Aufgaben:Exercise 4.5: Coaxial Cable - Impulse Response - Revision history

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	The  [[Signal_Representation/~~Fourier_Transform_and_Its_Inverse~~#~~Fouriertransformation~~\|Fourier transform]]  of  $H_{\rm K}(f)$  yields the impulse response  $h_{\rm K}(t)$,  which can be specified in closed-analytic form for a coaxial cable using the approximations described here.  For a binary system holds:		The  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#Fourier_transform\|Fourier transform]]  of  $H_{\rm K}(f)$  yields the impulse response  $h_{\rm K}(t)$,  which can be specified in closed-analytic form for a coaxial cable using the approximations described here.  For a binary system holds:
	:$$h_{\rm K}(t) = \frac{ {a}_{\rm \star}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot{\rm e}^{ - {{a}_{\rm \star}^2}/(2 \hspace{0.05cm} \pi \cdot \hspace{0.05cm} t/T)}\hspace{0.4cm}{\rm with}\hspace{0.2cm}{a}_{\rm \star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}\hspace{0.05cm}.$$		:$$h_{\rm K}(t) = \frac{ {a}_{\rm \star}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot{\rm e}^{ - {{a}_{\rm \star}^2}/(2 \hspace{0.05cm} \pi \cdot \hspace{0.05cm} t/T)}\hspace{0.4cm}{\rm with}\hspace{0.2cm}{a}_{\rm \star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}\hspace{0.05cm}.$$
	Subtask  '''(5)'''  refers to the basic reception pulse  $g_r(t) = g_s(t) \star h_{\rm K}(t)$, where the basic transmission pulse  $g_s(t)$  is assumed to be a rectangle  $($height  $s_0$,  duration  $T)$.		Subtask  '''(5)'''  refers to the basic reception pulse  $g_r(t) = g_s(t) \star h_{\rm K}(t)$, where the basic transmission pulse  $g_s(t)$  is assumed to be a rectangle  $($height  $s_0$,  duration  $T)$.
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	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.2 Coaxial Cable^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^4.2 Coaxial Cable^]]
	[[de:Aufgaben:4.~~5_Koaxialkabel_–_Impulsantwort~~]]		[[de:Aufgaben:Aufgabe 4.5: Koaxialkabel – Impulsantwort]]

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	'''(3)'''  Substituting the result into the given equation, we get (for simplicity, we use "${a}$"instead of "${a}_{\rm \star}$"):		'''(3)'''  Substituting the result into the given equation, we get (for simplicity, we use "${a}$" instead of "${a}_{\rm \star}$"):
	:$$h_{\rm K}(t_{\rm max}) = \frac{1}{T} \cdot \frac{ {a}}{ \sqrt{2 \pi^2 \cdot {{a}^6}/{(3\pi)^3}}}\hspace{0.1cm} \cdot		:$$h_{\rm K}(t_{\rm max}) = \frac{1}{T} \cdot \frac{ {a}}{ \sqrt{2 \pi^2 \cdot {{a}^6}/{(3\pi)^3}}}\hspace{0.1cm} \cdot{\rm exp} \left[ - \frac{{a}^2}{2\pi} \cdot\frac{3\pi}{{\rm 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} \approx \frac{1}{T} \cdot \frac{1.453}{{a}^2}\hspace{0.05cm}.$$
	{\rm exp} \left[ - \frac{{a}^2}{2\pi} \cdot		*With  $a = 6.9$  we obtain the final result:
	\frac{3\pi}{{\rm a}^2}\hspace{0.1cm}\right]		:$${\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}.$$
	= \frac{1}{T} \cdot \frac{1}{{a}^2}\cdot
	\sqrt{\frac{27 \pi		'''(4)'''  Using the result from  '''(3)''' , the appropriate equation of determination is:
	}{2}} \cdot {\rm e}^{-3/2} \approx \frac{1}{T} \cdot \frac{1.453}{{a}^2}		:$$\frac{ {a}/T}{ \sqrt{2 \pi^2 \cdot (t_{5\%}/T)^3}}= 0.05 \cdot 0.03 {1}/{T}\hspace{0.15cm}{= 0.0015 \cdot {1}/{T}}\hspace{0.2cm} \Rightarrow \hspace{0.2cm} (t_{5\%}/T)^{3/2} = \frac{a}{\sqrt{2} \cdot \pi \cdot 0.0015}\approx 1036 \hspace{0.3cm}\Rightarrow\hspace{0.3cm}\hspace{0.15cm}\underline{t_{5\%}/T \approx 103.5} \hspace{0.05cm}.$$
	\hspace{0.05cm}.$$*With  $a = 6.9$  we obtain 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 from  '''(3)''' , the appropriate equation of determination is::$$\frac{ {a}/T}{ \sqrt{2 \pi^2 \cdot (t_{5\%}/T)^3}}= 0.05 \cdot 0.03 {1}/{T}
	\hspace{0.15cm}{= 0.0015 \cdot {1}/{T}}
	\hspace{0.2cm} \Rightarrow \hspace{0.2cm} (t_{5\%}/T)^{3/2} = \frac{a}{\sqrt{2} \cdot \pi \cdot
	0.0015}\approx 1036 \hspace{0.3cm}\Rightarrow
	\hspace{0.3cm}
	\hspace{0.15cm}\underline{t_{5\%}/T \approx 103.5} \hspace{0.05cm}.$$
	*This value is a bit too large, because the second term  ${\rm e}^{-0.05}\approx 0.95$  was neglected.		*This value is a bit too large, because the second term  ${\rm e}^{-0.05}\approx 0.95$  was neglected.
	*The exact calculation gives  $t_{\rm 5\%}/T \approx 97$.		*The exact calculation gives  $t_{\rm 5\%}/T \approx 97$.

