Aufgaben:Exercise 2.3: Yet Another Multi-Path Channel - Revision history

Javier: Text replacement - "Dirac impulse" to "Dirac delta"

2021-05-26T07:04:56Z

Text replacement - "Dirac impulse" to "Dirac delta"

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

2021-03-23T12:37:42Z

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

Guenter at 15:52, 11 January 2021

2021-01-11T15:52:10Z

Guenter at 15:42, 11 January 2021

2021-01-11T15:42:23Z

Guenter at 15:27, 11 January 2021

2021-01-11T15:27:56Z

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

2020-12-16T13:16:40Z

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 - "Zum_Rechnen_mit_komplexen_Zahlen" to "Calculating_With_Complex_Numbers"

2020-09-01T08:34:11Z

Text replacement - "Zum_Rechnen_mit_komplexen_Zahlen" to "Calculating_With_Complex_Numbers"

Javier: Text replacement - "transmit signal" to "transmitted signal"

2020-08-25T09:37:45Z

Text replacement - "transmit signal" to "transmitted signal"

Javier: Text replacement - "[[Signaldarstellung/" to "[[Signal_Representation/"

2020-07-09T09:40:41Z

Text replacement - "[[Signaldarstellung/" to "[[Signal_Representation/"

@@ Line 42: / Line 42: @@
 $\tau_3 \ = \ ${ 10 3% } $\ \ \rm &micro; s$
-{What are the weights of the first three Dirac impulses?
+{What are the weights of the first three Dirac deltas?
 |type="{}"}
 $k_1 \ = \ ${ 0.75 3% }

← Older revision		Revision as of 12:37, 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 98: / Line 98: @@
-&#8658; $H(f)$ is thus periodic with $f_0 \ \underline {= 500 \ \ \rm kHz}$.
+&#8658; &nbsp; $H(f)$&nbsp; is thus periodic with&nbsp; $f_0 \ \underline {= 500 \ \ \rm kHz}$.
@@ Line 111: / Line 111: @@
 \frac {3}{16} \cdot \left [  {\rm e}^{{\rm j}B} + {\rm e}^{-{\rm j}B}\right ]- \frac {1}{8} \cdot \left [  {\rm e}^{{\rm j}(B-A)} + {\rm e}^{-{\rm j}(B-A)}\right ]\hspace{0.05cm}.$$
-Using&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Darstellung_nach_Betrag_und_Phase|Euler's theorem]], with consideration of the frequency periodicity, this results in
+Using&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Representation_by_Amplidute_and_Phase|Euler's theorem]], with consideration of the frequency periodicity, this results in
 :$$|H(f)|= \sqrt{\frac {7}{8 }- \frac {3}{4} \cdot \cos( 2 \pi f \tau_2) +
 \frac {3}{8} \cdot \cos( 2 \pi f \tau_3)- \frac {1}{4} \cdot \cos( 2 \pi f (\tau_3 - \tau_2))}$$
@@ Line 120: / Line 120: @@
-'''(6)'''&nbsp; At $f = 250 \ \rm kHz$, the frequency response is
+'''(6)'''&nbsp; At&nbsp; $f = 250 \ \rm kHz$, the frequency response is
 [[File:P_ID2162__Mob_A_2_3e.png|right|frame|Amplitude frequency response for three-way channel]]
-$$H(f = 250\,{\rm kHz})= k_1 + k_2 \cdot {\rm e}^{-{\rm j}\cdot \pi}+ k_3 \cdot {\rm e}^{-{\rm j}\cdot 5\pi} = k_1 - k_2 - k_3
+:$$H(f = 250\,{\rm kHz})= k_1 + k_2 \cdot {\rm e}^{-{\rm j}\cdot \pi}+ k_3 \cdot {\rm e}^{-{\rm j}\cdot 5\pi} = k_1 - k_2 - k_3
   \hspace{0.05cm}.$$
 If you now substitute &nbsp;
 :$$k_3 = k_1 - k_2 = 0.75 + 0.50\hspace{0.1cm} \underline {= 1.25}\hspace{0.05cm},$$
-the result is $|H(f = 250 \ \rm kHz)| = 0$ and thus the most unfavorable value for this signal frequency.
+the result is&nbsp; $|H(f = 250 \ \rm kHz)| = 0$&nbsp; and thus the most unfavorable value for this signal frequency.
-The graph shows $|H(f)|$ in the range between $0$ and $500 \ \rm kHz$:
+The graph shows&nbsp; $|H(f)|$&nbsp; in the range between&nbsp; $0$&nbsp; and&nbsp; $500 \ \rm kHz$:
-*The blue curve corresponds to $k_3 = 0.25$ according to the specifications of subtask '''(4)'''.
+*The blue curve corresponds to&nbsp; $k_3 = 0.25$&nbsp; according to the specifications of task&nbsp; '''(4)'''.
-*The red curve corresponds to $k_3 = 1.25$, the most unfavourable value for $f = 250 \ \rm kHz$.
+*The red curve corresponds to&nbsp; $k_3 = 1.25$, the most unfavourable value for&nbsp; $f = 250 \ \rm kHz$.
 {{ML-Fuß}}

