Aufgaben:Exercise 1.2: Coaxial Cable - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:55:37Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:55, 16 March 2026
Line 5:		Line 5:
	[[File: P_ID787__LZI_A_1_2.png \| Various coaxial cable types\|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  $l$  (with a diameter of $2.6 \ \text{mm}$  of the inner conductor and  an external diameter of $9.5 \ \text{mm}$)  is for frequencies  $f > 0$:		The frequency response of a normal coaxial cable of length  $l$  (with a diameter of $2.6 \ \text{mm}$  of the inner conductor and  an external diameter of $9.5 \ \text{mm}$)  is for frequencies  $f > 0$:
	:$$H(f) = {\rm e}^{-\alpha_{0\hspace{0.02cm}} \hspace{0.05cm} \cdot \hspace{0.05cm} l}\cdot {\rm e}^{-(\alpha_1 + {\rm j}\hspace{0.05cm} \cdot\hspace{0.05cm} \beta_1)\hspace{0.05cm} \cdot \hspace{0.05cm} f\cdot \hspace{0.05cm} l}\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm e}^{-(\alpha_2 + {\~~rmj~~}\hspace{0.05cm} \cdot \hspace{0.05cm}\beta_2) \hspace{0.05cm} \cdot\hspace{0.05cm} \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm} l}.$$		:$$H(f) = {\rm e}^{-\alpha_{0\hspace{0.02cm}} \hspace{0.05cm} \cdot \hspace{0.05cm} l}\cdot {\rm e}^{-(\alpha_1 + {\rm j}\hspace{0.05cm} \cdot\hspace{0.05cm} \beta_1)\hspace{0.05cm} \cdot \hspace{0.05cm} f\cdot \hspace{0.05cm} l}\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm e}^{-(\alpha_2 + {\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}\beta_2) \hspace{0.05cm} \cdot\hspace{0.05cm} \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm} l}.$$

	*The first term in this equation, arising from the ohmic losses, is described by the  "kilometric damping"  $α_0 = 0.00162\, \text{Np/km}$.		*The first term in this equation, arising from the ohmic losses, is described by the  "kilometric damping"  $α_0 = 0.00162\, \text{Np/km}$.
Line 105:		Line 105:

	[[Category:Linear and Time-Invariant Systems: Exercises\|^1.1 System Description in Frequency Domain^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^1.1 System Description in Frequency Domain^]]
	[[de:Aufgaben:1.~~2_Koaxialkabel~~]]		[[de:Aufgaben:Aufgabe 1.2: Koaxialkabel]]

Maintenance script: Join 1 split formula blocks

2026-03-16T14:15:56Z

Join 1 split formula blocks

← Older revision		Revision as of 16:15, 16 March 2026
Line 91:		Line 91:

	'''(4)'''  For the frequency response it holds:		'''(4)'''  For the frequency response it holds:
	:$$\begin{align*}H(f) & = K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot		:$$\begin{align}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} } = K \cdot {\rm e}^{- \sqrt{ f/f_0} }\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} \cdot\hspace{0.05cm}\sqrt{ f/f_0} } \hspace{0.05 cm} \Rightarrow \hspace{0.05 cm} \|H(f)\|^2 = K^2 \cdot{\rm e}^{- 2\hspace{0.05cm} \cdot\hspace{0.05cm}\sqrt{ f/f_0} }.\end{align}$$
	\hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot
	\hspace{0.05cm} f/f_0} } = K \cdot {\rm e}^{- \sqrt{ f/f_0} }
	\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} \cdot
	\hspace{0.05cm}\sqrt{ f/f_0} } \hspace{0.05 cm} \Rightarrow \hspace{0.05 cm} \|H(f)\|^2 = K^2 \cdot
	{\rm e}^{- 2\hspace{0.05cm} \cdot
	\hspace{0.05cm}\sqrt{ f/f_0} }.\end{align*}$$
	*For this,  $K^2 \cdot \rm e^{–2} ≈ 0.135$ is obtained for  $f = f_0$ . From this it follows that:		*For this,  $K^2 \cdot \rm e^{–2} ≈ 0.135$ is obtained for  $f = f_0$ . From this it follows that:
	:$$P_y = P_x \cdot \|H(f = f_0)\|^2 \hspace{0.15cm}\underline{\approx135\hspace{0.05cm}{\rm mW}}.$$		:$$P_y = P_x \cdot \|H(f = f_0)\|^2 \hspace{0.15cm}\underline{\approx135\hspace{0.05cm}{\rm mW}}.$$