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	[[File:P_ID1814__LZI_A_4_5.png\|right\|frame\|Impulse response of a coaxial cable]]		[[File:P_ID1814__LZI_A_4_5.png\|right\|frame\|Impulse response of a coaxial cable]]
	The frequency response of a coaxial cable  (German:  "Koaxialkabel"   ⇒   subscipt  "K")  of length  $l$  can be represented by the following formula:		The frequency response of a coaxial cable  (German:  "Koaxialkabel"   ⇒   subscipt  "K")  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{\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}.$$
	\cdot
	{\rm e}^{- (\alpha_1 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1) \hspace{0.05cm}\cdot f \hspace{0.05cm}\cdot \hspace{0.05cm}l} \cdot
	{\rm e}^{- (\alpha_2 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2) \hspace{0.05cm}\cdot \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm}l}
	\hspace{0.05cm}.$$
	The first term of this equation is due to the ohmic losses,  the second term to the transverse losses.  Dominant,  however,  is the skin effect,  which is expressed by the third term.		The first term of this equation is due to the ohmic losses,  the second term to the transverse losses.  Dominant,  however,  is the skin effect,  which is expressed by the third term.

	With the coefficients valid for the  "standard coaxial cable"  $\text{(2.6 mm}$  core diameter,  $\text{9.5 mm}$  outer diameter$)$		With the coefficients valid for the  "standard coaxial cable"  $\text{(2.6 mm}$  core diameter,  $\text{9.5 mm}$  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{0.2cm} \beta_2 = 0.2722 \hspace{0.15cm}\frac {\rm rad}{\rm km \cdot \sqrt{\rmMHz}}\hspace{0.05cm}$$
	\hspace{0.05cm},
	\hspace{0.2cm} \beta_2 = 0.2722 \hspace{0.15cm}\frac {\rm rad}{\rm km \cdot \sqrt{\rm
	~~MHz~~}}\hspace{0.05cm}$$
	this frequency response can also be represented as follows:		this 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}0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km}\hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}}}\hspace{0.05cm}.$$
	\hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}} } \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}
	0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km}
	\hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}}}
	\hspace{0.05cm}.$$
	That means:		That means:

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	If one defines the characteristic cable attenuation  ${a}_{\rm *}$  at half the bit rate  $($i.e., at  $R/2)$  and normalizes the frequency to  $R$,  one can treat digital systems of different bit rate and length uniformly:		If one defines the characteristic cable attenuation  ${a}_{\rm *}$  at half the bit rate  $($i.e., at  $R/2)$  and normalizes the frequency to  $R$,  one can treat digital systems of different bit rate and length uniformly:
	:$${a}_{\rm \star} = {a}_{\rm K}(f ={R}/{2})		:$${a}_{\rm \star} = {a}_{\rm K}(f ={R}/{2})\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm K}(f) = {\rm e}^{-{a}_{\rm \star} \cdot \sqrt{2f/R}}\cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} {a}_{\star} \cdot \sqrt{2f/R}}\hspace{0.4cm}{\rm with}\hspace{0.2cm}{a}_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}\hspace{0.05cm}.$$
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm K}(f) = {\rm e}^{-
	{a}_{\rm \star} \cdot \sqrt{2f/R}}\cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} {a}_{\star} \cdot \sqrt{2f/R}}\hspace{0.4cm}{\rm with}\hspace{0.2cm}{a}_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}
	\hspace{0.05cm}.$$
	*The corresponding  $\rm dB$  value is greater by a factor of  $8.686$.		*The corresponding  $\rm dB$  value is greater by a factor of  $8.686$.
	*For a binary system,  $R = 1/T$ holds,  so that the characteristic cable attenuation refers to the frequency  $f = 1/(2T)$.		*For a binary system,  $R = 1/T$ holds,  so that the characteristic cable attenuation refers to the frequency  $f = 1/(2T)$.
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	The  [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Fouriertransformation\|Fourier transform]]  of  $H_{\rm K}(f)$  yields the impulse response  $h_{\rm K}(t)$,  which can be specified in closed-analytic form for a coaxial cable using the approximations described here.  For a binary system holds:		The  [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Fouriertransformation\|Fourier transform]]  of  $H_{\rm K}(f)$  yields the impulse response  $h_{\rm K}(t)$,  which can be specified in closed-analytic form for a coaxial cable using the approximations described here.  For a binary system holds:
	:$$h_{\rm K}(t) = \frac{ {a}_{\rm \star}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot		:$$h_{\rm K}(t) = \frac{ {a}_{\rm \star}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot{\rm e}^{ - {{a}_{\rm \star}^2}/(2 \hspace{0.05cm} \pi \cdot \hspace{0.05cm} t/T)}\hspace{0.4cm}{\rm with}\hspace{0.2cm}{a}_{\rm \star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}\hspace{0.05cm}.$$
	{\rm e}^{ - {{a}_{\rm \star}^2}/(2 \hspace{0.05cm} \pi \cdot \hspace{0.05cm} t/T)}
	\hspace{0.4cm}{\rm with}\hspace{0.2cm}{a}_{\rm \star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}
	\hspace{0.05cm}.$$
	Subtask  '''(5)'''  refers to the basic reception pulse  $g_r(t) = g_s(t) \star h_{\rm K}(t)$, where the basic transmission pulse  $g_s(t)$  is assumed to be a rectangle  $($height  $s_0$,  duration  $T)$.		Subtask  '''(5)'''  refers to the basic reception pulse  $g_r(t) = g_s(t) \star h_{\rm K}(t)$, where the basic transmission pulse  $g_s(t)$  is assumed to be a rectangle  $($height  $s_0$,  duration  $T)$.

