Difference between revisions of "Aufgaben:Exercise 1.2: Coaxial Cable"

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{{quiz-Header|Buchseite=Lineare zeitinvariante Systeme/Systembeschreibung im Frequenzbereich}}
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{{quiz-Header|Buchseite=Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain}}
  
[[File: P_ID787__LZI_A_1_2.png | Verschiedene Koaxialkabel (Aufgabe A1.2) | right]]
+
 
Der Frequenzgang eines Normalkoaxialkabels (mit 2.6 mm Durchmesser des Innenleiters und  9.5 mm Außendurchmesser) der Länge $l$ lautet für Frequenzen $f$ > 0:  
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[[File: P_ID787__LZI_A_1_2.png | Various coaxial cable types|right|frame]]
$$H(f) = {\rm e}^{-\alpha_{0\hspace{0.02cm}} \hspace{0.05cm} \cdot \hspace{0.05cm} l}
+
The frequency response of a normal coaxial cable of length  $l$  (with a diameter of $2.6 \ \text{mm}$  of the inner conductor and  an external diameter of $9.5 \ \text{mm}$)  is for frequencies  $f > 0$:  
 +
:$$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
 
\cdot {\rm e}^{-(\alpha_1  + {\rm j}\hspace{0.05cm} \cdot
 
\hspace{0.05cm} \beta_1)\hspace{0.05cm} \cdot \hspace{0.05cm} f
 
\hspace{0.05cm} \beta_1)\hspace{0.05cm} \cdot \hspace{0.05cm} f
Line 11: 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}.$$
  
*Der erste, von den Ohmschen Verlusten herrührende Term in dieser Gleichung wird durch die so genannte kilometrische Dämpfung $α_0 = 0.00162\, \text{Np/km}$ beschrieben.  
+
*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 frequency-proportional damping component   ⇒   $α_1 · f · l$  mit  $α_1 = 0.000435 \,\text{Np/(km · MHz)}$   is due to the lateral losses. This only becomes noticeable at very high frequencies and is neglected in the following.
 +
 
 +
 
 +
*Also, the frequency-proportional phase  $β_1 · f · l$  with  $β_1 = 21.78 \,\text{rad/(km · MHz)}$  is left out of consideration because this only leads to an equal transit time for all frequencies.
 +
 
 +
 
 +
Hence, for frequencies between  $200 \ \text{kHz}$  and  $400 \ \text{MHz}$  the frequency response of the coaxial cable is predominantly determined by the influence of
 +
*the damping constant  $α_2 = 0.2722 \,\text{Np/(km · MHz}^{0.5})$, and
 +
*the phase constant  $β_2 = 0.2722 \,\text{rad/(km · MHz}^{0.5})$,
 +
 
 +
 
 +
which are due to the so-called  "skin effect".  For positive frequencies the following applies:
 +
:$$H(f) = K \cdot {\rm e}^{-(\alpha_2  + {\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \beta_2)\hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{f} \hspace{0.05cm} \cdot \hspace{0.05cm} l}.$$
 +
Because of the same numerical values of  $α_2$  and  $β_2$  this can be expressed as follows:
 +
:$$H(f) = K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot \hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f/f_0} },$$
 +
where the parameter  $f_0$  equally accounts for the two constants  $α_2$  and  $β_2$  and the cable length  $l$ .
 +
 
 +
 
 +
 
  
*Der frequenzproportionale Dämpfungsanteil  ⇒  $α_1 · f · l$ mit $α_1 = 0.000435 \,\text{Np/(km · MHz)}$ geht auf die Querverluste zurück. Dieser macht sich erst bei sehr hohen Frequenzen bemerkbar und wird im Folgenden vernachlässigt.
 
  
*Auch die frequenzproportionale Phase $β_1 · f · l$ mit $β_1 = 21.78 \,\text{rad/(km · MHz)}$ wird außer Acht gelassen werden, da diese nur eine für alle Frequenzen gleiche Laufzeit zur Folge hat.
 
  
  
Der Koaxialkabel–Frequenzgang wird deshalb für Frequenzen zwischen 200 kHz und 400 MHz im Wesentlichen durch den Einfluss der Dämpfungskonstanten $α_2 = 0.2722 \,\text{Np/(km · MHz}^{0.5})$ und der Phasenkonstanten $β_2 = 0.2722 \,\text{rad/(km · MHz}^{0.5})$ bestimmt, die auf den so genannten Skineffekt zurückzuführen sind. Für positive Frequenzen gilt:
+
''Please note:''
$$H(f) = K \cdot {\rm e}^{-(\alpha_2 + {\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \beta_2)\hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{f} \hspace{0.05cm} \cdot \hspace{0.05cm} l}.$$
+
*This exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain | System Description in Frequency Domain]].
Aufgrund der gleichen Zahlenwerte von $α_2$ und $β_2$ kann hierfür auch geschrieben werden:
+
   
$$H(f) = K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot \hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f/f_0} },$$
 
wobei der Parameter $f_0$ die beiden Konstanten $α_2$ und $β_2$ sowie die Kabellänge $l$ gleichermaßen berücksichtigt.
 
