Aufgaben:Exercise 2.5: Scatter Function - Revision history

Javier: Text replacement - "Category:Exercises for Mobile Communications" to "Category:Mobile Communications: Exercises"

2021-03-23T12:37:53Z

Text replacement - "Category:Exercises for Mobile Communications" to "Category:Mobile Communications: Exercises"

Guenter at 12:57, 14 January 2021

2021-01-14T12:57:58Z

Guenter at 16:42, 13 January 2021

2021-01-13T16:42:26Z

Guenter at 16:18, 13 January 2021

2021-01-13T16:18:48Z

Javier: Text replacement - "===Sample solution===" to "===Solution==="

2020-12-16T13:16:19Z

Text replacement - "===Sample solution===" to "===Solution==="

Javier: Text replacement - "Mobile Kommunikation/Das GWSSUS–Kanalmodell" to "Mobile_Communications/The_GWSSUS_Channel_Model"

2020-12-15T14:01:17Z

Text replacement - "Mobile Kommunikation/Das GWSSUS–Kanalmodell" to "Mobile_Communications/The_GWSSUS_Channel_Model"

Javier: Text replacement - "Mobile_Kommunikation/" to "Mobile_Communications/"

2020-07-09T09:13:27Z

Text replacement - "Mobile_Kommunikation/" to "Mobile_Communications/"

Javier at 15:43, 20 May 2020

2020-05-20T15:43:40Z

Javier at 11:07, 22 April 2020

2020-04-22T11:07:42Z

Javier at 11:01, 22 April 2020

2020-04-22T11:01:38Z

← Older revision		Revision as of 12:37, 23 March 2021
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−	[[Category:~~Exercises for~~ Mobile Communications\|^2.3 The GWSSUS Channel Model^]]	+	[[Category:Mobile Communications: Exercises\|^2.3 The GWSSUS Channel Model^]]

@@ Line 33: / Line 33: @@
 ''Notes:''
-* This task should clarify the subject matter of the chapter&nbsp; [[Mobile_Communications/The_GWSSUS_Channel_Model| The GWSSUS Channel Model]].
+* This exercise should clarify the subject matter of the chapter&nbsp; [[Mobile_Communications/The_GWSSUS_Channel_Model| The GWSSUS Channel Model]].
 * The relationship between the individual system functions is given in the&nbsp; [[Mobile_Communications/The_GWSSUS_Channel_Model#Generalized_system_functions_of_time_variant_systems|graph on the first page]]&nbsp; of this chapter.
 *Note that the magnitude function&nbsp; $|\eta_{\rm VD}(\tau, \hspace{0.05cm} f_{\rm D})|$&nbsp; is shown above, so negative weights of the Dirac functions cannot be recognized.

