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

Guenter at 14:14, 31 May 2022

2022-05-31T14:14:07Z

Guenter at 13:11, 31 May 2022

2022-05-31T13:11:28Z

Hwang at 12:45, 17 May 2022

2022-05-17T12:45:54Z

Hwang at 15:53, 16 May 2022

2022-05-16T15:53:49Z

Guenter at 14:49, 7 May 2022

2022-05-07T14:49:28Z

Guenter: Guenter moved page Aufgaben:Aufgabe 3.1Z: Frequenzgang des Koaxialkabels to Aufgaben:Exercise 3.1Z: Frequency Response of the Coaxial Cable

2022-05-07T14:46:56Z

Guenter moved page Aufgabe 3.1Z: Frequenzgang des Koaxialkabels to Exercise 3.1Z: Frequency Response of the Coaxial Cable

Hwang at 09:38, 25 April 2022

2022-04-25T09:38:10Z

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Javier: Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

2021-03-23T12:50:52Z

Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

Javier: Text replacement - "[[Lineare_zeitinvariante_Systeme/" to "[[Linear_and_Time_Invariant_Systems/"

2020-07-09T09:45:21Z

Text replacement - "[[Lineare_zeitinvariante_Systeme/" to "[[Linear_and_Time_Invariant_Systems/"

Guenter at 09:56, 18 February 2019

2019-02-18T09:56:10Z

@@ Line 83: / Line 83: @@
 $a_{\rm K}(f = 70\,{\rm MHz})\ = \ $ { 4.555 3% } $\ {\rm Np} $
-{What is the formula for the conversion between &nbsp;$\rm Np$&nbsp; and &nbsp;$\rm dB$?&nbsp; What is the &nbsp;$\rm dB$&nbsp; value for the attenuation calculated in subtask&nbsp; '''(4)'''?
+{What is the formula for the conversion between &nbsp;$\rm Np$&nbsp; and &nbsp;$\rm dB$?&nbsp; What is the &nbsp;$\rm dB$&ndash;value for the attenuation calculated in subtask&nbsp; '''(4)'''?
 |type="{}"}
 $a_{\rm K}(f = 70\,{\rm MHz})\ = \ $ { 39.57 3% } $\ {\rm dB} $
@@ Line 98: / Line 98: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>Solutions 1 and 4</u> are correct:
+'''(1)'''&nbsp; <u>Solutions 1 and 4</u>&nbsp; are correct:
-*The $\alpha_0$&ndash;term causes only frequency-independent attenuation and the $\beta_1$&ndash;term (linear phase) causes frequency-independent delay.
+*The&nbsp; $\alpha_0$&ndash;term causes only frequency-independent attenuation and the&nbsp; $\beta_1$&ndash;term&nbsp; (linear phase)&nbsp; causes frequency-independent delay.
-*All other terms contribute to the (linear) distortions.
+*All other terms contribute to the&nbsp; (linear)&nbsp; distortions.
-'''(2)'''&nbsp; With $a_0 = \alpha_0 \cdot l$, the following equation must be satisfied:
+'''(2)'''&nbsp; With&nbsp; $a_0 = \alpha_0 \cdot l$,&nbsp; the following equation must be satisfied:
 :$${\rm e}^{- a_0 }  \ge 0.99
    \hspace{0.3cm} \Rightarrow \hspace{0.3cm}a_0 < {\rm ln}
@@ Line 110: / Line 110: @@
    \hspace{0.05cm}.$$
-This gives the following for the maximum cable length