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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The characteristic cable attenuation  ${a}_{\rm \star} = 60 \ \rm dB$  corresponds approximately to  $6.9\ \rm Np$. Therefore the following must hold:		'''(1)'''  The characteristic cable attenuation  ${a}_{\rm \star} = 60 \ \rm dB$  corresponds approximately to  $6.9\ \rm Np$. Therefore the following must hold:
	:$$\alpha_2 \cdot l \cdot {R}/{2} = 6.9\,\,{\rm		:$$\alpha_2 \cdot l \cdot {R}/{2} = 6.9\,\,{\rmNp}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\rmNp}}{0.2722 \,\, {\rm Np}/({\rm km \cdot \sqrt{\rm MHz}})\cdot \sqrt{70\,\,{\rm MHz}}}\hspace{0.15cm}\underline{ \approx 3\,\,{\rm km}}\hspace{0.05cm}.$$
	Np}
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\rm
	Np}}{0.2722 \,\, {\rm Np}/({\rm km \cdot \sqrt{\rm MHz}})
	\cdot \sqrt{70\,\,{\rm MHz}}}\hspace{0.15cm}\underline{ \approx 3\,\,{\rm km}}
	\hspace{0.05cm}.$$


	'''(2)'''  With the substitutions		'''(2)'''  With the substitutions
	:$$x = \frac{ t}{ T}, \hspace{0.2cm} K_1 = \frac{ {a}_{\rm \star}/T}{\sqrt{2 \pi^2 }}, \hspace{0.2cm}		:$$x = \frac{ t}{ T}, \hspace{0.2cm} K_1 = \frac{ {a}_{\rm \star}/T}{\sqrt{2 \pi^2 }}, \hspace{0.2cm}K_2 = \frac{ {a}_{\rm \star}^2}{2 \pi}$$
	K_2 = \frac{ {a}_{\rm \star}^2}{2 \pi}$$
	the impulse response can be described as follows:		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}.$$
	\hspace{0.05cm}.$$
	*By setting the derivative to zero:		*By setting the derivative to zero:
	:$$- {3}/{2} \cdot K_1 \cdot x^{-5/2}\cdot {\rm e}^{-K_2/x}+ K_1 \cdot x^{-3/2}\cdot {\rm		:$$- {3}/{2} \cdot K_1 \cdot x^{-5/2}\cdot {\rm e}^{-K_2/x}+ K_1 \cdot x^{-3/2}\cdot {\rme}^{-K_2/x}\cdot (-K_2) \cdot (-x^{-2})= 0\hspace{0.05cm}.$$
	e}^{-K_2/x}\cdot (-K_2) \cdot (-x^{-2})= 0		:$$\Rightarrow \hspace{0.3cm} {3}/{2} \cdot x^{-5/2} = K_2 \cdotx^{-7/2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}x_{\rm max} = {2}/{3} \cdot K_2 = \frac{{a}_{\rm \star}^2}{3 \pi}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm} {3}/{2} \cdot x^{-5/2} = K_2 \~~cdot~~
	x^{-7/2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	x_{\rm max} = {2}/{3} \cdot K_2 = \frac{{a}_{\rm \star}^2}{3 \pi}
	\hspace{0.05cm}.$$
	*This results in the following for  $60 \ \rm dB$  cable attenuation  $({a}_{\rm \star} \approx 6.9 \ \rm Np)$:		*This results in the following for  $60 \ \rm dB$  cable attenuation  $({a}_{\rm \star} \approx 6.9 \ \rm Np)$:
	:$$x_{\rm max} = { t_{\rm max}}/{ T}= { 6.9^2}/{(3\pi)}\hspace{0.15cm}\underline{ \approx 5 }\hspace{0.05cm}.$$		:$$x_{\rm max} = { t_{\rm max}}/{ T}= { 6.9^2}/{(3\pi)}\hspace{0.15cm}\underline{ \approx 5 }\hspace{0.05cm}.$$
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	\sqrt{\frac{27 \pi		\sqrt{\frac{27 \pi
	}{2}} \cdot {\rm e}^{-3/2} \approx \frac{1}{T} \cdot \frac{1.453}{{a}^2}		}{2}} \cdot {\rm e}^{-3/2} \approx \frac{1}{T} \cdot \frac{1.453}{{a}^2}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$*With  $a = 6.9$  we obtain the final result::$${\rm Max}\,[h_{\rm K}(t)] = \frac{1.453}{{6.9\,}^2} \cdot {1}/{T}\hspace{0.15cm}\underline{\approx 0.03 \cdot {1}/{T}}
	*With  $a = 6.9$  we obtain the final result:		\hspace{0.05cm}.$$'''(4)'''  Using the result from  '''(3)''' , the appropriate equation of determination is::$$\frac{ {a}/T}{ \sqrt{2 \pi^2 \cdot (t_{5\%}/T)^3}}= 0.05 \cdot 0.03 {1}/{T}
	:$${\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}.$$


