Aufgaben:Exercise 1.4Z: On the Doppler Effect - Revision history

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 The equations given in the theoretical section for the reception frequency are
-* taking into account the theory of relativity (briefly referred to as "relativistic&rdquo;):
+* taking into account the theory of relativity (briefly referred to as "relativistic"):
 :$${\rm equation \hspace{0.15cm}(1):}\hspace{0.2cm}f_{\rm E} = f_{\rm S} \cdot \frac{\sqrt{1 - (v/c)^2}}{1 - v/c \cdot \cos(\alpha)}  \hspace{0.05cm},$$
-* without consideration of relativistic properties (referred to as "conventional&rdquo;):
+* without consideration of relativistic properties (referred to as "conventional"):
 :$${\rm equation \hspace{0.15cm}(2):}\hspace{0.2cm}f_{\rm E} = f_{\rm S} \cdot \big [ 1 + {v}/{c} \cdot \cos(\alpha) \big ] \hspace{0.05cm}.$$

@@ Line 14: / Line 14: @@
 The equations given in the theoretical section for the reception frequency are
-* taking into account the theory of relativity (briefly referred to as &bdquo;relativistic&rdquo;):
+* taking into account the theory of relativity (briefly referred to as "relativistic&rdquo;):
 :$${\rm equation \hspace{0.15cm}(1):}\hspace{0.2cm}f_{\rm E} = f_{\rm S} \cdot \frac{\sqrt{1 - (v/c)^2}}{1 - v/c \cdot \cos(\alpha)}  \hspace{0.05cm},$$
-* without consideration of relativistic properties (referred to as &bdquo;conventional&rdquo;):
+* without consideration of relativistic properties (referred to as "conventional&rdquo;):
 :$${\rm equation \hspace{0.15cm}(2):}\hspace{0.2cm}f_{\rm E} = f_{\rm S} \cdot \big [ 1 + {v}/{c} \cdot \cos(\alpha) \big ] \hspace{0.05cm}.$$

← Older revision		Revision as of 12:38, 23 March 2021
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−	[[Category:~~Exercises for~~ Mobile Communications\|^1.3 Rayleigh Fading with Memory^]]	+	[[Category:Mobile Communications: Exercises\|^1.3 Rayleigh Fading with Memory^]]

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 </quiz>
-===Solutions===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; With the driving direction&nbsp; $\rm (A)$, the receiver approaches the transmitter at an angle&nbsp; $\alpha = 0$.&nbsp; This gives '''(1)''' according to the relativistic equation:

← Older revision		Revision as of 13:57, 15 December 2020
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−	{{quiz-Header\|Buchseite=~~Mobile Kommunikation~~/~~Statistische Bindungen innerhalb des Rayleigh-Prozesses~~}}	+	{{quiz-Header\|Buchseite=Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process}}

	[[File:P_ID2118__Mob_Z_1_4.png\|right\|frame\|Directions of movement  $\rm (A)$, ...]]		[[File:P_ID2118__Mob_Z_1_4.png\|right\|frame\|Directions of movement  $\rm (A)$, ...]]

