Aufgaben:Exercise 1.2: Coaxial Cable - Revision history

Guenter at 15:40, 11 July 2021

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Oezer at 23:12, 28 June 2021

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← Older revision		Revision as of 15:40, 11 July 2021
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	''Please note:''		''Please note:''
−	*This ~~task~~ belongs to the chapter  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain \| System Description in Frequency Domain]].	+	*This exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain \| System Description in Frequency Domain]].

@@ Line 75: / Line 75: @@
 </quiz>
-===Sample solution===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; For the direct signal transmission factor the following holds:
@@ Line 99: / Line 99: @@
 \sqrt{2\hspace{0.05cm} \cdot
 \hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f/f_0} }.$$
-*By comparing coefficients with the equation given above the following is obtained:
+*By comparing the coefficients with the equation given above the following is obtained:
 :$${1}/{f_0} = \alpha_2^2 \hspace{0.05cm} \cdot \hspace{0.05cm} l^2 = \big ( \frac { {\rm 0.272} }{\rm km \hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{MHz} }\big )^2 \cdot ({\rm 5 \hspace{0.05cm} km})^2 =  \frac{1.855}{ {\rm MHz} }\hspace{0.2cm} \Rightarrow \hspace{0.2cm} f_0 \hspace{0.15cm}\rm \underline{= 0.540 \: MHz}.$$
-'''(4)'''&nbsp; For the frequency response it holds that:
+'''(4)'''&nbsp; For the frequency response it holds:
 :$$\begin{align*}H(f) & = K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot
 \hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot

@@ Line 3: / Line 3: @@
-[[File: P_ID787__LZI_A_1_2.png | Various coaxial cables|right|frame]]
+[[File: P_ID787__LZI_A_1_2.png | Various coaxial cable types|right|frame]]
 The frequency response of a normal coaxial cable of length&nbsp; $l$&nbsp; (with a diameter of&nbsp;$2.6 \ \text{mm}$&nbsp; of the inner conductor and&nbsp; an external diameter of $9.5 \ \text{mm}$)&nbsp; is for frequencies&nbsp; $f > 0$:
 :$$H(f) = {\rm e}^{-\alpha_{0\hspace{0.02cm}} \hspace{0.05cm} \cdot \hspace{0.05cm} l}
@@ Line 12: / Line 12: @@
 \hspace{0.05cm} \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm} l}.$$
-*The first term in this equation, arising from the ohmic losses, is described by the so-called kilometric damping&nbsp; $α_0 = 0.00162\, \text{Np/km}$&nbsp;.
+*The first term in this equation, arising from the ohmic losses, is described by the&nbsp; "kilometric damping"&nbsp; $α_0 = 0.00162\, \text{Np/km}$.
@@ Line 26: / Line 26: @@
-which are due to the so-called skin effect. For positive frequencies the following applies:
+which are due to the so-called&nbsp; "skin effect".&nbsp; For positive frequencies the following applies:
 :$$H(f) = K \cdot {\rm e}^{-(\alpha_2  + {\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \beta_2)\hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{f} \hspace{0.05cm} \cdot \hspace{0.05cm} l}.$$
 Because of the same numerical values of&nbsp; $α_2$&nbsp; and&nbsp; $β_2$&nbsp; this can be expressed as follows:
@@ Line 53: / Line 53: @@
-{What length&nbsp; $l_{\rm max}$&nbsp; could a cable have such that a direct signal is not damped by no more than&nbsp; $3\%$&nbsp;?
+{What length&nbsp; $l_{\rm max}$&nbsp; could a cable have such that a direct&nbsp; (DC)&nbsp; signal is not damped by no more than&nbsp; $3\%$&nbsp;?
 |type="{}"}
 $l_{\rm max} \ = \ $ { 18.8 5%  } &nbsp;$\text{ km}$
-{What is the characteristic frequency&nbsp; $f_0$&nbsp; for a cable length of&nbsp; $l = 5\,\text{ km}$. Consider the relation&nbsp; $\rm \sqrt{2j} = 1 + j$.
+{What is the characteristic frequency&nbsp; $f_0$&nbsp; for a cable length of&nbsp; $l = 5\,\text{ km}$.&nbsp; Consider the relation&nbsp; $\rm \sqrt{2j} = 1 + j$.
 |type="{}"}
 $f_0\ = \ $ { 0.54 5%  } &nbsp;$\text{ MHz}$
-{A cosine signal of frequency&nbsp; $f_x = f_0$&nbsp; is applied to the cable input with a power of&nbsp; $P_x = \,\text{1 W}$. What is the output power&nbsp; $P_y$?
+{A cosine signal of frequency&nbsp; $f_x = f_0$&nbsp; is applied to the cable input with a power of&nbsp; $P_x = \,\text{1 W}$.&nbsp; What is the output power&nbsp; $P_y$?
 |type="{}"}
 $P_y \ = \ $ { 135 5%  } &nbsp;$\text{ mW}$