@@ Line 80: / Line 80: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The time-variant impulse response $h(\tau, \hspace{0.05cm} t) = \eta_{\rm VZ}(\tau, \hspace{0.05cm} t)$ is the inverse Fourier transform of the delay&ndash;Doppler function $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D}) = s(\tau, \hspace{0.05cm} f_{\rm D})$:
+'''(1)'''&nbsp; The time-variant impulse response&nbsp; $h(\tau, \hspace{0.05cm} t) = \eta_{\rm VZ}(\tau, \hspace{0.05cm} t)$&nbsp; is the inverse Fourier transform of the delay&ndash;Doppler function&nbsp; $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D}) = s(\tau, \hspace{0.05cm} f_{\rm D})$:
 :$$\eta_{\rm VZ}(\tau, \hspace{0.05cm} t)
   \hspace{0.2cm}  \stackrel{t, \hspace{0.02cm}f_{\rm D}}{\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet} \hspace{0.2cm} \eta_{\rm VD}(\tau, f_{\rm D})\hspace{0.05cm}.$$
+'''(2)'''&nbsp; For the delay&nbsp; $\tau = 0$, the scatter function&nbsp; $\eta_{\rm VD}$&nbsp; consists of a single Dirac at $f_{\rm D} = 100 \ \rm Hz$.
 *According to the second Fourier integral, the desired time-domain function satisfies:
 :$$\eta_{\rm VZ}(\tau = 0, t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \int\limits_{-\infty}^{+\infty} \delta (f_{\rm D} - 100\,{\rm Hz}) \cdot {\rm e}^{{\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm} 2 \pi f_{\rm D} t}\hspace{0.15cm}{\rm d}f_{\rm D} =\frac{1}{\sqrt{2}} \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi  t \hspace{0.05cm}\cdot \hspace{0.05cm}100\,{\rm Hz}} .$$
-*The correct solution is therefore <u>solution 1</u>.
+*Correct is <u>solution 1</u>.
-'''(3)'''&nbsp; For the delay $\tau = 1 \ \ \rm &micro; s$ the delay&ndash;Doppler function consists of two Dirac functions at $&plusmn;50 \ \rm Hz$, each with weight $-0.5$.
+'''(3)'''&nbsp; For the delay&nbsp;  $\tau = 1 \ \ \rm &micro; s$&nbsp; the delay&ndash;Doppler function consists of two Dirac functions at&nbsp; $&plusmn;50 \ \rm Hz$, each with weight&nbsp; $-0.5$.
-*The time function is then
+*The time function is&nbsp;  $\eta_{\rm VZ}(\tau = 1\,{\rm \mu s}, t) = - \cos( 2 \pi t \cdot 50\,{\rm Hz})\hspace{0.05cm}.$
-*This function can be represented with $A = -1$ and $f_0 = 50 \ \rm Hz$ according to <u>solution 2</u>.
+*This function can be represented with&nbsp; $A = -1$&nbsp; and&nbsp; $f_0 = 50 \ \rm Hz$&nbsp; according to <u>solution 2</u>.
-'''(4)'''&nbsp; The three Dirac functions $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$ are at the Doppler frequencies $+100 \ \rm Hz$, $+50 \ \rm Hz$ and $-50 \ \rm Hz$.
+'''(4)'''&nbsp; The three Dirac functions&nbsp; $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$&nbsp; are at the Doppler frequencies&nbsp; $+100 \ \rm Hz$, $+50 \ \rm Hz$&nbsp; and&nbsp; $-50 \ \rm Hz$.
-*For all other Doppler frequencies, therefore, we must have $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D}) \equiv 0$.
+*For all other Doppler frequencies, we must have&nbsp; $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D}) \equiv 0$.
-*<u>Solution 2</u> is therefore correct.
+*<u>Solution 2</u> is correct.
-'''(5)'''&nbsp; If one looks at the scatter function $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$ in the direction of the $\tau$&ndash;axis, there is one Dirac function at each of the Doppler frequencies $100 \ \rm Hz$ and $&plusmn;50 \ \rm Hz$.
