Linear and Time Invariant Systems/Properties of Coaxial Cables - Revision history

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2026-03-16T13:47:02Z

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	The cable frequency response  $H_{\rm K}(f)$  results from the cable length  $l$  and the complex propagation function $($per unit length$)$		The cable frequency response  $H_{\rm K}(f)$  results from the cable length  $l$  and the complex propagation function $($per unit length$)$
	:$$\gamma(f) = \alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot \sqrt {f}+ {\rm j}\cdot (\beta_1 \cdot f + \beta_2 \cdot \sqrt {f})\hspace{0.05cm}\hspace{0.3cm}		:$$\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}.$$
	\Rightarrow \hspace{0.3cm}H_{\rm K}(f) = {\rm e}^{-\gamma(f)\hspace{0.05cm} \cdot \hspace{0.05cm} l} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\|H_{\rm K}(f)\| = {\rm e}^{-\alpha(f)\hspace{0.05cm} \cdot \hspace{0.05cm} l}\hspace{0.05cm}.$$

	The cable specific constants for the  »'''standard coaxial cable'''«  $\text{(2.6/9.5 mm)}$  are:		The cable specific constants for the  »'''standard coaxial cable'''«  $\text{(2.6/9.5 mm)}$  are:
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	Accordingly,  the kilometric attenuation and phase constants for the  »'''small coaxial cable'''«  $\text{(1.2/4.4 mm)}$:		Accordingly,  the kilometric attenuation and phase constants for the  »'''small coaxial cable'''«  $\text{(1.2/4.4 mm)}$:
	:$$\begin{align*}\alpha_0 & = 0.00783\, \frac{ {\rm Np} }{ {\rm km} }\hspace{0.05cm}, \hspace{0.2cm}		:$$\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.:  Dämpfung, Phase und Laufzeiten bei Weitverkehrs–Koaxialpaaren.  Frequenz 31, S. 23-28, 1977.</ref>.  They apply to a temperature of  $20^\circ\ \text{C (293 K)}$  and frequencies greater than  $\text{200 kHz}$.		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.:  Dämpfung, Phase und Laufzeiten bei Weitverkehrs–Koaxialpaaren.  Frequenz 31, S. 23-28, 1977.</ref>.  They apply to a temperature of  $20^\circ\ \text{C (293 K)}$  and frequencies greater than  $\text{200 kHz}$.
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	*Sometimes,  we assume the  »'''simplified attenuation function'''«:		*Sometimes,  we assume the  »'''simplified attenuation function'''«:
	:$${\rm a}_{\rm K}(f) = \alpha_2 \cdot \sqrt {f} \cdot l = {\rm a}_{\rm \star}\cdot \sqrt		:$${\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}.$$
	{ {2f}/{R}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\|H_{\rm K}(f)\| = {\rm e}^{- {\rm a}_{\rm K}(f)}\hspace{0.05cm}, \hspace{0.2cm} {\rm a}_{\rm K}(f)\hspace{0.15cm}{\rm in }\hspace{0.15cm}{\rm Np}\hspace{0.05cm}.$$

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

	The time domain representation is obtained by applying the  [[Signal_Representation/~~Fourier_Transform_and_its_Inverse~~#The_second_Fourier_integral\|»inverse Fourier transform«]]  and the  [[Signal_Representation/The_Convolution_Theorem_and_Operation\|»convolution theorem«]]:		The time domain representation is obtained by applying the  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_second_Fourier_integral\|»inverse Fourier transform«]]  and the  [[Signal_Representation/The_Convolution_Theorem_and_Operation\|»convolution theorem«]]:
	:$$h_{\rm K}(t) = \mathcal{F}^{-1} \left \{ {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} ~~b_1~~		:$$h_{\rm K}(t) = \mathcal{F}^{-1} \left \{ {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} b_1f}\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}.$$
	f}\right \} \star\mathcal{F}^{-1} \left \{ {\rm e}^{-2{\rm a}_{\rm \star}\hspace{0.03cm}\cdot \hspace{0.03cm} \sqrt{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}f/R} }\right \} \hspace{0.05cm}.$$