	'''(4)'''  Using the result from  '''(3)''' , the appropriate equation of determination is:
	:$$\frac{ {a}/T}{ \sqrt{2 \pi^2 \cdot (t_{5\%}/T)^3}}= 0.05 \cdot 0.03 {1}/{T}
	\hspace{0.15cm}{= 0.0015 \cdot {1}/{T}}		\hspace{0.15cm}{= 0.0015 \cdot {1}/{T}}
	\hspace{0.2cm} \Rightarrow \hspace{0.2cm} (t_{5\%}/T)^{3/2} = \frac{a}{\sqrt{2} \cdot \pi \cdot		\hspace{0.2cm} \Rightarrow \hspace{0.2cm} (t_{5\%}/T)^{3/2} = \frac{a}{\sqrt{2} \cdot \pi \cdot
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	'''(5)'''  <u>The second solution</u> is correct:		'''(5)'''  <u>The second solution</u> is correct:
	*In general:		*In general:
	:$$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 .$$
	*Since the channel pulse response $h_{\rm K}(t)$ changes only insignificantly within a symbol duration, it can also be written:		*Since the channel pulse response $h_{\rm K}(t)$ changes only insignificantly within a symbol duration, it can also be written:
	:$$g_r(t) = h_{\rm K}(t) \cdot s_0 \cdot T.$$		:$$g_r(t) = h_{\rm K}(t) \cdot s_0 \cdot T.$$

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.2 Coaxial Cable^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^4.2 Coaxial Cable^]]
			[[de:Aufgaben:4.5_Koaxialkabel_–_Impulsantwort]]

Hwang at 08:42, 18 November 2021

2021-11-18T08:42:28Z

← Older revision		Revision as of 10:42, 18 November 2021
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	<quiz display=simple>		<quiz display=simple>
	{~~Wie groß ist die Länge~~  $l$  ~~eines Normalkoaxialkabels, wenn sich für die Bitrate  $R = 140 \ \rm Mbit/s$  die charakteristische Kabeldämpfung~~  ${a}_{\rm \star} = 60 \ \rm dB$  ~~ergibt~~?		{What is the length  $l$  of a standard coaxial cable if the characteristic cable attenuation is  ${a}_{\rm \star} = 60 \ \rm dB$  for the bit rate  $R = 140 \ \rm Mbit/s$ ?
	\|type="{}"}		\|type="{}"}
	$l \ =\ $ { 3 3% } $\ \rm km$		$l \ =\ $ { 3 3% } $\ \rm km$


	{~~Zu welcher Zeit~~  $t_{\rm max}$  ~~besitzt~~  $h_{\rm K}(t)$  ~~sein Maximum~~? ~~Es gelte weiter~~  ${a}_{\rm \star} = 60 \ \rm dB$.		{At which time  $t_{\rm max}$  does  $h_{\rm K}(t)$  have its maximum? Let  ${a}_{\rm \star} = 60 \ \rm dB$.
	\|type="{}"}		\|type="{}"}
	$t_{\rm max} \ = \ $ { 5 3% } $\ \cdot T$		$t_{\rm max} \ = \ $ { 5 3% } $\ \cdot T$


	{~~Wie groß ist der Maximalwert der Impulsantwort~~?		{What is the maximum value of the impulse response?
	\|type="{}"}		\|type="{}"}
	${\rm Max}\, \big [h_{\rm K}(t)\big ] \ = \ $ { 0.03 3% } $\ \cdot 1/T$		${\rm Max}\, \big [h_{\rm K}(t)\big ] \ = \ $ { 0.03 3% } $\ \cdot 1/T$


	{~~Ab welcher Zeit~~  $t_{\rm 5\%}$  ~~ist~~  $h_{\rm K}(t)$  ~~kleiner als~~  $5\%$  ~~des Maximums~~?  ~~Berücksichtigen Sie als Näherung nur den ersten Term der angegebenen Formel~~.		{From which time  $t_{\rm 5\%}$  is  $h_{\rm K}(t)$  less than  $5\%$  of the maximum?  As an approximation, consider only the first term of the given formula.
	\|type="{}"}		\|type="{}"}
	$t_{\rm 5\%} \ = \ $ { 103.5 3% } $\ \cdot T$		$t_{\rm 5\%} \ = \ $ { 103.5 3% } $\ \cdot T$


	{~~Welche Aussagen treffen für den Empfangsgrundimpuls~~  $g_r(t)$  zu?		{Which statements are true for the basic reception pulse  $g_r(t)$ ?
	\|type="[]"}		\|type="[]"}
	- $g_r(t)$  ~~ist doppelt so breit wie~~  $h_{\rm K}(t)$.		- $g_r(t)$  is twice as wide as  $h_{\rm K}(t)$.
	+ ~~Es gilt näherungsweise~~  $g_r(t) = s_0 \cdot T \cdot h_{\rm K}(t)$.		+ It is approximated  $g_r(t) = s_0 \cdot T \cdot h_{\rm K}(t)$.
	- $g_r(t)$  ~~kann durch einen Gaußimpuls angenähert werden~~.		- $g_r(t)$  can be approximated by a Gaussian pulse.


