Aufgaben:Exercise 3.1: Impulse Response of the Coaxial Cable - Revision history

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	If one defines the characteristic cable attenuation  $a_*$  at half the bit rate  $(R_{\rm B}/2)$,  one can treat digital systems of different bit rate and length uniformly:		If one defines the characteristic cable attenuation  $a_*$  at half the bit rate  $(R_{\rm B}/2)$,  one can treat digital systems of different bit rate and length uniformly:
	:$$a_{\star} = a_{\rm K}(f = {R_{\rm B}}/{2})\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm K}(f) = {\rm e}^{-a_{\star} \cdot \sqrt{2f/R_{\~~rmB~~}}}\cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} a_{\star} \cdot \sqrt{2f/R_{\~~rmB~~}}}\hspace{0.4cm}{\rm with}\hspace{0.2cm}a_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}\hspace{0.05cm}.$$		:$$a_{\star} = a_{\rm K}(f = {R_{\rm B}}/{2})\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm K}(f) = {\rm e}^{-a_{\star} \cdot \sqrt{2f/R_{\rm B}}}\cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} a_{\star} \cdot \sqrt{2f/R_{\rm B}}}\hspace{0.4cm}{\rm with}\hspace{0.2cm}a_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}\hspace{0.05cm}.$$

	The corresponding  $\rm dB$  value is larger by a factor of  $8.688$.  For a binary system,  $R_{\rm B} = 1/T$  applies,  so that the characteristic cable attenuation is then related to the frequency  $f = 1/(2T)$.		The corresponding  $\rm dB$  value is larger by a factor of  $8.688$.  For a binary system,  $R_{\rm B} = 1/T$  applies,  so that the characteristic cable attenuation is then related to the frequency  $f = 1/(2T)$.
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	[[Category:Digital Signal Transmission: Exercises\|^3.1 Intersymbol Interference^]]		[[Category:Digital Signal Transmission: Exercises\|^3.1 Intersymbol Interference^]]
	[[de:Aufgaben:3.~~1_Impulsantwort_des_Koaxialkabels~~]]		[[de:Aufgaben:Aufgabe 3.1: Impulsantwort des Koaxialkabels]]

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	'''(1)'''  The characteristic cable attenuation  $a_* = 60 \ \rm dB$  corresponds to about  $6.9 \ \rm Np$.  Therefore,  it must hold:		'''(1)'''  The characteristic cable attenuation  $a_* = 60 \ \rm dB$  corresponds to about  $6.9 \ \rm Np$.  Therefore,  it must hold:
	:$$\alpha_2 \cdot l \cdot {R_{\rm B}}/{2} = 6.9\,\,{\~~rmNp~~}$$		:$$\alpha_2 \cdot l \cdot {R_{\rm B}}/{2} = 6.9\,\,{\rm Np}$$
	:$$\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\~~rmNp~~}}{0.2722\,\,\frac{\rm Np}{{\rm km} \cdot \sqrt{\rm MHz}}\cdot \sqrt{70\,\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 3\,\,{\rm km}}\hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\rm Np}}{0.2722\,\,\frac{\rm Np}{{\rm km} \cdot \sqrt{\rm MHz}}\cdot \sqrt{70\,\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 3\,\,{\rm km}}\hspace{0.05cm}.$$


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	*By setting the derivative to zero,  it follows:		*By setting the derivative to zero,  it follows:
	:$$- {3}/{2} \cdot K_1 \cdot x^{-5/2}\cdot {\rm e}^{-K_2/x}+ K_1 \cdot x^{-3/2}\cdot {\~~rme~~}^{-K_2/x}\cdot (-K_2) \cdot (-x^{-2})= 0\hspace{0.05cm}$$		:$$- {3}/{2} \cdot K_1 \cdot x^{-5/2}\cdot {\rm e}^{-K_2/x}+ K_1 \cdot x^{-3/2}\cdot {\rm e}^{-K_2/x}\cdot (-K_2) \cdot (-x^{-2})= 0\hspace{0.05cm}$$
	:$$\Rightarrow \hspace{0.3cm} {3}/{2} \cdot x^{-5/2} = K_2 \~~cdotx~~^{-7/2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}x_{\rm max} = {2}/{3} \cdot K_2 = { a_{\star}^2}/({3 \pi})\hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} {3}/{2} \cdot x^{-5/2} = K_2 \cdot x^{-7/2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}x_{\rm max} = {2}/{3} \cdot K_2 = { a_{\star}^2}/({3 \pi})\hspace{0.05cm}.$$