+'''(5)'''&nbsp; If you look at the scatter function&nbsp; $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$&nbsp; in the direction of the&nbsp; $\tau$&ndash;axis, there is one Dirac function at each of the Doppler frequencies&nbsp; $100 \ \rm Hz$&nbsp; and&nbsp; $&plusmn;50 \ \rm Hz$.
-*Here, depending on $f$, complex exponential oscillations with constant magnitude result in each case (from which it follows that the <u>solution 1</u> is correct):
+*Here, depending on $f$,&nbsp; complex exponential oscillations with constant magnitude result in each case&nbsp; (from which it follows that <u>solution 1</u> is correct):
 :$$|\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D} = 100\,{\rm Hz})| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{\sqrt{2}} = {\rm const.}$$
 :$$| \eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D}= \pm 50\,{\rm Hz})| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.5 = {\rm const.}$$
@@ Line 120: / Line 118: @@
 [[File:P_ID2168__Mob_A_2_5e_neu.png|right|frame|Relationships between all system functions]]
-'''(6)'''&nbsp; As can be seen from the given [[Mobile_Communications/Das_GWSSUS%E2%80%93Kanalmodell#Verallgemeinerte_Systemfunktionen_zeitvarianter_Systeme|graph]], <u>solutions 2 and 3</u> are correct.
+'''(6)'''&nbsp; As can be seen from the given [[Mobile_Communications/The_GWSSUS_Channel_Model#Generalized_system_functions_of_time_variant_systems|graph]], <u>solutions 2 and 3</u> are correct.
 *The graph shows all system functions.
@@ Line 129: / Line 127: @@
 ''Note:''
-Compare the time-variant transfer function $|\eta_{\rm FZ}(f, \hspace{0.05cm} t)|$ in the bottom right figure with the corresponding graph for [[Aufgaben:Exercise_2.4:_2-D_Transfer_Function| Exercise 2.4]]:
+Compare the time-variant transfer function&nbsp; $|\eta_{\rm FZ}(f, \hspace{0.05cm} t)|$&nbsp; in the bottom right figure with the corresponding graph in&nbsp; [[Aufgaben:Exercise_2.4:_2-D_Transfer_Function| Exercise 2.4]]:
-*The respective magnitude functions differ significantly, although $|\eta_{\rm VZ}(\tau, t)|$ is the same in both cases.
+*The respective magnitude functions differ significantly, although&nbsp; $|\eta_{\rm VZ}(\tau, t)|$&nbsp; is the same in both cases.
-*In Exercise 2.4, a cosine was implicitly assumed for $\eta_{\rm VZ}(\tau = 1 \ {\rm &micro; s}, t)$; here we have a negative cosine function.
+*In Exercise 2.4, a cosine was implicitly assumed for&nbsp; $\eta_{\rm VZ}(\tau = 1 \ {\rm &micro; s}, t)$;&nbsp; here we have a negative cosine function.
 *The (not explicitly) specified delay&ndash;Doppler function for Exercise 2.4 was
 :$$\eta_{\rm VD}(\tau, f_{\rm D}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \delta (\tau) \cdot \delta (f_{\rm D} - 100\,{\rm Hz})+$$
@@ Line 138: / Line 136: @@
   \hspace{0.05cm}.$$
-*Comparison with the equation in this task shows that only the signs of the Diracs have changed at $\tau = 1 \ \rm &micro; s$.
+*Comparison with the equation in this task shows that only the signs of the Diracs have changed at&nbsp; $\tau = 1 \ \rm &micro; s$.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 [[File:P_ID2164__Mob_A_2_5.png|right|frame|Delay-Doppler profile]]