+*This gives the following for the maximum cable length:
 :$$l_{\rm max} = \frac{a_0 }{\alpha_0 }  = \frac{0.01\,\,{\rm Np}}{0.00162\,\,{\rm Np/km}}\hspace{0.15cm}\underline {\approx 6.173\,\,{\rm km}}
    \hspace{0.05cm}.$$
-'''(3)'''&nbsp; For the attenuation curve, considering all terms:
+'''(3)'''&nbsp; For the attenuation curve,&nbsp; considering all terms:
 :$$a_{\rm K}(f) \ = \ \big[\alpha_0 + \alpha_1  \cdot f + \alpha_2  \cdot
    \sqrt{f}\hspace{0.05cm}\big] \cdot l = \big [0.00162 + 0.000435  \cdot 70 + 0.2722   \cdot \sqrt{70}\hspace{0.05cm}\big]\, \frac{\rm Np}{\rm km} \cdot 2\,{\rm km} = \hspace{0.15cm}\underline {= 4.619\, {\rm Np}}\hspace{0.05cm}.$$
-'''(4)'''&nbsp; According to the calculation at point (3), the attenuation value $\underline {4.555\,{\rm Np}}$ is obtained here.
+'''(4)'''&nbsp; According to the calculation at point&nbsp; '''(3)''',&nbsp; the attenuation value&nbsp; $\underline {4.555\,{\rm Np}}$&nbsp; is obtained here.
-'''(5)'''&nbsp; For any positive quantity $x$ holds:
+'''(5)'''&nbsp; For any positive quantity&nbsp; $x$&nbsp; holds:
 :$$x_{\rm Np} = {\rm ln} \hspace{0.10cm} x =  \frac{{\rm lg} \hspace{0.10cm} x}{{\rm lg} \hspace{0.10cm} {\rm e}}
    =   \frac{1}{{20 \cdot \rm lg} \hspace{0.10cm} {\rm e}} \cdot
    (20 \cdot {\rm lg} \hspace{0.10cm} x) = 0.1151 \cdot x_{\rm dB}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{\rm dB} = 8.686 \cdot x_{\rm
   Np}\hspace{0.05cm}.$$
-Thus, the attenuation value $4.555\,{\rm Np}$ is identical to $\underline{39.57\,{\rm dB} }$.
+*Thus,&nbsp; the attenuation value&nbsp; $4.555\,{\rm Np}$&nbsp; is identical to&nbsp; $\underline{39.57\,{\rm dB} }$.
-'''(6)'''&nbsp; <u>Solutions 1, 4 and 5</u> are correct:
+'''(6)'''&nbsp; <u>Solutions 1, 4 and 5</u>&nbsp; are correct:
-*With the restriction to the attenuation term with $\alpha_2$, the following holds for the frequency response:
+*With the restriction to the attenuation term with&nbsp; $\alpha_2$,&nbsp; the following holds for the frequency response:
 :$$H_{\rm K}(f)   =
    {\rm e}^{- \alpha_2  \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
@@ Line 139: / Line 139: @@
    {\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2  \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
    \sqrt{f}}   \hspace{0.05cm}.$$
-*If we omit the $\beta_1$ phase term, nothing changes with respect to the distortions. Only the phase and group delay would become smaller (both equal) by the value $\tau_1 = (\beta_1 \cdot l)/2\pi$.
+*If we omit the&nbsp; $\beta_1$&nbsp; phase term,&nbsp; nothing changes with respect to the distortions.&nbsp; Only the phase and group delay would become smaller&nbsp; (both equal)&nbsp; by the value&nbsp; $\tau_1 = (\beta_1 \cdot l)/2\pi$.