	It must be taken into account here:		It must be taken into account here:
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	:$$g_r(t) = g_s(t) \star h_{\rm K}(t)\hspace{0.05cm}.$$		:$$g_r(t) = g_s(t) \star h_{\rm K}(t)\hspace{0.05cm}.$$
	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:		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:
	:$$g_r(t) = 2 s_0 \cdot \left [ {\rm Q} \left (\frac {{\rm a}_{\rm \star}/\sqrt {\pi}}{ \sqrt{ (t/T - 0.5)}}\right ) -		:$$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}.$$
	{\rm Q} \left (\frac {{\rm a}_{\rm \star}/\sqrt {\pi}}{ \sqrt{ (t/T + 0.5)}}\right ) \right ]\hspace{0.05cm}.$$
	Here  $\rm a_∗$  denotes the  »characteristic cable attenuation«  $($in Neper$)$  and  ${\rm Q}(x)$  the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability\|»complementary Gaussian error function«]].		Here  $\rm a_∗$  denotes the  »characteristic cable attenuation«  $($in Neper$)$  and  ${\rm Q}(x)$  the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability\|»complementary Gaussian error function«]].

Maintenance script: Add German interlanguage link

2026-03-16T12:27:48Z

Add German interlanguage link

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	{{Display}}		{{Display}}
			[[de:Lineare_zeitinvariante_Systeme/Eigenschaften_von_Koaxialkabeln]]

Guenter at 17:37, 24 November 2023

2023-11-24T17:37:16Z

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	==Special features of coaxial cable systems==		==Special features of coaxial cable systems==
	<br>		<br>
	Assuming binary transmission with non-return-to-zero  (NRZ)  rectangular pulses  $($~~symbol~~ duration $T)$  ~~and a coaxial transmission channel,  the following system model is obtained.  In particular, it should be noted:~~		[[File:EN_LZI_4_2_S5_v2.png \|right\|frame\| Binary transmission system with coaxial cable]]
			Assuming binary transmission

			*with non-return-to-zero  $\rm (NRZ)$  rectangular pulses  $($duration $T)$

	~~[[File:EN_LZI_4_2_S5_v2.png \|right\|frame\| Binary~~ transmission ~~system with coaxial cable<br><br><br>]]~~		*and a coaxial transmission channel,

	*In a simulation, the phase delay time of the coaxial cable is conveniently left out of consideration.  Then the basic receiver pulse  $g_r(t)$  (with ${\rm a}_{\rm \star}$ in Neper)~~  is approximated by~~
			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$)$:
	:$$g_r(t) \approx s_0 \cdot T \cdot h_{\rm K}(t) = \frac {s_0 \cdot {\rm a}_{\rm \star}/\pi}{ \sqrt{2 \cdot(t/T)^3}}\cdot {\rm e}^{ -{{\rm a}_{\rm \star}^2}/( {2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}t/T}) } \hspace{0.05cm}.$$		:$$g_r(t) \approx s_0 \cdot T \cdot h_{\rm K}(t) = \frac {s_0 \cdot {\rm a}_{\rm \star}/\pi}{ \sqrt{2 \cdot(t/T)^3}}\cdot {\rm e}^{ -{{\rm a}_{\rm \star}^2}/( {2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}t/T}) } \hspace{0.05cm}.$$
	*Because of the good shielding of coaxial cables against other impairments,  the  [[Aufgaben:Exercise_1.3Z:_Thermal_Noise\|~~$\text{~~thermal noise}$]]  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  $N_0/2$.
	*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".		$(2)$  Because of the good shielding of coaxial cables against other impairments,  the  [[Aufgaben:Exercise_1.3Z:_Thermal_Noise\|»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  $N_0/2$.
	*This noise addition point is also useful because the frequency response  $H_{\rm K}(f)$  decisively attenuates all noise accumulated along the cable.
	*Then the received signal is with the amplitude coefficients  $a_{\nu}$:		$(3)$  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  $a_{\nu}$:
	:$$r(t) = \sum_{\nu = - \infty}^{+ \infty}a_{\nu}\cdot g_r(t - \nu \cdot T)+ n(t) \hspace{0.05cm} .$$		:$$r(t) = \sum_{\nu = - \infty}^{+ \infty}a_{\nu}\cdot g_r(t - \nu \cdot T)+ n(t) \hspace{0.05cm} .$$

Guenter at 16:57, 24 November 2023

2023-11-24T16:57:57Z

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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Conclusion:}$  When simulating and optimizing transmission systems,		$\text{Conclusion:}$  When simulating and optimizing transmission systems,  <br>'''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'''.}}
	*one usually omits the phase term ~~with~~  $b_1 = β_1 · l$,

	*since this results exclusively in a  $($often not disturbing$)$  phase delay,  '''but no signal ~~distortion~~'''.}}