	This gives for  $60 \ \rm dB$  cable attenuation  $(a_ ≈ 6.9 \ \rm Np)$:		This gives for  $60 \ \rm dB$  cable attenuation  $(a_ ≈ 6.9 \ \rm Np)$:

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	If one defines the characteristic cable attenuation  $a_*$  at half the bit rate  $(R_{\rm B}/2)$,  one can treat digital systems of different bit rate and length uniformly:		If one defines the characteristic cable attenuation  $a_*$  at half the bit rate  $(R_{\rm B}/2)$,  one can treat digital systems of different bit rate and length uniformly:
	:$$a_{\star} = a_{\rm K}(f = {R_{\rm B}}/{2})		:$$a_{\star} = a_{\rm K}(f = {R_{\rm B}}/{2})\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm K}(f) = {\rm e}^{-a_{\star} \cdot \sqrt{2f/R_{\rmB}}}\cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} a_{\star} \cdot \sqrt{2f/R_{\rmB}}}\hspace{0.4cm}{\rm with}\hspace{0.2cm}a_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}\hspace{0.05cm}.$$
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm K}(f) = {\rm e}^{-
	a_{\star} \cdot \sqrt{2f/R_{\rm
	B}}}\cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} a_{\star} \cdot \sqrt{2f/R_{\rm
	B}}}\hspace{0.4cm}{\rm with}\hspace{0.2cm}a_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}
	\hspace{0.05cm}.$$

	The corresponding  $\rm dB$  value is larger by a factor of  $8.688$.  For a binary system,  $R_{\rm B} = 1/T$  applies,  so that the characteristic cable attenuation is then related to the frequency  $f = 1/(2T)$.		The corresponding  $\rm dB$  value is larger by a factor of  $8.688$.  For a binary system,  $R_{\rm B} = 1/T$  applies,  so that the characteristic cable attenuation is then related to the frequency  $f = 1/(2T)$.
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	'''(3)'''  Substituting the result of  '''(2)'''  into the given equation,  we obtain  (using  $a$  instead of  $a_*$):		'''(3)'''  Substituting the result of  '''(2)'''  into the given equation,  we obtain  (using  $a$  instead of  $a_*$):
	:$$h_{\rm K}(t_{\rm max}) \ = \ \frac{1}{T} \cdot \frac{ a}{ \sqrt{2 \pi^2 \cdot \frac{a^6}{(3\pi)^3}}}\hspace{0.1cm} \cdot		:$$h_{\rm K}(t_{\rm max}) \ = \ \frac{1}{T} \cdot \frac{ a}{ \sqrt{2 \pi^2 \cdot \frac{a^6}{(3\pi)^3}}}\hspace{0.1cm} \cdot{\rm exp} \left[ - \frac{a^2}{2\pi} \cdot\frac{3\pi}{a^2}\hspace{0.1cm}\right]= \frac{1}{T} \cdot \frac{1}{a^2}\cdot\sqrt{\frac{27 \pi}{2}} \cdot {\rm e}^{-3/2}\hspace{0.15cm}\approx \frac{1}{T} \cdot \frac{1.453}{a^2}\hspace{0.05cm}.$$
	{\rm exp} \left[ - \frac{a^2}{2\pi} \cdot
	\frac{3\pi}{a^2}\hspace{0.1cm}\right]= \frac{1}{T} \cdot \frac{1}{a^2}\cdot
	\sqrt{\frac{27 \pi
	}{2}} \cdot {\rm e}^{-3/2}\hspace{0.15cm}\approx \frac{1}{T} \cdot \frac{1.453}{a^2}
	\hspace{0.05cm}.$$

	*Thus,  with  $a = 6.9$,  we arrive at the final result:		*Thus,  with  $a = 6.9$,  we arrive at the final result:

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	[[File:P_ID1370__Dig_A_3_1.png\|right\|frame\|Impulse response of a coaxial cable]]		[[File:P_ID1370__Dig_A_3_1.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 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 first term of this equation is due to the ohmic losses.
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	With the coefficients valid for a so-called  "standard coaxial cable"  $(2.6 \ \rm mm$  core diameter and  $9.5 \ \rm mm$  outer diameter$)$		With the coefficients valid for a so-called  "standard coaxial cable"  $(2.6 \ \rm mm$  core diameter and  $9.5 \ \rm mm$  outer diameter$)$
	:$$\alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km \cdot \sqrt{\rm MHz}}		:$$\alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km \cdot \sqrt{\rm MHz}}\hspace{0.05cm},\hspace{0.2cm} \beta_2 = 0.2722 \hspace{0.15cm}\frac{\rm rad}{\rm km \cdot \sqrt{\rm MHz}}\hspace{0.05cm},$$
	\hspace{0.05cm},
	\hspace{0.2cm} \beta_2 = 0.2722 \hspace{0.15cm}\frac{\rm rad}{\rm km \cdot \sqrt{\rm MHz}}\hspace{0.05cm},$$

	the frequency response can also be represented as follows:		the frequency response can also be represented as follows:
	:$$H_{\rm K}(f) \approx {\rm e}^{- 0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km}		:$$H_{\rm K}(f) \approx {\rm e}^{- 0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km}\hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}} } \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}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: Attenuation curve  $a_{\rm K}(f)$  and phase curve  $b_{\rm K}(f)$  are identical except for the pseudo units  "$\rm Np$"  and  "$\rm rad$".		That means: Attenuation curve  $a_{\rm K}(f)$  and phase curve  $b_{\rm K}(f)$  are identical except for the pseudo units  "$\rm Np$"  and  "$\rm rad$".
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	The Fourier transform of  $H_{\rm K}(f)$  yields the impulse response  $h_{\rm K}(t)$,  which can be given in closed-analytic form for a coaxial cable with the approximations described here.  For a binary system holds:		The Fourier transform of  $H_{\rm K}(f)$  yields the impulse response  $h_{\rm K}(t)$,  which can be given in closed-analytic form for a coaxial cable with the approximations described here.  For a binary system holds:
	:$$h_{\rm K}(t) = \frac{ a_{\star}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot		:$$h_{\rm K}(t) = \frac{ a_{\star}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot{\rm exp} \left[ - \frac{a_{\star}^2}{2 \pi \cdot t/T}\hspace{0.1cm}\right]\hspace{0.4cm}{\rm with}\hspace{0.2cm}a_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}\hspace{0.05cm}.$$
	{\rm exp} \left[ - \frac{a_{\star}^2}{2 \pi \cdot t/T}\hspace{0.1cm}\right]
	\hspace{0.4cm}{\rm with}\hspace{0.2cm}a_{\star}\hspace{0.2cm}{\rm in}\hspace{0.2cm}{\rm Np}
	\hspace{0.05cm}.$$

	Subtask  '''(5)'''  is related to the basic receiver pulse  $g_r(t) = g_s(t) * h_K(t)$,  where  $g_s(t)$  should be assumed to be a rectangular pulse with height  $s_0$  and duration  $T$.		Subtask  '''(5)'''  is related to the basic receiver pulse  $g_r(t) = g_s(t) * h_K(t)$,  where  $g_s(t)$  should be assumed to be a rectangular pulse with height  $s_0$  and duration  $T$.
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The characteristic cable attenuation  $a_* = 60 \ \rm dB$  corresponds to about  $6.9 \ \rm Np$.  Therefore,  it must hold:		'''(1)'''  The characteristic cable attenuation  $a_* = 60 \ \rm dB$  corresponds to about  $6.9 \ \rm Np$.  Therefore,  it must hold:
	:$$\alpha_2 \cdot l \cdot {R_{\rm B}}/{2} = 6.9\,\,{\rm		:$$\alpha_2 \cdot l \cdot {R_{\rm B}}/{2} = 6.9\,\,{\rmNp}$$
	Np}$$		:$$\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\rmNp}}{0.2722\,\,\frac{\rm Np}{{\rm km} \cdot \sqrt{\rm MHz}}\cdot \sqrt{70\,\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 3\,\,{\rm km}}\hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\rm
	Np}}{0.2722\,\,\frac{\rm Np}{{\rm km} \cdot \sqrt{\rm MHz}}
	\cdot \sqrt{70\,\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 3\,\,{\rm km}}
	\hspace{0.05cm}.$$


	'''(2)'''  With the substitutions		'''(2)'''  With the substitutions
	:$$x = { t}/{ T}, \hspace{0.2cm} K_1 = \frac{a_*/T}{\sqrt{2\pi^2 }}, \hspace{0.2cm}		:$$x = { t}/{ T}, \hspace{0.2cm} K_1 = \frac{a_/T}{\sqrt{2\pi^2 }}, \hspace{0.2cm}K_2 = \frac{a_^2}{2\pi}$$
	K_2 = \frac{a_*^2}{2\pi}$$

