equation modified in this way is also a good approximation for other basic transmission pulses as long as &nbsp;$\rm a_∗≥ 60 \ \rm dB$&nbsp; is sufficiently large.&nbsp; $Δt_s$&nbsp; then indicates the&nbsp;[[Signal_Representation/Special_Cases_of_Pulses#Gaussian_pulse|&raquo;equivalent pulse duration&laquo;]]&nbsp; of &nbsp;$g_r(t)$.}}
  
==Empfangsgrundimpuls==
 
Mit dem Sendegrundimpuls $g_s(t)$ und der Impulsantwort $h_{\rm K}(t)$ ergibt sich für den Empfangsgrundimpuls:
 
$$g_r(t) = g_s(t) \star h_{\rm K}(t)\hspace{0.05cm}.$$
 
Verwendet man am Sender einen NRZ–Rechteckimpuls $g_s(t)$ mit der Amplitude $s_0$ und Dauer $Δt_s = T$, so ergibt sich für den Grundimpuls am Ausgang des Koaxialkabels:
 
$$g_r(t) = 2 s_0 \cdot \left [ {\rm Q} \left (\frac {{\rm a}_{\rm \star}/\sqrt {\pi}}{ \sqrt{ (t/T - 0.5)}}\right  ) -
 
{\rm Q} \left (\frac {{\rm a}_{\rm \star}/\sqrt {\pi}}{ \sqrt{ (t/T + 0.5)}}\right  ) \right ]\hspace{0.05cm},$$
 
mit $a_∗$: charakteristische Kabeldämpfung in Neper, $Q(x)$: komplemantäre Gaußsche Fehlerfunktion.
 
  
 +
==Special features of coaxial cable systems==
 +
<br>
 +
[[File:EN_LZI_4_2_S5_v2.png |right|frame| Binary transmission system with coaxial cable]]
 +
Assuming binary transmission
 +
 +
*with non-return-to-zero&nbsp; $\rm (NRZ)$&nbsp; rectangular pulses&nbsp; $($duration $T)$&nbsp;
  
{{Beispiel}}
+
*and a coaxial transmission channel,&nbsp;
Die Abbildung zeigt die normierte Koaxialkabelimpulsantwort $T · h_{\rm K}(t)$ und den auf die Sendeamplitude $s_0$ normierten Empfangsgrundimpuls $g_r(t)$ für die charakteristischen Kabeldämpfungen $a_∗ =$ 40 dB, 60 dB, 80 dB und 100 dB. Kleinere Werte von $a_∗$ sind für die Praxis nicht relevant.
 
  
[[File:P_ID1804__LZI_4_2_S4_90.png | Impulsantwort und Rechteckantwort (Empfangsgrundimpuls) des Koaxialkabels]]
 
  
 +
the following system model is obtained.&nbsp; In particular,&nbsp; it should be noted:
  
Man erkennt aus dieser Darstellung:
+
$(1)$&nbsp; In a simulation,&nbsp; the phase delay time of the coaxial cable is conveniently left out of consideration.&nbsp; Then the basic receiver pulse  &nbsp;$g_r(t)$&nbsp; is approximated by&nbsp; $($with ${\rm a}_{\rm \star}$&nbsp;in Neper$)$:
*Mit $a_∗ =$ 40 dB ist $g_r(t)/s_0$ an der Spitze geringfügig (etwa um den Faktor 0.95) kleiner als die normierte Impulsantwort $T · h_{\rm K}(t)$.
+
:$$g_r(t) \approx s_0 \cdot T \cdot h_{\rm K}(t) =  \frac {s_0 \cdot {\rm a}_{\rm \star}/\pi}{ \sqrt{2    \cdot(t/T)^3}}\cdot {\rm e}^{  -{{\rm a}_{\rm \star}^2}/( {2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}t/T}) } \hspace{0.05cm}.$$
*Dagegen sind für den Fall $a_∗ ≥$ 60 dB die Rechteckantwort und die Impulsantwort innerhalb der Zeichengenauigkeit nicht zu unterscheiden.
 