	==Basic receiver pulse==		==Basic receiver pulse==

	<br>		<br>
	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:		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:
	:$$g_r(t) = g_s(t) \star h_{\rm K}(t)\hspace{0.05cm}.$$		:$$g_r(t) = g_s(t) \star h_{\rm K}(t)\hspace{0.05cm}.$$
	If a non-return-to-zero (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:		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:
	:$$g_r(t) = 2 s_0 \cdot \left [ {\rm Q} \left (\frac {{\rm a}_{\rm \star}/\sqrt {\pi}}{ \sqrt{ (t/T - 0.5)}}\right ) -		:$$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}.$$		{\rm Q} \left (\frac {{\rm a}_{\rm \star}/\sqrt {\pi}}{ \sqrt{ (t/T + 0.5)}}\right ) \right ]\hspace{0.05cm}.$$
	Here  $\rm a_∗$  denotes the characteristic cable attenuation  (in Neper)  and  ${\rm Q}(x)$  the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability\|~~$\text{~~complementary Gaussian error function}$]].		Here  $\rm a_∗$  denotes the  »characteristic cable attenuation«  $($in Neper$)$  and  ${\rm Q}(x)$  the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability\|»complementary Gaussian error function«]].

	~~[[File:EN_LZI_4_2_S4.png\|right\|frame\|Impulse response of the coaxial cable and basic receiver pulse]]~~
	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
	$\text{Example 2:}$		$\text{Example 2:}$
	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 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 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.
			*the basic receiver pulse  $($»rectangular response«$)$  $g_r(t)$  normalized to the transmission amplitude  $s_0$   ⇒   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:		One recognizes the following from this diagram:
	*With  $\rm a_∗ = 40 \ \rm dB$,  the normalized ~~rectangular response~~  $g_r(t)/s_0$  is slightly  (about a factor of $0.95)$  smaller ~~at the peak~~ than the normalized impulse response  $T · h_{\rm K}(t)$.  Here is a small difference between impulse response and basic receiver pulse .		#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 .
	*In contrast,  for ~~the case~~  $a_∗ ≥ 60 \ \rm dB$,  the ~~rectangular response~~ and the impulse response are indistinguishable within the drawing accuracy.		#In contrast,  for  $a_∗ ≥ 60 \ \rm dB$,  the basic receiver pulse and the impulse response are indistinguishable within the drawing accuracy.
	*For a return-to-zero  (RZ)  pulse,  the above equation for ~~the basic receiver pulse~~ would still need to be multiplied by the ~~duty cycle~~  $Δt_s/T$~~ ~~.  In this case  $g_r(t)/s_0$  is smaller than  $T · h_{\rm K}(t)$  by at least this factor.		#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.
	*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 [[Signal_Representation/Special_Cases_of_Pulses#Gaussian_pulse\|~~$\text{~~equivalent pulse duration}$]]  of  $g_r(t)$.}}		#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 [[Signal_Representation/Special_Cases_of_Pulses#Gaussian_pulse\|»equivalent pulse duration«]]  of  $g_r(t)$.}}


	We draw your attention to the  (German language)  interactive SWF applet  [[Applets:Zeitverhalten_von_Kupferkabeln\|"Zeitverhalten von Kupferkabeln"]]  ⇒   "Time behavior of copper cables",  which deals with the topic discussed here.

Guenter at 13:35, 24 November 2023

2023-11-24T13:35:32Z

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Guenter at 17:48, 23 November 2023

2023-11-23T17:48:26Z

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Guenter at 16:24, 16 January 2023

2023-01-16T16:24:34Z

← Older revision		Revision as of 18:24, 16 January 2023
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	Assuming binary transmission with non-return-to-zero  (NRZ)  rectangular pulses  $($symbol duration $T)$  and a coaxial transmission channel,  the following system model is obtained.  In particular, it should be noted:		Assuming binary transmission with non-return-to-zero  (NRZ)  rectangular pulses  $($symbol duration $T)$  and a coaxial transmission channel,  the following system model is obtained.  In particular, it should be noted:

	[[File:~~EN_LZI_4_2_S5~~.png \|right\|frame\| Binary transmission system with coaxial cable ~~'''KORREKTUR: receiver filter, noise faults'''~~<br><br><br>]]		[[File:EN_LZI_4_2_S5_v2.png \|right\|frame\| Binary transmission system with coaxial cable<br><br><br>]]

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

Guenter at 16:16, 12 January 2023

2023-01-12T16:16:56Z

← Older revision		Revision as of 18:16, 12 January 2023
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	[[File:EN_LZI_4_2_S5.png \|right\|frame\| Binary transmission system with coaxial cable '''KORREKTUR: receiver filter, noise faults'''<br><br><br>]]		[[File:EN_LZI_4_2_S5.png \|right\|frame\| Binary transmission system with coaxial cable '''KORREKTUR: receiver filter, noise faults'''<br><br><br>]]