	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,  it follows:		*By setting the derivative to zero,  it follows:
	:$$- {3}/{2} \cdot K_1 \cdot x^{-5/2}\cdot {\rm e}^{-K_2/x}+ K_1 \cdot x^{-3/2}\cdot {\rm		:$$- {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 = { a_{\star}^2}/({3 \pi})\hspace{0.05cm}.$$
	\hspace{0.05cm}$$
	:$$\Rightarrow \hspace{0.3cm} {3}/{2} \cdot x^{-5/2} = K_2 \~~cdot~~
	x^{-7/2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	x_{\rm max} = {2}/{3} \cdot K_2 = { a_{\star}^2}/({3 \pi})
	\hspace{0.05cm}.$$

	This gives for  $60 \ \rm dB$  cable attenuation  $(a_ ≈ 6.9 \ \rm Np)$:		This gives for  $60 \ \rm dB$  cable attenuation  $(a_ ≈ 6.9 \ \rm Np)$:
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	*Thus,  with  $a = 6.9$,  we arrive at the final result:		*Thus,  with  $a = 6.9$,  we arrive at the final result:
	:$${\rm Max} \ [h_{\rm K}(t)] = \frac{1.453}{{6.9\,}^2} \cdot {1}/{T}\hspace{0.15cm}\underline {\approx 0.03 \cdot {1}/{T}}		:$${\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)'''  Using the result of subtask  '''(3)''',  the determining equation is:		'''(4)'''  Using the result of subtask  '''(3)''',  the determining equation is:
	:$$\frac{ a/T}{ \sqrt{2 \pi^2 \cdot (t_{5\%}/T)^3}}= 0.05 \cdot 0.03 \cdot {1}/{T}= 0.0015 \cdot {1}/{T} \hspace{0.3cm}		:$$\frac{ a/T}{ \sqrt{2 \pi^2 \cdot (t_{5\%}/T)^3}}= 0.05 \cdot 0.03 \cdot {1}/{T}= 0.0015 \cdot {1}/{T} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} (t_{5\%}/T)^{3/2} = \frac{a}{\sqrt{2} \cdot \pi \cdot0.0015}\approx 1036 \hspace{0.3cm}\Rightarrow\hspace{0.3cm}t_{5\%}/T \hspace{0.15cm}\underline {\approx 103.5}\hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm} (t_{5\%}/T)^{3/2} = \frac{a}{\sqrt{2} \cdot \pi \~~cdot~~
	0.0015}\approx 1036 \hspace{0.3cm}\Rightarrow
	\hspace{0.3cm}t_{5\%}/T \hspace{0.15cm}\underline {\approx 103.5}\hspace{0.05cm}.$$

	*This value is slightly too large because the second term  ${\rm e}^{\rm – 0.05} ≈ 0.95$  was neglected.		*This value is slightly too large because the second term  ${\rm e}^{\rm – 0.05} ≈ 0.95$  was neglected.
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	'''(5)'''  The <u>second solution</u>  is correct.  In general:		'''(5)'''  The <u>second solution</u>  is correct.  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 impulse response  $h_{\rm K}(t)$  changes only insignificantly within a symbol duration,  it can also be written for this purpose:		*Since the channel impulse response  $h_{\rm K}(t)$  changes only insignificantly within a symbol duration,  it can also be written for this purpose:

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	[[Category:Digital Signal Transmission: Exercises\|^3.1 Intersymbol Interference^]]		[[Category:Digital Signal Transmission: Exercises\|^3.1 Intersymbol Interference^]]
			[[de:Aufgaben:3.1_Impulsantwort_des_Koaxialkabels]]

Guenter: Guenter moved page Aufgaben:Exercise 3.1: Impulse response of the Coaxial Cable to Aufgaben:Exercise 3.1: Impulse Response of the Coaxial Cable

2022-05-31T12:59:51Z

Guenter moved page Aufgaben:Exercise 3.1: Impulse response of the Coaxial Cable to Aufgaben:Exercise 3.1: Impulse Response of the Coaxial Cable

← Older revision	Revision as of 14:59, 31 May 2022
(No difference)

