View source for Linear and Time Invariant Systems/Properties of Coaxial Cables

 
{{Header
|Untermenü=Properties of Electrical Cables
|Vorherige Seite=Some Results from Line Transmission Theory
|Nächste Seite=Properties of Balanced Copper Pairs
}}
==Complex propagation function of coaxial cables==
<br>
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: 
*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; $\text{4.4 mm}$.


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

The cable specific constants for the&nbsp; &raquo;'''standard coaxial cable'''&laquo;&nbsp; $\text{(2.6/9.5 mm)}$&nbsp; are:
:$$\begin{align*}\alpha_0  & = 0.00162\, \frac{ {\rm Np} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_1 = 0.000435\, \frac{ {\rm Np} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \alpha_2 = 0.2722\, \frac{ {\rm Np} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}, \\ \beta_1 & = 21.78\, \frac{ {\rm rad} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \beta_2 = 0.2722\, \frac{ {\rm rad} }{ {\rm km \cdot \sqrt{MHz} } } \hspace{0.05cm}.\end{align*}$$

Accordingly,&nbsp; the kilometric attenuation and phase constants for the&nbsp; &raquo;'''small coaxial cable'''&laquo;&nbsp; $\text{(1.2/4.4 mm)}$:
:$$\begin{align*}\alpha_0  & = 0.00783\, \frac{ {\rm Np} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm}
 \alpha_1 = 0.000443\, \frac{ {\rm Np} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm}  \alpha_2 = 0.5984\, \frac{ {\rm Np} }{ {\rm km \cdot \sqrt{MHz} } }\hspace{0.05cm}, \\ \beta_1 & = 22.18\, \frac{ {\rm rad} }{ {\rm km \cdot MHz} }\hspace{0.05cm}, \hspace{0.2cm} \beta_2 = 0.5984\, \frac{ {\rm rad} }{ {\rm km \cdot \sqrt{MHz} } } \hspace{0.05cm}.\end{align*}$$

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;  

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;]]:
#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. 
#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.  
#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.

==Characteristic cable attenuation==
<br>
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; $\text{500 MHz}$:
[[File:EN_LZI_4_2_S2_neu.png |right|frame| Attenuation function and characteristic attenuation of coaxial cables]]
:$${\alpha}_{\rm K}(f)  \hspace{-0.05cm} =  \alpha_0  \hspace{-0.05cm}+ \hspace{-0.05cm} \alpha_1 \cdot f  \hspace{-0.05cm}+ \hspace{-0.05cm} \alpha_2  \hspace{-0.05cm}\cdot  \hspace{-0.05cm}\sqrt {f} \hspace{0.01cm}  \hspace{0.01cm}.$$
:$$\Rightarrow \hspace{0.3cm}{\rm a}_{\rm K}(f) =\alpha_{\rm K}(f) \cdot l $$

'''Notes on the representation chosen here''' 

#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; ${a}_{\rm K}(f)$. 
#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;:
:$${\rm a}_{\rm K}(f)  = \alpha_2 \cdot \sqrt {f} \cdot l =  {\rm a}_{\rm \star}\cdot \sqrt
 { {2f}/{R}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}|H_{\rm K}(f)|  = {\rm e}^{- {\rm a}_{\rm K}(f)}\hspace{0.05cm}, \hspace{0.2cm}  {\rm a}_{\rm K}(f)\hspace{0.15cm}{\rm in }\hspace{0.15cm}{\rm Np}\hspace{0.05cm}.$$

{{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; $(l)$.}} 


'''Interpretation of the right graph'''

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 
*the normal coaxial cable&nbsp; $\text{(2.6/9.5 mm)}$ &nbsp; &rArr; &nbsp; left ordinate labeling,&nbsp; and

*for the small coaxial cable&nbsp; $\text{(1.2/4.4 mm)}$ &nbsp; &rArr; &nbsp; right ordinate labeling.


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;$4$. 


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;.