-*On the other hand, if the $\beta_2$&ndash;term is omitted, completely different ratios result:
+*On the other hand,&nbsp; if the&nbsp; $\beta_2$&ndash;term is omitted,&nbsp; completely different ratios result:
-:: The frequency response $H_{\rm K}(f)$ now no longer satisfies the requirement of a causal system; for such a system $H_{\rm K}(f)$ must be minimum-phase.
+# The frequency response&nbsp; $H_{\rm K}(f)$&nbsp; now no longer satisfies the requirement of a causal system;&nbsp; for such a system&nbsp; $H_{\rm K}(f)$&nbsp; must be minimum-phase.
-:: The impulse response $h_{\rm K}(t)$ is symmetrical about $t = 0$ with real frequency response, which does not correspond to the conditions.
+# The impulse response&nbsp; $h_{\rm K}(t)$&nbsp; is symmetrical about&nbsp; $t = 0$&nbsp; with real frequency response,&nbsp; which does not correspond to the conditions.
 *Therefore as an approximation for the coaxial cable frequency response is allowed:
 :$$a_{\rm K}(f) = \alpha_2  \cdot l \cdot
    \sqrt{f}, \hspace{0.2cm}b_{\rm K}(f) = a_{\rm K}(f) \cdot
    \frac{\rm rad}{\rm Np}\hspace{0.05cm}.$$
-*That means: $a_{\rm K}(f)$ and $b_{\rm K}(f)$ of a coaxial cable are identical in form and differ only in their units.
+*That means:&nbsp; $a_{\rm K}(f)$&nbsp; and&nbsp; $b_{\rm K}(f)$&nbsp; of a coaxial cable are identical in form and differ only in their units.
-*For a digital system with bit rate $R_{\rm B} = 140\,{\rm Mbit/s}$ &#8658; $R_{\rm B}/2 = 70\,{\rm Mbit/s}$ and cable length $l = 2\,{\rm km}$, $a_* &asymp; 40\,{\rm dB}$ is indeed valid &ndash; see solution to '''(5)'''.
+*For a digital system with bit rate&nbsp; $R_{\rm B} = 140\,{\rm Mbit/s}$ &nbsp; &#8658; &nbsp; $R_{\rm B}/2 = 70\,{\rm Mbit/s}$&nbsp; and cable length&nbsp; $l = 2\,{\rm km}$:&nbsp; <br> &nbsp; &nbsp; &nbsp; $a_* &asymp; 40\,{\rm dB}$ is indeed valid &ndash; see solution to subtask&nbsp; '''(5)'''.
-*A system with four times the bit rate ($R_{\rm B}/2 = 280\,{\rm Mbit/s}$) and half the length ($l = 1\,{\rm km}$) results in the same characteristic cable attenuation.
+*A system with four times the bit rate&nbsp; $(R_{\rm B}/2 = 280\,{\rm Mbit/s})$&nbsp; and half the length&nbsp; $(l = 1\,{\rm km})$&nbsp; results in the same characteristic cable attenuation.
-*In contrast, for a system with $R_{\rm B}/2 = 35\,{\rm Mbit/s}$ and $l = 2\,{\rm km}$ to
+*In contrast,&nbsp; for a system with&nbsp; $R_{\rm B}/2 = 35\,{\rm Mbit/s}$&nbsp; and&nbsp; $l = 2\,{\rm km}$&nbsp; holds:
 :$$a_{\rm dB} = 0.2722 \frac{\rm Np}{\rm km\cdot \sqrt{\rm MHz}}  \cdot 2 \ \rm km \cdot \sqrt{35 \ \rm MHz} \cdot 8.6859 \frac{\rm dB}{\rm Np} &asymp; 28 \ \rm dB.$$