-For the mobile radio channel as a time-variant system, there are a total of four system functions that are linked with each other via the Fourier transform. With the nomenclature from our learning tutorial, these are:
+For the mobile radio channel as a time-variant system, there are a total of four system functions that are linked with each other via the Fourier transform.&nbsp; With the nomenclature from our tutorial, these are:
-* the time-variant impulse response&nbsp; $h(\tau, \hspace{0.05cm}t)$, which we also denote here as&nbsp; $\eta_{\rm VZ}(\tau,\hspace{0.05cm} t)$&nbsp;,
+* the time-variant impulse response&nbsp; $h(\tau, \hspace{0.05cm}t)$, which we also denote here as&nbsp; $\eta_{\rm VZ}(\tau,\hspace{0.05cm} t)$,
 * the delay-Doppler function&nbsp; $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$,
 * the frequency-Doppler function&nbsp; $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D})$,
@@ Line 10: / Line 10: @@
-The indices represent the delay (<b>V</b>) $\tau$, the time (<b>Z</b>) &nbsp; $t$, the frequency (<b>F</b>)&nbsp; $f$&nbsp; and the Doppler frequency (<b>D</b>)&nbsp; $f_{\rm D}$.
+The four possible system functions are uniformly denoted by&nbsp; $\boldsymbol{\eta}_{12}$&nbsp;.<br>
-The delay&ndash;Doppler function&nbsp; $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$&nbsp; is shown in the top plot:
+*The first subindex is either a&nbsp; $\boldsymbol{\rm V}$&nbsp; $($because of German&nbsp; $\rm V\hspace{-0.05cm}$erzögerung &nbsp; &rArr; &nbsp; delay time &nbsp;$\tau)$&nbsp; or&nbsp; a&nbsp; $\boldsymbol{\rm F}$&nbsp; $($frequency&nbsp; $f)$.<br>
 :$$\eta_{\rm VD}(\tau, f_{\rm D}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \delta (\tau) \cdot \delta (f_{\rm D} - 100\,{\rm Hz})-$$
 :$$\hspace{1.75cm} \ - \ \hspace{-0.1cm} \frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} - 50\,{\rm Hz})-
@@ Line 18: / Line 23: @@
   \hspace{0.05cm}.$$
-In the literature,&nbsp; $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$&nbsp; is often also called <i>scatter function</i> and denoted with&nbsp; $s(\tau, \hspace{0.05cm}f_{\rm D})$&nbsp;.
+In the literature,&nbsp; $\eta_{\rm VD}(\tau, \hspace{0.05cm}f_{\rm D})$&nbsp; is often also called&nbsp; '''scatter function'''&nbsp; and denoted with&nbsp; $s(\tau, \hspace{0.05cm}f_{\rm D})$&nbsp;.
 In this task, the associated delay&ndash;time function&nbsp; $\eta_{\rm VZ}(\tau, \hspace{0.05cm}t)$&nbsp; and the frequency&ndash;Doppler function&nbsp; $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D})$&nbsp; are to be determined.
 ''Notes:''
-* This task should clarify the subject matter of the chapter&nbsp; [[Mobile_Communications/Das_GWSSUS%E2%80%93Kanalmodell| Das GWSSUS&ndash;Kanalmodell]].
+* This task should clarify the subject matter of the chapter&nbsp; [[Mobile_Communications/The_GWSSUS_Channel_Model| The GWSSUS Channel Model]].
-* The relationship between the individual system functions is given in the&nbsp; [[Mobile_Communications/Das_GWSSUS%E2%80%93Kanalmodell#Verallgemeinerte_Systemfunktionen_zeitvarianter_Systeme|graph on the first page]]&nbsp; of this chapter.
+* The relationship between the individual system functions is given in the&nbsp; [[Mobile_Communications/The_GWSSUS_Channel_Model#Generalized_system_functions_of_time_variant_systems|graph on the first page]]&nbsp; of this chapter.
 *Note that the magnitude function&nbsp; $|\eta_{\rm VD}(\tau, \hspace{0.05cm} f_{\rm D})|$&nbsp; is shown above, so negative weights of the Dirac functions cannot be recognized.
@@ Line 33: / Line 41: @@
 ===Questionnaire===
 <quiz display=simple>
-{At which values of &nbsp; $\tau$ are the components of 2D impulse response&nbsp; $\eta_{\rm VZ}(\tau, \hspace{0.05cm}t)$&nbsp;?
+{At which values of&nbsp; $\tau$&nbsp; there are the components of 2D impulse response&nbsp; $\eta_{\rm VZ}(\tau, \hspace{0.05cm}t)$&nbsp;?
 |type="[]"}
 + $\tau = 0$,
@@ Line 39: / Line 47: @@
 - other $\tau$&ndash;values.
-{Calculate&nbsp; $|\eta_{\rm VZ}(\tau = 0,\hspace{0.05cm}t)|$. Which of the following statements are true?
+{Calculate&nbsp; $|\eta_{\rm VZ}(\tau = 0,\hspace{0.05cm}t)|$.&nbsp; Which of the following statements are true?
 |type="()"}
 + $|\eta_{\rm VZ}(\tau = 0,\hspace{0.05cm} t)|$&nbsp; is independent of&nbsp; $t$.
@@ Line 45: / Line 53: @@
 - &nbsp; $\eta_{\rm VZ}(\tau = 0, \hspace{0.05cm}t) = A \cdot \sin {(2\pi f_0 t)}$.
-{Calculate&nbsp; $|\eta_{\rm VZ}(\tau = 1 \ {\rm &micro; s},\hspace{0.05cm} t)|$. Which of the following statements are true?
+{Calculate&nbsp; $|\eta_{\rm VZ}(\tau = 1 \ {\rm &micro; s},\hspace{0.05cm} t)|$.&nbsp; Which of the following statements are true?
 |type="()"}
 - $|\eta_{\rm VZ}(\tau = 1 \ {\rm &micro; s},\hspace{0.05cm} t)|$&nbsp; is independent of&nbsp; $t$.
@@ Line 51: / Line 59: @@
 - &nbsp; $\eta_{\rm VZ}(\tau = 1 \ {\rm &micro; s}, \hspace{0.05cm}t) = A \cdot \sin {(2\pi f_0 t)}$.
-{Consider now the frequency&ndash;Doppler representation&nbsp; $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D})$. For which values of &nbsp; $f_{\rm D}$ is this function <b>not</b> equal to zero?
+{Consider the frequency&ndash;Doppler representation&nbsp; $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D})$.&nbsp; For which values of&nbsp; $f_{\rm D}$ is this function <b>not</b> equal to zero?
 |type="[]"}
 - $f_{\rm D} = 0$,
@@ Line 59: / Line 67: @@
 {Which of the following statements are true for&nbsp; $\eta_{\rm FD}(f,\hspace{0.05cm} f_{\rm D})$?
 |type="()"}
-+ $|\eta_{\rm FD}(f,\hspace{0.05cm} f_{\rm D} = 100 \ \rm Hz)|$&nbsp; is independent of&nbsp; $f_{\rm D}$.
++ $|\eta_{\rm FD}(f,\hspace{0.05cm} f_{\rm D} = 100 \ \rm Hz)|$&nbsp; is independent of $f_{\rm D}$.
 - &nbsp; $\eta_{\rm FD}(f, \hspace{0.05cm} f_{\rm D} = 50 \ {\rm Hz}) = A \cdot \cos {(2\pi t_0 f)}$.
 - &nbsp; $\eta_{\rm FD}(f, \hspace{0.05cm}f_{\rm D} = 50 \ {\rm Hz}) = A \cdot \sin {(2\pi t_0 f)}$.
@@ Line 65: / Line 73: @@
 {How do you get the time-variant transfer function&nbsp; $\eta_{\rm FZ}(f, \hspace{0.05cm}t)$?
 |type="[]"}
-- By Fourier transformation of&nbsp; $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$&nbsp; with respect to &nbsp; $\tau$.
+- By Fourier transformation of&nbsp; $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D})$&nbsp; with respect to&nbsp; $\tau$.
 + By Fourier transformation of&nbsp; $\eta_{\rm VZ}(\tau, \hspace{0.05cm}t)$&nbsp; with respect to&nbsp; $\tau$.
-+ By Fourier inverse transformation of&nbsp; $\eta_{\rm FD}(f,\hspace{0.05cm} f_{\rm D})$&nbsp; with respect to&nbsp; $f_{\rm D}$.
++ By inverse Fourier transformation of&nbsp; $\eta_{\rm FD}(f,\hspace{0.05cm} f_{\rm D})$&nbsp; with respect to $f_{\rm D}$.
 </quiz>