Guenter at 12:58, 31 May 2022

2022-05-31T12:58:50Z

← Older revision		Revision as of 14:58, 31 May 2022
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	*Since the channel impulse response  $h_{\rm K}(t)$  changes only insignificantly within a symbol duration,  it can also be written for this purpose:		*Since the channel impulse response  $h_{\rm K}(t)$  changes only insignificantly within a symbol duration,  it can also be written for this purpose:
	:$$g_r(t) = h_{\rm K}(t) \cdot s_0 \cdot T .$$		:$$g_r(t) \approx h_{\rm K}(t) \cdot s_0 \cdot T .$$
	{{ML-Fuß}}		{{ML-Fuß}}


	[[Category:Digital Signal Transmission: Exercises\|^3.1 Intersymbol Interference^]]		[[Category:Digital Signal Transmission: Exercises\|^3.1 Intersymbol Interference^]]

Guenter at 12:38, 31 May 2022

2022-05-31T12:38:51Z

← Older revision		Revision as of 14:38, 31 May 2022
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The characteristic cable attenuation $a_* = 60 \ \rm dB$ corresponds to about $6.9 \ \rm Np$. Therefore, it must hold:		'''(1)'''  The characteristic cable attenuation  $a_* = 60 \ \rm dB$  corresponds to about  $6.9 \ \rm Np$.  Therefore,  it must hold:
	:$$\alpha_2 \cdot l \cdot {R_{\rm B}}/{2} = 6.9\,\,{\rm		:$$\alpha_2 \cdot l \cdot {R_{\rm B}}/{2} = 6.9\,\,{\rm
	Np}		Np}$$
	~~\hspace{0.3cm}~~\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\rm		:$$\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\rm
	Np}}{0.2722\,\,\frac{\rm Np}{{\rm km} \cdot \sqrt{\rm MHz}}		Np}}{0.2722\,\,\frac{\rm Np}{{\rm km} \cdot \sqrt{\rm MHz}}
	\cdot \sqrt{70\,\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 3\,\,{\rm km}}		\cdot \sqrt{70\,\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 3\,\,{\rm km}}
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	By setting the derivative to zero, it follows:		*By setting the derivative to zero,  it follows:
	:$$- {3}/{2} \cdot K_1 \cdot x^{-5/2}\cdot {\rm e}^{-K_2/x}+ K_1 \cdot x^{-3/2}\cdot {\rm		:$$- {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
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	This gives for $60 \ \rm dB$ cable attenuation $(a_* ≈ 6.9 \ \rm Np)$:		This gives for  $60 \ \rm dB$  cable attenuation  $(a_ ≈ 6.9 \ \rm Np)$:
	:$$x_{\rm max} = { t_{\rm max}}/{ T} \hspace{-0.1cm}: \hspace{0.2cm} { t_{\rm max}}/{ T} = { 6.9^2}/({3\pi})\hspace{0.15cm}\underline {\approx 5} \hspace{0.2cm}.$$		:$$x_{\rm max} = { t_{\rm max}}/{ T} \hspace{-0.1cm}: \hspace{0.2cm} { t_{\rm max}}/{ T} = { 6.9^2}/({3\pi})\hspace{0.15cm}\underline {\approx 5} \hspace{0.2cm}.$$


