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

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 Exercise 3.1: Impulse response of the Coaxial Cable to Exercise 3.1: Impulse Response of the Coaxial Cable

Guenter at 12:58, 31 May 2022

2022-05-31T12:58:50Z

Guenter at 12:38, 31 May 2022

2022-05-31T12:38:51Z

Guenter at 12:30, 31 May 2022

2022-05-31T12:30:14Z

Hwang at 15:53, 16 May 2022

2022-05-16T15:53:03Z

Guenter at 14:48, 7 May 2022

2022-05-07T14:48:57Z

Guenter: Guenter moved page Aufgaben:Aufgabe 3.1: Impulsantwort des Koaxialkabels to Aufgaben:Exercise 3.1: Impulse response of the Coaxial Cable

2022-05-07T14:45:49Z

Guenter moved page Aufgabe 3.1: Impulsantwort des Koaxialkabels to Exercise 3.1: Impulse response of the Coaxial Cable

Hwang at 09:09, 25 April 2022

2022-04-25T09:09:43Z

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2021-05-28T14:29:51Z

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2021-05-28T14:28:25Z

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@@ Line 147: / Line 147: @@
 *Since the channel impulse response&nbsp; $h_{\rm K}(t)$&nbsp; changes only insignificantly within a symbol duration,&nbsp; it can also be written for this purpose:
-:$$g_r(t) = h_{\rm K}(t) \cdot s_0 \cdot T .$$
+:$$g_r(t) \approx h_{\rm K}(t) \cdot s_0 \cdot T .$$
 {{ML-Fuß}}
 [[Category:Digital Signal Transmission: Exercises|^3.1 Intersymbol Interference^]]