	*In a simulation, the phase delay time of the coaxial cable is conveniently left out of consideration.  Then the basic ~~reception pulse '''KORREKTUR: received/~~receiver pulse~~'''~~  $g_r(t)$  (with ${\rm a}_{\rm \star}$ in Neper)  is approximated by		*In a simulation, the phase delay time of the coaxial cable is conveniently left out of consideration.  Then the basic receiver pulse  $g_r(t)$  (with ${\rm a}_{\rm \star}$ in Neper)  is approximated by
	:$$g_r(t) \approx s_0 \cdot T \cdot h_{\rm K}(t) = \frac {s_0 \cdot {\rm a}_{\rm \star}/\pi}{ \sqrt{2 \cdot(t/T)^3}}\cdot {\rm e}^{ -{{\rm a}_{\rm \star}^2}/( {2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}t/T}) } \hspace{0.05cm}.$$		:$$g_r(t) \approx s_0 \cdot T \cdot h_{\rm K}(t) = \frac {s_0 \cdot {\rm a}_{\rm \star}/\pi}{ \sqrt{2 \cdot(t/T)^3}}\cdot {\rm e}^{ -{{\rm a}_{\rm \star}^2}/( {2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}t/T}) } \hspace{0.05cm}.$$
	*Because of the good shielding of coaxial cables against other impairments,  the  [[Aufgaben:Exercise_1.3Z:_Thermal_Noise\|$\text{thermal noise}$]]  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  $N_0/2$.		*Because of the good shielding of coaxial cables against other impairments,  the  [[Aufgaben:Exercise_1.3Z:_Thermal_Noise\|$\text{thermal noise}$]]  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  $N_0/2$.

Guenter at 16:11, 12 January 2023

2023-01-12T16:11:27Z

Guenter at 16:06, 12 January 2023

2023-01-12T16:06:30Z

← Older revision		Revision as of 18:06, 12 January 2023
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	<br>		<br>
	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  $\text{500 MHz}$:		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  $\text{500 MHz}$:
	[[File:EN_LZI_4_2_S2_neu.png \|right\|frame\| Attenuation function and characteristic attenuation of coaxial cables ~~<br>'''Korrektur:''' alpha, Np/km~~]]		[[File:EN_LZI_4_2_S2_neu.png \|right\|frame\| Attenuation function and characteristic attenuation of coaxial cables]]
	:$${\alpha}_{\rm K}(f) \hspace{-0.05cm} = \alpha_0 \hspace{-0.05cm}+ \hspace{-0.05cm} \alpha_1 \cdot f \hspace{-0.05cm}+ \hspace{-0.05cm} \alpha_2 \hspace{-0.05cm}\cdot \hspace{-0.05cm}\sqrt {f} \hspace{0.01cm} \hspace{0.01cm}.$$		:$${\alpha}_{\rm K}(f) \hspace{-0.05cm} = \alpha_0 \hspace{-0.05cm}+ \hspace{-0.05cm} \alpha_1 \cdot f \hspace{-0.05cm}+ \hspace{-0.05cm} \alpha_2 \hspace{-0.05cm}\cdot \hspace{-0.05cm}\sqrt {f} \hspace{0.01cm} \hspace{0.01cm}.$$
	:$$\Rightarrow \hspace{0.3cm}{\rm a}_{\rm K}(f) =\alpha_{\rm K}(f) \cdot l $$		:$$\Rightarrow \hspace{0.3cm}{\rm a}_{\rm K}(f) =\alpha_{\rm K}(f) \cdot l $$
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	$\text{Example 1:}$		$\text{Example 1:}$
	The results of this section are illustrated by the following graph as an example.		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  $\rm a_∗ = 60 \ dB$~~<br>'''Korrektur:''' Rahmen~~]]		[[File:EN_LZI_4_2_S3_neu.png\|right\|frame\| Impulse response of a coaxial cable with  $\rm a_∗ = 60 \ dB$]]
	*The normalized impulse response  $T · h_{\rm K}(t)$  of a coaxial cable with  $\rm a_∗ = 60 \ dB \ \ (6.9\ Np)$ is shown.		*The normalized impulse response  $T · h_{\rm K}(t)$  of a coaxial cable with  $\rm a_∗ = 60 \ dB \ \ (6.9\ Np)$ is shown.
	*The attenuation coefficients  $α_0$  and  $α_1$  can thus be neglected,  as shown in the last section.		*The attenuation coefficients  $α_0$  and  $α_1$  can thus be neglected,  as shown in the last section.

