Difference between revisions of "Aufgaben:Exercise 3.1: Impulse Response of the Coaxial Cable"

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{{quiz-Header|Buchseite=Digitalsignalübertragung/Ursachen_und_Auswirkungen_von_Impulsinterferenzen
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{{quiz-Header|Buchseite=Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference
 
}}
 
}}
  
[[File:P_ID1370__Dig_A_3_1.png|right|frame|Impulsantwort eines Koaxialkabels]]
+
[[File:P_ID1370__Dig_A_3_1.png|right|frame|Impulse response of a coaxial cable]]
Der Frequenzgang eines Koaxialkabels der Länge $l$ ist durch folgende Formel darstellbar:
+
The frequency response of a coaxial cable of length  $l$  can be represented by the following formula:
 
:$$H_{\rm K}(f) \ = \ {\rm e}^{- \alpha_0 \hspace{0.05cm} \cdot \hspace{0.05cm} l}
 
:$$H_{\rm K}(f) \ = \ {\rm e}^{- \alpha_0 \hspace{0.05cm} \cdot \hspace{0.05cm} l}
   \cdot $$
+
   \cdot  
:$$\ \cdot \
 
 
   {\rm e}^{- (\alpha_1 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1) \hspace{0.05cm}\cdot f \hspace{0.05cm}\cdot \hspace{0.05cm}l}  \cdot
 
   {\rm e}^{- (\alpha_1 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1) \hspace{0.05cm}\cdot f \hspace{0.05cm}\cdot \hspace{0.05cm}l}  \cdot
  $$
+
  \ {\rm e}^{- (\alpha_2 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2) \hspace{0.05cm}\cdot \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm}l}
:$$\ \cdot \ {\rm e}^{- (\alpha_2 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2) \hspace{0.05cm}\cdot \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm}l}
 
 
     \hspace{0.05cm}.$$
 
     \hspace{0.05cm}.$$
  
Der erste Term dieser Gleichung ist auf die Ohmschen Verluste zurückzuführen und der zweite Term auf die Querverluste. Dominant ist jedoch der Skineffekt, der durch den dritten Term ausgedrückt wird.
+
*The first term of this equation is due to the ohmic losses.
Mit den für ein so genanntes Normalkoaxialkabel ($2.6 \ \rm mm$ Kerndurchmesser und $9.5 \ \rm mm$ Außendurchmesser) gültigen Koeffizienten
+
* The second term is due to the transverse losses.
 +
*Dominant,  however,  is the skin effect,  which is expressed by the third term.
 +
 
 +
 
 +
With the coefficients valid for a so-called  "standard coaxial cable"  $(2.6 \ \rm mm$  core diameter and  $9.5 \ \rm mm$  outer diameter$)$
 
:$$\alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km \cdot \sqrt{\rm MHz}}
 
:$$\alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km \cdot \sqrt{\rm MHz}}
 
   \hspace{0.05cm},
 
   \hspace{0.05cm},
   \hspace{0.2cm} \beta_2 = 0.2722 \hspace{0.15cm}\frac{\rm rad}{\rm km \cdot \sqrt{\rm MHz}}\hspace{0.05cm}.$$
+
   \hspace{0.2cm} \beta_2 = 0.2722 \hspace{0.15cm}\frac{\rm rad}{\rm km \cdot \sqrt{\rm MHz}}\hspace{0.05cm},$$
  
lässt sich dieser Frequenzgang auch wie folgt darstellen:
+
the frequency response can also be represented as follows:
 
:$$H_{\rm K}(f) \approx {\rm e}^{- 0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km}
 
:$$H_{\rm K}(f) \approx {\rm e}^{- 0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km}
 
   \hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}} } \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}
 
   \hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}} } \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}
Line 26: Line 28:
 
     \hspace{0.05cm}.$$
 
     \hspace{0.05cm}.$$
  
Das heißt, der Dämpfungsverlauf $a_K(f)$ und der Phasenverlauf $b_K(f)$ sind bis auf die Pseudoeinheiten „$\rm Np$” bzw. „$\rm rad$” identisch.
+
That means: Attenuation curve  $a_{\rm K}(f)$  and phase curve  $b_{\rm K}(f)$  are identical except for the pseudo units  "$\rm Np$"  and  "$\rm rad$".
  