@@ Line 65: / Line 65: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Here we have $r(\tau) = s(\tau) &#8727; h(\tau)$, where $s(\tau)$ denotes a rectangular pulse of duration $T = 5 \ \ \rm &micro; s$ and the impulse response $h(\tau)$ is made up of $M$ weighted Dirac functions at $\tau_1, \tau_2, \ \text{...} \ , \tau_M$
+'''(1)'''&nbsp; Here we have&nbsp; $r(\tau) = s(\tau) &#8727; h(\tau)$, where&nbsp; $s(\tau)$&nbsp; denotes a rectangular pulse of duration&nbsp; $T = 5 \ \ \rm &micro; s$&nbsp; and the impulse response&nbsp; $h(\tau)$&nbsp; is made up of&nbsp; $M$&nbsp; weighted Dirac functions at&nbsp; $\tau_1, \tau_2, \ \text{...} \ , \tau_M$.
-The sketched output signal $r(\tau)$ can only result if
+The sketched output signal&nbsp; $r(\tau)$&nbsp; can only result if
-* $\tau_1 = 0$ (otherwise $r(\tau)$ would not start at $\tau = 0$),
+* $\tau_1 = 0$&nbsp; $($otherwise $r(\tau)$&nbsp; would not start at $\tau = 0)$,
-* $\tau_M = 10 \ \rm &micro; s$ is (this results in the rectangular section between $10 \ \rm &micro; s$ and $15 \ \ \rm &micro; s$),
+* $\tau_M = 10 \ \rm &micro; s$&nbsp; $($this results in the rectangular section between&nbsp; $10 \ \rm &micro; s$ and $15 \ \ \rm &micro; s)$,
-* there is another Dirac function at $\tau_2 = 2 \ \rm &micro; s$ between the two.
+* there is another Dirac function at&nbsp; $\tau_2 = 2 \ \rm &micro; s$&nbsp; between the two.
-That means: &nbsp; The impulse response here consists of $\underline {M = 3}$ Dirac functions.
+That means: &nbsp; The impulse response here consists of&nbsp; $\underline {M = 3}$&nbsp; Dirac functions.
@@ Line 80: / Line 80: @@
-'''(3)'''&nbsp; If you compare input $s(\tau)$ and output $r(\tau)$, you will get the following results:
+'''(3)'''&nbsp; If you compare input&nbsp; $s(\tau)$&nbsp; and output&nbsp; $r(\tau)$, you will get the following results:
-* Interval $0 < \tau < 2 \ {\rm &micro; s} \text{:} \, s(\tau) = s_0, \hspace{1cm} r(\tau) = 0.75 \cdot s_0 \,\,\Rightarrow\,\, k_1 \ \underline {= 0.75}$,
+* Interval&nbsp; $0 < \tau < 2 \ {\rm &micro; s} \text{:} \, s(\tau) = s_0, \hspace{1cm} r(\tau) = 0.75 \cdot s_0 \,\,\Rightarrow\,\, k_1 \ \underline {= 0.75}$,
-* Interval $2 \ {\rm &micro; s} < \tau < 5 \ {\rm &micro; s} \text{:} \, \hspace{2.4cm} r(\tau) =(k_1 + k_2) \cdot s_0 = 0.25 \cdot s_0 \Rightarrow k_2 \ \underline {= \, &ndash;0.50}$,