@@ Line 3: / Line 3: @@
 [[File:P_ID1371__Dig_Z_3_1.png|right|frame|Some coaxial cable types]]
-A so-called standard coaxial cable with core diameter &nbsp;$2.6 \ \rm mm$, outer diameter &nbsp;$9.5 \ \rm mm$&nbsp; and length &nbsp;$l$&nbsp; has the following frequency response:
+A so-called&nbsp; "standard coaxial cable"
 :$$H_{\rm K}(f)   \ = \ {\rm e}^{- \alpha_0 \hspace{0.05cm} \cdot \hspace{0.05cm} l}  \cdot
    {\rm e}^{- \alpha_1  \hspace{0.05cm}\cdot \hspace{0.05cm}l \hspace{0.05cm}\cdot \hspace{0.05cm}f}  \cdot
@@ Line 12: / Line 18: @@
    \sqrt{f}}   \hspace{0.05cm}.$$
-The attenuation parameters &nbsp;$\alpha_0$, &nbsp;$\alpha_1$&nbsp; and &nbsp;$\alpha_2$&nbsp; are to be entered in Neper &nbsp;$(\rm Np)$, the phase parameters &nbsp;$\beta_1$&nbsp; and &nbsp;$\beta_2$&nbsp; in Radian &nbsp;$(\rm rad)$.&nbsp;
+The attenuation parameters &nbsp;$\alpha_0$, &nbsp;$\alpha_1$&nbsp; and &nbsp;$\alpha_2$&nbsp; are to be entered in&nbsp; "Neper" &nbsp;$(\rm Np)$,&nbsp; the phase parameters &nbsp;$\beta_1$&nbsp; and &nbsp;$\beta_2$&nbsp; in&nbsp; "radian" &nbsp;$(\rm rad)$.&nbsp;
 The following numerical values apply:
 :$$\alpha_0 = 0.00162 \hspace{0.15cm}\frac{\rm Np}{\rm km} \hspace{0.05cm},\hspace{0.2cm}
@@ Line 19: / Line 25: @@
    \alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km\cdot\sqrt{\rm MHz}} \hspace{0.05cm}.$$
-Often, to describe a &nbsp;[[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Frequenzbereich|"linear time-invariant system"]]&nbsp; (LZI) in terms of system theory, one uses
+Often,&nbsp; to describe a &nbsp;[[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Frequenzbereich|"linear time-invariant system"]]&nbsp; $\rm (LTI)$&nbsp; in terms of system theory,&nbsp; one uses
-* the attenuation function $($in &nbsp;$\rm Np$&nbsp; or &nbsp;$\rm dB)$:
+* the attenuation function&nbsp; $($in &nbsp;$\rm Np$&nbsp; or &nbsp;$\rm dB)$:
 :$$a_{\rm K}(f) = - {\rm ln} \hspace{0.10cm}|H_{\rm K}(f)|= - 20 \cdot {\rm lg} \hspace{0.10cm}|H_{\rm K}(f)|
      \hspace{0.05cm},$$
-* the phase function $($in &nbsp;$\rm rad$ or $\rm degrees)$:
+* the phase function&nbsp; $($in &nbsp;$\rm rad$&nbsp; or&nbsp; $\rm degrees)$:
 :$$b_{\rm K}(f) = - {\rm arc} \hspace{0.10cm}H_{\rm K}(f)
      \hspace{0.05cm}.$$
@@ Line 32: / Line 38: @@
    \sqrt{f}}  \cdot
    {\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2  \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
-   \sqrt{f}}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} a_{\rm K}(f) = \alpha_2 \cdot l \cdot
+   \sqrt{f}}$$
    \sqrt{f}, \hspace{0.2cm}b_{\rm K}(f) = a_{\rm K}(f) \cdot
    \frac{\rm rad}{\rm Np}\hspace{0.05cm}.$$
