Aufgaben:Exercise 2.4: 2D Transfer Function - Revision history

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

2021-03-23T12:38:04Z

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

Guenter at 16:12, 11 January 2021

2021-01-11T16:12:27Z

Guenter at 16:04, 11 January 2021

2021-01-11T16:04:08Z

Guenter: Guenter moved page Aufgaben:Exercise 2.4: 2-D Transfer Function to Aufgaben:Exercise 2.4: 2D Transfer Function

2021-01-11T15:57:07Z

Guenter moved page Exercise 2.4: 2-D Transfer Function to Exercise 2.4: 2D Transfer Function

Guenter at 15:56, 11 January 2021

2021-01-11T15:56:51Z

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

2020-12-16T13:16:41Z

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

Javier: Text replacement - "Mobile Kommunikation/Mehrwegeempfang beim Mobilfunk" to "Mobile_Communications/Multipath_Reception_in_Mobile_Communications"

2020-12-15T14:00:22Z

Text replacement - "Mobile Kommunikation/Mehrwegeempfang beim Mobilfunk" to "Mobile_Communications/Multipath_Reception_in_Mobile_Communications"

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

2020-07-09T09:13:05Z

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

Javier at 15:43, 20 May 2020

2020-05-20T15:43:03Z

Javier at 14:42, 15 April 2020

2020-04-15T14:42:49Z

← Older revision		Revision as of 12:38, 23 March 2021
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−	[[Category:~~Exercises for~~ Mobile Communications\|^2.2 Multi-Path Reception in Wireless Systems^]]	+	[[Category:Mobile Communications: Exercises\|^2.2 Multi-Path Reception in Wireless Systems^]]

@@ Line 53: / Line 53: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The period can be read from the given graph. If the magnitude representation is taken into account, the result is $T_0 \ \underline {= 20 \ \ \rm ms}$.
+'''(1)'''&nbsp; The period can be read from the given graph.&nbsp; If the magnitude representation is taken into account, the result is&nbsp; $T_0 \ \underline {= 20 \ \ \rm ms}$.
-'''(2)'''&nbsp; At time $t_1 \ \underline {= 5 \ \ \rm ms}$ we have $h(\tau = 1 \ {\rm &micro; s}, t_1) = 0$. Accordingly, the following applies
+'''(2)'''&nbsp; At time&nbsp; $t_1 \ \underline {= 5 \ \ \rm ms}$&nbsp; we have&nbsp; $h(\tau = 1 \ {\rm &micro; s}, t_1) = 0$.&nbsp; Accordingly, the following applies
 :$$h(\tau = 1\,{\rm &micro; s},\hspace{0.05cm}t_1) = \frac{1}{ \sqrt{2}} \cdot \delta(\tau)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
   H(f,\hspace{0.05cm}t_1) = \frac{1}{ \sqrt{2}} = {\rm const.}$$
-The same applies to $t_2 \ \underline {= 15 \ \ \rm ms}$:
+The same applies to&nbsp; $t_2 \ \underline {= 15 \ \ \rm ms}$:
 :$$h(\tau = 1\,{\rm &micro; s},\hspace{0.05cm}t_2) = \frac{1}{ \sqrt{2}} \cdot \delta(\tau)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
   H(f,\hspace{0.05cm}t_2) = \frac{1}{ \sqrt{2}} = {\rm const.}$$
@@ Line 66: / Line 66: @@
-'''(3)'''&nbsp; At time $t = 0$ the impulse response with $\tau_1 = 1 \ \ \rm &micro; s$ is
+'''(3)'''&nbsp; At time&nbsp; $t = 0$&nbsp; the impulse response with&nbsp; $\tau_1 = 1 \ \ \rm &micro; s$&nbsp; is
 :$$h(\tau,\hspace{0.05cm}t = 0) = \frac{1}{ \sqrt{2}} \cdot \delta(\tau)+ \delta(\tau - \tau_1)\hspace{0.05cm}.$$
@@ Line 76: / Line 76: @@
 Therefore,
-* $H_0(f)$ is periodic with $1/\tau_1 = 1 \ \rm MHz$.
+* $H_0(f)$&nbsp; is periodic with&nbsp; $1/\tau_1 = 1 \ \rm MHz$.
 * For the maximum and minimum values, we have
 :$${\rm Max}\, \left [ \, |H_0(f)|\, \right ] \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt {  1.5 + { \sqrt{2}} } \approx 1.707 \hspace{0.05cm},\hspace{0.5cm}{\rm Min}\, \left [ \, |H_0(f)|\, \right ] \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt {  1.5 - { \sqrt{2}} } \approx 0.293 \hspace{0.05cm}. $$
-* At $f = 0$, $|H_0(f)|$ has a maximum.
+* At $f = 0$,&nbsp; $|H_0(f)|$ has a maximum.
-Therefore, <u>all three solution suggestions</u> are correct.
+Therefore, <u>all three solution suggestions</u>&nbsp; are correct.
 '''(4)'''&nbsp; For the time $t = 10 \ \rm ms$ the following equations apply:
@@ Line 93: / Line 93: @@
 [[File:P_ID2163__Mob_A_2_4d.png|right|frame|2D impulse response $|h(\tau, \hspace{0.05cm}t)|$ and 2D transfer function $|H(f, \hspace{0.05cm}t)|$]]
 <u>Solutions 1 and 2</u> are correct:
-*The frequency period does not change from $t = 0$.
+*The frequency period does not change compaired to&nbsp; $t = 0$.
-*The maximum value is still $1,707$ and the minimum value $0,293$ does not change compared to the subtask '''(3)'''.
+*The maximum value is still&nbsp; $1.707$&nbsp; and the minimum value&nbsp; $0.293$&nbsp; does not change compared to the subtask&nbsp; '''(3)'''.
-*For $f = 0$ there is now a minimum and not a maximum.
+*For $f = 0$&nbsp; there is now a minimum and not a maximum.
-The graph on the right shows the magnitude $|H(f, t)|$ of the 2D transfer function.
+The graph on the right shows the magnitude&nbsp; $|H(f, t)|$&nbsp; of the 2D transfer function.

@@ Line 4: / Line 4: @@
 [[File:P_ID2161__Mob_A_2_4.png|right|frame|2D impulse response&nbsp; $|h(\tau, \hspace{0.05cm}t)|$]]
 The graph shows the two-dimensional impulse response&nbsp; $h(\tau, \hspace{0.05cm}t)$&nbsp; of a mobile radio system in magnitude representation.
-*It can be seen that the 2D&ndash;impulse response only has components at delays&nbsp; $\tau = 0$&nbsp; and&nbsp; $\tau = 1 \ \rm &micro; s$&nbsp;.
+*It can be seen that the 2D impulse response only has components at delays&nbsp; $\tau = 0$&nbsp; and&nbsp; $\tau = 1 \ \rm &micro; s$&nbsp;.
 *At these times:
-$$h(\tau = 0\,{\rm &micro; s},\hspace{0.05cm}t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{ \sqrt{2}} = {\rm const.}$$
+:$$h(\tau = 0\,{\rm &micro; s},\hspace{0.05cm}t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{ \sqrt{2}} = {\rm const.}$$
-$$h(\tau = 1\,{\rm &micro; s},\hspace{0.05cm}t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}   \cos(2\pi \cdot {t}/{ T_0})\hspace{0.05cm}.$$
+:$$h(\tau = 1\,{\rm &micro; s},\hspace{0.05cm}t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}   \cos(2\pi \cdot {t}/{ T_0})\hspace{0.05cm}.$$
-For all other values of &nbsp; $\tau$, we have &nbsp; $h(\tau, \hspace{0.05cm}t) \equiv 0$.
+For all other values of&nbsp; $\tau$,&nbsp; we have&nbsp; $h(\tau, \hspace{0.05cm}t) \equiv 0$.
 We want to obtain the two-dimensional transfer function&nbsp; $H(f, \hspace{0.05cm} t)$&nbsp; as the Fourier transform of&nbsp; $h(\tau, t)$&nbsp; with respect to the delay &nbsp; $\tau$:
@@ Line 15: / Line 15: @@
   \hspace{0.2cm}  \stackrel {f,\hspace{0.05cm}\tau}{\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} h(\tau,\hspace{0.05cm}t)
   \hspace{0.05cm}.$$
 ''Notes:''
 * This task belongs to chapter&nbsp; [[Mobile_Communications/Multi-Path_Reception_in_Mobile_Communications| Multi&ndash;Path Reception in Mobile Communications]].
-* A similar problem is treated in&nbsp; [[Aufgaben:Exercise_2.5:_Scatter_Function| Task 2.5]]&nbsp; but with a different nomenclature.
+* A similar problem is treated in&nbsp; [[Aufgaben:Exercise_2.5:_Scatter_Function| Exercize 2.5]]&nbsp; but with a different nomenclature.
@@ Line 27: / Line 29: @@
 ===Questionnaire===
 <quiz display=simple>
-{What is the period &nbsp; $T_0$&nbsp; of the function&nbsp; $h(\tau = 1 \ {\rm &micro; s},\hspace{0.05cm} t)$? Note that the graph shows the <u>magnitude</u>&nbsp; $|h(\tau, \hspace{0.05cm}t)|$&nbsp;.
+{What is the period &nbsp; $T_0$&nbsp; of the function&nbsp; $h(\tau = 1 \ {\rm &micro; s},\hspace{0.05cm} t)$?&nbsp; Note that the graph shows the <u>magnitude</u>&nbsp; $|h(\tau, \hspace{0.05cm}t)|$&nbsp;.
 |type="{}"}
 $T_0 \ = \ ${ 20 3% } $\ \ \rm ms$
-{At what times&nbsp; $t_1$&nbsp; $($between&nbsp; $0$&nbsp; and&nbsp; $10 \ \rm ms)$&nbsp; and&nbsp; $t_2$&nbsp; $($between&nbsp; $10 \ \ \rm ms$&nbsp; and&nbsp; $20 \ \ \rm ms)$&nbsp; is&nbsp; $H(f, \hspace{0.05cm}t)$&nbsp; constant in &nbsp; $f$&nbsp; ?
+{At what times&nbsp; $t_1$&nbsp; $($between&nbsp; $0$&nbsp; and&nbsp; $10 \ \rm ms)$&nbsp; and&nbsp; $t_2$&nbsp; $($between&nbsp; $10 \ \ \rm ms$&nbsp; and&nbsp; $20 \ \ \rm ms)$&nbsp; is&nbsp; $H(f, \hspace{0.05cm}t)$&nbsp; constant in $f$&nbsp; ?
 |type="{}"}
 $t_1 \ = \ ${ 5 3% } $\ \ \rm ms$
 $t_2 \ = \ ${ 15 3% } $\ \ \rm ms$
-{Calculate&nbsp; $H_0(f) = H(f, \hspace{0.05cm}t = 0)$. Which statements are true?
+{Calculate&nbsp; $H_0(f) = H(f, \hspace{0.05cm}t = 0)$.&nbsp; Which statements are true?
 |type="[]"}
-+ We have &nbsp; $H_0(f) = H_0(f + i \cdot 1 \ {\rm MHz}), \ i = &plusmn;1, &plusmn;2, \ \ \text{...}$
++ We have&nbsp; $H_0(f) = H_0(f + i \cdot 1 \ {\rm MHz}), \ i = &plusmn;1, &plusmn;2, \ \ \text{...}$
 + We have approximately&nbsp; $0.293 &#8804; |H_0(f)| &#8804; 1.707$.
-+ $|H_0(f)|$&nbsp; has a maximum at&nbsp; $f = 0$&nbsp;.
++ $|H_0(f)|$&nbsp; has a maximum at $f = 0$&nbsp;.
-{Calculate&nbsp; $H_{10}(f) = H(f, \hspace{0.05cm}t = 10 \ \rm ms)$. Which statements are true?
+{Calculate&nbsp; $H_{10}(f) = H(f, \hspace{0.05cm}t = 10 \ \rm ms)$.&nbsp; Which statements are true?
 |type="[]"}
-+ We have &nbsp; $H_{10}(f) = H_{10}(f + i \cdot 1 \ {\rm MHz}),\ i = &plusmn;1, &plusmn;2, \ \ \text{...}$
++ We have&nbsp; $H_{10}(f) = H_{10}(f + i \cdot 1 \ {\rm MHz}),\ i = &plusmn;1, &plusmn;2, \ \ \text{...}$
 + We have approximately&nbsp; $0.293 &#8804; H_{10}(f) &#8804; 1.707$.
-- $|H_{10}(f)|$&nbsp; has a maximum at&nbsp; $f = 0$&nbsp;.
+- $|H_{10}(f)|$&nbsp; has a maximum at $f = 0$&nbsp;.
 </quiz>