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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Die charakteristische Kabeldämpfung~~  ${a}_{\rm \star} = 60 \ \rm dB$  ~~entspricht in etwa~~  $6.9\ \rm Np$. ~~Deshalb muss gelten~~:		'''(1)'''  The characteristic cable attenuation  ${a}_{\rm \star} = 60 \ \rm dB$  corresponds approximately to  $6.9\ \rm Np$. Therefore the following must hold:
	:$$\alpha_2 \cdot l \cdot {R}/{2} = 6.9\,\,{\rm		:$$\alpha_2 \cdot l \cdot {R}/{2} = 6.9\,\,{\rm
	Np}		Np}
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	'''(2)'''  ~~Mit den Substitutionen~~		'''(2)'''  With the substitutions
	:$$x = \frac{ t}{ T}, \hspace{0.2cm} K_1 = \frac{ {a}_{\rm \star}/T}{\sqrt{2 \pi^2 }}, \hspace{0.2cm}		:$$x = \frac{ t}{ T}, \hspace{0.2cm} K_1 = \frac{ {a}_{\rm \star}/T}{\sqrt{2 \pi^2 }}, \hspace{0.2cm}
	K_2 = \frac{ {a}_{\rm \star}^2}{2 \pi}$$		K_2 = \frac{ {a}_{\rm \star}^2}{2 \pi}$$
	~~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:
	:$$- {3}/{2} \cdot K_1 \cdot x^{-5/2}\cdot {\rm e}^{-K_2/x}+ K_1 \cdot x^{-3/2}\cdot {\rm		:$$- {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 111:		Line 111:
	x_{\rm max} = {2}/{3} \cdot K_2 = \frac{{a}_{\rm \star}^2}{3 \pi}		x_{\rm max} = {2}/{3} \cdot K_2 = \frac{{a}_{\rm \star}^2}{3 \pi}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	*~~Daraus ergibt sich für~~  $60 \ \rm dB$  ~~Kabeldämpfung~~  $({a}_{\rm \star} \approx 6.9 \ \rm Np)$:		*This results in the following for  $60 \ \rm dB$  cable attenuation  $({a}_{\rm \star} \approx 6.9 \ \rm Np)$:
	:$$x_{\rm max} = { t_{\rm max}}/{ T}= { 6.9^2}/{(3\pi)}\hspace{0.15cm}\underline{ \approx 5 }\hspace{0.05cm}.$$		:$$x_{\rm max} = { t_{\rm max}}/{ T}= { 6.9^2}/{(3\pi)}\hspace{0.15cm}\underline{ \approx 5 }\hspace{0.05cm}.$$



	'''(3)'''  ~~Setzt man das Ergebnis in die vorgegebene Gleichung ein~~, ~~so erhält man~~ (~~zur Vereinfachung verwenden wir~~ "${a}$"~~anstelle von~~ "${a}_{\rm \star}$"):		'''(3)'''  Substituting the result into the given equation, we get (for simplicity, we use "${a}$"instead of "${a}_{\rm \star}$"):
	:$$h_{\rm K}(t_{\rm max}) = \frac{1}{T} \cdot \frac{ {a}}{ \sqrt{2 \pi^2 \cdot {{a}^6}/{(3\pi)^3}}}\hspace{0.1cm} \cdot		:$$h_{\rm K}(t_{\rm max}) = \frac{1}{T} \cdot \frac{ {a}}{ \sqrt{2 \pi^2 \cdot {{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
Line 124:		Line 124:
	}{2}} \cdot {\rm e}^{-3/2} \approx \frac{1}{T} \cdot \frac{1.453}{{a}^2}		}{2}} \cdot {\rm e}^{-3/2} \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~~:		*With  $a = 6.9$  we obtain 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)'''  ~~Mit dem Ergebnis aus~~  '''(3)'''  ~~lautet die geeignete Bestimmungsgleichung~~:		'''(4)'''  Using the result from  '''(3)''' , the appropriate equation of determination is:
	:$$\frac{ {a}/T}{ \sqrt{2 \pi^2 \cdot (t_{5\%}/T)^3}}= 0.05 \cdot 0.03 {1}/{T}		:$$\frac{ {a}/T}{ \sqrt{2 \pi^2 \cdot (t_{5\%}/T)^3}}= 0.05 \cdot 0.03 {1}/{T}
	\hspace{0.15cm}{= 0.0015 \cdot {1}/{T}}		\hspace{0.15cm}{= 0.0015 \cdot {1}/{T}}
Line 136:		Line 136:
	\hspace{0.3cm}		\hspace{0.3cm}
	\hspace{0.15cm}\underline{t_{5\%}/T \approx 103.5} \hspace{0.05cm}.$$		\hspace{0.15cm}\underline{t_{5\%}/T \approx 103.5} \hspace{0.05cm}.$$
	*~~Dieser Wert ist etwas zu groß~~, ~~da der zweite Term~~  ${\rm e}^{-0.05}\approx 0.95$  ~~vernachlässigt wurde~~.		*This value is a bit too large, because the second term  ${\rm e}^{-0.05}\approx 0.95$  was neglected.
	*~~Die exakte Berechnung liefert~~  $t_{\rm 5\%}/T \approx 97$.		*The exact calculation gives  $t_{\rm 5\%}/T \approx 97$.



	'''(5)'''  ~~Richtig ist~~ <u>~~der zweite Lösungsvorschlag~~</u>:		'''(5)'''  <u>The second solution</u> is correct:
	*~~Allgemein gilt~~:		*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~~ $h_{\rm K}(t)$ ~~innerhalb einer Symboldauer nur unwesentlich ändert~~, ~~kann auch geschrieben werden~~:		*Since the channel pulse response $h_{\rm K}(t)$ changes only insignificantly within a symbol duration, it can also be written:
	:$$g_r(t) = h_{\rm K}(t) \cdot s_0 \cdot T.$$		:$$g_r(t) = h_{\rm K}(t) \cdot s_0 \cdot T.$$
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 17:27, 12 November 2021

2021-11-12T17:27:37Z

← Older revision		Revision as of 19:27, 12 November 2021
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	[[File:P_ID1814__LZI_A_4_5.png\|right\|frame\|Impulse response of a coaxial cable]]		[[File:P_ID1814__LZI_A_4_5.png\|right\|frame\|Impulse response of a coaxial cable]]
	The frequency response of a coaxial cable of length  $l$  can be represented by the following formula:		The frequency response of a coaxial cable  (German:  "Koaxialkabel"   ⇒   subscipt  "K")  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
Line 9:		Line 9:
	{\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}		{\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}.$$		\hspace{0.05cm}.$$
	The first term of this equation is due to the ohmic losses, the second term to the transverse losses. Dominant, however, is the skin effect, which is expressed by the third term.		The first term of this equation is due to the ohmic losses,  the second term to the transverse losses.  Dominant,  however,  is the skin effect,  which is expressed by the third term.