@@ Line 89: / Line 89: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The characteristic cable attenuation $a_* = 60 \ \rm dB$ corresponds to about $6.9 \ \rm Np$. Therefore, it must hold:
+'''(1)'''&nbsp; The characteristic cable attenuation&nbsp; $a_* = 60 \ \rm dB$&nbsp; corresponds to about&nbsp; $6.9 \ \rm Np$.&nbsp; Therefore,&nbsp; it must hold:
 :$$\alpha_2 \cdot l \cdot {R_{\rm B}}/{2} = 6.9\,\,{\rm
-Np}
+Np}$$
-\hspace{0.3cm}\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\rm
+:$$\Rightarrow \hspace{0.3cm} l = \frac{6.9\,\,{\rm
 Np}}{0.2722\,\,\frac{\rm Np}{{\rm km} \cdot \sqrt{\rm MHz}}
 \cdot \sqrt{70\,\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 3\,\,{\rm km}}
@@ Line 106: / Line 106: @@
    \hspace{0.05cm}.$$
-By setting the derivative to zero, it follows:
+*By setting the derivative to zero,&nbsp; it follows:
 :$$- {3}/{2} \cdot  K_1 \cdot x^{-5/2}\cdot {\rm e}^{-K_2/x}+ K_1 \cdot x^{-3/2}\cdot {\rm
    e}^{-K_2/x}\cdot (-K_2) \cdot (-x^{-2})= 0
@@ Line 115: / Line 115: @@
    \hspace{0.05cm}.$$
-This gives for $60 \ \rm dB$ cable attenuation $(a_* &asymp; 6.9 \ \rm Np)$:
+*This gives for&nbsp; $60 \ \rm dB$&nbsp; cable attenuation&nbsp; $(a_* &asymp; 6.9 \ \rm Np)$:
 :$$x_{\rm max} =   { t_{\rm max}}/{  T} \hspace{-0.1cm}: \hspace{0.2cm} { t_{\rm max}}/{  T} = { 6.9^2}/({3\pi})\hspace{0.15cm}\underline {\approx 5} \hspace{0.2cm}.$$
-'''(3)'''&nbsp; Substituting the result of (2) into the given equation, we obtain (we use $a$ instead of $a_*$):
+'''(3)'''&nbsp; Substituting the result of&nbsp; '''(2)'''&nbsp; into the given equation,&nbsp; we obtain&nbsp; (using&nbsp; $a$&nbsp; instead of&nbsp; $a_*$):
 :$$h_{\rm K}(t_{\rm max})  \ = \  \frac{1}{T} \cdot \frac{ a}{  \sqrt{2  \pi^2 \cdot \frac{a^6}{(3\pi)^3}}}\hspace{0.1cm} \cdot
    {\rm exp} \left[ - \frac{a^2}{2\pi} \cdot
@@ Line 127: / Line 127: @@
   \hspace{0.05cm}.$$
-Thus, with $a = 6.9$, we arrive at the final result:
+*Thus,&nbsp; with&nbsp; $a = 6.9$,&nbsp; we arrive at the final result:
 :$${\rm Max} \ [h_{\rm K}(t)]  = \frac{1.453}{{6.9\,}^2} \cdot {1}/{T}\hspace{0.15cm}\underline {\approx  0.03 \cdot {1}/{T}}
    \hspace{0.05cm}.$$
-'''(4)'''&nbsp; Using the result of subtask (3), the determining equation is:
+'''(4)'''&nbsp; Using the result of subtask&nbsp; '''(3)''',&nbsp; the determining equation is:
 :$$\frac{ a/T}{  \sqrt{2  \pi^2 \cdot (t_{5\%}/T)^3}}= 0.05 \cdot 0.03 \cdot {1}/{T}= 0.0015  \cdot {1}/{T} \hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}  (t_{5\%}/T)^{3/2} = \frac{a}{\sqrt{2} \cdot \pi \cdot
@@ Line 138: / Line 138: @@
    \hspace{0.3cm}t_{5\%}/T \hspace{0.15cm}\underline {\approx 103.5}\hspace{0.05cm}.$$
-This value is slightly too large because the second term ${\rm e}^{\rm &ndash; 0.05} &asymp; 0.95$ was neglected. The exact calculation gives $t_{\rm 5\%}/T &asymp; 97$.
+*This value is slightly too large because the second term&nbsp; ${\rm e}^{\rm &ndash; 0.05} &asymp; 0.95$&nbsp; was neglected.
-'''(5)'''&nbsp; The <u>second solution</u> is correct. In general:
+'''(5)'''&nbsp; The <u>second solution</u>&nbsp; is correct.&nbsp; In general:
 :$$g_r(t) = g_s(t) \star h_{\rm K}(t) = s_0 \cdot
 \int_{t-T/2}^{t+T/2} h_{\rm K}(\tau) \,{\rm d} \tau .$$
-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&nbsp; $h_{\rm K}(t)$&nbsp; changes only insignificantly within a symbol duration,&nbsp; it can also be written for this purpose:
 :$$g_r(t) = h_{\rm K}(t) \cdot s_0 \cdot T .$$
 {{ML-Fuß}}