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:47:26Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:47, 16 March 2026
Line 5:		Line 5:
	[[File: P_ID787__LZI_A_1_2.png \| Various coaxial cable types\|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  $l$  (with a diameter of $2.6 \ \text{mm}$  of the inner conductor and  an external diameter of $9.5 \ \text{mm}$)  is for frequencies  $f > 0$:		The frequency response of a normal coaxial cable of length  $l$  (with a diameter of $2.6 \ \text{mm}$  of the inner conductor and  an external diameter of $9.5 \ \text{mm}$)  is for frequencies  $f > 0$:
	:$$H(f) = {\rm e}^{-\alpha_{0\hspace{0.02cm}} \hspace{0.05cm} \cdot \hspace{0.05cm} l}		:$$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\hspace{0.05cm} \beta_1)\hspace{0.05cm} \cdot \hspace{0.05cm} f\cdot \hspace{0.05cm} l}\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm e}^{-(\alpha_2 + {\rmj}\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}.$$
	\cdot {\rm e}^{-(\alpha_1 + {\rm j}\hspace{0.05cm} \cdot
	\hspace{0.05cm} \beta_1)\hspace{0.05cm} \cdot \hspace{0.05cm} f
	\cdot \hspace{0.05cm} l}\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm e}^{-(\alpha_2 + {\rm
	j}\hspace{0.05cm} \cdot \hspace{0.05cm}\beta_2) \hspace{0.05cm} \cdot
	\hspace{0.05cm} \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm} l}.$$

	*The first term in this equation, arising from the ohmic losses, is described by the  "kilometric damping"  $α_0 = 0.00162\, \text{Np/km}$.		*The first term in this equation, arising from the ohmic losses, is described by the  "kilometric damping"  $α_0 = 0.00162\, \text{Np/km}$.
Line 78:		Line 73:
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  For the direct signal transmission factor the following holds:		'''(1)'''  For the direct signal transmission factor the following holds:
	:$$K = H(f=0) = {\rm e}^{-\alpha_0 \hspace{0.05cm}\cdot		:$$K = H(f=0) = {\rm e}^{-\alpha_0 \hspace{0.05cm}\cdot\hspace{0.05cm} l} = {\rm e}^{-0.00162 \hspace{0.05cm} \cdot\hspace{0.05cm} 5}\hspace{0.15cm}\underline{ \approx 0.992}.$$
	\hspace{0.05cm} l} = {\rm e}^{-0.00162 \hspace{0.05cm} \cdot
	\hspace{0.05cm} 5}\hspace{0.15cm}\underline{ \approx 0.992}.$$



	'''(2)'''  With  ${\rm a_0 } = α_0 · l$  the following equation should be satisfied:		'''(2)'''  With  ${\rm a_0 } = α_0 · l$  the following equation should be satisfied:
	:$${\rm e}^{\rm -a_0 } \ge 0.97		:$${\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}.$$
	\hspace{0.2cm} \Rightarrow \hspace{0.2cm} {\rm a_0 } < \ln \frac{1}{0.97
	} \approx 0.0305\,{\rm Np}.$$
	*Thus, the maximum length is   $l_{\rm max} = 0.0305 \ \text{Np}/0.00162 \ \text{Np/km} \rm \underline{\: ≈ \: 18.8 \: km}$.		*Thus, the maximum length is   $l_{\rm max} = 0.0305 \ \text{Np}/0.00162 \ \text{Np/km} \rm \underline{\: ≈ \: 18.8 \: km}$.

Line 93:		Line 84:

	'''(3)'''  Because of  $β_2 = α_2$  and the given relation  $\rm 1 + j = \sqrt{2j}$  the frequency response can be formulated as follows:		'''(3)'''  Because of  $β_2 = α_2$  and the given relation  $\rm 1 + j = \sqrt{2j}$  the frequency response can be formulated as follows:
	:$$H(f) = K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot		:$$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 \hspace{0.05cm} \cdot \hspace{0.05cm} {\alpha_2}^2\hspace{0.05cm} \cdot \hspace{0.05cm} l^2} }= 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} }.$$
	\hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot
	\hspace{0.05cm} f \hspace{0.05cm} \cdot \hspace{0.05cm} {\alpha_2}^2
	\hspace{0.05cm} \cdot \hspace{0.05cm} l^2} }= K \cdot {\rm e}^{-
	\sqrt{2\hspace{0.05cm} \cdot
	\hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f/f_0} }.$$
	*By comparing the coefficients with the equation given above the following is obtained:		*By comparing the coefficients with the equation given above the following is obtained:
	:$${1}/{f_0} = \alpha_2^2 \hspace{0.05cm} \cdot \hspace{0.05cm} l^2 = \big ( \frac { {\rm 0.272} }{\rm km \hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{MHz} }\big )^2 \cdot ({\rm 5 \hspace{0.05cm} km})^2 = \frac{1.855}{ {\rm MHz} }\hspace{0.2cm} \Rightarrow \hspace{0.2cm} f_0 \hspace{0.15cm}\rm \underline{= 0.540 \: MHz}.$$		:$${1}/{f_0} = \alpha_2^2 \hspace{0.05cm} \cdot \hspace{0.05cm} l^2 = \big ( \frac { {\rm 0.272} }{\rm km \hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{MHz} }\big )^2 \cdot ({\rm 5 \hspace{0.05cm} km})^2 = \frac{1.855}{ {\rm MHz} }\hspace{0.2cm} \Rightarrow \hspace{0.2cm} f_0 \hspace{0.15cm}\rm \underline{= 0.540 \: MHz}.$$
Line 118:		Line 104:

	'''(5)'''  With the higher frequency  $f_x = 10\:\text{MHz}$  the output power is significantly smaller compared to  $f_x = 0.54\:\text{MHz}$ :		'''(5)'''  With the higher frequency  $f_x = 10\:\text{MHz}$  the output power is significantly smaller compared to  $f_x = 0.54\:\text{MHz}$ :
	:$$P_y = P_x \cdot {\rm e}^{- 2\hspace{0.05cm} \cdot		:$$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}}.$$
	\hspace{0.05cm}\sqrt{ 10/0.54} }\approx P_x \cdot {\rm e}^{- 8.6 } \hspace{0.15cm}\underline{\approx 0.184 \hspace{0.1cm}{\rm mW}}.$$

	{{ML-Fuß}}		{{ML-Fuß}}

Maintenance script: Add German interlanguage link

2026-03-16T12:30:16Z

Add German interlanguage link

← Older revision		Revision as of 14:30, 16 March 2026
Line 126:		Line 126:

	[[Category:Linear and Time-Invariant Systems: Exercises\|^1.1 System Description in Frequency Domain^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^1.1 System Description in Frequency Domain^]]
			[[de:Aufgaben:1.2_Koaxialkabel]]

Guenter at 15:40, 11 July 2021

2021-07-11T15:40:45Z

← Older revision		Revision as of 17:40, 11 July 2021
Line 39:		Line 39:

	''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]].

Guenter at 15:40, 9 July 2021

2021-07-09T15:40:10Z

← Older revision		Revision as of 17:40, 9 July 2021
Line 75:		Line 75:
	</quiz>		</quiz>

	===~~Sample solution~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  For the direct signal transmission factor the following holds:		'''(1)'''  For the direct signal transmission factor the following holds:
Line 99:		Line 99:
	\sqrt{2\hspace{0.05cm} \cdot		\sqrt{2\hspace{0.05cm} \cdot
	\hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f/f_0} }.$$		\hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f/f_0} }.$$
	*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:
	:$${1}/{f_0} = \alpha_2^2 \hspace{0.05cm} \cdot \hspace{0.05cm} l^2 = \big ( \frac { {\rm 0.272} }{\rm km \hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{MHz} }\big )^2 \cdot ({\rm 5 \hspace{0.05cm} km})^2 = \frac{1.855}{ {\rm MHz} }\hspace{0.2cm} \Rightarrow \hspace{0.2cm} f_0 \hspace{0.15cm}\rm \underline{= 0.540 \: MHz}.$$		:$${1}/{f_0} = \alpha_2^2 \hspace{0.05cm} \cdot \hspace{0.05cm} l^2 = \big ( \frac { {\rm 0.272} }{\rm km \hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{MHz} }\big )^2 \cdot ({\rm 5 \hspace{0.05cm} km})^2 = \frac{1.855}{ {\rm MHz} }\hspace{0.2cm} \Rightarrow \hspace{0.2cm} f_0 \hspace{0.15cm}\rm \underline{= 0.540 \: MHz}.$$



	'''(4)'''  For the frequency response it holds ~~that~~:		'''(4)'''  For the frequency response it holds:
	:$$\begin{align*}H(f) & = K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot		:$$\begin{align*}H(f) & = K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot
	\hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot		\hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot

Guenter at 15:36, 9 July 2021

2021-07-09T15:36:38Z

Oezer at 23:12, 28 June 2021

2021-06-28T23:12:59Z

← Older revision		Revision as of 01:12, 29 June 2021
Line 112:		Line 112:
	{\rm e}^{- 2\hspace{0.05cm} \cdot		{\rm e}^{- 2\hspace{0.05cm} \cdot
	\hspace{0.05cm}\sqrt{ f/f_0} }.\end{align*}$$		\hspace{0.05cm}\sqrt{ f/f_0} }.\end{align*}$$
	*For this  $K^2 \cdot \rm e^{–2} ≈ 0.135$ is obtained for  $f = f_0$ . From this it follows that:		*For this,  $K^2 \cdot \rm e^{–2} ≈ 0.135$ is obtained for  $f = f_0$ . From this it follows that:
	:$$P_y = P_x \cdot \|H(f = f_0)\|^2 \hspace{0.15cm}\underline{\approx135\hspace{0.05cm}{\rm mW}}.$$		:$$P_y = P_x \cdot \|H(f = f_0)\|^2 \hspace{0.15cm}\underline{\approx135\hspace{0.05cm}{\rm mW}}.$$

Oezer at 23:07, 28 June 2021

2021-06-28T23:07:18Z

← Older revision		Revision as of 01:07, 29 June 2021
Line 4:		Line 4:

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

Oezer at 23:06, 28 June 2021

2021-06-28T23:06:58Z

← Older revision		Revision as of 01:06, 29 June 2021
Line 4:		Line 4:

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

← Older revision		Revision as of 17:36, 9 July 2021
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  $l$  (with a diameter of $2.6 \ \text{mm}$  of the inner conductor and  an external diameter of $9.5 \ \text{mm}$)  is for frequencies  $f > 0$:		The frequency response of a normal coaxial cable of length  $l$  (with a diameter of $2.6 \ \text{mm}$  of the inner conductor and  an external diameter of $9.5 \ \text{mm}$)  is for frequencies  $f > 0$:
	:$$H(f) = {\rm e}^{-\alpha_{0\hspace{0.02cm}} \hspace{0.05cm} \cdot \hspace{0.05cm} l}		:$$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}.$$		\hspace{0.05cm} \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm} l}.$$

	*The first term in this equation, arising from the ohmic losses, is described by the ~~so-called~~ kilometric damping  $α_0 = 0.00162\, \text{Np/km}$~~ ~~.		*The first term in this equation, arising from the ohmic losses, is described by the  "kilometric damping"  $α_0 = 0.00162\, \text{Np/km}$.


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  "skin effect".  For positive frequencies the following applies:
	:$$H(f) = K \cdot {\rm e}^{-(\alpha_2 + {\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \beta_2)\hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{f} \hspace{0.05cm} \cdot \hspace{0.05cm} l}.$$		:$$H(f) = K \cdot {\rm e}^{-(\alpha_2 + {\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \beta_2)\hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{f} \hspace{0.05cm} \cdot \hspace{0.05cm} l}.$$
	Because of the same numerical values of  $α_2$  and  $β_2$  this can be expressed as follows:		Because of the same numerical values of  $α_2$  and  $β_2$  this can be expressed as follows:
Line 53:		Line 53:


	{What length  $l_{\rm max}$  could a cable have such that a direct signal is not damped by no more than  $3\%$ ?		{What length  $l_{\rm max}$  could a cable have such that a direct  (DC)  signal is not damped by no more than  $3\%$ ?
	\|type="{}"}		\|type="{}"}
	$l_{\rm max} \ = \ $ { 18.8 5% }  $\text{ km}$		$l_{\rm max} \ = \ $ { 18.8 5% }  $\text{ km}$


	{What is the characteristic frequency  $f_0$  for a cable length of  $l = 5\,\text{ km}$. Consider the relation  $\rm \sqrt{2j} = 1 + j$.		{What is the characteristic frequency  $f_0$  for a cable length of  $l = 5\,\text{ km}$.  Consider the relation  $\rm \sqrt{2j} = 1 + j$.
	\|type="{}"}		\|type="{}"}
	$f_0\ = \ $ { 0.54 5% }  $\text{ MHz}$		$f_0\ = \ $ { 0.54 5% }  $\text{ MHz}$


	{A cosine signal of frequency  $f_x = f_0$  is applied to the cable input with a power of  $P_x = \,\text{1 W}$. What is the output power  $P_y$?		{A cosine signal of frequency  $f_x = f_0$  is applied to the cable input with a power of  $P_x = \,\text{1 W}$.  What is the output power  $P_y$?
	\|type="{}"}		\|type="{}"}
	$P_y \ = \ $ { 135 5% }  $\text{ mW}$		$P_y \ = \ $ { 135 5% }  $\text{ mW}$