	With the coefficients valid for a "standard coaxial cable"  $\text{(2.6 mm}$  core diameter,  $\text{9.5 mm}$  outer diameter$)$		With the coefficients valid for the  "standard coaxial cable"  $\text{(2.6 mm}$  core diameter,  $\text{9.5 mm}$  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},
Line 22:		Line 22:
	\hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}}}		\hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	~~⇒~~   Attenuation curve  ${a}_{\rm K}(f)$  and phase curve  $b_{\rm K}(f)$  are identical except for the pseudo units "Np" and "rad", respectively.		That means:

			Attenuation curve  ${a}_{\rm K}(f)$  and phase curve  $b_{\rm K}(f)$  are identical except for the pseudo units  "Np"  and  "rad",  respectively.


Line 28:		Line 30:
	:$${a}_{\rm \star} = {a}_{\rm K}(f ={R}/{2})		:$${a}_{\rm \star} = {a}_{\rm K}(f ={R}/{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}_{\rm \star} \cdot \sqrt{2f/R}}\cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} {a}_{\star} \cdot \sqrt{2f/R}}\hspace{0.4cm}{\rm ~~mit~~}\hspace{0.2cm}{a}_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}		{a}_{\rm \star} \cdot \sqrt{2f/R}}\cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} {a}_{\star} \cdot \sqrt{2f/R}}\hspace{0.4cm}{\rm with}\hspace{0.2cm}{a}_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	*The corresponding  $\rm dB$&~~ndash~~;value is greater by a factor of  $8.686$~~  größer~~.~~ ~~		*The corresponding  $\rm dB$  value is greater by a factor of  $8.686$.
	*For a binary system,  $R = 1/T$ holds, so that the characteristic cable attenuation refers to the frequency  $f = 1/(2T)$~~ ~~.		*For a binary system,  $R = 1/T$ holds,  so that the characteristic cable attenuation refers to the frequency  $f = 1/(2T)$.


	The  [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Fouriertransformation\|Fourier transform]]  of  $H_{\rm K}(f)$  yields the impulse response  $h_{\rm K}(t)$, which can be specified in closed-analytic form for a coaxial cable using the approximations described here. For a binary system holds:		The  [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Fouriertransformation\|Fourier transform]]  of  $H_{\rm K}(f)$  yields the impulse response  $h_{\rm K}(t)$,  which can be specified in closed-analytic form for a coaxial cable using the approximations described here.  For a binary system holds:
	:$$h_{\rm K}(t) = \frac{ {a}_{\rm \star}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot		:$$h_{\rm K}(t) = \frac{ {a}_{\rm \star}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot
	{\rm e}^{ - {{a}_{\rm \star}^2}/(2 \hspace{0.05cm} \pi \cdot \hspace{0.05cm} t/T)}		{\rm e}^{ - {{a}_{\rm \star}^2}/(2 \hspace{0.05cm} \pi \cdot \hspace{0.05cm} t/T)}
	\hspace{0.4cm}{\rm with}\hspace{0.2cm}{a}_{\rm \star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}		\hspace{0.4cm}{\rm with}\hspace{0.2cm}{a}_{\rm \star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	Subtask  '''(5)'''  refers to the basic ~~receiver~~  $g_r(t) = g_s(t) \star h_{\rm K}(t)$, where  $g_s(t)$  is assumed to be a rectangle ~~with~~ height  $s_0$  ~~and~~ duration  $T$~~ ~~.		Subtask  '''(5)'''  refers to the basic reception pulse  $g_r(t) = g_s(t) \star h_{\rm K}(t)$, where the basic transmission pulse  $g_s(t)$  is assumed to be a rectangle  $($height  $s_0$,  duration  $T)$.







	''Notes:''		Notes:
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Koaxialkabeln\|Properties of Coaxial Cables]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Koaxialkabeln\|Properties of Coaxial Cables]].

	*You can use the interactive applet  [[Applets:Zeitverhalten_von_Kupferkabeln\|Zeitverhalten von Kupferkabeln]]  to check your results.		*You can use the  (German language)  interactive SWF applet  [[Applets:Zeitverhalten_von_Kupferkabeln\|"Zeitverhalten von Kupferkabeln"]]   ⇒   "Time behavior of copper cables"  to check your results.

Hwang at 14:59, 8 November 2021

2021-11-08T14:59:29Z

← Older revision		Revision as of 16:59, 8 November 2021
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	===~~Fragebogen~~===		===Questions===

	<quiz display=simple>		<quiz display=simple>
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	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Die charakteristische Kabeldämpfung  ${a}_{\rm \star} = 60 \ \rm dB$  entspricht in etwa  $6.9\ \rm Np$. Deshalb muss gelten:		'''(1)'''  Die charakteristische Kabeldämpfung  ${a}_{\rm \star} = 60 \ \rm dB$  entspricht in etwa  $6.9\ \rm Np$. Deshalb muss gelten:

Hwang at 14:58, 8 November 2021

2021-11-08T14:58:15Z

← Older revision		Revision as of 16:58, 8 November 2021
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			{{quiz-Header\|Buchseite=Linear_and_Time_Invariant_Systems/Properties_of_Coaxial_Cables
	{{quiz-Header\|Buchseite=~~Lineare zeitinvariante Systeme~~/~~Koaxialkabel~~
	}}		}}

	[[File:P_ID1814__LZI_A_4_5.png\|right\|frame\|~~Impulsantwort eines Koaxialkabels~~]]		[[File:P_ID1814__LZI_A_4_5.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
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	{\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}		{\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~~, ~~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, the second term to the transverse losses. Dominant, however, is the skin effect, which is expressed by the third term.