*Bei einem RZ–Impuls ist die obige Gleichung für den Empfangsgrundimpuls noch mit dem Tastverhältnis $Δts/T$ zu multiplizieren. In diesem Fall ist $g_r(t)/s_0$ deutlich kleiner als $T · h_{\rm K}(t)$.
 
*Die so modifizierte Gleichung stellt auch eine gute Näherung für andere Sendegrundimpulse dar, so lange $a_∗$ hinreichend groß ist (≥ 60 dB). $Δt_s$ gibt dann die äquivalente Sendeimpulsdauer an.
 
{{end}}
 
  
 +
$(2)$&nbsp; Because of the good shielding of coaxial cables against other impairments,&nbsp; the&nbsp; [[Aufgaben:Exercise_1.3Z:_Thermal_Noise|&raquo;thermal noise&laquo;]]&nbsp;  is the dominant stochastic perturbation.&nbsp; In this case,&nbsp; the signal &nbsp;$n(t)$&nbsp; is Gaussian and white.&nbsp; It can described by the&nbsp; $($two-sided$)$&nbsp; noise power density &nbsp;$N_0/2$.
  
Wir möchten Sie auf ein Interaktionsmodul hinweisen, das die hier behandelte Thematik zum Inhalt hat:
+
$(3)$&nbsp; By far the largest noise component arises in the input stage of the receiver,&nbsp; so that it is expedient to add the noise signal &nbsp;$n(t)$&nbsp; at the&nbsp; interface&nbsp; &raquo;cable ⇒ receiver&laquo;.&nbsp; This noise addition point is also useful because the frequency response &nbsp;$H_{\rm K}(f)$&nbsp; decisively attenuates all noise accumulated along the cable.
$$\href{http://en.lntwww.de/cgi-bin/extern/uni.pl?uno=hyperlink&due=block&b_id=2430&hyperlink_typ=block_verweis&hyperlink_fenstergroesse=blockverweis_gross&session_id=7761LSYGTN1463149483}{Zeitverhalten von Kupferkabeln}$$
+
 +
$(3)$&nbsp; Then the received signal is with the amplitude coefficients &nbsp;$a_{\nu}$:
 +
:$$r(t) = \sum_{\nu = - \infty}^{+ \infty}a_{\nu}\cdot g_r(t - \nu \cdot T)+ n(t) \hspace{0.05cm} .$$
  
==Einige Bemerkungen zu Koaxialkabelsystemen==
 
Geht man von binärer Übertragung mit NRZ–Rechteckimpulsen (Symboldauer $T$) und einem koaxialen Übertragungskanal aus, so ergibt sich das folgende Systemmodell:
 
  
 +
==Exercises for the chapter==
 +
<br>
 +
[[Aufgaben:Exercise_4.4:_Coaxial_Cable_-_Frequency_Response| Exercise 4.4: Coaxial Cable - Frequency Response]]
  
[[File:P_ID1805__LZI_4_2_S5_neu.png | Binäres Übertragungssystem mit Koaxialkabel]]
+
[[Aufgaben:Exercise_4.5:_Coaxial_Cable_-_Impulse_Response| Exercise 4.5: Coaxial Cable - Impulse Response]]
  
 +
[[Aufgaben:Exercise_4.5Z:_Impulse_Response_once_again|Exercise 4.5Z: Impulse Response once again]]
  
Insbesondere ist zu beachten:
 
*Bei einer Simulation lässt man zweckmäßigerweise die Laufzeit des Koaxialkabels außer Betracht. Dann gilt für den Empfangsgrundimpuls näherungsweise:
 
$$g_r(t) \approx s_0 \cdot T \cdot h_{\rm K}(t)  =  \frac {s_0 \cdot {\rm a}_{\rm \star}/\pi}{ \sqrt{2    \hspace{0.05cm}(t/T)^3}}\cdot {\rm exp} \left [ -\frac {{\rm a}_{\rm \star}^2}{ {2\pi \hspace{0.05cm}t/T}} \right ] \hspace{0.05cm}, \hspace{0.2cm} \hspace{0.15cm} {\rm mit}\hspace{0.15cm}{\rm a}_{\rm \star}\hspace{0.15cm} {\rm in}\hspace{0.15cm} {\rm Neper}\hspace{0.05cm}.$$
 