	'''(3)'''  Substituting the result of (2) into the given equation, we obtain (~~we use~~ $a$ instead of $a_*$):		'''(3)'''  Substituting the result of  '''(2)'''  into the given equation,  we obtain  (using  $a$  instead of  $a_*$):
	:$$h_{\rm K}(t_{\rm max}) \ = \ \frac{1}{T} \cdot \frac{ a}{ \sqrt{2 \pi^2 \cdot \frac{a^6}{(3\pi)^3}}}\hspace{0.1cm} \cdot		:$$h_{\rm K}(t_{\rm max}) \ = \ \frac{1}{T} \cdot \frac{ a}{ \sqrt{2 \pi^2 \cdot \frac{a^6}{(3\pi)^3}}}\hspace{0.1cm} \cdot
	{\rm exp} \left[ - \frac{a^2}{2\pi} \cdot		{\rm exp} \left[ - \frac{a^2}{2\pi} \cdot
Line 127:		Line 127:
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	Thus, with $a = 6.9$, we arrive at the final result:		*Thus,  with  $a = 6.9$,  we arrive at the final result:
	:$${\rm Max} \ [h_{\rm K}(t)] = \frac{1.453}{{6.9\,}^2} \cdot {1}/{T}\hspace{0.15cm}\underline {\approx 0.03 \cdot {1}/{T}}		:$${\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)'''  Using the result of subtask (3), the determining equation is:		'''(4)'''  Using the result of subtask  '''(3)''',  the determining equation is:
	:$$\frac{ a/T}{ \sqrt{2 \pi^2 \cdot (t_{5\%}/T)^3}}= 0.05 \cdot 0.03 \cdot {1}/{T}= 0.0015 \cdot {1}/{T} \hspace{0.3cm}		:$$\frac{ a/T}{ \sqrt{2 \pi^2 \cdot (t_{5\%}/T)^3}}= 0.05 \cdot 0.03 \cdot {1}/{T}= 0.0015 \cdot {1}/{T} \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} (t_{5\%}/T)^{3/2} = \frac{a}{\sqrt{2} \cdot \pi \cdot		\Rightarrow \hspace{0.3cm} (t_{5\%}/T)^{3/2} = \frac{a}{\sqrt{2} \cdot \pi \cdot
Line 138:		Line 138:
	\hspace{0.3cm}t_{5\%}/T \hspace{0.15cm}\underline {\approx 103.5}\hspace{0.05cm}.$$		\hspace{0.3cm}t_{5\%}/T \hspace{0.15cm}\underline {\approx 103.5}\hspace{0.05cm}.$$

	This value is slightly too large because the second term ${\rm e}^{\rm – 0.05} ≈ 0.95$ was neglected. The exact calculation gives $t_{\rm 5\%}/T ≈ 97$.		*This value is slightly too large because the second term  ${\rm e}^{\rm – 0.05} ≈ 0.95$  was neglected.
			*The exact calculation gives  $t_{\rm 5\%}/T ≈ 97$.


	'''(5)'''  The <u>second solution</u> is correct. In general:		'''(5)'''  The <u>second solution</u>  is correct.  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 impulse response $h_{\rm K}(t)$ changes only insignificantly within a symbol duration, it can also be written for this purpose:		*Since the channel impulse response  $h_{\rm K}(t)$  changes only insignificantly within a symbol duration,  it can also be written for this purpose:
	:$$g_r(t) = h_{\rm K}(t) \cdot s_0 \cdot T .$$		:$$g_r(t) = h_{\rm K}(t) \cdot s_0 \cdot T .$$
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 12:30, 31 May 2022

2022-05-31T12:30:14Z

← Older revision		Revision as of 14:30, 31 May 2022
Line 11:		Line 11:
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	The first term of this equation is due to the ohmic losses~~, and the~~ second term is due 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 is due to the transverse losses.
			*Dominant,  however,  is the skin effect,  which is expressed by the third term.

	With the coefficients valid for a so-called standard coaxial cable  $(2.6 \ \rm mm$  core diameter and  $9.5 \ \rm mm$  outer diameter$)$
			With the coefficients valid for a so-called  "standard coaxial cable"  $(2.6 \ \rm mm$  core diameter and  $9.5 \ \rm mm$  outer diameter$)$
	:$$\alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km \cdot \sqrt{\rm MHz}}		:$$\alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km \cdot \sqrt{\rm MHz}}
	\hspace{0.05cm},		\hspace{0.05cm},
	\hspace{0.2cm} \beta_2 = 0.2722 \hspace{0.15cm}\frac{\rm rad}{\rm km \cdot \sqrt{\rm MHz}}\hspace{0.05cm}.$$		\hspace{0.2cm} \beta_2 = 0.2722 \hspace{0.15cm}\frac{\rm rad}{\rm km \cdot \sqrt{\rm MHz}}\hspace{0.05cm},$$

	~~this~~ frequency response can also be represented as follows:		the frequency response can also be represented as follows:
	:$$H_{\rm K}(f) \approx {\rm e}^{- 0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km}		:$$H_{\rm K}(f) \approx {\rm e}^{- 0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km}
	\hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}} } \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}		\hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}} } \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}
Line 25:		Line 28:
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

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

	If one defines the characteristic cable attenuation  $a_*$  at half the bit rate  $(R_{\rm B}/2)$, one can treat digital systems of different bit rate and length uniformly:		If one defines the characteristic cable attenuation  $a_*$  at half the bit rate  $(R_{\rm B}/2)$,  one can treat digital systems of different bit rate and length uniformly:
	:$$a_{\star} = a_{\rm K}(f = {R_{\rm B}}/{2})		:$$a_{\star} = a_{\rm K}(f = {R_{\rm B}}/{2})
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm K}(f) = {\rm e}^{-		\hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm K}(f) = {\rm e}^{-
Line 35:		Line 38:
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