Definiert man die charakteristische Kabeldämpfung $a_*$ bei der halben Bitrate ($R_B/2$), so kann man Digitalsysteme unterschiedlicher Bitrate und Länge einheitlich behandeln:
+
If one defines the characteristic cable attenuation  $a_*$  at half the bit rate  $(R_{\rm B}/2)$,  one can treat digital systems of different bit rate and length uniformly:
 
:$$a_{\star} = a_{\rm K}(f = {R_{\rm B}}/{2})
 
:$$a_{\star} = a_{\rm K}(f = {R_{\rm B}}/{2})
 
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm K}(f) = {\rm e}^{-
 
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm K}(f) = {\rm e}^{-
 
a_{\star} \cdot \sqrt{2f/R_{\rm
 
a_{\star} \cdot \sqrt{2f/R_{\rm
 
   B}}}\cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} a_{\star} \cdot \sqrt{2f/R_{\rm
 
   B}}}\cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} a_{\star} \cdot \sqrt{2f/R_{\rm
   B}}}\hspace{0.4cm}{\rm mit}\hspace{0.2cm}a_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}
+
   B}}}\hspace{0.4cm}{\rm with}\hspace{0.2cm}a_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}
 
   \hspace{0.05cm}.$$
 
   \hspace{0.05cm}.$$
  
Der entsprechende $\rm dB$–Wert ist um den Faktor $8.688$ größer. Bei einem Binärsystem gilt $R_B = 1/T$, so dass sich dann die charakteristische Kabeldämpfung auf die Frequenz $f = 1/(2T)$ bezieht. Die Fouriertransformierte von $H_K(f)$ liefert die Impulsantwort $h_K(t)$, die für ein Koaxialkabel mit den hier beschriebenen Näherungen in geschlossen-analytischer Form angebbar ist. Für ein Binärsystem gilt:  
+
The corresponding  $\rm dB$  value is larger by a factor of  $8.688$.  For a binary system,  $R_{\rm B} = 1/T$  applies,  so that the characteristic cable attenuation is then related to the frequency  $f = 1/(2T)$. 
 +
 
 +
 
 +
The Fourier transform of  $H_{\rm K}(f)$  yields the impulse response  $h_{\rm K}(t)$,  which can be given in closed-analytic form for a coaxial cable with the approximations described here.  For a binary system holds:
 
:$$h_{\rm K}(t) =  \frac{ a_{\star}/T}{  \sqrt{2  \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot
 
:$$h_{\rm K}(t) =  \frac{ a_{\star}/T}{  \sqrt{2  \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot
 
   {\rm exp} \left[ - \frac{a_{\star}^2}{2  \pi  \cdot t/T}\hspace{0.1cm}\right]
 
   {\rm exp} \left[ - \frac{a_{\star}^2}{2  \pi  \cdot t/T}\hspace{0.1cm}\right]
   \hspace{0.4cm}{\rm mit}\hspace{0.2cm}a_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}
+
   \hspace{0.4cm}{\rm with}\hspace{0.2cm}a_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}
 
\hspace{0.05cm}.$$
 
\hspace{0.05cm}.$$
  
Die Teilaufgabe e) bezieht sich auf den Empfangsgrundimpuls $g_r(t) = g_s(t) * h_K(t)$, wobei für $g_s(t)$ ein Rechteckimpuls mit der Höhe $s_0$ und der Dauer $T$ angenommen werden soll.  
+
Subtask  '''(5)'''  is related to the basic receiver pulse  $g_r(t) = g_s(t) * h_K(t)$,  where  $g_s(t)$  should be assumed to be a rectangular pulse with height  $s_0$  and duration  $T$. 
 +
 
 +
 
 +
 
 +
 
 +
 
 +
Notes:
 +
*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference|"Causes and Effects of Intersymbol Interference"]].
  