*Das $\href{http://en.lntwww.de/cgi-bin/extern/uni.pl?uno=hyperlink&due=block&b_id=1231&hyperlink_typ=block_verweis&hyperlink_fenstergroesse=blockverweis_gross&session_id=7761LSYGTN1463149483}{thermische Rauschen}$  ist wegen der sehr guten Abschirmung der Koaxialkabel gegenüber anderen Störungen die dominante Störursache. $n(t)$ ist gaußverteilt und weiß und wird durch die (zweiseitige) Rauschleistungsdichte $N_0/2$ beschrieben.
 
*Der weitaus größte Rauschanteil entsteht in der Eingangsstufe des Empfängers, so dass man $n(t)$ zweckmäßigerweise an der Schnittstelle Kabel–Empfänger addiert:
 
$$r(t) = \sum_{\nu = - \infty}^{+ \infty}a_{\nu}\cdot g_r(t - \nu \cdot T)+ n(t) \hspace{0.05cm} .$$
 
*Dieser Punkt ist auch deshalb sinnvoll, da alle entlang des Kabels akkumulierten Rauschstörungen durch den Kabelfrequenzgang $H_{\rm K}(f)$ entscheidend gedämpft werden.
 
  
==Quellenverzeichnis==
+
==References==
 
<references/>
 
<references/>
  
 
{{Display}}
 
{{Display}}

Latest revision as of 18:37, 24 November 2023

Complex propagation function of coaxial cables


Coaxial cables consist of an inner conductor and – separated by a dielectric – an outer conductor.  Two different types of cable have been standardized,  with the diameters of the inner and outer conductors mentioned for identification purposes:

  • the  »standard coaxial cable«  whose inner conductor has a diameter of  $\text{2.6 mm}$  and whose outer diameter is  $\text{9.5 mm}$,
  • the  »small coaxial cable«  with diameters  $\text{1.2 mm}$  and  $\text{4.4 mm}$.


The cable frequency response  $H_{\rm K}(f)$  results from the cable length  $l$  and the complex propagation function $($per unit length$)$

$$\gamma(f) = \alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot \sqrt {f}+ {\rm j}\cdot (\beta_1 \cdot f + \beta_2 \cdot \sqrt {f})\hspace{0.05cm}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}H_{\rm K}(f) = {\rm e}^{-\gamma(f)\hspace{0.05cm} \cdot \hspace{0.05cm} l} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}|H_{\rm K}(f)| = {\rm e}^{-\alpha(f)\hspace{0.05cm} \cdot \hspace{0.05cm} l}\hspace{0.05cm}.$$

The cable specific constants for the  »standard coaxial cable«  $\text{(2.6/9.5 mm)}$  are:

$$\begin{align*}\alpha_0 & = 0.00162\, \frac{ {\rm Np} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_1 = 0.000435\, \frac{ {\rm Np} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_2 = 0.2722\, \frac{ {\rm Np} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}, \\ \beta_1 & = 21.78\, \frac{ {\rm rad} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \beta_2 = 0.2722\, \frac{ {\rm rad} }{ {\rm km \cdot \sqrt{MHz} } } \hspace{0.05cm}.\end{align*}$$

Accordingly,  the kilometric attenuation and phase constants for the  »small coaxial cable«  $\text{(1.2/4.4 mm)}$:

$$\begin{align*}\alpha_0 & = 0.00783\, \frac{ {\rm Np} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_1 = 0.000443\, \frac{ {\rm Np} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_2 = 0.5984\, \frac{ {\rm Np} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}, \\ \beta_1 & = 22.18\, \frac{ {\rm rad} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \beta_2 = 0.5984\, \frac{ {\rm rad} }{ {\rm km \cdot \sqrt{MHz} } } \hspace{0.05cm}.\end{align*}$$

These values can be calculated from the geometric dimensions of the cables and have been confirmed by measurements at the Fernmeldetechnisches Zentralamt in Darmstadt - see [Wel77][1].  They apply to a temperature of  $20^\circ\ \text{C (293 K)}$  and frequencies greater than  $\text{200 kHz}$. 