@@ Line 59: / Line 59: @@
 ===Solutions===
 {{ML-Kopf}}
-'''(1)'''&nbsp; With the direction of travel (A), the receiver approaches the transmitter at an angle&nbsp; $\alpha = 0$.&nbsp; This gives (1) according to the relativistic equation:
+'''(1)'''&nbsp; With the driving direction&nbsp; $\rm (A)$, the receiver approaches the transmitter at an angle&nbsp; $\alpha = 0$.&nbsp; This gives '''(1)''' according to the relativistic equation:
 :$$f_{\rm E} = f_{\rm S} \cdot \frac{\sqrt{1 - (v/c)^2}}{1 - v/c }
   \hspace{0.3cm} \Rightarrow \hspace{0.3cm} f_{\rm D} = f_{\rm E} - f_{\rm S}  = f_{\rm S} \cdot \left [  \frac{\sqrt{1 - (v/c)^2}}{1 - v/c } - 1 \right ]\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}{f_{\rm D}}/{f_{\rm S}} =   \frac{\sqrt{1 - (v/c)^2}}{1 - v/c } - 1 \hspace{0.05cm}.$$
-*With $\upsilon_1/c = 0.6$ you get
+*With&nbsp; $v_1/c = 0.6$&nbsp; you get
 :$${f_{\rm D}}/{f_{\rm S}} =   \frac{\sqrt{1 - 0.6^2}}{1 - 0.6 } - 1 = \frac{0.8}{0.4 } - 1 \hspace{0.15cm} \underline{ = 1}
 \hspace{0.3cm}\Rightarrow\hspace{0.3cm} {f_{\rm E}}/{f_{\rm S}} = 2
   \hspace{0.05cm}.$$
-*Correspondingly with $\upsilon_2/c = 10^{\rm &ndash;5}$:
+*Correspondingly with&nbsp; $v_2/c = 10^{\rm -5}$:
 :$${f_{\rm D}}/{f_{\rm S}} =   \frac{\sqrt{1 - (10^{-5})^2}}{1 - (10^{-5}) } - 1  \approx 1 + 10^{-5} - 1 \hspace{0.15cm} \underline{ = 10^{-5}}
 \hspace{0.3cm}\Rightarrow\hspace{0.3cm} {f_{\rm E}}/{f_{\rm S}} = 1.00001
@@ Line 75: / Line 75: @@
-'''(2)'''&nbsp; Now the receiver moves away from the transmitter ($\alpha = 180^°$).
+'''(2)'''&nbsp; Now the receiver moves away from the transmitter&nbsp; $(\alpha = 180^°)$.
-*The receive frequency $f_{\rm E}$ is lower than the transmit frequency $f_{\rm S}$ and the Doppler frequency $f_{\rm D}$ is negative. With ${\rm cos}(\alpha) = \ &ndash;1$ you now get
+*The reception frequency&nbsp; $f_{\rm E}$&nbsp; is lower than the transmission frequency&nbsp; $f_{\rm S}$&nbsp; and the Doppler frequency&nbsp; $f_{\rm D}$&nbsp; is negative.&nbsp; With&nbsp; ${\rm cos}(\alpha) = \ -1$&nbsp; you now get
 :$${f_{\rm D}}/{f_{\rm S}} =   \frac{\sqrt{1 - (v/c)^2}}{1 + v/c } - 1 =
 \left\{ \begin{array}{c}  \hspace{0.15cm} \underline{  -0.5} \\ \\
@@ Line 84: / Line 84: @@
 \hspace{0.05cm}.$$
-*Converting to $f_{\rm E}/f_{\rm S}$ results in:
+*Converting to&nbsp; $f_{\rm E}/f_{\rm S}$&nbsp; results in:
 :$${f_{\rm E}}/{f_{\rm S}} =
 \left\{ \begin{array}{c}  \hspace{0.15cm} {  0.5} \\ \\
@@ Line 99: / Line 99: @@
 This results in the following numerical values:
-* Direction (A), $v_1  = 1.8 \cdot 10^8 \ {\rm m/s}\text{:}\hspace{0.4cm} f_{\rm D}/f_{\rm S} \ \underline {= \ 0.6} \ \ \ &#8658; \ \ \ f_{\rm E}/f_{\rm S} = 1.6,$
+* Direction&nbsp; $\rm (A)$,&nbsp; $v_1  = 1.8 \cdot 10^8 \ {\rm m/s}\text{:}\hspace{0.4cm} f_{\rm D}/f_{\rm S} \ \underline {= \ 0.6} \ \ \ &#8658; \ \ \ f_{\rm E}/f_{\rm S} = 1.6,$
-* Direction (A), $v_2  = 3.0 \cdot 10^3 \ {\rm m/s}\text{:}\hspace{0.4cm} f_{\rm D}/f_{\rm S} \ \underline {= \ 10^{\rm &ndash;5}} \ \ \ &#8658; \ \ \ f_{\rm E}/f_{\rm S} = 1.00001,$