 +
*Reference is made in particular to the section  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces|"Signals, Basis Functions and Vector Spaces"]].
 +
  
''Hinweis:''
 
* Die Aufgabe bezieht sich auf das Kapitel [[Digitalsignal%C3%BCbertragung/Ursachen_und_Auswirkungen_von_Impulsinterferenzen|Ursachen und Auswirkungen von Impulsinterferenzen]] dieses Buches und auf das Kapitel [[Digitalsignal%C3%BCbertragung/Signale,_Basisfunktionen_und_Vektorr%C3%A4ume|Signale, Basisfunktionen und Vektorräume]] des Buches „Lineare zeitinvariante Systeme”.
 
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Wie groß ist die Länge $l$ eines Normalkoaxialkabels, wenn sich für die Bitrate $R_B = 140 \ \rm Mbit/s$ die charakteristische Kabeldämpfung zu $a_* = 60 \ \rm dB$ ergibt?
+
{What is the length &nbsp;$l$&nbsp; of a standard coaxial cable,&nbsp; if for the bit rate &nbsp;$R_{\rm B} = 140 \ \rm Mbit/s$&nbsp; the characteristic cable attenuation is &nbsp;$a_* = 60 \ \rm dB$?&nbsp;
 
|type="{}"}
 
|type="{}"}
$l$ = { 3 3% } $\rm km $
+
$l \ = \ $ { 3 3% } $\ \rm km $
  
{Zu welcher Zeit $t_{\rm max}$ besitzt $h_K(t)$ sein Maximum? Es gelte weiter $a_* = 60 \ \rm dB$.
+
{At what time &nbsp;$t_{\rm max}$&nbsp; does &nbsp;$h_{\rm K}(t)$&nbsp; have its maximum?&nbsp; Let &nbsp;$a_* = 60 \ \rm dB$ be further valid.
 
|type="{}"}
 
|type="{}"}
$t_{\rm max}/T$ = { 5 3% } $ $
+
$t_{\rm max}/T= \ $ { 5 3% }  
  
{Wie groß ist der Maximalwert der Impulsantwort?
+
{What is the maximum value of the impulse response?&nbsp; Let &nbsp;$a_* = 60 \ \rm dB$&nbsp; continue to hold.
 
|type="{}"}
 
|type="{}"}
${\rm Max} [h_K(t)]$ = { 0.03 3% } $1/T $
+
${\rm Max}\  \big [h_{\rm K}(t)\big ]= \ $ { 0.03 3% } $\ \cdot 1/T $
  
{Ab welcher Zeit $t_{\rm 5\%}$ ist $h_K(t)$ kleiner als $5\%$ des Maximums? Berücksichtigen Sie als Näherung nur den ersten Term der angegebenen Formel.
+
{At what time &nbsp;$t_{\rm 5\%}$&nbsp; is &nbsp;$h_{\rm K}(t)$&nbsp; less than &nbsp;$5\%$&nbsp; of the maximum?&nbsp; Consider only the first term of the given formula as an approximation.
 
|type="{}"}
 
|type="{}"}
$t_{\rm 5\%}/T$ = { 103.5 3% } $1/T $
+
$t_{\rm 5\%}/T= \ $ { 103.5 3% }
  
{Welche Aussagen treffen für den Empfangsgrundimpuls $g_r(t)$ zu?
+
{Which statements are true for the basic receiver pulse &nbsp;$g_r(t)$?&nbsp;
 
|type="[]"}
 
|type="[]"}
- $g_r(t)$ ist doppelt so breit wie $h_K(t)$.
+
- $g_r(t)$&nbsp; is twice as wide as &nbsp;$h_{\rm K}(t)$.
+ Es gilt näherungsweise $g_r(t) = h_K(t) \cdot s_0 \cdot T$.
+
+ It is approximately &nbsp;$g_r(t) = h_{\rm K}(t) \cdot s_0 \cdot T$.
- $g_r(t)$ kann durch einen Gaußimpuls angenähert werden.
+
- $g_r(t)$&nbsp; can be approximated by a Gaussian pulse.
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Die charakteristische Kabeldämpfung $a_* = 60 \ \rm dB$ entspricht etwa $6.9 \ \rmNp$ Deshalb muss gelten:
+
'''(1)'''&nbsp; The characteristic cable attenuation&nbsp; $a_* = 60 \ \rm dB$&nbsp; corresponds to about&nbsp; $6.9 \ \rm Np$.&nbsp; Therefore,&nbsp; it must hold:
 