There is the following connection to the  »primary line parameters«:

  1. The ohmic losses originating from the frequency-independent component  $R\hspace{0.05cm}'$  are modeled by the parameter  $α_0$  and cause a  $($for coaxial cables small$)$  frequency-independent attenuation.
  2. The component  $α_1 · f$  of the  »attenuation function (per unit length)«  is due to the derivation losses  $(G\hspace{0.08cm}’)$  and the frequency-proportional term  $β_1 · f$  causes only delay but no distortion.
  3. The components  $α_2$  and  $β_2$  are due to the  »skin effect«,  which causes the current density inside the conductor to be lower than at the surface in the case of higher-frequency alternating current.  As a result,  the  »serial resistance (per unit length)«   $R\hspace{0.05cm}’$  of an electric line increases with the square root of the frequency.

Characteristic cable attenuation


The graph shows the frequency-dependent attenuation curve for the normal coaxial cable and the small coaxial cable.  Shown on the left is the cable attenuation per unit length of the two coaxial cable types in the frequency range up to  $\text{500 MHz}$:

Attenuation function and characteristic attenuation of coaxial cables
$${\alpha}_{\rm K}(f) \hspace{-0.05cm} = \alpha_0 \hspace{-0.05cm}+ \hspace{-0.05cm} \alpha_1 \cdot f \hspace{-0.05cm}+ \hspace{-0.05cm} \alpha_2 \hspace{-0.05cm}\cdot \hspace{-0.05cm}\sqrt {f} \hspace{0.01cm} \hspace{0.01cm}.$$
$$\Rightarrow \hspace{0.3cm}{\rm a}_{\rm K}(f) =\alpha_{\rm K}(f) \cdot l $$

Notes on the representation chosen here

  1. To make the difference between the attenuation function per unit length  »alpha«  and the function  »a«  $($after multiplication by length$)$  more recognizable,  the attenuation function is written here as  ${\rm a}_{\rm K}(f)$  and not  $($italics$)$  as  ${a}_{\rm K}(f)$.
  2. The ordinate labeling is given here in  »Np/km«.  Often it is also done in  »dB/km«,  with the following conversion:
$$\ln(10)/20 = 0.11513\text{...} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 1 \ \rm dB = 0.11513\text{... Np.} $$

Interpretation of the left graph

  • It can be seen from the curves shown that the error is still tolerable when neglecting the frequency-independent component  $α_0$  and the frequency-proportional term  $(α_1\cdot f)$.
  • Sometimes,  we assume the  »simplified attenuation function«:
$${\rm a}_{\rm K}(f) = \alpha_2 \cdot \sqrt {f} \cdot l = {\rm a}_{\rm \star}\cdot \sqrt { {2f}/{R}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}|H_{\rm K}(f)| = {\rm e}^{- {\rm a}_{\rm K}(f)}\hspace{0.05cm}, \hspace{0.2cm} {\rm a}_{\rm K}(f)\hspace{0.15cm}{\rm in }\hspace{0.15cm}{\rm Np}\hspace{0.05cm}.$$

$\text{Definition:}$  We denote as  »characteristic cable attenuation«  $\rm a_∗$ 

  • the attenuation of a coaxial cable at half the bit rate
  • due to the  $α_2$ term alone   ⇒   »skin effect«,  thus neglecting the  $α_0$  and the  $α_1$ term:
$${\rm a}_{\rm \star} = {\rm a}_{\rm K}(f = {R}/{2}) = \alpha_2 \cdot \sqrt {{R}/{2}} \cdot l\hspace{0.05cm}.$$

This value is particularly suitable for comparing different conducted transmission systems with different

  1. coaxial cable types  $($normal or small coaxial cable$)$,  each identified by the parameter  $\alpha_2$,
  2. bit rates  $(R)$,  and
  3. cable lengths  $(l)$.