@@ Line 112: / Line 112: @@
 {\rm e}^{- 2\hspace{0.05cm} \cdot
 \hspace{0.05cm}\sqrt{ f/f_0} }.\end{align*}$$
-*For this&nbsp; $K^2 \cdot \rm e^{–2} ≈ 0.135$ is obtained for&nbsp; $f = f_0$&nbsp;. From this it follows that:
+*For this,&nbsp; $K^2 \cdot \rm e^{–2} ≈ 0.135$ is obtained for&nbsp; $f = f_0$&nbsp;. From this it follows that:
 :$$P_y = P_x \cdot |H(f = f_0)|^2 \hspace{0.15cm}\underline{\approx135\hspace{0.05cm}{\rm mW}}.$$

@@ Line 4: / Line 4: @@
 [[File: P_ID787__LZI_A_1_2.png | Various coaxial cables|right|frame]]
-The frequency response of a normal coaxial cable of length&nbsp; $l$&nbsp; (with a diameter of&nbsp; $2.6 \ \text{mm}$&nbsp; of the inner conductor and&nbsp; an external diameter of $9.5 \ \text{mm}$)&nbsp; is for frequencies&nbsp; $f > 0$:
+The frequency response of a normal coaxial cable of length&nbsp; $l$&nbsp; (with a diameter of&nbsp;$2.6 \ \text{mm}$&nbsp; of the inner conductor and&nbsp; an external diameter of $9.5 \ \text{mm}$)&nbsp; is for frequencies&nbsp; $f > 0$:
 :$$H(f) = {\rm e}^{-\alpha_{0\hspace{0.02cm}} \hspace{0.05cm} \cdot \hspace{0.05cm} l}
 \cdot {\rm e}^{-(\alpha_1  + {\rm j}\hspace{0.05cm} \cdot