:$$\alpha_2 \cdot l \cdot {R_{\rm B}}/{2} = 6.9\,\,{\rm
 
:$$\alpha_2 \cdot l \cdot {R_{\rm B}}/{2} = 6.9\,\,{\rm
Np}
+
Np}$$
\hspace{0.3cm}\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\rm
+
:$$\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\rm
Np}}{0.2722\,\,\frac{Np}{{\rm km} \cdot \sqrt{MHz}}
+
Np}}{0.2722\,\,\frac{\rm Np}{{\rm km} \cdot \sqrt{\rm MHz}}
 
\cdot \sqrt{70\,\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 3\,\,{\rm km}}
 
\cdot \sqrt{70\,\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 3\,\,{\rm km}}
 
   \hspace{0.05cm}.$$
 
   \hspace{0.05cm}.$$
  
  
'''(2)'''&nbsp; Mit den Substitutionen
+
'''(2)'''&nbsp; With the substitutions
 
:$$x =  { t}/{  T}, \hspace{0.2cm} K_1 = \frac{a_*/T}{\sqrt{2\pi^2 }}, \hspace{0.2cm}
 
:$$x =  { t}/{  T}, \hspace{0.2cm} K_1 = \frac{a_*/T}{\sqrt{2\pi^2 }}, \hspace{0.2cm}
 
   K_2 = \frac{a_*^2}{2\pi}$$
 
   K_2 = \frac{a_*^2}{2\pi}$$
  
kann die Impulsantwort wie folgt beschrieben werden
+
the impulse response can be described as follows:
 
:$$h_{\rm K}(x) =  K_1 \cdot x^{-3/2}\cdot {\rm e}^{-K_2/x}
 
:$$h_{\rm K}(x) =  K_1 \cdot x^{-3/2}\cdot {\rm e}^{-K_2/x}
 
   \hspace{0.05cm}.$$
 
   \hspace{0.05cm}.$$
  
Durch Nullsetzen der Ableitung folgt daraus:
+
*By setting the derivative to zero,&nbsp; it follows:
 
:$$- {3}/{2} \cdot  K_1 \cdot x^{-5/2}\cdot {\rm e}^{-K_2/x}+ K_1 \cdot x^{-3/2}\cdot {\rm
 
:$$- {3}/{2} \cdot  K_1 \cdot x^{-5/2}\cdot {\rm e}^{-K_2/x}+ K_1 \cdot x^{-3/2}\cdot {\rm
 
   e}^{-K_2/x}\cdot (-K_2) \cdot (-x^{-2})= 0
 
   e}^{-K_2/x}\cdot (-K_2) \cdot (-x^{-2})= 0
Line 100: Line 112:
 
:$$\Rightarrow \hspace{0.3cm} {3}/{2} \cdot x^{-5/2} = K_2 \cdot
 
:$$\Rightarrow \hspace{0.3cm} {3}/{2} \cdot x^{-5/2} = K_2 \cdot
 
  x^{-7/2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
 
  x^{-7/2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
  x_{\rm max} = {2}/{3} \cdot  K_2 = { a_{\star}^2}/({{3  \pi}})
+
  x_{\rm max} = {2}/{3} \cdot  K_2 = { a_{\star}^2}/({3  \pi})
 
   \hspace{0.05cm}.$$
 
   \hspace{0.05cm}.$$
  
Daraus ergibt sich für $60 \ \rm dB$ Kabeldämpfung($a_* &asymp; 6.9 \ \rm Np$)
+
*This gives for&nbsp; $60 \ \rm dB$&nbsp; cable attenuation&nbsp; $(a_* &asymp; 6.9 \ \rm Np)$:
 