@@ Line 70: / Line 70: @@
 </quiz>
-===Sample solution===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; The time-variant impulse response $h(\tau, \hspace{0.05cm} t) = \eta_{\rm VZ}(\tau, \hspace{0.05cm} t)$ is the inverse Fourier transform of the delay&ndash;Doppler function $\eta_{\rm VD}(\tau,\hspace{0.05cm} f_{\rm D}) = s(\tau, \hspace{0.05cm} f_{\rm D})$:

← Older revision		Revision as of 14:01, 15 December 2020
Line 1:		Line 1:

−	{{quiz-Header\|Buchseite=~~Mobile Kommunikation~~/~~Das GWSSUS–Kanalmodell~~}}	+	{{quiz-Header\|Buchseite=Mobile_Communications/The_GWSSUS_Channel_Model}}

	[[File:P_ID2164__Mob_A_2_5.png\|right\|frame\|Delay-Doppler profile]]		[[File:P_ID2164__Mob_A_2_5.png\|right\|frame\|Delay-Doppler profile]]

@@ Line 111: / Line 111: @@
-[[File:P_ID2168__Mob_A_2_5e_new.png|right|frame|interrelation of all system functions]]
+[[File:P_ID2168__Mob_A_2_5e_neu.png|right|frame|Relationships between all system functions]]
-'''(6)'''&nbsp; As can be seen from the given [[Mobile_Communications/The_GWSSUS%E2%80%93Channel Model#Generalized_System Functions_Time-Variant_Systems|Graphics]], the <u>solution alternatives 2 and 3</u> are applicable.
+'''(6)'''&nbsp; As can be seen from the given [[Mobile_Kommunikation/Das_GWSSUS%E2%80%93Kanalmodell#Verallgemeinerte_Systemfunktionen_zeitvarianter_Systeme|graph]], <u>solutions 2 and 3</u> are correct.
-*The graphic shows all system functions.
+*The graph shows all system functions.
 *The Fourier correspondences (shown in green) illustrate the relationships between these system functions.
@@ Line 121: / Line 121: @@
 ''Note:''
-Compare the time-variant transfer function $|\eta_{\rm FZ}(f, \hspace{0.05cm} t)|$ in the figure below right with the corresponding graphic for [[Tasks:Task_2.4:_2D-Transfer Function| Task 2.4]]:
+Compare the time-variant transfer function $|\eta_{\rm FZ}(f, \hspace{0.05cm} t)|$ in the bottom right figure with the corresponding graph for [[Aufgaben:Exercise_2.4:_2-D_Transfer_Function| Exercise 2.4]]:
-*The respective amount functions shown differ significantly, although $|\eta_{\rm VZ}(\tau, t)|$ is the same in both cases.
+*The respective magnitude functions differ significantly, although $|\eta_{\rm VZ}(\tau, t)|$ is the same in both cases.
-*In task 2.4, a cosine was implicitly assumed for $\eta_{\rm VZ}(\tau = 1 \ {\rm &micro; s}, t)$, here a minus&ndash;cosine function.
+*In Exercise 2.4, a cosine was implicitly assumed for $\eta_{\rm VZ}(\tau = 1 \ {\rm &micro; s}, t)$; here we have a negative cosine function.
-*The (not explicitly) specified delay&ndash;Doppler function for task 2.4 was
+*The (not explicitly) specified delay&ndash;Doppler function for Exercise 2.4 was
-$$\eta_{\rm VD}(\tau, f_{\rm D}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2} \cdot \delta (\tau) \cdot \delta (f_{\rm D} - 100\,{\rm Hz})+$
+:$$\eta_{\rm VD}(\tau, f_{\rm D}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2}} \cdot \delta (\tau) \cdot \delta (f_{\rm D} - 100\,{\rm Hz})+$$
-$$\hspace{2cm}+\hspace{0.22cm}\frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} - 50\,{\rm Hz})+ $
+:$$\hspace{2cm}+\hspace{0.22cm}\frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} - 50\,{\rm Hz})+ $$
-$$\hspace{2cm}+\hspace{0.22cm} \frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} + 50\,{\rm Hz})
+:$$\hspace{2cm}+\hspace{0.22cm} \frac{1}{2} \cdot \delta (\tau- 1\,{\rm \mu s}) \cdot \delta (f_{\rm D} + 50\,{\rm Hz})
   \hspace{0.05cm}.$$
-*Comparison with the equation on the [[Tasks:2.5_Scatter-Function|Specifications]] shows that only the signs of the Diracs have changed at $\tau = 1 \ \rm &micro; s$.
+*Comparison with the equation in this task shows that only the signs of the Diracs have changed at $\tau = 1 \ \rm &micro; s$.
 {{ML-Fuß}}