Interpretation of the right graph

The right–hand diagram above shows the characteristic cable attenuation  $\rm a_∗$  in  »Neper«  $\rm (Np)$  as a function of the bit rate  $R$  and the cable length  $l$  for

  • the normal coaxial cable  $\text{(2.6/9.5 mm)}$   ⇒   left ordinate labeling,  and
  • for the small coaxial cable  $\text{(1.2/4.4 mm)}$   ⇒   right ordinate labeling.


This diagram shows the PCM systems of hierarchy levels   $3$  to  $5$  proposed by the  $\text{ITU-T}$  $($»ITU Telecommunication Standardization Sector«$)$  in the 1970s.  One recognizes:

  1. For all these systems for PCM speech transmission,  the characteristic cable attenuation assumes values between  $7 \ \rm Np \ \ (≈ 61 \ dB)$  and  $10.6 \ \rm Np \ \ (≈ 92 \ dB)$ .
  2. The system  $\text{PCM 480}$  – designed for 480 simultaneous telephone calls - with bit rate  $R ≈ 35 \ \rm Mbit/s$  was specified for both the normal coaxial cable  $($with  $l = 9.3 \ \rm km)$  and the small coaxial cable  $($with  $l = 4 \ \rm km)$.  The  $\rm a_∗$–values  $10.4\ \rm Np$  resp.  $9.9\ \rm Np$  are in the same order of magnitude.
  3. The system  $\text{PCM 1920}$  of the fourth hierarchy level  $($specified for the normal coaxial cable$)$  with  $R ≈ 140 \ \rm Mbit/s$  and  $l = 4.65 \ \rm km$  is parameterized by  $\rm a_∗ = 10.6 \ \rm Np$  or  $10.6 \ {\rm Np}· 8.688 \ \rm dB/Np ≈ 92\ \rm dB$ .
  4. Although the system  $\text{PCM 7680}$  has a four times greater bit rate  $R ≈ 560 \ \rm Mbit/s$ ,  the characteristic cable attenuation of   $\rm a_∗ ≈ 61 \ dB$  is due to the better medium  »normal coaxial cable« and the shorter cable sections $(l = 1. 55 \ \rm km)$  by a factor of  $3$  significantly lower .
  5. These numerical values also show that for coaxial cable systems,  the cable length  $l$  is more critical than the bit rate  $R$.  If one wants to double the cable length,  one has to reduce the bit rate by a factor  $4$.


You can view the topic described here with the interactive HTLM 5/JavaScript applet  »Attenuation of Copper Cables« .

Impulse response of a coaxial cable


To calculate the impulse response,  the first two of the in total five  components of the complex propagation function  $($per unit length$)$  can be neglected  $($the reasoning can be found in the previous section$)$.  So we start from the following equation:

$$\gamma(f) = \alpha_0 + \alpha_1 \cdot f + {\rm j} \cdot \beta_1 \cdot f +\alpha_2 \cdot \sqrt {f}+ {\rm j}\cdot \beta_2 \cdot \sqrt {f} \approx {\rm j} \cdot \beta_1 \cdot f +\alpha_2 \cdot \sqrt {f}+ {\rm j}\cdot \beta_2 \cdot \sqrt {f} \hspace{0.05cm}.$$

Considering

  1. the cable length  $l$,
  2. the characteristic cable attenuation  $\rm a_∗$,  and
  3. that  $α_2$  $($in  »Np«$)$  and  $β_2$  (in  »rad«$)$  are numerically equal,


thus applies to the  »frequency response of the coaxial cable«:

$$H_{\rm K}(f) = {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1 f} \cdot {\rm e}^{-{\rm a}_{\rm \star}\hspace{0.05cm}\cdot \hspace{0.05cm} \sqrt{2f/R} }\cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}{\rm a}_{\rm \star}\hspace{0.05cm}\cdot \hspace{0.02cm} \sqrt{2f/R}}= {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1 f} \cdot {\rm e}^{-2{\rm a}_{\rm \star}\hspace{0.03cm}\cdot \hspace{0.03cm} \sqrt{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}f/R}} \hspace{0.05cm}.$$