+* Direction&nbsp; $\rm (A)$,&nbsp; $v_2  = 3.0 \cdot 10^3 \ {\rm m/s}\text{:}\hspace{0.4cm} f_{\rm D}/f_{\rm S} \ \underline {= \ 10^{\rm &ndash;5}} \ \ \ &#8658; \ \ \ f_{\rm E}/f_{\rm S} = 1.00001,$
-* Direction (B), $v_1  = 1.8 \cdot 10^8 \ {\rm m/s}\text{:}\hspace{0.4cm} f_{\rm D}/f_{\rm S} \ \underline {= \ &ndash;0.6} \ \ \ &#8658; \ \ \ f_{\rm E}/f_{\rm S} = 0.4,$
+* Direction&nbsp; $\rm (B)$,&nbsp; $v_1  = 1.8 \cdot 10^8 \ {\rm m/s}\text{:}\hspace{0.4cm} f_{\rm D}/f_{\rm S} \ \underline {= \ &ndash;0.6} \ \ \ &#8658; \ \ \ f_{\rm E}/f_{\rm S} = 0.4,$
-* Direction (B), $v_2  = 3.0 \cdot 10^3 \ {\rm m/s}\text{:}\hspace{0.4cm} f_{\rm D}/f_{\rm S} \ \underline {= \ &ndash;10^{\rm &ndash;5}} \ \ \ &#8658; \ \ \ f_{\rm E}/f_{\rm S} = 0.99999.$
+* Direction&nbsp; $\rm (B)$,&nbsp; $v_2  = 3.0 \cdot 10^3 \ {\rm m/s}\text{:}\hspace{0.4cm} f_{\rm D}/f_{\rm S} \ \underline {= \ &ndash;10^{\rm &ndash;5}} \ \ \ &#8658; \ \ \ f_{\rm E}/f_{\rm S} = 0.99999.$
 You can tell:
-*For realistic speeds &ndash; including $v \ \approx \ 10000 \ {\rm km/h}$ &ndash; the conventional equation (2) gives the same result as the relativistic equation (1) up to the accuracy of a pocket calculator.
+*For realistic speeds &ndash; including&nbsp; $v \ \approx \ 10000 \ {\rm km/h}$&nbsp; &ndash; the conventional equation&nbsp; '''(2)'''&nbsp; gives the same result as the relativistic equation&nbsp; '''(1)'''&nbsp; up to the accuracy of a pocket calculator.
-*With the approximation, the angles $\alpha = 0^°$ and $\alpha = 180^\circ$ result in the same absolute value for the Doppler frequency.
+*With the approximation, the angles&nbsp; $\alpha = 0^°$&nbsp; and&nbsp; $\alpha = 180^\circ$&nbsp; result in the same absolute value for the Doppler frequency.
 *The approximations differ only in the sign.
-*In the relativistic equation this symmetry is no longer present. See subtasks (1) and (2).
+*In the relativistic equation this symmetry is no longer present.&nbsp; See subtasks '''(1)''' and '''(2)'''.
-'''(4)'''&nbsp; Equation (2) leads here to the result:
+'''(4)'''&nbsp; Equation '''(2)''' leads here to the result:
-$$f_{\rm D} = f_{\rm E} - f_{\rm S} = f_{\rm S} \cdot {v_3}/{c} \cdot \cos(\alpha)
+:$$f_{\rm D} = f_{\rm E} - f_{\rm S} = f_{\rm S} \cdot {v_3}/{c} \cdot \cos(\alpha)
 \hspace{0.05cm}.$$
-* The direction of travel (C) is perpendicular ($\alpha = 90^\circ$) to the connection line transmitter&ndash;receiver. In this case, no Doppler shift occurs:
+* The driving direction&nbsp; $\rm (C)$&nbsp; is perpendicular&nbsp; $(\alpha = 90^\circ)$&nbsp; to the connection line transmitter&ndash;receiver.&nbsp; In this case, no Doppler shift occurs:
 :$$f_{\rm D} \ \ \underline {= \ 0}.$$
-* The direction of movement (D) is characterized by $\alpha = \ &ndash;135^\circ$. As a result:
+* The driving direction&nbsp; $\rm (D)$&nbsp; is characterized by&nbsp; $\alpha = \ &ndash;135^\circ$.&nbsp; As a result:
 :$$f_{\rm D} =  2 \cdot 10^{9}\,\,{\rm Hz} \cdot  \frac{30\,\,{\rm m/s}}{3 \cdot 10^{8}\,\,{\rm m/s}} \cdot  \cos(-135^{\circ}) \hspace{0.15cm} \underline{ \approx -141\,\,{\rm Hz}}  \hspace{0.05cm}.$$
 {{ML-Fuß}}