  
'''Hinweis:''' Diese Aufgabe bezieht sich auf den Theorieteil von [[Lineare_zeitinvariante_Systeme/Systembeschreibung_im_Frequenzbereich | Systembeschreibung im Frequenzbereich]].
 
  
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
  
{Wie groß ist die Frequenzgang–Konstante $K$ für die Kabellänge $l = 5\,\text{ km}$?  
+
{What is the frequency response constant&nbsp; $K$&nbsp; for a cable length of&nbsp; $l = 5\,\text{ km}$?  
 
|type="{}"}
 
|type="{}"}
$K =$ { 0.992 5%  }
+
$K \ = \ $ { 0.992 5%  }
  
  
{Welche Länge $l_{\rm max}$ könnte ein Kabel besitzen, damit ein Gleichsignal um nicht mehr als 3% gedämpft wird?  
+
{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="{}"}
 
|type="{}"}
$l_{\rm max} =$ { 18.8 5%  } km
+
$l_{\rm max} \ = \ $ { 18.8 5%  } &nbsp;$\text{ km}$
  
  
{Wie groß ist die charakteristische Frequenz $f_0$ für die Kabellänge $l = 5\,\text{ km}$. Berücksichtigen Sie die Beziehung $\rm \sqrt{2j} = 1 + j$.  
+
{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="{}"}
 
|type="{}"}
$f_0 =$ { 0.54 5%  } MHz
+
$f_0\ = \ $ { 0.54 5%  } &nbsp;$\text{ MHz}$
  
  
{Wie groß ist die Ausgangsleistung $P_y$, wenn man am Kabeleingang ein Cosinussignal der Frequenz $f_0$ mit Leistung $P_x = \,\text{1 W}$ anlegt?  
+
{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="{}"}
 
|type="{}"}
$f = f_0$: &nbsp; $P_y =$ { 135 5%  } mW
+
$P_y \ = \ $ { 135 5%  } &nbsp;$\text{ mW}$
  
  
{Welche Ausgangsleistung erhält man mit der Signalfrequenz $f_x =$ 10 MHz?
+
{What output power is obtained with the signal frequency&nbsp; $f_x = 10 \ \rm MHz$?
 
|type="{}"}
 
|type="{}"}
$f = 10\,\text{MHz}$: &nbsp; $P_y =$ { 0.184 5%  } mW
+
$P_y \ = \ $ { 0.184 5%  } &nbsp;$\text{ mW}$
  
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''1.''' Für den Gleichsignalübertragungsfaktor gilt:  
+
'''(1)'''&nbsp; For the direct signal transmission factor the following holds:  
$$K = H(f=0) = {\rm e}^{-\alpha_0 \hspace{0.05cm}\cdot
+
:$$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} l} = {\rm e}^{-0.00162 \hspace{0.05cm} \cdot
 
\hspace{0.05cm} 5}\hspace{0.15cm}\underline{ \approx 0.992}.$$
 
\hspace{0.05cm} 5}\hspace{0.15cm}\underline{ \approx 0.992}.$$
  
  
'''2.''' Mit $a_0 = α_0 · l$ müsste folgende Gleichung erfüllt sein:
+
 
$${\rm e}^{\rm -a_0 } \ge 0.97
+
'''(2)'''&nbsp; With&nbsp; ${\rm a_0 } = α_0 · l$&nbsp; the following equation should be satisfied:
 +
:$${\rm e}^{\rm -a_0 } \ge 0.97
 
\hspace{0.2cm} \Rightarrow \hspace{0.2cm} {\rm a_0 } < \ln \frac{1}{0.97
 
\hspace{0.2cm} \Rightarrow \hspace{0.2cm} {\rm a_0 } < \ln \frac{1}{0.97
 
} \approx 0.0305\,{\rm Np}.$$
 
} \approx 0.0305\,{\rm Np}.$$
Damit erhält man für die maximale Länge $l_{\rm max} =$ 0.0305 Np/0.00162 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.''' Wegen $β_2 = α_2$ und der angegebenen Beziehung $\rm 1 + j = (2j)^{0.5}$ 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:
$$H(f) = K \cdot {\rm e}^{- \sqrt{2{\rm j}\hspace{0.05cm} \cdot
+
:$$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} f \hspace{0.05cm} \cdot \hspace{0.05cm} {\alpha_2}^2
 