+* Interval&nbsp; $2 \ {\rm &micro; s} < \tau < 5 \ {\rm &micro; s} \text{:} \, \hspace{2.45cm} r(\tau) =(k_1 + k_2) \cdot s_0 = 0.25 \cdot s_0 \Rightarrow k_2 \ \underline {= \, -0.50}$,
-* Interval $10 \ {\rm &micro; s} < \tau < 15 \ {\rm &micro; s} \text{:} \, \hspace{1.95cm} r(\tau) =k_3 \cdot s_0 = 0.25 \cdot s_0 \,\Rightarrow\, k_3 \ \underline {= 0.25}$.
+* Interval&nbsp; $10 \ {\rm &micro; s} < \tau < 15 \ {\rm &micro; s} \text{:} \, \hspace{1.99cm} r(\tau) =k_3 \cdot s_0 = 0.25 \cdot s_0 \,\Rightarrow\, k_3 \ \underline {= 0.25}$.
-'''(4)'''&nbsp; Using the time-shifting property, the Fourier transform of the impulse response $h(\tau)$ is:
 :$$h(\tau) = k_1 \cdot \delta( \tau) + k_2 \cdot \delta( \tau - \tau_2)+ k_3 \cdot \delta( \tau - \tau_3) \hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}H(f) = k_1 + k_2 \cdot {\rm e}^{- {\rm j}\cdot 2 \pi f \tau_2}+  k_3 \cdot {\rm e}^{- {\rm j}\cdot 2 \pi f \tau_3}
@@ Line 91: / Line 93: @@
 Analysis of the individual contributions leads to the following conclusion:
-* The first part is constant (any period is valid).
+* The first part is constant &nbsp; &#8658; &nbsp; periode&nbsp; $f_1 &#8594; &#8734;$..
 * The second part is periodic with $f_2 = 1/\tau_2 = 500 \ \rm kHz$.
 * The third part is periodic with $f_3 = 1/\tau_3 = 100 \ \rm kHz$.
@@ Line 99: / Line 101: @@
-'''(5)'''&nbsp; With $A = 2\pi f \cdot \tau_2$ and $B = 2\pi f \cdot \tau_3$ you get
+'''(5)'''&nbsp; With&nbsp; $A = 2\pi f \cdot \tau_2$&nbsp; and&nbsp; $B = 2\pi f \cdot \tau_3$&nbsp; you get
 :$$|H(f)|^2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} H(f) \cdot H^{\star}(f)=
   \left [ \frac {3}{4} - \frac {1}{2} \cdot {\rm e}^{-{\rm j}A} + \frac {1}{4} \cdot {\rm e}^{-{\rm j}B}\right ]
@@ Line 109: / Line 111: @@
 \frac {3}{16} \cdot \left [  {\rm e}^{{\rm j}B} + {\rm e}^{-{\rm j}B}\right ]- \frac {1}{8} \cdot \left [  {\rm e}^{{\rm j}(B-A)} + {\rm e}^{-{\rm j}(B-A)}\right ]\hspace{0.05cm}.$$
-Using [[Signal_Representation/Calculating_With_Complex_Numbers#Darstellung_nach_Betrag_und_Phase|Euler's theorem]], with consideration of the frequency periodicity, this results in
+Using&nbsp; [[Signal_Representation/Calculating_With_Complex_Numbers#Darstellung_nach_Betrag_und_Phase|Euler's theorem]], with consideration of the frequency periodicity, this results in
 :$$|H(f)|= \sqrt{\frac {7}{8 }- \frac {3}{4} \cdot \cos( 2 \pi f \tau_2) +
 \frac {3}{8} \cdot \cos( 2 \pi f \tau_3)- \frac {1}{4} \cdot \cos( 2 \pi f (\tau_3 - \tau_2))}$$