@@ Line 18: / Line 18: @@
 ''Notes:''
-* This task belongs to chapter&nbsp; [[Mobile_Communications/Mehrwegeempfang_beim_Mobilfunk| Mehrwegeempfang beim Mobilfunk]].
+* This task belongs to chapter&nbsp; [[Mobile_Communications/Multi-Path_Reception_in_Mobile_Communications| Multi&ndash;Path Reception in Mobile Communications]].
 * A similar problem is treated in&nbsp; [[Aufgaben:Exercise_2.5:_Scatter_Function| Task 2.5]]&nbsp; but with a different nomenclature.

@@ Line 49: / Line 49: @@
 </quiz>
-===Sample solution===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; The period can be read from the given graph. If the magnitude representation is taken into account, the result is $T_0 \ \underline {= 20 \ \ \rm ms}$.

← Older revision		Revision as of 14:00, 15 December 2020
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−	{{quiz-Header\|Buchseite=~~Mobile Kommunikation~~/~~Mehrwegeempfang beim Mobilfunk~~}}	+	{{quiz-Header\|Buchseite=Mobile_Communications/Multipath_Reception_in_Mobile_Communications}}

	[[File:P_ID2161__Mob_A_2_4.png\|right\|frame\|2D impulse response  $\|h(\tau, \hspace{0.05cm}t)\|$]]		[[File:P_ID2161__Mob_A_2_4.png\|right\|frame\|2D impulse response  $\|h(\tau, \hspace{0.05cm}t)\|$]]

@@ Line 75: / Line 75: @@
 Therefore,
 * $H_0(f)$ is periodic with $1/\tau_1 = 1 \ \rm MHz$.
-* For the maximum value or minimum value, we have
+* For the maximum and minimum values, we have
 :$${\rm Max}\, \left [ \, |H_0(f)|\, \right ] \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt {  1.5 + { \sqrt{2}} } \approx 1.707 \hspace{0.05cm},\hspace{0.5cm}{\rm Min}\, \left [ \, |H_0(f)|\, \right ] \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt {  1.5 - { \sqrt{2}} } \approx 0.293 \hspace{0.05cm}. $$
 * At $f = 0$, $|H_0(f)|$ has a maximum.
@@ Line 83: / Line 83: @@
 '''(4)'''&nbsp; For the time $t = 10 \ \rm ms$ the following equations apply:
-$$h(\tau,\hspace{0.05cm}t = 10\,{\rm ms}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{ \sqrt{2} \cdot \delta(\tau)- \delta(\tau - \tau_1)\hspace{0.05cm},$$
+:$$h(\tau,\hspace{0.05cm}t = 10\,{\rm ms}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{ \sqrt{2}} \cdot \delta(\tau)- \delta(\tau - \tau_1)\hspace{0.05cm},$$
-$$H_{10}(f) = H(f,\hspace{0.05cm}t = 10\,{\rm ms}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}
+:$$H_{10}(f) = H(f,\hspace{0.05cm}t = 10\,{\rm ms}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}
-  \frac{1}{ \sqrt{2}} - \cos( 2 \pi f \tau_1)+ {\rm j}\cdot \sin( 2 \pi f \tau_1)\hspace{0.05cm},$$
+  \frac{1}{ \sqrt{2}} -  \cos( 2 \pi f \tau_1)+ {\rm j}\cdot \sin( 2 \pi f \tau_1)\hspace{0.05cm},$$
 :$$ |H_{10}(f)| \hspace{-0.1cm} \ = \ \hspace{-0.1cm}
-  \sqrt { 1.5 - { \sqrt{2}} \cdot \cos( 2 \pi f \tau_1)}\hspace{0.05cm}.$
+  \sqrt { 1.5 -  { \sqrt{2}} \cdot \cos( 2 \pi f \tau_1)}\hspace{0.05cm}.$$
 [[File:P_ID2163__Mob_A_2_4d.png|right|frame|2D impulse response $|h(\tau, \hspace{0.05cm}t)|$ and 2D transfer function $|H(f, \hspace{0.05cm}t)|$]]
-Correct are the <u>solutions 1 and 2</u>:
+<u>Solutions 1 and 2</u> are correct:
-*The frequency period does not change from $t = $0.
+*The frequency period does not change from $t = 0$.
-*The maximum value is still $1,707$ and the minimum value $0,293$ does not change compared to the subtask '''(3)''.
+*The maximum value is still $1,707$ and the minimum value $0,293$ does not change compared to the subtask '''(3)'''.
-*For $f = 0$ there is now a minimum and no maximum.
+*For $f = 0$ there is now a minimum and not a maximum.
-The graph on the right shows the amount $|H(f, t)|$ of the 2D&ndash;transfer function.
+The graph on the right shows the magnitude $|H(f, t)|$ of the 2D transfer function.