:$$x_{\rm max} =  { t_{\rm max}}/{  T} \hspace{-0.1cm}: \hspace{0.2cm} { t_{\rm max}}/{  T} = { 6.9^2}/({3\pi})\hspace{0.15cm}\underline {\approx 5} \hspace{0.2cm}.$$
 
:$$x_{\rm max} =  { t_{\rm max}}/{  T} \hspace{-0.1cm}: \hspace{0.2cm} { t_{\rm max}}/{  T} = { 6.9^2}/({3\pi})\hspace{0.15cm}\underline {\approx 5} \hspace{0.2cm}.$$
  
  
'''(3)'''&nbsp; Setzt man das Ergebnis von b) in die vorgegebene Gleichung ein, so erhält man (zur Vereinfachung verwenden wir $a$ anstelle von $a_*$):
+
'''(3)'''&nbsp; Substituting the result of&nbsp; '''(2)'''&nbsp; into the given equation,&nbsp; we obtain&nbsp; (using&nbsp; $a$&nbsp; instead of&nbsp; $a_*$):
 
:$$h_{\rm K}(t_{\rm max})  \ = \  \frac{1}{T} \cdot \frac{ a}{  \sqrt{2  \pi^2 \cdot \frac{a^6}{(3\pi)^3}}}\hspace{0.1cm} \cdot
 
:$$h_{\rm K}(t_{\rm max})  \ = \  \frac{1}{T} \cdot \frac{ a}{  \sqrt{2  \pi^2 \cdot \frac{a^6}{(3\pi)^3}}}\hspace{0.1cm} \cdot
 
   {\rm exp} \left[ - \frac{a^2}{2\pi} \cdot
 
   {\rm exp} \left[ - \frac{a^2}{2\pi} \cdot
   \frac{3\pi}{a^2}\hspace{0.1cm}\right]$$
+
   \frac{3\pi}{a^2}\hspace{0.1cm}\right]= \frac{1}{T} \cdot \frac{1}{a^2}\cdot
:$$ \ = \frac{1}{T} \cdot \frac{1}{a^2}\cdot
 
 
   \sqrt{\frac{27 \pi
 
   \sqrt{\frac{27 \pi
 
  }{2}} \cdot {\rm e}^{-3/2}\hspace{0.15cm}\approx  \frac{1}{T} \cdot \frac{1.453}{a^2}
 
  }{2}} \cdot {\rm e}^{-3/2}\hspace{0.15cm}\approx  \frac{1}{T} \cdot \frac{1.453}{a^2}
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
Mit $a = 6.9$ kommt man somit zum Endergebnis:
+
*Thus,&nbsp; with&nbsp; $a = 6.9$,&nbsp; we arrive at the final result:
:$${\rm Max}[h_{\rm K}(t)]  = \frac{1.453}{{6.9\,}^2} \cdot {1}/{T}\hspace{0.15cm}\underline {\approx  0.03 \cdot {1}/{T}}
+
:$${\rm Max} \ [h_{\rm K}(t)]  = \frac{1.453}{{6.9\,}^2} \cdot {1}/{T}\hspace{0.15cm}\underline {\approx  0.03 \cdot {1}/{T}}
 
   \hspace{0.05cm}.$$
 
   \hspace{0.05cm}.$$
  
  
'''(4)'''&nbsp; Mit dem Ergebnis aus 3) lautet die Bestimmungsgleichung:
+
'''(4)'''&nbsp; Using the result of subtask&nbsp; '''(3)''',&nbsp; the determining equation is:
:$$\frac{ a/T}{  \sqrt{2  \pi^2 \cdot (t_{5\%}/T)^3}}= 0.05 \cdot 0.03 \cdot {1}/{T}= 0.0015  \cdot {1}/{T}$$
+
:$$\frac{ a/T}{  \sqrt{2  \pi^2 \cdot (t_{5\%}/T)^3}}= 0.05 \cdot 0.03 \cdot {1}/{T}= 0.0015  \cdot {1}/{T} \hspace{0.3cm}
:$$\Rightarrow \hspace{0.3cm}  (t_{5\%}/T)^{3/2} = \frac{a}{\sqrt{2} \cdot \pi \cdot
+
\Rightarrow \hspace{0.3cm}  (t_{5\%}/T)^{3/2} = \frac{a}{\sqrt{2} \cdot \pi \cdot
 