@@ Line 4: / Line 4: @@
 [[File:P_ID2160__Mob_A_2_3.png|right|frame|Given piecewise constant impulse response]]
 We consider a multipath channel, which is characterized by the following impulse response:
-$$h(\tau, \hspace{0.05cm} t) = h(\tau) = \sum_{m = 1}^{M} k_m \cdot \delta( \tau - \tau_m)
+:$$h(\tau, \hspace{0.05cm} t) = h(\tau) = \sum_{m = 1}^{M} k_m \cdot \delta( \tau - \tau_m)
   \hspace{0.05cm}.$$
-All coefficients&nbsp; $k_{m}$&nbsp; are real (positive or negative). Furthermore, we note
+All coefficients&nbsp; $k_{m}$&nbsp; are real (positive or negative).&nbsp; Furthermore, we note:
-* From the specification&nbsp; $h(\tau, \hspace{0.05cm}t) = h(\tau)$&nbsp; you can see that the channel is time invariant.
+* From the specification&nbsp; $h(\tau, \hspace{0.05cm}t) = h(\tau)$&nbsp; you can see that the channel is time&ndash;invariant.
-* Generally, the channel has $M$ paths. The&nbsp; value of $M$ should be determined from the graph.
+* Generally, the channel has&nbsp; $M$&nbsp; paths.&nbsp; The value of&nbsp; $M$&nbsp; should be determined from the graph.
-* The following relations apply to the delay times:&nbsp; $\tau_1 < \tau_2 < \tau_3 < \ \ text{...}$
+* The following relations apply to the delay times:&nbsp; $\tau_1 < \tau_2 < \tau_3 < \ \text{...}$.
-The graph shows the output signal&nbsp; $r(\tau)$&nbsp; of the channel when the following transmitted signal is present at the input (shown in the equivalent low-pass range):
+The graph shows the output signal&nbsp; $r(\tau)$&nbsp; of the channel when the following transmitted signal is present at the input&nbsp; (shown in the equivalent low-pass range):
 :$$s(\tau) = \left\{ \begin{array}{c} s_0\\
    \end{array} \right.\quad
@@ Line 24: / Line 24: @@
 ''Notes:''
-* This task refers to the chapter&nbsp; [[Mobile_Communications/Mehrwegeempfang_beim_Mobilfunk| Mehrwegeempfang beim Mobilfunk]].
+* This task refers to the chapter&nbsp; [[Mobile_Communications/Multi-Path_Reception_in_Mobile_Communications| Multi&ndash;Path Reception in Mobile Communications]].
-* For the solution of subtask '''(1)''' assume that the impulse response&nbsp; $h(\tau)$&nbsp; has a span of 5 microseconds.
+* For the solution of subtask&nbsp; '''(1)'''&nbsp; assume that the impulse response&nbsp; $h(\tau)$&nbsp; has a span of five microseconds.
@@ Line 32: / Line 32: @@
 ===Questionnaire===
 <quiz display=simple>
-{What is the impulse response&nbsp; $h(\tau)$? How many paths&nbsp; $(M)$&nbsp; are there?
+{What is the impulse response&nbsp; $h(\tau)$?&nbsp; How many paths&nbsp; $(M)$&nbsp; are there?
 |type="{}"}
 $M \ = \ ${ 3 }
-{Specify the first three delays&nbsp; $\tau_m$&nbsp;.
+{Specify the first three delays&nbsp; $\tau_m$.
 |type="{}"}
 $\tau_1 \ = \ ${ 0. } $\ \ \rm &micro; s$
@@ Line 48: / Line 48: @@
 $k_3 \ = \ ${ 0.25 3% }
-{Calculate the frequency response&nbsp; $H(f)$. What is the frequency period&nbsp; $f_0$? <br><i>Note:</i> &nbsp; With integer&nbsp; $i$, it must hold that &nbsp; $H(f + i \cdot f_0) = H(f)$&nbsp;.
+{Calculate the frequency response&nbsp; $H(f)$.&nbsp; What is the frequency period&nbsp; $f_0$? <br><i> &nbsp; &nbsp;Note:</i> &nbsp; With integer&nbsp; $i$, it must hold that &nbsp; $H(f + i \cdot f_0) = H(f)$&nbsp;.
 |type="{}"}
 $f_0 \ = \ ${ 500 3% } $\ \ \rm kHz$
-{Calculate the magnitude of the frequency response. Which values result for the frequencies&nbsp; $f = 0$,&nbsp; $f = 250 \ \rm kHz$&nbsp; and&nbsp; $f = 500 \ \rm kHz$?
+{Calculate the magnitude of the frequency response.&nbsp; Which values result for the frequencies&nbsp; $f = 0$,&nbsp; $f = 250 \ \rm kHz$&nbsp; and&nbsp; $f = 500 \ \rm kHz$?
 |type="{}"}
 $|H(f = 0)| \ = \ ${ 0.5 3% }
@@ Line 58: / Line 58: @@
 $|H(f = 500 \ \rm kHz)| \ = \ ${ 0.5 3% }
-{What is the worst value&nbsp; $({\rm worst \ case})$&nbsp; for&nbsp; $k_3$&nbsp; at frequency&nbsp; $f = 250 \ \rm kHz$?
+{What is the worst value&nbsp; $({\rm worst \ case})$&nbsp; for&nbsp; $k_3$&nbsp; at frequency&nbsp; $f = 250 \ \rm kHz$&nbsp;?
 |type="{}"}
 $k_3 \ = \ ${ 1.25 3% }

← 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_ID2160__Mob_A_2_3.png\|right\|frame\|Given piecewise constant impulse response]]		[[File:P_ID2160__Mob_A_2_3.png\|right\|frame\|Given piecewise constant impulse response]]