	~~Mit den für ein~~ "~~Normalkoaxialkabel~~"  $\text{(2.6 mm}$  ~~Kerndurchmesser~~,  $\text{9.5 mm}$  ~~Außendurchmesser~~$)$  ~~gültigen Koeffizienten~~		With the coefficients valid for a "standard coaxial cable"  $\text{(2.6 mm}$  core diameter,  $\text{9.5 mm}$  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		\hspace{0.2cm} \beta_2 = 0.2722 \hspace{0.15cm}\frac {\rm rad}{\rm km \cdot \sqrt{\rm
	MHz}}\hspace{0.05cm}$$		MHz}}\hspace{0.05cm}$$
	~~lässt sich dieser Frequenzgang auch wie folgt darstellen~~:		this 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 23:		Line 22:
	\hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}}}		\hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	⇒   ~~Dämpfungsverlauf~~  ${a}_{\rm K}(f)$  ~~und Phasenverlauf~~  $b_{\rm K}(f)$  ~~sind bis auf die Pseudoeinheiten~~ "Np" ~~bzw.~~ "rad" ~~identisch~~.		⇒   Attenuation curve  ${a}_{\rm K}(f)$  and phase curve  $b_{\rm K}(f)$  are identical except for the pseudo units "Np" and "rad", respectively.


	~~Definiert man die charakteristische Kabeldämpfung~~  ${a}_{\rm *}$  ~~bei der halben Bitrate~~  $($~~also bei~~  $R/2)$  ~~und normiert die Frequenz auf~~  $R$,  ~~so kann man Digitalsysteme unterschiedlicher Bitrate und Länge einheitlich behandeln~~:		If one defines the characteristic cable attenuation  ${a}_{\rm *}$  at half the bit rate  $($i.e., at  $R/2)$  and normalizes the frequency to  $R$,  one can treat digital systems of different bit rate and length uniformly:
	:$${a}_{\rm \star} = {a}_{\rm K}(f ={R}/{2})		:$${a}_{\rm \star} = {a}_{\rm K}(f ={R}/{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}_{\rm \star} \cdot \sqrt{2f/R}}\cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} {a}_{\star} \cdot \sqrt{2f/R}}\hspace{0.4cm}{\rm mit}\hspace{0.2cm}{a}_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}		{a}_{\rm \star} \cdot \sqrt{2f/R}}\cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} {a}_{\star} \cdot \sqrt{2f/R}}\hspace{0.4cm}{\rm mit}\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~~  $8.686$  größer.		*The corresponding  $\rm dB$–value is greater by a factor of  $8.686$  größer.
	*~~Bei einem Binärsystem gilt~~  $R = 1/T$, so ~~dass sich die charakteristische Kabeldämpfung auf die Frequenz~~  $f = 1/(2T)$  ~~bezieht~~.		*For a binary system,  $R = 1/T$ holds, so that the characteristic cable attenuation refers to the frequency  $f = 1/(2T)$ .


	~~Die~~  [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Fouriertransformation\|~~Fouriertransformierte~~]]  ~~von~~  $H_{\rm K}(f)$  ~~liefert die Impulsantwort~~  $h_{\rm K}(t)$, ~~die für ein Koaxialkabel mit den hier beschriebenen Näherungen~~ in ~~geschlossen–analytischer Form angebbar ist~~. ~~Für ein Binärsystem gilt~~:		The  [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Fouriertransformation\|Fourier transform]]  of  $H_{\rm K}(f)$  yields the impulse response  $h_{\rm K}(t)$, which can be specified in closed-analytic form for a coaxial cable using the approximations described here. For a binary system holds:
	:$$h_{\rm K}(t) = \frac{ {a}_{\rm \star}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot		:$$h_{\rm K}(t) = \frac{ {a}_{\rm \star}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot
	{\rm e}^{ - {{a}_{\rm \star}^2}/(2 \hspace{0.05cm} \pi \cdot \hspace{0.05cm} t/T)}		{\rm e}^{ - {{a}_{\rm \star}^2}/(2 \hspace{0.05cm} \pi \cdot \hspace{0.05cm} t/T)}
	\hspace{0.4cm}{\rm ~~mit~~}\hspace{0.2cm}{a}_{\rm \star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}		\hspace{0.4cm}{\rm with}\hspace{0.2cm}{a}_{\rm \star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	~~Die Teilaufgabe~~  '''(5)'''  ~~bezieht sich auf den Empfangsgrundimpuls~~  $g_r(t) = g_s(t) \star h_{\rm K}(t)$, ~~wobei für~~  $g_s(t)$  ~~ein Rechteck mit Höhe~~  $s_0$  ~~und Dauer~~  $T$  ~~angenommen wird~~.		Subtask  '''(5)'''  refers to the basic receiver  $g_r(t) = g_s(t) \star h_{\rm K}(t)$, where  $g_s(t)$  is assumed to be a rectangle with height  $s_0$  and duration  $T$ .


Line 48:		Line 47:


	''~~Hinweise~~:''		''Notes:''
	*~~Die Aufgabe gehört zum Kapitel~~  [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Koaxialkabeln\|~~Eigenschaften von Koaxialkabeln~~]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Koaxialkabeln\|Properties of Coaxial Cables]].