   0.0015}\approx 1036 \hspace{0.3cm}\Rightarrow
 
   0.0015}\approx 1036 \hspace{0.3cm}\Rightarrow
 
   \hspace{0.3cm}t_{5\%}/T \hspace{0.15cm}\underline {\approx 103.5}\hspace{0.05cm}.$$
 
   \hspace{0.3cm}t_{5\%}/T \hspace{0.15cm}\underline {\approx 103.5}\hspace{0.05cm}.$$
  
Dieser Wert ist etwas zu groß, da der zweite Term $e^{\rm &ndash; 0.05} &asymp; 0.95$ vernachlässigt wurde. Die exakte Berechnung liefert $t_{\rm 5\%}/T &asymp; 97$.
+
*This value is slightly too large because the second term&nbsp; ${\rm e}^{\rm &ndash; 0.05} &asymp; 0.95$&nbsp; was neglected.  
 +
*The exact calculation gives&nbsp; $t_{\rm 5\%}/T &asymp; 97$.
  
  
'''(5)'''&nbsp; Richtig ist der <u>zweite Lösungsvorschlag</u>. Allgemein gilt:
+
'''(5)'''&nbsp; The <u>second solution</u>&nbsp; is correct.&nbsp; In general:
 
:$$g_r(t) = g_s(t) \star h_{\rm K}(t) = s_0 \cdot
 
:$$g_r(t) = g_s(t) \star h_{\rm K}(t) = s_0 \cdot
 
\int_{t-T/2}^{t+T/2} h_{\rm K}(\tau) \,{\rm d} \tau .$$
 
\int_{t-T/2}^{t+T/2} h_{\rm K}(\tau) \,{\rm d} \tau .$$
  
Da sich die Kanalimpulsantwort $h_K(t)$ innerhalb einer Symboldauer nur unwesentlich ändert, kann hierfür auch geschrieben werden:
+
*Since the channel impulse response&nbsp; $h_{\rm K}(t)$&nbsp; changes only insignificantly within a symbol duration,&nbsp; it can also be written for this purpose:
:$$g_r(t) = h_{\rm K}(t) \cdot s_0 \cdot T .$$
+
:$$g_r(t) \approx h_{\rm K}(t) \cdot s_0 \cdot T .$$
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
[[Category:Aufgaben zu Digitalsignalübertragung|^3.1 Auswirkungen von Impulsinterferenzen^]]
+
[[Category:Digital Signal Transmission: Exercises|^3.1 Intersymbol Interference^]]

Latest revision as of 13:59, 31 May 2022

Impulse response of a coaxial cable

The frequency response of a coaxial cable of length  $l$  can be represented by the following formula:

$$H_{\rm K}(f) \ = \ {\rm e}^{- \alpha_0 \hspace{0.05cm} \cdot \hspace{0.05cm} l} \cdot {\rm e}^{- (\alpha_1 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1) \hspace{0.05cm}\cdot f \hspace{0.05cm}\cdot \hspace{0.05cm}l} \cdot \ {\rm e}^{- (\alpha_2 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2) \hspace{0.05cm}\cdot \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm}l} \hspace{0.05cm}.$$
  • The first term of this equation is due to the ohmic losses.
  • The second term is due to the transverse losses.
  • Dominant,  however,  is the skin effect,  which is expressed by the third term.