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


	The Fourier transform of  $H_{\rm K}(f)$  yields the impulse response  $h_{\rm K}(t)$, which can be given in closed-analytic form for a coaxial cable with the approximations described here. For a binary system holds:		The Fourier transform of  $H_{\rm K}(f)$  yields the impulse response  $h_{\rm K}(t)$,  which can be given in closed-analytic form for a coaxial cable with the approximations described here.  For a binary system holds:
	:$$h_{\rm K}(t) = \frac{ a_{\star}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot		:$$h_{\rm K}(t) = \frac{ a_{\star}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot
	{\rm exp} \left[ - \frac{a_{\star}^2}{2 \pi \cdot t/T}\hspace{0.1cm}\right]		{\rm exp} \left[ - \frac{a_{\star}^2}{2 \pi \cdot t/T}\hspace{0.1cm}\right]
Line 44:		Line 47:
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	Subtask '''(5)''' is related to the basic receiver pulse  $g_r(t) = g_s(t) * h_K(t)$, where  $g_s(t)$  should be assumed to be a rectangular pulse with height  $s_0$  and duration  $T$.		Subtask  '''(5)'''  is related to the basic receiver pulse  $g_r(t) = g_s(t) * h_K(t)$,  where  $g_s(t)$  should be assumed to be a rectangular pulse with height  $s_0$  and duration  $T$.


Line 50:		Line 53:


			Notes:
			*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference\|"Causes and Effects of Intersymbol Interference"]].

	~~''Notes:''~~
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference\|"Causes and Effects of Intersymbol Interference"]].
	*Reference is made in particular to the section  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces\|"Signals, Basis Functions and Vector Spaces"]].		*Reference is made in particular to the section  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces\|"Signals, Basis Functions and Vector Spaces"]].

Line 61:		Line 64:

	<quiz display=simple>		<quiz display=simple>
	{What is the length  $l$  of a standard coaxial cable, if for the bit rate  $R_{\rm B} = 140 \ \rm Mbit/s$  the characteristic cable attenuation is  $a_* = 60 \ \rm dB$?		{What is the length  $l$  of a standard coaxial cable,  if for the bit rate  $R_{\rm B} = 140 \ \rm Mbit/s$  the characteristic cable attenuation is  $a_* = 60 \ \rm dB$?
	\|type="{}"}		\|type="{}"}
	$l \ = \ $ { 3 3% } $\ \rm km $		$l \ = \ $ { 3 3% } $\ \rm km $

	{At what time  $t_{\rm max}$  does  $~~h_K~~(t)$  have its maximum? Let  $a_* = 60 \ \rm dB$ be further valid.		{At what time  $t_{\rm max}$  does  $h_{\rm K}(t)$  have its maximum?  Let  $a_* = 60 \ \rm dB$ be further valid.
	\|type="{}"}		\|type="{}"}
	$t_{\rm max}/T= \ $ { 5 3% }		$t_{\rm max}/T= \ $ { 5 3% }

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

	{At what time  $t_{\rm 5\%}$  is  $h_{\rm K}(t)$  less than  $5\%$  of the maximum? Consider only the first term of the given formula as an approximation.		{At what time  $t_{\rm 5\%}$  is  $h_{\rm K}(t)$  less than  $5\%$  of the maximum?  Consider only the first term of the given formula as an approximation.
	\|type="{}"}		\|type="{}"}
	$t_{\rm 5\%}/T= \ $ { 103.5 3% }		$t_{\rm 5\%}/T= \ $ { 103.5 3% }

Hwang at 15:53, 16 May 2022

2022-05-16T15:53:03Z

← Older revision		Revision as of 17:53, 16 May 2022
Line 52:		Line 52:

	''Notes:''		''Notes:''
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference\|Causes and Effects of Intersymbol Interference]].		*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference\|"Causes and Effects of Intersymbol Interference"]].
	*Reference is made in particular to the section  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces\|Signals, Basis Functions and Vector Spaces]].		*Reference is made in particular to the section  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces\|"Signals, Basis Functions and Vector Spaces"]].