==Impulse response of a coaxial cable==
<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:
:$$\gamma(f)  = \alpha_0 + \alpha_1 \cdot f +  {\rm j} \cdot \beta_1 \cdot f  +\alpha_2 \cdot \sqrt {f}+ {\rm j}\cdot \beta_2 \cdot \sqrt {f} \approx    {\rm j} \cdot \beta_1 \cdot f  +\alpha_2 \cdot \sqrt {f}+ {\rm j}\cdot \beta_2 \cdot \sqrt {f} \hspace{0.05cm}.$$

Considering
#the cable length &nbsp;$l$,
#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, 


thus applies to the&nbsp; &raquo;frequency response of the coaxial cable&laquo;:
:$$H_{\rm K}(f)    = {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1 f} \cdot {\rm e}^{-{\rm a}_{\rm \star}\hspace{0.05cm}\cdot \hspace{0.05cm} \sqrt{2f/R} }\cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}{\rm a}_{\rm \star}\hspace{0.05cm}\cdot \hspace{0.02cm} \sqrt{2f/R}}= {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1 f} \cdot {\rm e}^{-2{\rm a}_{\rm \star}\hspace{0.03cm}\cdot \hspace{0.03cm} \sqrt{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}f/R}} \hspace{0.05cm}.$$
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&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;]]:
:$$h_{\rm K}(t)  = \mathcal{F}^{-1} \left \{ {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1
 f}\right \} \star\mathcal{F}^{-1} \left \{ {\rm e}^{-2{\rm a}_{\rm \star}\hspace{0.03cm}\cdot \hspace{0.03cm} \sqrt{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}f/R} }\right \} \hspace{0.05cm}.$$

It must be taken into account here:
*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;.
  
*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;:
:$$h_{\rm K}(t\hspace{0.05cm}' + \tau_{\rm P}\hspace{0.05cm} ')  = \frac {1}{T} \cdot \frac {{\rm a}_{\rm \star}}{\pi \cdot \sqrt{2   \cdot t\hspace{0.05cm}'\hspace{0.05cm}^3}}\cdot {\rm exp} \left [ -\frac {{\rm a}_{\rm \star}^2}{ {2\pi \cdot t\hspace{0.05cm}'}} \right ]\hspace{0.05cm},\hspace{0.2cm}{\rm a}_{\rm \star}\hspace{0.15cm}{\rm in\hspace{0.15cm} Np}\hspace{0.05cm}.$$
:Here&nbsp; $T = 1/R$&nbsp; denotes&nbsp; &raquo;the symbol duration of a binary system&laquo;&nbsp; and it holds &nbsp;$τ_{\rm P} \hspace{0.05cm}' = τ_{\rm P}/T$.

{{GraueBox|TEXT=  
$\text{Example 1:}$&nbsp; 
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.


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 
#for systems with small or normal coaxial cable,&nbsp; 
#different lengths, 
#different bit rates.&nbsp; 


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;$T = 30 \ \rm  ns)$:
:$$\tau_{\rm B}= \frac {\beta_1 \cdot l}{2\pi} =\frac {22.18\, { {\rm rad} }/{ {(\rm km \cdot MHz)} }\cdot 2.8\,{\rm km} }{2\pi} = 9.9\,{\rm &micro; s}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\tau_{\rm B}\hspace{0.05cm}' ={\tau_{\rm B} }/{T} \approx 330\hspace{0.05cm}.$$

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;. }}


{{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==

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


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;$g_r(t)$.}}


==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; 

*and a coaxial transmission channel,&nbsp; 


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

$(1)$&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$)$: 
:$$g_r(t) \approx s_0 \cdot T \cdot h_{\rm K}(t)  =  \frac {s_0 \cdot {\rm a}_{\rm \star}/\pi}{ \sqrt{2    \cdot(t/T)^3}}\cdot {\rm e}^{  -{{\rm a}_{\rm \star}^2}/( {2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}t/T}) } \hspace{0.05cm}.$$

$(2)$&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;$N_0/2$.

$(3)$&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.
 
$(3)$&nbsp; Then the received signal is with the amplitude coefficients &nbsp;$a_{\nu}$:
:$$r(t) = \sum_{\nu = - \infty}^{+ \infty}a_{\nu}\cdot g_r(t - \nu \cdot T)+ n(t) \hspace{0.05cm} .$$


==Exercises for the chapter==
<br>
[[Aufgaben:Exercise_4.4:_Coaxial_Cable_-_Frequency_Response| Exercise 4.4: Coaxial Cable - Frequency Response]]

[[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]]


==References==
<references/>

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