With the coefficients valid for a so-called  "standard coaxial cable"  $(2.6 \ \rm mm$  core diameter and  $9.5 \ \rm mm$  outer diameter$)$

$$\alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km \cdot \sqrt{\rm MHz}} \hspace{0.05cm}, \hspace{0.2cm} \beta_2 = 0.2722 \hspace{0.15cm}\frac{\rm rad}{\rm km \cdot \sqrt{\rm MHz}}\hspace{0.05cm},$$

the frequency response can also be represented as follows:

$$H_{\rm K}(f) \approx {\rm e}^{- 0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km} \hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}} } \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} 0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km} \hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}}} \hspace{0.05cm}.$$

That means: Attenuation curve  $a_{\rm K}(f)$  and phase curve  $b_{\rm K}(f)$  are identical except for the pseudo units  "$\rm Np$"  and  "$\rm rad$".

If one defines the characteristic cable attenuation  $a_*$  at half the bit rate  $(R_{\rm B}/2)$,  one can treat digital systems of different bit rate and length uniformly:

$$a_{\star} = a_{\rm K}(f = {R_{\rm B}}/{2}) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm K}(f) = {\rm e}^{- a_{\star} \cdot \sqrt{2f/R_{\rm B}}}\cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} a_{\star} \cdot \sqrt{2f/R_{\rm B}}}\hspace{0.4cm}{\rm with}\hspace{0.2cm}a_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np} \hspace{0.05cm}.$$

The corresponding  $\rm dB$  value is larger by a factor of  $8.688$.  For a binary system,  $R_{\rm B} = 1/T$  applies,  so that the characteristic cable attenuation is then related to the frequency  $f = 1/(2T)$. 


The Fourier transform of  $H_{\rm K}(f)$  yields the impulse response  $h_{\rm K}(t)$,  which can be given in closed-analytic form for a coaxial cable with the approximations described here.  For a binary system holds:

$$h_{\rm K}(t) = \frac{ a_{\star}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot {\rm exp} \left[ - \frac{a_{\star}^2}{2 \pi \cdot t/T}\hspace{0.1cm}\right] \hspace{0.4cm}{\rm with}\hspace{0.2cm}a_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np} \hspace{0.05cm}.$$

Subtask  (5)  is related to the basic receiver pulse  $g_r(t) = g_s(t) * h_K(t)$,  where  $g_s(t)$  should be assumed to be a rectangular pulse with height  $s_0$  and duration  $T$. 



Notes:



Questions

1

What is the length  $l$  of a standard coaxial cable,  if for the bit rate  $R_{\rm B} = 140 \ \rm Mbit/s$  the characteristic cable attenuation is  $a_* = 60 \ \rm dB$? 

$l \ = \ $

$\ \rm km $

2

At what time  $t_{\rm max}$  does  $h_{\rm K}(t)$  have its maximum?  Let  $a_* = 60 \ \rm dB$ be further valid.

$t_{\rm max}/T= \ $

3

What is the maximum value of the impulse response?  Let  $a_* = 60 \ \rm dB$  continue to hold.

${\rm Max}\ \big [h_{\rm K}(t)\big ]= \ $

$\ \cdot 1/T $

4

At what time  $t_{\rm 5\%}$  is  $h_{\rm K}(t)$  less than  $5\%$  of the maximum?  Consider only the first term of the given formula as an approximation.

$t_{\rm 5\%}/T= \ $

5

Which statements are true for the basic receiver pulse  $g_r(t)$? 

$g_r(t)$  is twice as wide as  $h_{\rm K}(t)$.
It is approximately  $g_r(t) = h_{\rm K}(t) \cdot s_0 \cdot T$.
$g_r(t)$  can be approximated by a Gaussian pulse.