	*~~Sie können zur Überprüfung Ihrer Ergebnisse das interaktive Applet~~  [[Applets:Zeitverhalten_von_Kupferkabeln\|Zeitverhalten von Kupferkabeln]]  ~~benutzen~~.		*You can use the interactive applet  [[Applets:Zeitverhalten_von_Kupferkabeln\|Zeitverhalten von Kupferkabeln]]  to check your results.

Guenter at 14:51, 29 October 2021

2021-10-29T14:51:37Z

← Older revision		Revision as of 16:51, 29 October 2021
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	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.2 ~~Koaxialkabel~~^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^4.2 Coaxial Cable^]]

← Older revision		Revision as of 16:03, 16 March 2026
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	With the coefficients valid for the  "standard coaxial cable"  $\text{(2.6 mm}$  core diameter,  $\text{9.5 mm}$  outer diameter$)$		With the coefficients valid for the  "standard coaxial cable"  $\text{(2.6 mm}$  core diameter,  $\text{9.5 mm}$  outer diameter$)$
	:$$\alpha_2 = 0.2722 \hspace{0.15cm}\frac {\rm Np}{\rm km \cdot \sqrt{\rm MHz}}\hspace{0.05cm},\hspace{0.2cm} \beta_2 = 0.2722 \hspace{0.15cm}\frac {\rm rad}{\rm km \cdot \sqrt{\~~rmMHz~~}}\hspace{0.05cm}$$		:$$\alpha_2 = 0.2722 \hspace{0.15cm}\frac {\rm Np}{\rm km \cdot \sqrt{\rm MHz}}\hspace{0.05cm},\hspace{0.2cm} \beta_2 = 0.2722 \hspace{0.15cm}\frac {\rm rad}{\rm km \cdot \sqrt{\rm MHz}}\hspace{0.05cm}$$
	this frequency response can also be represented as follows:		this frequency response can also be represented as follows:
	:$$H_{\rm K}(f) \approx {\rm e}^{- 0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km}\hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}} } \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km}\hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}}}\hspace{0.05cm}.$$		:$$H_{\rm K}(f) \approx {\rm e}^{- 0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km}\hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}} } \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km}\hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}}}\hspace{0.05cm}.$$
Line 72:		Line 72:
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The characteristic cable attenuation  ${a}_{\rm \star} = 60 \ \rm dB$  corresponds approximately to  $6.9\ \rm Np$. Therefore the following must hold:		'''(1)'''  The characteristic cable attenuation  ${a}_{\rm \star} = 60 \ \rm dB$  corresponds approximately to  $6.9\ \rm Np$. Therefore the following must hold:
	:$$\alpha_2 \cdot l \cdot {R}/{2} = 6.9\,\,{\~~rmNp~~}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\~~rmNp~~}}{0.2722 \,\, {\rm Np}/({\rm km \cdot \sqrt{\rm MHz}})\cdot \sqrt{70\,\,{\rm MHz}}}\hspace{0.15cm}\underline{ \approx 3\,\,{\rm km}}\hspace{0.05cm}.$$		:$$\alpha_2 \cdot l \cdot {R}/{2} = 6.9\,\,{\rm Np}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\rm Np}}{0.2722 \,\, {\rm Np}/({\rm km \cdot \sqrt{\rm MHz}})\cdot \sqrt{70\,\,{\rm MHz}}}\hspace{0.15cm}\underline{ \approx 3\,\,{\rm km}}\hspace{0.05cm}.$$


Line 80:		Line 80:
	:$$h_{\rm K}(x) = K_1 \cdot x^{-3/2}\cdot {\rm e}^{-K_2/x}\hspace{0.05cm}.$$		:$$h_{\rm K}(x) = K_1 \cdot x^{-3/2}\cdot {\rm e}^{-K_2/x}\hspace{0.05cm}.$$
	*By setting the derivative to zero:		*By setting the derivative to zero:
	:$$- {3}/{2} \cdot K_1 \cdot x^{-5/2}\cdot {\rm e}^{-K_2/x}+ K_1 \cdot x^{-3/2}\cdot {\~~rme~~}^{-K_2/x}\cdot (-K_2) \cdot (-x^{-2})= 0\hspace{0.05cm}.$$		:$$- {3}/{2} \cdot K_1 \cdot x^{-5/2}\cdot {\rm e}^{-K_2/x}+ K_1 \cdot x^{-3/2}\cdot {\rm e}^{-K_2/x}\cdot (-K_2) \cdot (-x^{-2})= 0\hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm} {3}/{2} \cdot x^{-5/2} = K_2 \~~cdotx~~^{-7/2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}x_{\rm max} = {2}/{3} \cdot K_2 = \frac{{a}_{\rm \star}^2}{3 \pi}\hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} {3}/{2} \cdot x^{-5/2} = K_2 \cdot x^{-7/2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}x_{\rm max} = {2}/{3} \cdot K_2 = \frac{{a}_{\rm \star}^2}{3 \pi}\hspace{0.05cm}.$$
	*This results in the following for  $60 \ \rm dB$  cable attenuation  $({a}_{\rm \star} \approx 6.9 \ \rm Np)$:		*This results in the following for  $60 \ \rm dB$  cable attenuation  $({a}_{\rm \star} \approx 6.9 \ \rm Np)$:
	:$$x_{\rm max} = { t_{\rm max}}/{ T}= { 6.9^2}/{(3\pi)}\hspace{0.15cm}\underline{ \approx 5 }\hspace{0.05cm}.$$		:$$x_{\rm max} = { t_{\rm max}}/{ T}= { 6.9^2}/{(3\pi)}\hspace{0.15cm}\underline{ \approx 5 }\hspace{0.05cm}.$$