The following abbreviations are used here:

$$b_1\hspace{0.1cm}{(\rm in }\hspace{0.15cm}{\rm rad)}= \beta_1 \cdot l \hspace{0.05cm}, \hspace{0.8cm} {\rm a}_{\rm \star}\hspace{0.1cm}{(\rm in }\hspace{0.15cm}{\rm Np)}= \alpha_2 \cdot \sqrt {R/2} \cdot l \hspace{0.05cm}.$$

The time domain representation is obtained by applying the  »inverse Fourier transform«  and the  »convolution theorem«:

$$h_{\rm K}(t) = \mathcal{F}^{-1} \left \{ {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1 f}\right \} \star\mathcal{F}^{-1} \left \{ {\rm e}^{-2{\rm a}_{\rm \star}\hspace{0.03cm}\cdot \hspace{0.03cm} \sqrt{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}f/R} }\right \} \hspace{0.05cm}.$$

It must be taken into account here:

  • The first term yields the Dirac delta function  $δ(t - τ_{\rm P})$  shifted by the phase delay  $τ_{\rm P} = b_1/2π$ .
  • The second term can be given analytically closed.  We write  $h_{\rm K}(t + τ_P)$,  so that the phase delay  $τ_{\rm P}$  need not be considered further.
$$h_{\rm K}(t + \tau_{\rm P}) = \frac {{\rm a}_{\rm \star}}{\pi \cdot \sqrt{2 \cdot R \cdot t^3}}\cdot {\rm exp} \left [ -\frac {{\rm a}_{\rm \star}^2}{ {2\pi \cdot R\cdot t}} \right ]\hspace{0.05cm},\hspace{0.2cm}{\rm a}_{\rm \star}\hspace{0.15cm}{\rm in\hspace{0.15cm} Np}\hspace{0.05cm}.$$
  • Since the bit rate  $R$  also has already been considered in the $\rm a_∗$  definition;  this equation can be easily represented with the normalized time  $t\hspace{0.05cm}' = t/T$ :
$$h_{\rm K}(t\hspace{0.05cm}' + \tau_{\rm P}\hspace{0.05cm} ') = \frac {1}{T} \cdot \frac {{\rm a}_{\rm \star}}{\pi \cdot \sqrt{2 \cdot t\hspace{0.05cm}'\hspace{0.05cm}^3}}\cdot {\rm exp} \left [ -\frac {{\rm a}_{\rm \star}^2}{ {2\pi \cdot t\hspace{0.05cm}'}} \right ]\hspace{0.05cm},\hspace{0.2cm}{\rm a}_{\rm \star}\hspace{0.15cm}{\rm in\hspace{0.15cm} Np}\hspace{0.05cm}.$$
Here  $T = 1/R$  denotes  »the symbol duration of a binary system«  and it holds  $τ_{\rm P} \hspace{0.05cm}' = τ_{\rm P}/T$.

$\text{Example 1:}$  The results of this section are illustrated by the following graph as an example.

Impulse response of a coaxial cable with  $\rm a_∗ = 60 \ dB$
  1. The normalized impulse response  $T · h_{\rm K}(t)$  of a coaxial cable with  $\rm a_∗ = 60 \ dB \ \ (6.9\ Np)$ is shown.
  2. The attenuation coefficients  $α_0$  and  $α_1$  can thus be neglected,  as shown in the last section.
  3. For the left graph, the parameter  $β_1 = 0$  was also set.


Because of the parameterization by the coefficient  $\rm a_∗$,  and the time normalization to the symbol duration  $T$  the left curve is equally valid

  1. for systems with small or normal coaxial cable, 
  2. different lengths,
  3. different bit rates. 