\hspace{0.05cm} \cdot \hspace{0.05cm} l^2} }= K \cdot {\rm e}^{-
 
\hspace{0.05cm} \cdot \hspace{0.05cm} l^2} }= K \cdot {\rm e}^{-
\sqrt{2{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f/f_0} }.$$
+
\sqrt{2\hspace{0.05cm} \cdot
Durch Koeffizientenvergleich mit der vorne angegebenen Gleichung erhält man:  
+
\hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f/f_0} }.$$
$$\frac{1}{f_0} = \alpha_2^2 \hspace{0.05cm} \cdot \hspace{0.05cm} l^2 = ( \frac { {\rm 0.272} }{\rm km \hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{MHz} })^2 \cdot ({\rm 5 \hspace{0.05cm} km})^2 =  \frac{1.855}{ {\rm MHz} }\hspace{0.2cm} \Rightarrow \hspace{0.2cm} f_0 \hspace{0.15cm}\rm \underline{= 0.540 \: MHz}$$
+
*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.''' Für den Frequenzgang gilt:  
+
'''(4)'''&nbsp; For the frequency response it holds:  
$$\begin{align*}H(f) & = K \cdot {\rm e}^{- \sqrt{2{\rm j}\hspace{0.05cm} \cdot
+
:$$\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} }
 
\hspace{0.05cm} f/f_0} } = K \cdot {\rm e}^{- \sqrt{ f/f_0} }
\cdot {\rm e}^{- {\rm j}\sqrt{ f/f_0} } \\ & \Rightarrow \hspace{0.05 cm}  |H(f)|^2 = K^2 \cdot
+
\cdot {\rm e}^{- {\rm j}\hspace{0.05cm} \cdot
{\rm e}^{- 2\sqrt{ f/f_0} }.\end{align*}$$
+
\hspace{0.05cm}\sqrt{ f/f_0} } \hspace{0.05 cm} \Rightarrow \hspace{0.05 cm}  |H(f)|^2 = K^2 \cdot
Für $f = f_0$ erhält man hierfür $\rm e^{–2}$ ≈ 0.135. Daraus folgt weiter:  
+
{\rm e}^{- 2\hspace{0.05cm} \cdot
$$P_y = P_x \cdot |H(f = f_0)|^2 \hspace{0.15cm}\underline{\approx135\hspace{0.05cm}{\rm mW}}.$$
+
\hspace{0.05cm}\sqrt{ f/f_0} }.\end{align*}$$
 +
*For this,&nbsp; $K^2 \cdot \rm e^{–2} ≈ 0.135$ is obtained for&nbsp; $f = f_0$&nbsp;. From this it follows that:  
 +
:$$P_y = P_x \cdot |H(f = f_0)|^2 \hspace{0.15cm}\underline{\approx135\hspace{0.05cm}{\rm mW}}.$$
 +
 
  
  
'''5.''' In diesem Fall gilt:  
+
'''(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;:  
$$P_y = P_x \cdot {\rm e}^{- 2\sqrt{ 10/0.54} }\approx P_x \cdot {\rm e}^{- 8.6 } \hspace{0.15cm}\underline{\approx 0.184 \hspace{0.1cm}{\rm mW}}.$$
+
:$$P_y = P_x \cdot {\rm e}^{- 2\hspace{0.05cm} \cdot
 +
\hspace{0.05cm}\sqrt{ 10/0.54} }\approx P_x \cdot {\rm e}^{- 8.6 } \hspace{0.15cm}\underline{\approx 0.184 \hspace{0.1cm}{\rm mW}}.$$
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
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[[Category:Aufgaben zu Lineare zeitinvariante Systeme|^1.1 Systembeschreibung im Frequenzbereich^]]
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[[Category:Linear and Time-Invariant Systems: Exercises|^1.1 System Description in Frequency Domain^]]

Latest revision as of 16:40, 11 July 2021


Various coaxial cable types

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  $f > 0$:

$$H(f) = {\rm e}^{-\alpha_{0\hspace{0.02cm}} \hspace{0.05cm} \cdot \hspace{0.05cm} l} \cdot {\rm e}^{-(\alpha_1 + {\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \beta_1)\hspace{0.05cm} \cdot \hspace{0.05cm} f \cdot \hspace{0.05cm} l}\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm e}^{-(\alpha_2 + {\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm}\beta_2) \hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm} l}.$$
  • 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 frequency-proportional damping component   ⇒   $α_1 · f · l$  mit  $α_1 = 0.000435 \,\text{Np/(km · MHz)}$  is due to the lateral losses. This only becomes noticeable at very high frequencies and is neglected in the following.