@@ Line 77: / Line 77: @@
 ===Sample solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Für den Gleichsignalübertragungsfaktor gilt:
+'''(1)'''&nbsp; For the direct signal transmission factor the following holds:
 :$$K = H(f=0) = {\rm e}^{-\alpha_0 \hspace{0.05cm}\cdot
 \hspace{0.05cm} l} = {\rm e}^{-0.00162 \hspace{0.05cm} \cdot
@@ Line 84: / Line 84: @@
-'''(2)'''&nbsp; Mit&nbsp; ${\rm a_0 } = α_0 · l$&nbsp; müsste folgende Gleichung erfüllt sein:
+'''(2)'''&nbsp; With&nbsp; ${\rm a_0 } = α_0 · l$&nbsp; the following equation should be satisfied:
 :$${\rm e}^{\rm -a_0 } \ge 0.97
 \hspace{0.2cm} \Rightarrow \hspace{0.2cm} {\rm a_0 } < \ln \frac{1}{0.97
 } \approx 0.0305\,{\rm Np}.$$
-*Damit erhält man für die maximale Länge&nbsp; $l_{\rm max} = 0.0305 \ \text{Np}/0.00162 \ \text{Np/km} \rm \underline{\: ≈ \: 18.8 \: km}$.
+*Thus, the maximum length is &nbsp; $l_{\rm max} = 0.0305 \ \text{Np}/0.00162 \ \text{Np/km} \rm \underline{\: ≈ \: 18.8 \: km}$.
-'''(3)'''&nbsp; Wegen&nbsp; $β_2 = α_2$&nbsp; und der angegebenen Beziehung&nbsp; $\rm 1 + j = \sqrt{2j}$&nbsp; kann für den Frequenzgang auch geschrieben werden:
+'''(3)'''&nbsp; Because of&nbsp; $β_2 = α_2$&nbsp; and  the given relation&nbsp; $\rm 1 + j = \sqrt{2j}$&nbsp; the frequency response can be formulated as follows:
 :$$H(f) = K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot
 \hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot
@@ Line 99: / Line 99: @@
 \sqrt{2\hspace{0.05cm} \cdot
 \hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f/f_0} }.$$
-*Durch Koeffizientenvergleich mit der vorne angegebenen Gleichung erhält man:
+*By comparing coefficients with the equation given above the following is obtained:
 :$${1}/{f_0} = \alpha_2^2 \hspace{0.05cm} \cdot \hspace{0.05cm} l^2 = \big ( \frac { {\rm 0.272} }{\rm km \hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{MHz} }\big )^2 \cdot ({\rm 5 \hspace{0.05cm} km})^2 =  \frac{1.855}{ {\rm MHz} }\hspace{0.2cm} \Rightarrow \hspace{0.2cm} f_0 \hspace{0.15cm}\rm \underline{= 0.540 \: MHz}.$$
-'''(4)'''&nbsp; Für den Frequenzgang gilt:
+'''(4)'''&nbsp; For the frequency response it holds that:
 :$$\begin{align*}H(f) & = K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot
 \hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot
@@ Line 112: / Line 112: @@
 {\rm e}^{- 2\hspace{0.05cm} \cdot
 \hspace{0.05cm}\sqrt{ f/f_0} }.\end{align*}$$
-*Für&nbsp; $f = f_0$&nbsp; erhält man hierfür&nbsp; $K^2 \cdot \rm e^{–2} ≈ 0.135$. Daraus folgt weiter:
+*For this&nbsp; $K^2 \cdot \rm e^{–2} ≈ 0.135$ is obtained for&nbsp; $f = f_0$&nbsp;. From this it follows that:
 :$$P_y = P_x \cdot |H(f = f_0)|^2 \hspace{0.15cm}\underline{\approx135\hspace{0.05cm}{\rm mW}}.$$
-'''(5)'''&nbsp; Mit der höheren Frequenz&nbsp; $f_x = 10\:\text{MHz}$&nbsp; ist die Ausgangsleistung gegenüber&nbsp; $f_x = 0.54\:\text{MHz}$&nbsp; signifikant kleiner:
+'''(5)'''&nbsp; With the higher frequency&nbsp; $f_x = 10\:\text{MHz}$&nbsp; the output power is significantly smaller compared to&nbsp; $f_x = 0.54\:\text{MHz}$&nbsp;:
 :$$P_y = P_x \cdot {\rm e}^{- 2\hspace{0.05cm} \cdot
 \hspace{0.05cm}\sqrt{ 10/0.54} }\approx P_x \cdot {\rm e}^{- 8.6 } \hspace{0.15cm}\underline{\approx 0.184 \hspace{0.1cm}{\rm mW}}.$$