@@ Line 11: / Line 11: @@
      \hspace{0.05cm}.$$
-The first term of this equation is due to the ohmic losses, and the second term is due to the transverse losses. Dominant, however, is the skin effect, which is expressed by the third term.
+*The first term of this equation is due to the ohmic losses.
 With the coefficients valid for a so-called standard coaxial cable &nbsp;$(2.6 \ \rm mm$&nbsp; core diameter and &nbsp;$9.5 \ \rm mm$&nbsp; outer diameter$)$
 :$$\alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km \cdot \sqrt{\rm MHz}}
    \hspace{0.05cm},
-   \hspace{0.2cm} \beta_2 = 0.2722 \hspace{0.15cm}\frac{\rm rad}{\rm km \cdot \sqrt{\rm MHz}}\hspace{0.05cm}.$$
+   \hspace{0.2cm} \beta_2 = 0.2722 \hspace{0.15cm}\frac{\rm rad}{\rm km \cdot \sqrt{\rm MHz}}\hspace{0.05cm},$$
-this frequency response can also be represented as follows:
+the frequency response can also be represented as follows:
 :$$H_{\rm K}(f) \approx {\rm e}^{- 0.2722 \hspace{0.05cm}\cdot \hspace{0.05cm}l/{\rm km}
    \hspace{0.05cm}\cdot \sqrt{f/{\rm MHz}} } \cdot {\rm e}^{- {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}
@@ Line 25: / Line 28: @@
      \hspace{0.05cm}.$$
-That means: &nbsp; attenuation curve &nbsp;$a_{\rm K}(f)$&nbsp; and phase curve &nbsp;$b_{\rm K}(f)$&nbsp; are identical except for the pseudo units "$\rm Np$" and "$\rm rad$".
+That means: Attenuation curve &nbsp;$a_{\rm K}(f)$&nbsp; and phase curve &nbsp;$b_{\rm K}(f)$&nbsp; are identical except for the pseudo units&nbsp; "$\rm Np$"&nbsp; and&nbsp; "$\rm rad$".
-If one defines the characteristic cable attenuation &nbsp;$a_*$&nbsp; at half the bit rate &nbsp;$(R_{\rm B}/2)$, one can treat digital systems of different bit rate and length uniformly:
+If one defines the characteristic cable attenuation &nbsp;$a_*$&nbsp; at half the bit rate &nbsp;$(R_{\rm B}/2)$,&nbsp; one can treat digital systems of different bit rate and length uniformly:
 :$$a_{\star} = a_{\rm K}(f = {R_{\rm B}}/{2})
 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}H_{\rm K}(f) = {\rm e}^{-
@@ Line 35: / Line 38: @@
    \hspace{0.05cm}.$$
-The corresponding &nbsp;$\rm dB$ value is larger by a factor of &nbsp;$8.688$.&nbsp; For a binary system, &nbsp;$R_{\rm B} = 1/T$ applies, so that the characteristic cable attenuation is then related to the frequency &nbsp;$f = 1/(2T)$.&nbsp;
+The corresponding &nbsp;$\rm dB$&nbsp; value is larger by a factor of &nbsp;$8.688$.&nbsp; For a binary system, &nbsp;$R_{\rm B} = 1/T$&nbsp; applies,&nbsp; so that the characteristic cable attenuation is then related to the frequency &nbsp;$f = 1/(2T)$.&nbsp;
-The Fourier transform of &nbsp;$H_{\rm K}(f)$&nbsp; yields the impulse response &nbsp;$h_{\rm K}(t)$, 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 &nbsp;$H_{\rm K}(f)$&nbsp; yields the impulse response &nbsp;$h_{\rm K}(t)$,&nbsp; which can be given in closed-analytic form for a coaxial cable with the approximations described here.&nbsp; For a binary system holds:
 :$$h_{\rm K}(t) =   \frac{ a_{\star}/T}{  \sqrt{2  \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot
    {\rm exp} \left[ - \frac{a_{\star}^2}{2  \pi  \cdot t/T}\hspace{0.1cm}\right]
@@ Line 44: / Line 47: @@
 \hspace{0.05cm}.$$
-Subtask '''(5)''' is related to the basic receiver pulse &nbsp;$g_r(t) = g_s(t) * h_K(t)$, where &nbsp;$g_s(t)$&nbsp; should be assumed to be a rectangular pulse with height &nbsp;$s_0$&nbsp; and duration &nbsp;$T$.&nbsp;
+Subtask&nbsp; '''(5)'''&nbsp; is related to the basic receiver pulse &nbsp;$g_r(t) = g_s(t) * h_K(t)$,&nbsp; where &nbsp;$g_s(t)$&nbsp; should be assumed to be a rectangular pulse with height &nbsp;$s_0$&nbsp; and duration &nbsp;$T$.&nbsp;
@@ Line 50: / Line 53: @@
 *Reference is made in particular to the section&nbsp; [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces|"Signals, Basis Functions and Vector Spaces"]].
@@ Line 61: / Line 64: @@
 <quiz display=simple>
-{What is the length &nbsp;$l$&nbsp; of a standard coaxial cable, if for the bit rate &nbsp;$R_{\rm B} = 140 \ \rm Mbit/s$&nbsp; the characteristic cable attenuation is &nbsp;$a_* = 60 \ \rm dB$?&nbsp;
+{What is the length &nbsp;$l$&nbsp; of a standard coaxial cable,&nbsp; if for the bit rate &nbsp;$R_{\rm B} = 140 \ \rm Mbit/s$&nbsp; the characteristic cable attenuation is &nbsp;$a_* = 60 \ \rm dB$?&nbsp;
 |type="{}"}
 $l \ = \ $ { 3 3% } $\ \rm km $
-{At what time &nbsp;$t_{\rm max}$&nbsp; does &nbsp;$h_K(t)$&nbsp; have its maximum? Let &nbsp;$a_* = 60 \ \rm dB$ be further valid.
+{At what time &nbsp;$t_{\rm max}$&nbsp; does &nbsp;$h_{\rm K}(t)$&nbsp; have its maximum?&nbsp; Let &nbsp;$a_* = 60 \ \rm dB$ be further valid.
 |type="{}"}
 $t_{\rm max}/T= \ $ { 5 3% }
-{What is the maximum value of the impulse response? Let &nbsp;$a_* = 60 \ \rm dB$ continue to hold.
+{What is the maximum value of the impulse response?&nbsp; Let &nbsp;$a_* = 60 \ \rm dB$&nbsp; continue to hold.
 |type="{}"}
 ${\rm Max}\  \big [h_{\rm K}(t)\big ]= \ $ { 0.03 3% } $\ \cdot 1/T $
-{At what time &nbsp;$t_{\rm 5\%}$&nbsp; is &nbsp;$h_{\rm K}(t)$&nbsp; less than &nbsp;$5\%$&nbsp; of the maximum? Consider only the first term of the given formula as an approximation.
+{At what time &nbsp;$t_{\rm 5\%}$&nbsp; is &nbsp;$h_{\rm K}(t)$&nbsp; less than &nbsp;$5\%$&nbsp; of the maximum?&nbsp; Consider only the first term of the given formula as an approximation.
 |type="{}"}
 $t_{\rm 5\%}/T= \ $ { 103.5 3% }