@@ Line 24: / Line 24: @@
 ''Notes:''
-* This task belongs to the topic of&nbsp; [[Mobile_Communications/Statistical_bindings_within_the_Rayleigh_process|Statistical bindings within the Rayleigh process]].
+* This task belongs to the topic of&nbsp; [[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process|Statistical bindings within the Rayleigh process]].
 * $c = 3 \cdot 10^8 \ \ \rm m/s$&nbsp; is the speed of light.
 * To check your results you can use the applet&nbsp; [[Applets:The_Doppler_Effect|The Doppler Effect]].

@@ Line 13: / Line 13: @@
-The equations given in the theory section for the reception frequency are
+The equations given in the theoretical section for the reception frequency are
 * taking into account the theory of relativity (briefly referred to as &bdquo;relativistic&rdquo;):
-$${\rm equation \hspace{0.15cm}(1):}\hspace{0.2cm}f_{\rm E} = f_{\rm S} \cdot \frac{\sqrt{1 - (v/c)^2}}{1 - v/c \cdot \cos(\alpha)}  \hspace{0.05cm},$$
+:$${\rm equation \hspace{0.15cm}(1):}\hspace{0.2cm}f_{\rm E} = f_{\rm S} \cdot \frac{\sqrt{1 - (v/c)^2}}{1 - v/c \cdot \cos(\alpha)}  \hspace{0.05cm},$$
 * without consideration of relativistic properties (referred to as &bdquo;conventional&rdquo;):
-$${\rm equation \hspace{0.15cm}(2):}\hspace{0.2cm}f_{\rm E} = f_{\rm S} \cdot \big [ 1 + {v}/{c} \cdot \cos(\alpha) \big ] \hspace{0.05cm}.$$
+:$${\rm equation \hspace{0.15cm}(2):}\hspace{0.2cm}f_{\rm E} = f_{\rm S} \cdot \big [ 1 + {v}/{c} \cdot \cos(\alpha) \big ] \hspace{0.05cm}.$$
@@ Line 34: / Line 34: @@
 ===Questions===
 <quiz display=simple>
-{Which Doppler frequencies result for the speeds&nbsp; $v_1$&nbsp; and&nbsp; $v_2$&nbsp; in direction of travel &nbsp;$\rm (A)$&nbsp; with <b> equation (1)</b>?
+{Which Doppler frequencies result for the speeds&nbsp; $v_1$&nbsp; and&nbsp; $v_2$&nbsp; in driving direction &nbsp;$\rm (A)$&nbsp; with <b> equation (1)</b>?
 |type="{}"}
 $v_1\text{:} \hspace{0.4cm} f_{\rm D}/f_{\rm S} \ = \ $ { 1 3% }
 $v_2\text{:} \hspace{0.4cm} f_{\rm D}/f_{\rm S} \ = \ $ { 1 3% } $\cdot \ 10^{-5}$
-{Which Doppler frequencies are obtained for the opposite direction of travel &nbsp;$\rm (B)$&nbsp; with <b>equation (1)</b>.
+{Which Doppler frequencies are obtained for the opposite driving direction &nbsp;$\rm (B)$&nbsp; with <b>equation (1)</b>.
 |type="{}"}
 $v_1\text{:} \hspace{0.4cm} f_{\rm D}/f_{\rm S} \ = \ $ { -0.515--0.485 }
@@ Line 51: / Line 51: @@
 $\hspace{2.96cm} v_2\text{:} \hspace{0.4cm} f_{\rm D}/f_{\rm S}\ = \ $ { -1.03--0.97 } $\cdot \ 10^{\rm &ndash;5}$
-{Let $f_{\rm S} = 2 \rm GHz$. Which Doppler frequencies result for the direction of travel &nbsp;$\rm (C)$&nbsp; and &nbsp;$\rm (D)$&nbsp; with <b> equation (2)</b>?
+{Let $f_{\rm S} = 2 \ \rm GHz$.&nbsp; Which Doppler frequencies result for the driving direction &nbsp;$\rm (C)$&nbsp; and &nbsp;$\rm (D)$&nbsp; with&nbsp; <b> equation (2)</b>?
 |type="{}"}
 ${\rm direction \ (C)}, \ \ v_3\text{:} \hspace{0.4cm} f_{\rm D} \ = \ $ { 0. } $\ \ \rm Hz$
@@ Line 59: / Line 59: @@
 ===Solutions===
 {{ML-Kopf}}
-'''(1)'''&nbsp; With the direction of travel (A), the receiver approaches the transmitter at an angle $\alpha = 0$. This gives (1) according to the relativistic equation:
+'''(1)'''&nbsp; With the direction of travel (A), the receiver approaches the transmitter at an angle&nbsp; $\alpha = 0$.&nbsp; This gives (1) according to the relativistic equation:
 :$$f_{\rm E} = f_{\rm S} \cdot \frac{\sqrt{1 - (v/c)^2}}{1 - v/c }
   \hspace{0.3cm} \Rightarrow \hspace{0.3cm} f_{\rm D} = f_{\rm E} - f_{\rm S}  = f_{\rm S} \cdot \left [  \frac{\sqrt{1 - (v/c)^2}}{1 - v/c } - 1 \right ]\hspace{0.3cm}