  • Also, the frequency-proportional phase  $β_1 · f · l$  with  $β_1 = 21.78 \,\text{rad/(km · MHz)}$  is left out of consideration because this only leads to an equal transit time for all frequencies.


Hence, for frequencies between  $200 \ \text{kHz}$  and  $400 \ \text{MHz}$  the frequency response of the coaxial cable is predominantly determined by the influence of

  • the damping constant  $α_2 = 0.2722 \,\text{Np/(km · MHz}^{0.5})$, and
  • the phase constant  $β_2 = 0.2722 \,\text{rad/(km · MHz}^{0.5})$,


which are due to the so-called  "skin effect".  For positive frequencies the following applies:

$$H(f) = K \cdot {\rm e}^{-(\alpha_2 + {\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} \beta_2)\hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{f} \hspace{0.05cm} \cdot \hspace{0.05cm} l}.$$

Because of the same numerical values of  $α_2$  and  $β_2$  this can be expressed as follows:

$$H(f) = K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot \hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f/f_0} },$$

where the parameter  $f_0$  equally accounts for the two constants  $α_2$  and  $β_2$  and the cable length  $l$ .




Please note:



Questions

1

What is the frequency response constant  $K$  for a cable length of  $l = 5\,\text{ km}$?

$K \ = \ $

2

What length  $l_{\rm max}$  could a cable have such that a direct  (DC)  signal is not damped by no more than  $3\%$ ?

$l_{\rm max} \ = \ $

 $\text{ km}$

3

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

$f_0\ = \ $

 $\text{ MHz}$

4

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  $P_y$?

$P_y \ = \ $

 $\text{ mW}$

5

What output power is obtained with the signal frequency  $f_x = 10 \ \rm MHz$?

$P_y \ = \ $

 $\text{ mW}$


Solution

(1)  For the direct signal transmission factor the following holds:

$$K = H(f=0) = {\rm e}^{-\alpha_0 \hspace{0.05cm}\cdot \hspace{0.05cm} l} = {\rm e}^{-0.00162 \hspace{0.05cm} \cdot \hspace{0.05cm} 5}\hspace{0.15cm}\underline{ \approx 0.992}.$$


(2)  With  ${\rm a_0 } = α_0 · l$  the following equation should be satisfied:

$${\rm e}^{\rm -a_0 } \ge 0.97 \hspace{0.2cm} \Rightarrow \hspace{0.2cm} {\rm a_0 } < \ln \frac{1}{0.97 } \approx 0.0305\,{\rm Np}.$$
  • Thus, the maximum length is   $l_{\rm max} = 0.0305 \ \text{Np}/0.00162 \ \text{Np/km} \rm \underline{\: ≈ \: 18.8 \: km}$.


(3)  Because of  $β_2 = α_2$  and the given relation  $\rm 1 + j = \sqrt{2j}$  the frequency response can be formulated as follows:

$$H(f) = K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot \hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f \hspace{0.05cm} \cdot \hspace{0.05cm} {\alpha_2}^2 \hspace{0.05cm} \cdot \hspace{0.05cm} l^2} }= K \cdot {\rm e}^{- \sqrt{2\hspace{0.05cm} \cdot \hspace{0.05cm}{\rm j}\hspace{0.05cm} \cdot \hspace{0.05cm} f/f_0} }.$$
  • By comparing the coefficients with the equation given above the following is obtained:
$${1}/{f_0} = \alpha_2^2 \hspace{0.05cm} \cdot \hspace{0.05cm} l^2 = \big ( \frac { {\rm 0.272} }{\rm km \hspace{0.05cm} \cdot \hspace{0.05cm} \sqrt{MHz} }\big )^2 \cdot ({\rm 5 \hspace{0.05cm} km})^2 = \frac{1.855}{ {\rm MHz} }\hspace{0.2cm} \Rightarrow \hspace{0.2cm} f_0 \hspace{0.15cm}\rm \underline{= 0.540 \: MHz}.$$


(4)  For the frequency response it holds:

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


(5)  With the higher frequency  $f_x = 10\:\text{MHz}$  the output power is significantly smaller compared to  $f_x = 0.54\:\text{MHz}$ :

$$P_y = P_x \cdot {\rm e}^{- 2\hspace{0.05cm} \cdot \hspace{0.05cm}\sqrt{ 10/0.54} }\approx P_x \cdot {\rm e}^{- 8.6 } \hspace{0.15cm}\underline{\approx 0.184 \hspace{0.1cm}{\rm mW}}.$$