← Older revision		Revision as of 18:11, 12 January 2023
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	*since this results exclusively in a  (often not disturbing)  phase delay,  but no signal distortion.}}		*since this results exclusively in a  (often not disturbing)  phase delay,  but no signal distortion.}}

	==Basic ~~reception~~ pulse==		==Basic receiver pulse==
	~~'''KORREKTUR: received/receiver pulse'''~~
	<br>		<br>
	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 ~~reception pulse '''KORREKTUR: received/~~receiver pulse~~'''~~  $g_r(t)$   ⇒   "basic pulse of the received signal  $r(t)$"  is:		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:
	:$$g_r(t) = g_s(t) \star h_{\rm K}(t)\hspace{0.05cm}.$$		:$$g_r(t) = g_s(t) \star h_{\rm K}(t)\hspace{0.05cm}.$$
	If a non-return-to-zero (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:		If a non-return-to-zero (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:
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	Here  $\rm a_∗$  denotes the characteristic cable attenuation  (in Neper)  and  ${\rm Q}(x)$  the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability\|$\text{complementary Gaussian error function}$]].		Here  $\rm a_∗$  denotes the characteristic cable attenuation  (in Neper)  and  ${\rm Q}(x)$  the  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables#Exceedance_probability\|$\text{complementary Gaussian error function}$]].

	[[File:EN_LZI_4_2_S4.png\|right\|frame\|Impulse response of the coaxial cable and basic ~~reception pulse '''KORREKTUR: received/~~receiver pulse~~''' ~~ ]]		[[File:EN_LZI_4_2_S4.png\|right\|frame\|Impulse response of the coaxial cable and basic receiver pulse]]
	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
	$\text{Example 2:}$		$\text{Example 2:}$
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	*the normalized coaxial cable impulse response  $T · h_{\rm K}(t)$   ⇒   solid curves,		*the normalized coaxial cable impulse response  $T · h_{\rm K}(t)$   ⇒   solid curves,
	*the basic ~~reception pulse '''KORREKTUR: received/~~receiver pulse~~'''~~  ("rectangular response")  $g_r(t)$  normalized to the transmission amplitude  $s_0$   ⇒   dotted line.		*the basic receiver pulse  ("rectangular response")  $g_r(t)$  normalized to the transmission amplitude  $s_0$   ⇒   dotted line.


	One recognizes the following from this diagram:		One recognizes the following from this diagram:
	*With  $\rm a_∗ = 40 \ \rm dB$,  the normalized rectangular response  $g_r(t)/s_0$  is slightly  (about a factor of $0.95)$  smaller at the peak than the normalized impulse response  $T · h_{\rm K}(t)$.  Here is a small difference between impulse response and basic ~~reception pulse '''KORREKTUR: received/~~receiver pulse~~'''~~.		*With  $\rm a_∗ = 40 \ \rm dB$,  the normalized rectangular response  $g_r(t)/s_0$  is slightly  (about a factor of $0.95)$  smaller at the peak than the normalized impulse response  $T · h_{\rm K}(t)$.  Here is a small difference between impulse response and basic receiver pulse .
	*In contrast,  for the case  $a_∗ ≥ 60 \ \rm dB$,  the rectangular response and the impulse response are indistinguishable within the drawing accuracy.		*In contrast,  for the case  $a_∗ ≥ 60 \ \rm dB$,  the rectangular response and the impulse response are indistinguishable within the drawing accuracy.
	*For a return-to-zero  (RZ)  pulse,  the above equation for the basic ~~reception impulse '''KORREKTUR: received/~~receiver pulse~~'''~~ would still need to be multiplied by the duty cycle  $Δt_s/T$ .  In this case  $g_r(t)/s_0$  is smaller than  $T · h_{\rm K}(t)$  by at least this factor.		*For a return-to-zero  (RZ)  pulse,  the above equation for the basic receiver pulse would still need to be multiplied by the duty cycle  $Δt_s/T$ .  In this case  $g_r(t)/s_0$  is smaller than  $T · h_{\rm K}(t)$  by at least this factor.
	*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 [[Signal_Representation/Special_Cases_of_Pulses#Gaussian_pulse\|$\text{equivalent pulse duration}$]]  of  $g_r(t)$.}}		*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 [[Signal_Representation/Special_Cases_of_Pulses#Gaussian_pulse\|$\text{equivalent pulse duration}$]]  of  $g_r(t)$.}}