-This is allowed because &nbsp;$\alpha_2$&nbsp; and &nbsp;$\beta_2$&nbsp; have exactly the same numerical value &ndash; just different pseudo units. Using the definition of the &nbsp;'''characteristic cable attenuation'''&nbsp; (in Neper or decibels)
+This is allowed because &nbsp;$\alpha_2$&nbsp; and &nbsp;$\beta_2$&nbsp; have exactly the same numerical value &ndash; just different pseudo units.&nbsp;
 :$$a_{\rm * (Np)} = a_{\rm K}(f = {R_{\rm B}}/{2}) = 0.1151 \cdot a_{\rm * (dB)}$$
@@ Line 44: / Line 53: @@
 Notes:
 *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"]].
@@ Line 56: / Line 63: @@
 <quiz display=simple>
-{Which terms of &nbsp;$H_{\rm K}(f)$&nbsp; do not lead to distortions? The
+{Which terms of &nbsp;$H_{\rm K}(f)$&nbsp; do not lead to distortions?&nbsp; The
 |type="[]"}
 + $\alpha_0$&ndash;term,
@@ Line 68: / Line 75: @@
 $l_{\rm max} \ = \ $  { 6.173 3% } $\ {\rm km} $
-{What is the attenuation $($in &nbsp;$\rm Np)$&nbsp; at the frequency &nbsp;$f = 70\,{\rm MHz}$, if the cable length is &nbsp;$l = 2\,{\rm km}$?&nbsp;
+{What is the attenuation&nbsp; $($in &nbsp;$\rm Np)$&nbsp; at the frequency &nbsp;$f = 70\,{\rm MHz}$,&nbsp; if the cable length is &nbsp;$l = 2\,{\rm km}$?&nbsp;
 |type="{}"}
 $a_{\rm K}(f = 70\,{\rm MHz})\ = \ $ { 4.619 3% } $\ {\rm Np} $
-{All other things being equal, what attenuation results when only the &nbsp;$\alpha_2$&ndash;term is considered?
+{All other things being equal,&nbsp; what attenuation results when only the &nbsp;$\alpha_2$&ndash;term is considered?
 |type="{}"}
 $a_{\rm K}(f = 70\,{\rm MHz})\ = \ $ { 4.555 3% } $\ {\rm Np} $
-{What is the formula for the conversion between &nbsp;$\rm Np$&nbsp; and &nbsp;$\rm dB$? What is the &nbsp;$\rm dB$ value for the attenuation calculated in '''(4)'''?
+{What is the formula for the conversion between &nbsp;$\rm Np$&nbsp; and &nbsp;$\rm dB$?&nbsp; What is the &nbsp;$\rm dB$&nbsp; value for the attenuation calculated in subtask&nbsp; '''(4)'''?
 |type="{}"}
 $a_{\rm K}(f = 70\,{\rm MHz})\ = \ $ { 39.57 3% } $\ {\rm dB} $
@@ Line 82: / Line 89: @@
 {Which of the statements are true provided that one restricts oneself to the &nbsp;$\alpha_2$&ndash;value with respect to the attenuation function?
 |type="[]"}
-+ One can also do without the phase term &nbsp;$\beta_1$&nbsp;
++ One can also do without the phase term &nbsp;$\beta_1$.&nbsp;
-- One can also do without the phase term &nbsp;$\beta_2$&nbsp;
+- One can also do without the phase term &nbsp;$\beta_2$.&nbsp;
 - $a_* &asymp; 40\,{\rm dB}$&nbsp; is valid for a system with &nbsp;$R_{\rm B} = 70\,{\rm Mbit/s}$&nbsp; and &nbsp;$l = 2\,{\rm km}$.
 + $a_* &asymp; 40\,{\rm dB}$&nbsp; is valid for a system with &nbsp;$R_{\rm B} = 140\,{\rm Mbit/s}$&nbsp; and &nbsp;$l = 2\,{\rm km}$.