For example for

  • normal coaxial cable  $\text{2.6/9.5 mm}$,  bit rate  $R = 140 \ \rm Mbit/s$,  cable length  $l = 3 \ \rm km$   ⇒   $\text{system A}$,
  • small coaxial cable  $\text{1.2/4.4 mm}$,  bit rate  $R = 35 \ \rm Mbit/s$,
    cable length  $l = 2.8 \ \rm km$   ⇒   $\text{system B}$.


It can be seen in the left–hand diagram that even at this moderate cable attenuation  $\rm a_∗ = 60 \ \rm dB$  the impulse response already extends over more than   $200$  symbol durations due to the skin effect  $(α_2 = β_2 ≠ 0)$.  Since the integral over  $h_{\rm K}(t)$  is equal to  $H_{\rm K}(f = 0) = 1$,  the maximum value becomes very small:

$${\rm Max}\big [h_{\rm K}(t)\big ] \approx 0.03.$$

In the diagram on the right,  the effects of the phase parameter  $β_1$  can be seen.  Note the different time scales of the left and the right diagram:

  • For  $\text{system A}$  $(β_1 = 21.78 \ \rm rad/(km · MHz)$,  $T = 7.14\ \rm ns)$   $β_1$  leads to a phase delay of
$$\tau_{\rm A}= \frac {\beta_1 \cdot l}{2\pi} =\frac {21.78\, { {\rm rad} }/{ {(\rm km \cdot MHz)} }\cdot 3\,{\rm km} }{2\pi} = 10.4\,{\rm \mu s}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\tau_{\rm A}\hspace{0.05cm}' = {\tau_{\rm A} }/{T} \approx 1457\hspace{0.05cm}.$$
  • On the other hand,  the following can be obtained for  $\text{system B}$  $(β_1 = 22.18 \ \rm rad/(km · MHz)$,  $T = 30 \ \rm ns)$:
$$\tau_{\rm B}= \frac {\beta_1 \cdot l}{2\pi} =\frac {22.18\, { {\rm rad} }/{ {(\rm km \cdot MHz)} }\cdot 2.8\,{\rm km} }{2\pi} = 9.9\,{\rm µ s}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\tau_{\rm B}\hspace{0.05cm}' ={\tau_{\rm B} }/{T} \approx 330\hspace{0.05cm}.$$

Although  $τ_{\rm A} ≈ τ_{\rm B}$  holds,  completely different ratios result because of the time normalization to  $T = 1/R$ .


$\text{Conclusion:}$  When simulating and optimizing transmission systems, 
one usually omits the linear phase term  $b_1 = β_1 · f$,  since this results exclusively in a  $($often not disturbing$)$  phase delay,  but no signal distortions.

Basic receiver pulse


With the  »basic transmission pulse«  $g_s(t)$   ⇒   »basic pulse of the transmitted signal«  $s(t)$  and the  »impulse response  $h_{\rm K}(t)$  of the channel,  the result for the  »basic receiver pulse«  $g_r(t)$   ⇒   »basic pulse of the received signal«  $r(t)$  is:

$$g_r(t) = g_s(t) \star h_{\rm K}(t)\hspace{0.05cm}.$$

If a non-return-to-zero  $\rm (NRZ)$  rectangular pulse  $g_s(t)$  with amplitude  $s_0$  and duration  $Δt_s = T$  is used at the transmitter,  the following results for the basic pulse at the receiver input:

$$g_r(t) = 2 s_0 \cdot \left [ {\rm Q} \left (\frac {{\rm a}_{\rm \star}/\sqrt {\pi}}{ \sqrt{ (t/T - 0.5)}}\right ) - {\rm Q} \left (\frac {{\rm a}_{\rm \star}/\sqrt {\pi}}{ \sqrt{ (t/T + 0.5)}}\right ) \right ]\hspace{0.05cm}.$$

Here  $\rm a_∗$  denotes the  »characteristic cable attenuation«  $($in Neper$)$  and  ${\rm Q}(x)$  the  »complementary Gaussian error function«.