@@ Line 52: / Line 52: @@
 ''Notes:''
-*The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference|Causes and Effects of Intersymbol Interference]].
+*The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference|"Causes and Effects of Intersymbol Interference"]].
-*Reference is made in particular to the section&nbsp; [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces|Signals, Basis Functions and Vector Spaces]].
+*Reference is made in particular to the section&nbsp; [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces|"Signals, Basis Functions and Vector Spaces"]].

← Older revision		Revision as of 14:48, 7 May 2022
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−	[[Category:Digital Signal Transmission: Exercises\|^3.1 ~~Impulsinterferenzen~~^]]	+	[[Category:Digital Signal Transmission: Exercises\|^3.1 Intersymbol Interference^]]

← Older revision	Revision as of 12:59, 31 May 2022
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← Older revision	Revision as of 14:45, 7 May 2022
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@@ Line 25: / Line 25: @@
      \hspace{0.05cm}.$$
-Das heißt: &nbsp; Dämpfungsverlauf &nbsp;$a_{\rm K}(f)$&nbsp; und Phasenverlauf &nbsp;$b_{\rm K}(f)$&nbsp; sind bis auf die Pseudoeinheiten "$\rm Np$&rdquo; bzw. "$\rm rad$&rdquo; identisch.
+Das heißt: &nbsp; Dämpfungsverlauf &nbsp;$a_{\rm K}(f)$&nbsp; und Phasenverlauf &nbsp;$b_{\rm K}(f)$&nbsp; sind bis auf die Pseudoeinheiten "$\rm Np$" bzw. "$\rm rad$" identisch.
 Definiert man die charakteristische Kabeldämpfung &nbsp;$a_*$&nbsp; bei der halben Bitrate &nbsp;$(R_{\rm B}/2)$, so kann man Digitalsysteme unterschiedlicher Bitrate und Länge einheitlich behandeln: