Aufgaben:Exercise 2.1Z: 2D-Frequency and 2D-Time Representations - 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 10:13, 16 February 2021

2021-02-16T10:13:17Z

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

2020-12-16T13:16:19Z

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

Javier: Text replacement - "Mobile Kommunikation/Allgemeine Beschreibung zeitvarianter Systeme" to "Mobile_Communications/General_Description_of_Time_Variant_Systems"

2020-12-15T13:59:26Z

Text replacement - "Mobile Kommunikation/Allgemeine Beschreibung zeitvarianter Systeme" to "Mobile_Communications/General_Description_of_Time_Variant_Systems"

Guenter at 16:58, 14 December 2020

2020-12-14T16:58:00Z

Guenter at 16:57, 14 December 2020

2020-12-14T16:57:05Z

Guenter at 16:48, 14 December 2020

2020-12-14T16:48:42Z

Guenter at 16:27, 14 December 2020

2020-12-14T16:27:15Z

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

2020-07-09T09:12:42Z

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

Javier at 15:40, 20 May 2020

2020-05-20T15:40:33Z

← Older revision		Revision as of 12:37, 23 March 2021
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−	[[Category:~~Exercises for~~ Mobile Communications\|^2.1 Description of Time-Variant Systems^]]	+	[[Category:Mobile Communications: Exercises\|^2.1 Description of Time-Variant Systems^]]

@@ Line 89: / Line 89: @@
 *For a two-way channel,&nbsp; $H(f, \ t)$&nbsp; is not a constant in&nbsp; $f$&nbsp; for any&nbsp; $t$.
 *Since in the&nbsp; $H(f, \ t)$&nbsp; graph the real and imaginary part have a non-zero mean &nbsp; &rArr; &nbsp; the Rayleigh&ndash;channel can also be excluded.
-*The data for the present task comes from a [[Mobile_Communications/Non-Frequency_Selective_Fading_With_Direct_Component| Rice channel]] with following parameters:
+*The data for the present task comes from a&nbsp; [[Mobile_Communications/Non-Frequency_Selective_Fading_With_Direct_Component| Rice channel]]&nbsp; with following parameters:
 :$$\sigma = {1}/{\sqrt{2}} \hspace{0.05cm},\hspace{0.2cm}
   x_0 = {1}/{\sqrt{2}} \hspace{0.05cm},\hspace{0.2cm}y_0 = -{1}/{\sqrt{2}} \hspace{0.05cm},\hspace{0.2cm}

@@ Line 64: / Line 64: @@
 </quiz>
-===Sample solution===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; As can be seen in the graph, the transfer function&nbsp; $H(f, \ t)$&nbsp; depends on&nbsp; $t$.&nbsp; Thus&nbsp; $h(\tau, \ t)$&nbsp; is also time-dependent.&nbsp; Correct is therefore <u>YES</u>.

← Older revision		Revision as of 13:59, 15 December 2020
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−	{{quiz-Header\|Buchseite=~~Mobile Kommunikation~~/~~Allgemeine Beschreibung zeitvarianter Systeme~~}}	+	{{quiz-Header\|Buchseite=Mobile_Communications/General_Description_of_Time_Variant_Systems}}

	[[File:P_ID2145__Mob_z_2_1.png\|right\|frame\|2D transfer function:  <br>real and imaginary parts]]		[[File:P_ID2145__Mob_z_2_1.png\|right\|frame\|2D transfer function:  <br>real and imaginary parts]]

@@ Line 89: / Line 89: @@
 *For a two-way channel,&nbsp; $H(f, \ t)$&nbsp; is not a constant in&nbsp; $f$&nbsp; for any&nbsp; $t$.
 *Since in the&nbsp; $H(f, \ t)$&nbsp; graph the real and imaginary part have a non-zero mean &nbsp; &rArr; &nbsp; the Rayleigh&ndash;channel can also be excluded.
-*The data for the present task comes from a [[Mobile_Communications/Nichtfrequenzselektives_Fading_mit_Direktkomponente#Kanalmodell_und_Rice.E2.80.93WDF| Rice channel]] with following parameters:
+*The data for the present task comes from a [[Mobile_Communications/Non-Frequency_Selective_Fading_With_Direct_Component| Rice channel]] with following parameters:
 :$$\sigma = {1}/{\sqrt{2}} \hspace{0.05cm},\hspace{0.2cm}
   x_0 = {1}/{\sqrt{2}} \hspace{0.05cm},\hspace{0.2cm}y_0 = -{1}/{\sqrt{2}} \hspace{0.05cm},\hspace{0.2cm}

@@ Line 66: / Line 66: @@
 ===Sample solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; As can be seen in the graph, the transfer function $H(f, t)$ is dependent on $t$. Thus $h(\tau, t)$ is also time-dependent. Correct is therefore <u>YES</u>.
+'''(1)'''&nbsp; As can be seen in the graph, the transfer function&nbsp; $H(f, \ t)$&nbsp; depends on&nbsp; $t$.&nbsp; Thus&nbsp; $h(\tau, \ t)$&nbsp; is also time-dependent.&nbsp; Correct is therefore <u>YES</u>.
-'''(2)'''&nbsp; If we look at a fixed point in time, for example $t = 2 \ \rm ms$, we obtain the following for the time-variant transfer function
+'''(2)'''&nbsp; If we look at a fixed point in time, for example&nbsp; $t = 2 \ \rm ms$, we obtain the following for the time-variant transfer function:
-$$H(f,\hspace{0.05cm} t = 2\, {\rm ms}) = - {\rm j} \cdot 1.3 \hspace{0.05cm} = {\rm const.}$$
+:$$H(f,\hspace{0.05cm} t = 2\, {\rm ms}) = - {\rm j} \cdot 1.3 \hspace{0.05cm} = {\rm const.}$$
-Thus the corresponding 2D&ndash;impulse response is
+*Thus the corresponding 2D&ndash;impulse response is
-$$h(\tau,\hspace{0.05cm} t = 2\, {\rm ms}) = - {\rm j} \cdot 1.3 \cdot \delta (\tau) \hspace{0.05cm}
+:$$h(\tau,\hspace{0.05cm} t = 2\, {\rm ms}) = - {\rm j} \cdot 1.3 \cdot \delta (\tau) \hspace{0.05cm}
    \hspace{0.3cm}\Rightarrow \hspace{0.3cm} M = 1 \hspace{0.05cm}.$$
-There is only one path ($M=1$). This means that the correct solution is <u>NO</u>.
+*There is only one path&nbsp; $(M=1)$.&nbsp; This means that the correct solution is <u>NO</u>.
@@ Line 86: / Line 87: @@
 '''(4)'''&nbsp; <u>Solution 4</u> is correct:
 *For the AWGN channel, no transfer function can be specified.
-*For a two-way channel, $H(f, t)$ is not constant in $f$ for any $t$.
+*For a two-way channel,&nbsp; $H(f, \ t)$&nbsp; is not a constant in&nbsp; $f$&nbsp; for any&nbsp; $t$.
-*Since in the $H(f, t)$ graph the real and imaginary part have a non-zero mean, the Rayleigh&ndash;channel can also be excluded.
+*Since in the&nbsp; $H(f, \ t)$&nbsp; graph the real and imaginary part have a non-zero mean &nbsp; &rArr; &nbsp; the Rayleigh&ndash;channel can also be excluded.
-*The data for the present task comes from a [[Mobile_Communications/Nichtfrequenzselektives_Fading_mit_Direktkomponente#Kanalmodell_und_Rice.E2.80.93WDF| Rice channel]] with the following parameters:
+*The data for the present task comes from a [[Mobile_Communications/Nichtfrequenzselektives_Fading_mit_Direktkomponente#Kanalmodell_und_Rice.E2.80.93WDF| Rice channel]] with following parameters:
 :$$\sigma = {1}/{\sqrt{2}} \hspace{0.05cm},\hspace{0.2cm}
   x_0 = {1}/{\sqrt{2}} \hspace{0.05cm},\hspace{0.2cm}y_0 = -{1}/{\sqrt{2}} \hspace{0.05cm},\hspace{0.2cm}