@@ Line 19: / Line 19: @@
    \alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km\cdot\sqrt{\rm MHz}} \hspace{0.05cm}.$$
-Often, to describe a &nbsp;[[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Frequenzbereich|linear time-invariant system]]&nbsp; (LZI) in terms of system theory, one uses
+Often, to describe a &nbsp;[[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Frequenzbereich|"linear time-invariant system"]]&nbsp; (LZI) in terms of system theory, one uses
 * the attenuation function $($in &nbsp;$\rm Np$&nbsp; or &nbsp;$\rm dB)$:
 :$$a_{\rm K}(f) = - {\rm ln} \hspace{0.10cm}|H_{\rm K}(f)|= - 20 \cdot {\rm lg} \hspace{0.10cm}|H_{\rm K}(f)|

@@ Line 47: / Line 47: @@
 ''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 14:49, 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 12:50, 23 March 2021
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−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^3.1 Impulsinterferenzen^]]	+	[[Category:Digital Signal Transmission: Exercises\|^3.1 Impulsinterferenzen^]]

@@ Line 94: / Line 94: @@
 *Der $\alpha_0$&ndash;Term bewirkt nur eine frequenzunabhängige Dämpfung und der $\beta_1$&ndash;Term (lineare Phase) eine frequenzunabhängige Laufzeit.
 *Alle anderen Terme tragen zu den (linearen) Verzerrungen bei.
@@ Line 105: / Line 106: @@
 :$$l_{\rm max} = \frac{a_0 }{\alpha_0 }  = \frac{0.01\,\,{\rm Np}}{0.00162\,\,{\rm Np/km}}\hspace{0.15cm}\underline {\approx 6.173\,\,{\rm km}}
    \hspace{0.05cm}.$$
 '''(3)'''&nbsp; Für den Dämpfungsverlauf gilt bei Berücksichtigung aller Terme:
-:$$a_{\rm K}(f) \ = \ [\alpha_0 + \alpha_1  \cdot f + \alpha_2  \cdot
+:$$a_{\rm K}(f) \ = \ \big[\alpha_0 + \alpha_1  \cdot f + \alpha_2  \cdot
-   \sqrt{f}\hspace{0.05cm}] \cdot l =  [0.00162 + 0.000435  \cdot 70 + 0.2722   \cdot \sqrt{70}\hspace{0.05cm}]\, \frac{\rm Np}{\rm km} \cdot 2\,{\rm km} = \hspace{0.15cm}\underline {= 4.619\, {\rm Np}}\hspace{0.05cm}.$$
+   \sqrt{f}\hspace{0.05cm}\big] \cdot l = \big [0.00162 + 0.000435  \cdot 70 + 0.2722   \cdot \sqrt{70}\hspace{0.05cm}\big]\, \frac{\rm Np}{\rm km} \cdot 2\,{\rm km} = \hspace{0.15cm}\underline {= 4.619\, {\rm Np}}\hspace{0.05cm}.$$
 '''(4)'''&nbsp; Entsprechend der Berechnung bei Punkt (3) erhält man hier den Dämpfungswert $\underline {4.555\,{\rm Np}}$.
 '''(5)'''&nbsp; Für eine jede positive Größe $x$ gilt:
@@ Line 117: / Line 121: @@
    (20 \cdot {\rm lg} \hspace{0.10cm} x) = 0.1151 \cdot x_{\rm dB}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{\rm dB} = 8.686 \cdot x_{\rm
   Np}\hspace{0.05cm}.$$
-Der Dämpfungswert $4.555\,{\rm Np}$ ist somit identisch mit <u>$39.57$</u>.
+Der Dämpfungswert $4.555\,{\rm Np}$ ist somit identisch mit $\underline{39.57\,{\rm dB} }$.
@@ Line 137: / Line 141: @@
    \frac{\rm rad}{\rm Np}\hspace{0.05cm}.$$
 *Das heißt: $a_{\rm K}(f)$ und $b_{\rm K}(f)$ eines Koaxialkabels sind formgleich und unterscheiden sich lediglich in ihren Einheiten.
-*Bei einem Digitalsystem mit Bitrate $R_{\rm B} = 140\,{\rm Mbit/s}$ &#8658; $R_{\rm B}/2 = 70\,{\rm Mbit/s}$ und Kabellänge $l = 2\,{\rm km}$ gilt tatsächlich $a_* &asymp; 40\,{\rm dB}$ (siehe Musterlösung zur letzten Teilaufgabe).
+*Bei einem Digitalsystem mit Bitrate $R_{\rm B} = 140\,{\rm Mbit/s}$ &#8658; $R_{\rm B}/2 = 70\,{\rm Mbit/s}$ und Kabellänge $l = 2\,{\rm km}$ gilt tatsächlich $a_* &asymp; 40\,{\rm dB}$ &ndash; siehe Musterlösung zu '''(5)'''.
 *Ein System mit vierfacher Bitrate ($R_{\rm B}/2 = 280\,{\rm Mbit/s}$) und halber Länge ($l = 1\,{\rm km}$) führt zur gleichen charakteristischen Kabeldämpfung.
-*Dagegen gilt für ein System mit $R_{\rm B}/2 = 35\,{\rm Mbit/s}$ und $l = 2\,{\rm km}$:
+*Dagegen gilt für ein System mit $R_{\rm B}/2 = 35\,{\rm Mbit/s}$ und $l = 2\,{\rm km}$ zu
 :$$a_{\rm dB} = 0.2722 \frac{\rm Np}{\rm km\cdot \sqrt{\rm MHz}}  \cdot 2 \ \rm km \cdot \sqrt{35 \ \rm MHz} \cdot 8.6859 \frac{\rm dB}{\rm Np} &asymp; 28 \ \rm dB.$$