@@ Line 21: / Line 21: @@
-Der Koaxialkabel–Frequenzgang wird deshalb für Frequenzen zwischen&nbsp; $200 \ \text{kHz}$&nbsp; und&nbsp; $400 \ \text{MHz}$&nbsp; vorwiegend durch den Einfluss
+Hence, for frequencies between&nbsp; $200 \ \text{kHz}$&nbsp; and&nbsp; $400 \ \text{MHz}$&nbsp; the frequency response of the coaxial cable is predominantly determined by the influence of
-*der Dämpfungskonstanten&nbsp; $α_2 = 0.2722 \,\text{Np/(km · MHz}^{0.5})$, und
+*the damping constant&nbsp; $α_2 = 0.2722 \,\text{Np/(km · MHz}^{0.5})$, and
-*der Phasenkonstanten&nbsp; $β_2 = 0.2722 \,\text{rad/(km · MHz}^{0.5})$
+*the phase constant&nbsp; $β_2 = 0.2722 \,\text{rad/(km · MHz}^{0.5})$,
-bestimmt, die auf den so genannten Skineffekt zurückzuführen sind. Für positive Frequenzen gilt:
+which are due to the so-called skin effect. For positive frequencies the following applies:
 :$$H(f) = K \cdot {\rm e}^{-(\alpha_2  + {\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \beta_2)\hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{f} \hspace{0.05cm} \cdot \hspace{0.05cm} l}.$$
-Aufgrund der gleichen Zahlenwerte von&nbsp; $α_2$&nbsp; und&nbsp; $β_2$&nbsp; kann hierfür auch geschrieben werden:
+Because of the same numerical values of&nbsp; $α_2$&nbsp; and&nbsp; $β_2$&nbsp; this can be expressed as follows:
 :$$H(f) = K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot \hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f/f_0} },$$
-wobei der Parameter&nbsp; $f_0$&nbsp; die beiden Konstanten&nbsp; $α_2$&nbsp; und&nbsp; $β_2$&nbsp; sowie die Kabellänge&nbsp; $l$&nbsp; gleichermaßen berücksichtigt.
+where the parameter&nbsp; $f_0$&nbsp; equally accounts for the two constants&nbsp; $α_2$&nbsp; and&nbsp; $β_2$&nbsp; and the cable length&nbsp; $l$&nbsp;.
@@ Line 48: / Line 48: @@
 <quiz display=simple>
-{Wie groß ist die Frequenzgang–Konstante&nbsp; $K$&nbsp; für die Kabellänge&nbsp; $l = 5\,\text{ km}$?
+{What is the frequency response constant&nbsp; $K$&nbsp; for a cable length of&nbsp; $l = 5\,\text{ km}$?
 |type="{}"}
 $K  \ = \ $ { 0.992 5%  }
-{Welche Länge&nbsp; $l_{\rm max}$&nbsp; könnte ein Kabel besitzen, damit ein Gleichsignal um nicht mehr als&nbsp; $3\%$&nbsp; gedämpft wird?
+{What length&nbsp; $l_{\rm max}$&nbsp; could a cable have such that a direct signal is not damped by no more than&nbsp; $3\%$&nbsp;?
 |type="{}"}
 $l_{\rm max} \ = \ $ { 18.8 5%  } &nbsp;$\text{ km}$
-{Wie groß ist die charakteristische Frequenz&nbsp; $f_0$&nbsp; für die Kabellänge&nbsp; $l = 5\,\text{ km}$. Berücksichtigen Sie die Beziehung&nbsp; $\rm \sqrt{2j} = 1 + j$.
+{What is the characteristic frequency&nbsp; $f_0$&nbsp; for a cable length of&nbsp; $l = 5\,\text{ km}$. Consider the relation&nbsp; $\rm \sqrt{2j} = 1 + j$.
 |type="{}"}
 $f_0\ = \ $ { 0.54 5%  } &nbsp;$\text{ MHz}$
-{Am Kabeleingang liegt ein Cosinussignal der Frequenz&nbsp; $f_x = f_0$&nbsp; mit Leistung&nbsp; $P_x = \,\text{1 W}$ an. Wie groß ist die Ausgangsleistung&nbsp; $P_y$?
+{A cosine signal of frequency&nbsp; $f_x = f_0$&nbsp; is applied to the cable input with a power of&nbsp; $P_x = \,\text{1 W}$. What is the output power&nbsp; $P_y$?
 |type="{}"}
 $P_y \ = \ $ { 135 5%  } &nbsp;$\text{ mW}$
-{Welche Ausgangsleistung erhält man mit der Signalfrequenz&nbsp; $f_x = 10 \ \rm MHz$?
+{What output power is obtained with the signal frequency&nbsp; $f_x = 10 \ \rm MHz$?
 |type="{}"}
 $P_y \ = \ $ { 0.184 5%  } &nbsp;$\text{ mW}$

@@ Line 12: / Line 12: @@
 \hspace{0.05cm} \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm} l}.$$
-*Der erste, von den Ohmschen Verlusten herrührende Term in dieser Gleichung wird durch die so genannte kilometrische Dämpfung&nbsp; $α_0 = 0.00162\, \text{Np/km}$&nbsp; beschrieben.
+*The first term in this equation, arising from the ohmic losses, is described by the so-called kilometric damping&nbsp; $α_0 = 0.00162\, \text{Np/km}$&nbsp;.
-*Der frequenzproportionale Dämpfungsanteil &nbsp; ⇒ &nbsp; $α_1 · f · l$&nbsp; mit&nbsp; $α_1 = 0.000435 \,\text{Np/(km · MHz)}$&nbsp; geht auf die Querverluste zurück. Dieser macht sich erst bei sehr hohen Frequenzen bemerkbar und wird im Folgenden vernachlässigt.
+*The frequency-proportional damping component &nbsp; ⇒ &nbsp; $α_1 · f · l$&nbsp; mit&nbsp; $α_1 = 0.000435 \,\text{Np/(km · MHz)}$&nbsp;  is due to the lateral losses. This only becomes noticeable at very high frequencies and is neglected in the following.
-*Auch die frequenzproportionale Phase&nbsp; $β_1 · f · l$&nbsp; mit&nbsp; $β_1 = 21.78 \,\text{rad/(km · MHz)}$&nbsp; wird außer Acht gelassen werden, da diese nur eine für alle Frequenzen gleiche Laufzeit zur Folge hat.
+*Also, the frequency-proportional phase&nbsp; $β_1 · f · l$&nbsp; with&nbsp; $β_1 = 21.78 \,\text{rad/(km · MHz)}$&nbsp; is left out of consideration because this only leads to an equal transit time for all frequencies.