Solution

(1)  The characteristic cable attenuation  $a_* = 60 \ \rm dB$  corresponds to about  $6.9 \ \rm Np$.  Therefore,  it must hold:

$$\alpha_2 \cdot l \cdot {R_{\rm B}}/{2} = 6.9\,\,{\rm Np}$$
$$\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\rm Np}}{0.2722\,\,\frac{\rm Np}{{\rm km} \cdot \sqrt{\rm MHz}} \cdot \sqrt{70\,\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 3\,\,{\rm km}} \hspace{0.05cm}.$$


(2)  With the substitutions

$$x = { t}/{ T}, \hspace{0.2cm} K_1 = \frac{a_*/T}{\sqrt{2\pi^2 }}, \hspace{0.2cm} K_2 = \frac{a_*^2}{2\pi}$$

the impulse response can be described as follows:

$$h_{\rm K}(x) = K_1 \cdot x^{-3/2}\cdot {\rm e}^{-K_2/x} \hspace{0.05cm}.$$
  • By setting the derivative to zero,  it follows:
$$- {3}/{2} \cdot K_1 \cdot x^{-5/2}\cdot {\rm e}^{-K_2/x}+ K_1 \cdot x^{-3/2}\cdot {\rm e}^{-K_2/x}\cdot (-K_2) \cdot (-x^{-2})= 0 \hspace{0.05cm}$$
$$\Rightarrow \hspace{0.3cm} {3}/{2} \cdot x^{-5/2} = K_2 \cdot x^{-7/2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} x_{\rm max} = {2}/{3} \cdot K_2 = { a_{\star}^2}/({3 \pi}) \hspace{0.05cm}.$$
  • This gives for  $60 \ \rm dB$  cable attenuation  $(a_* ≈ 6.9 \ \rm Np)$:
$$x_{\rm max} = { t_{\rm max}}/{ T} \hspace{-0.1cm}: \hspace{0.2cm} { t_{\rm max}}/{ T} = { 6.9^2}/({3\pi})\hspace{0.15cm}\underline {\approx 5} \hspace{0.2cm}.$$


(3)  Substituting the result of  (2)  into the given equation,  we obtain  (using  $a$  instead of  $a_*$):

$$h_{\rm K}(t_{\rm max}) \ = \ \frac{1}{T} \cdot \frac{ a}{ \sqrt{2 \pi^2 \cdot \frac{a^6}{(3\pi)^3}}}\hspace{0.1cm} \cdot {\rm exp} \left[ - \frac{a^2}{2\pi} \cdot \frac{3\pi}{a^2}\hspace{0.1cm}\right]= \frac{1}{T} \cdot \frac{1}{a^2}\cdot \sqrt{\frac{27 \pi }{2}} \cdot {\rm e}^{-3/2}\hspace{0.15cm}\approx \frac{1}{T} \cdot \frac{1.453}{a^2} \hspace{0.05cm}.$$
  • Thus,  with  $a = 6.9$,  we arrive at the final result:
$${\rm Max} \ [h_{\rm K}(t)] = \frac{1.453}{{6.9\,}^2} \cdot {1}/{T}\hspace{0.15cm}\underline {\approx 0.03 \cdot {1}/{T}} \hspace{0.05cm}.$$


(4)  Using the result of subtask  (3),  the determining equation is:

$$\frac{ a/T}{ \sqrt{2 \pi^2 \cdot (t_{5\%}/T)^3}}= 0.05 \cdot 0.03 \cdot {1}/{T}= 0.0015 \cdot {1}/{T} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} (t_{5\%}/T)^{3/2} = \frac{a}{\sqrt{2} \cdot \pi \cdot 0.0015}\approx 1036 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}t_{5\%}/T \hspace{0.15cm}\underline {\approx 103.5}\hspace{0.05cm}.$$
  • This value is slightly too large because the second term  ${\rm e}^{\rm – 0.05} ≈ 0.95$  was neglected.
  • The exact calculation gives  $t_{\rm 5\%}/T ≈ 97$.


(5)  The second solution  is correct.  In general:

$$g_r(t) = g_s(t) \star h_{\rm K}(t) = s_0 \cdot \int_{t-T/2}^{t+T/2} h_{\rm K}(\tau) \,{\rm d} \tau .$$
  • Since the channel impulse response  $h_{\rm K}(t)$  changes only insignificantly within a symbol duration,  it can also be written for this purpose:
$$g_r(t) \approx h_{\rm K}(t) \cdot s_0 \cdot T .$$