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Mobile Kommunikation/Allgemeine Beschreibung zeitvarianter Systeme}}
-[[File:P_ID2145__Mob_z_2_1.png|right|frame|2D transfer function:&nbsp; real and imaginary parts]]
+[[File:P_ID2145__Mob_z_2_1.png|right|frame|2D transfer function:&nbsp; <br>real and imaginary parts]]
-To describe a time-variant channel with several paths, the <i>two-dimensional impulse response</i> is used
+To describe a time-variant channel with several paths, the&nbsp; '''two-dimensional impulse response'''&nbsp; is used:
-$$h(\tau,\hspace{0.05cm}t) = \sum_{m = 1}^{M} z_m(t) \cdot {\rm \delta} (\tau - \tau_m)\hspace{0.05cm}.$$
+:$$h(\tau,\hspace{0.05cm}t) = \sum_{m = 1}^{M} z_m(t) \cdot {\rm \delta} (\tau - \tau_m)\hspace{0.05cm}.$$
-The first parameter&nbsp; $(\tau)$&nbsp; indicates the delay, the second &nbsp;$(t)$&nbsp; is related to the time variance of the channel.
+The first parameter&nbsp; $(\tau)$&nbsp; indicates the delay, the second parameter&nbsp;$(t)$&nbsp; is related to the time variance of the channel.
-The Fourier transform of&nbsp; $h(\tau, t)$&nbsp; in $\tau$ is the <i>time-variant transfer function</i>
+The Fourier transform of&nbsp; $h(\tau, \ t)$&nbsp; with respect to&nbsp; $\tau$&nbsp; is the &nbsp;'''time-variant transfer function''':
 :$$H(f,\hspace{0.05cm} t)
   \hspace{0.2cm}  \stackrel {f,\hspace{0.05cm}\tau}{\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} h(\tau,\hspace{0.05cm}t)
   \hspace{0.05cm}.$$
-In the graph,&nbsp; $H(f, t)$&nbsp; is displayed as a function of frequency, for different values of absolute time &nbsp;$t$&nbsp; in the range of $0 \ \text{...} \ 10 \ \rm ms$.
+*In the graph,&nbsp; $H(f, \ t)$&nbsp; is displayed as a function of frequency, for different values of absolute time &nbsp;$t$&nbsp; in the range of&nbsp; $0 \ \text{...} \ 10 \ \rm ms$.
-In general,&nbsp; $H(f, t)$&nbsp; is complex. The real part (top) and the imaginary part (bottom) are drawn separately.
+*In general,&nbsp; $H(f, \ t)$&nbsp; is complex.&nbsp; The real part (top) and the imaginary part (bottom) are drawn separately.
@@ Line 25: / Line 25: @@
 ''Notes:''
 * This task belongs to the chapter&nbsp; [[Mobile_Communications/General_Description_of_Time_Variant_Systems|General description of time&ndash;variant systems]].
-* In the above equation, an single-path channel is represented with the parameter&nbsp; $M = 1$&nbsp;.
+* In the above equation, an single-path channel is represented with parameter&nbsp; $M = 1$&nbsp;.
 * Here are some numerical values of the specified time-variant transfer function:
-$$H(f,\hspace{0.05cm} t \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0\, {\rm ms}) \approx 0.3 - {\rm j} \cdot 0.4 \hspace{0.05cm},\hspace{0.2cm}
+:$$H(f,\hspace{0.05cm} t \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0\, {\rm ms}) \approx 0.3 - {\rm j} \cdot 0.4 \hspace{0.05cm},\hspace{0.2cm}
   H(f,\hspace{0.05cm} t = 2\, {\rm ms}) \approx 0.0 - {\rm j} \cdot 1.3 \hspace{0.05cm},$$
-$$H(f,\hspace{0.05cm} t \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 4\, {\rm ms}) \approx 0.1 - {\rm j} \cdot 1.5 \hspace{0.05cm},\hspace{0.2cm}
+:$$H(f,\hspace{0.05cm} t \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 4\, {\rm ms}) \approx 0.1 - {\rm j} \cdot 1.5 \hspace{0.05cm},\hspace{0.2cm}
   H(f,\hspace{0.05cm} t = 6\, {\rm ms}) \approx 0.5 - {\rm j} \cdot 0.8 \hspace{0.05cm},$$
-$$H(f,\hspace{0.05cm} t \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 8\, {\rm ms}) \approx 0.9 - {\rm j} \cdot 0.1 \hspace{0.05cm},\hspace{0.2cm}
+:$$H(f,\hspace{0.05cm} t \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 8\, {\rm ms}) \approx 0.9 - {\rm j} \cdot 0.1 \hspace{0.05cm},\hspace{0.2cm}
   H(f,\hspace{0.05cm} t = 10\, {\rm ms}) \approx 1.4 \hspace{0.05cm}.$$
-* As can already be guessed from the above graph, neither the real nor the imaginary part of the 2D transfer function&nbsp; $H(f, t)$&nbsp; are zero-mean.
+* As can already be guessed from the above graph, neither the real nor the imaginary part of the 2D transfer function&nbsp; $H(f, \ t)$&nbsp; are zero-mean.
@@ Line 52: / Line 52: @@
 {How can the 2D impulse response be described here?
 |type="[]"}
-- $h(\tau, t) = A \cdot \delta(\tau) + B \cdot \delta(\tau \, &ndash;5 \, \rm &micro; s)$.
+- $h(\tau, \ t) = A \cdot \delta(\tau) + B \cdot \delta(\tau \, &ndash;5 \, \rm &micro; s)$.
-- $h(\tau, t) = A \cdot \delta(\tau)$.
+- $h(\tau, \ t) = A \cdot \delta(\tau)$.
-+ $h(\tau, t) = z(t) \cdot \delta(\tau)$.
++ $h(\tau, \ t) = z(t) \cdot \delta(\tau)$.
-{Estimate what type of channel the data was recorded for.
+{Estimate for which channel the data was recorded.
 |type="()"}
 - AWGN channel,