$\text{Example 2:}$  The figure shows for the characteristic cable attenuations  $\rm a_∗ = 40 \ \rm dB$, ... ,  $100 \ \rm dB$  $($smaller  $\rm a_∗$–values   are not relevant for practice$)$

  • the normalized coaxial cable impulse response  $T · h_{\rm K}(t)$   ⇒   solid curves,
  • the basic receiver pulse  $($»rectangular response«$)$  $g_r(t)$  normalized to the transmission amplitude  $s_0$   ⇒   dotted line.
Impulse response of the coaxial cable and basic receiver pulse


One recognizes the following from this diagram:

  1. With  $\rm a_∗ = 40 \ \rm dB$,  the normalized basic receiver pulse  $g_r(t)/s_0$  is at the peak slightly  $($about a factor of $0.95)$  smaller than the normalized impulse response  $T · h_{\rm K}(t)$.  Here is a small difference between impulse response and basic receiver pulse .
  2. In contrast,  for  $a_∗ ≥ 60 \ \rm dB$,  the basic receiver pulse and the impulse response are indistinguishable within the drawing accuracy.
  3. For a return-to-zero  $\rm (RZ)$  pulse,  the above equation for  $g_r(t)$  would still need to be multiplied by the factor  $Δt_s/T$.  In this case  $g_r(t)/s_0$  is smaller than  $T · h_{\rm K}(t)$  by at least this factor.
  4. The equation modified in this way is also a good approximation for other basic transmission pulses as long as  $\rm a_∗≥ 60 \ \rm dB$  is sufficiently large.  $Δt_s$  then indicates the »equivalent pulse duration«  of  $g_r(t)$.


Special features of coaxial cable systems


Binary transmission system with coaxial cable

Assuming binary transmission

  • with non-return-to-zero  $\rm (NRZ)$  rectangular pulses  $($duration $T)$ 
  • and a coaxial transmission channel, 


the following system model is obtained.  In particular,  it should be noted:

$(1)$  In a simulation,  the phase delay time of the coaxial cable is conveniently left out of consideration.  Then the basic receiver pulse  $g_r(t)$  is approximated by  $($with ${\rm a}_{\rm \star}$ in Neper$)$:

$$g_r(t) \approx s_0 \cdot T \cdot h_{\rm K}(t) = \frac {s_0 \cdot {\rm a}_{\rm \star}/\pi}{ \sqrt{2 \cdot(t/T)^3}}\cdot {\rm e}^{ -{{\rm a}_{\rm \star}^2}/( {2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}t/T}) } \hspace{0.05cm}.$$

$(2)$  Because of the good shielding of coaxial cables against other impairments,  the  »thermal noise«  is the dominant stochastic perturbation.  In this case,  the signal  $n(t)$  is Gaussian and white.  It can described by the  $($two-sided$)$  noise power density  $N_0/2$.

$(3)$  By far the largest noise component arises in the input stage of the receiver,  so that it is expedient to add the noise signal  $n(t)$  at the  interface  »cable ⇒ receiver«.  This noise addition point is also useful because the frequency response  $H_{\rm K}(f)$  decisively attenuates all noise accumulated along the cable.

$(3)$  Then the received signal is with the amplitude coefficients  $a_{\nu}$:

$$r(t) = \sum_{\nu = - \infty}^{+ \infty}a_{\nu}\cdot g_r(t - \nu \cdot T)+ n(t) \hspace{0.05cm} .$$


Exercises for the chapter


Exercise 4.4: Coaxial Cable - Frequency Response

Exercise 4.5: Coaxial Cable - Impulse Response

Exercise 4.5Z: Impulse Response once again


References

  1. Wellhausen, H. W.:  Dämpfung, Phase und Laufzeiten bei Weitverkehrs–Koaxialpaaren.  Frequenz 31, S. 23-28, 1977.