Aufgaben:Exercise 2.7Z: Coherence Bandwidth of the LTI Two-Path Channel: Unterschied zwischen den Versionen
Javier (Diskussion | Beiträge) (Die Seite wurde neu angelegt: „ {{quiz-Header|Buchseite=Mobile Kommunikation/Das GWSSUS–Kanalmodell}} right|frame|Zwei Zweiwegekanäle Zum GWSSUS&nd…“) |
Javier (Diskussion | Beiträge) |
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[[Datei:P_ID2178__Mob_Z_2_7.png|right|frame|Zwei Zweiwegekanäle]] | [[Datei:P_ID2178__Mob_Z_2_7.png|right|frame|Zwei Zweiwegekanäle]] | ||
− | + | For the GWSSUS–model, two parameters are given, which both statistically capture the resulting delay $\tau$ . More information on the topic „multipath propagation” can be found in section [[Mobile_Kommunikation/Das_GWSSUS%E2%80%93Kanalmodell#Simulation_gem.C3.A4.C3.9F_dem_GWSSUS.E2.80.93Modell| Simulation gemäß dem GWSSUS–Modell]] of the theory part. | |
− | * | + | * The <b>delay spread</b> $T_{\rm V}$ is by definition equal to the standard deviation of the random variable $\tau$. <br>This can be determined from the probability density $f_{\rm V}(\tau)$ . The PDF $f_{\rm V}(\tau)$ has the same shape as the delay power density spectrum ${\it \Phi}_{\rm V}(\tau)$. |
− | * | + | * The <b>coherence bandwidth</b> $B_{\rm K}$ describes the same situation in the frequency domain. <br> This is implicitly defined by the frequency–correlation function $\varphi_{\rm F}(\delta f)$ defined as the $\delta f$–value at which its amount first dropped to half: |
− | + | $$$\varphi_{\rm F}(\Delta f = B_{\rm K})| \stackrel {!}{=} {1}/{2} \cdot |\varphi_{\rm F}(\delta f = 0)| \hspace{0.05cm}.$$ | |
− | + | The connection between ${\it \Phi}_{\rm V}(\tau)$ and $\varphi_{\rm F}(\delta f)$ is given by the Fourier transform: | |
− | + | $$\varphi_{\rm F}(\delta f) | |
− | \hspace{0.2cm} | + | \{\hspace{0.2cm} {\bullet\!} {\hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$ |
− | * | + | *Both definitions are only partially suitable for a time invariant channel. |
− | * | + | *Often one uses for a time invariant two-way channel (i.e. with constant path weights according to the above graphic) as an approximation for the coherence bandwidth: |
− | :$$B_{\rm K}\hspace{0.01cm}' = \frac{1}{\ | + | :$$B_{\rm K}\hspace{0.01cm}' = \frac{1}{\frac_{\rm max} - \frac_{\rm min}} \hspace{0.05cm}.$$ |
− | In | + | In this task we want to clarify |
− | * | + | * why there are different definitions for the coherence band in the literature, |
− | * | + | * which connection exists between $B_{\rm K}$ and $B_{\rm K}\hspace{0.01cm}'$ and |
− | * | + | * which definitions make sense for which boundary conditions. |
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− | '' | + | ''Notes:'' |
− | * | + | *This task belongs to the chapter [[Mobile_Kommunikation/Das_GWSSUS%E2%80%93Kanalmodell| Das GWSSUS–Kanalmodell]]. |
− | * | + | *This task also refers to some theory pages in chapter [[Mobile_Kommunikation/Mehrwegeempfang_beim_Mobilfunk| Mehrwegeempfang beim Mobilfunk]]. |
− | === | + | ===Questionnaire=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {Which coherence bandwidth approximations $B_{\rm K}\hspace{0.01cm}'$ are there for channel $\rm A$ and $\rm B$? |
|type="{}"} | |type="{}"} | ||
− | + | Channel ${\rm A} \text \ \hspace{0.4cm} B_{\rm K}\hspace{0.01cm}' \ = \ ${ 1000 3% } $\ \ \rm kHz$ | |
− | + | Channel ${\rm B} \text \ \hspace{0.4cm} B_{\rm K}\hspace{0.01cm}' \ = \ ${ 1000 3% } $\ \ \rm kHz$ | |
− | { | + | {What is the WDF $f_{\rm V}(\tau)$? $G$ indicates the weight of the second path. |
|type="()"} | |type="()"} | ||
- $f_{\rm V}(\tau) = \delta(\tau) + G \cdot \delta(\tau \, –\tau_0)$, | - $f_{\rm V}(\tau) = \delta(\tau) + G \cdot \delta(\tau \, –\tau_0)$, | ||
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+ $f_{\rm V}(\tau) = 1/(1 + G^2) \cdot \delta(\tau) + G^2/(1 + G^2) \cdot \delta(\tau \, –\tau_0)$. | + $f_{\rm V}(\tau) = 1/(1 + G^2) \cdot \delta(\tau) + G^2/(1 + G^2) \cdot \delta(\tau \, –\tau_0)$. | ||
− | { | + | {Calculate the multipath propagation $ T_{\rm V}$. |
|type="{}"} | |type="{}"} | ||
− | + | Channel ${\rm A} \text \ \hspace{0.4cm} T_{\rm V} \ = \ ${ 0.5 3% } $\ \rm µ s$ | |
− | + | Channel ${\rm B} \text \ \hspace{0.4cm} T_{\rm V} \ = \ ${ 0.4 3% } $\ \rm µ s$ | |
− | { | + | {What is the coherence bandwidth $B_{\rm K}$ does the channel ${\rm A}$ have? |
|type="[]"} | |type="[]"} | ||
− | + | + | + It applies $B_{\rm K} = 333 \ \rm kHz$. |
− | - | + | - It is $B_{\rm K} = 500 \ \rm kHz$. |
− | - | + | - It applies $B_{\rm K} = 1 \ \rm MHz$. |
− | - $B_{\rm K}$ | + | - $B_{\rm K}$ cannot be specified according to this definition. |
− | { | + | {Which coherence bandwidth $B_{\rm K}$ does channel ${\rm B}$ have? |
|type="[]"} | |type="[]"} | ||
− | - | + | - It applies $B_{\rm K} = 333 \ \rm kHz$. |
− | - | + | - It is $B_{\rm K} = 500 \ \rm kHz$. |
− | - | + | - It applies $B_{\rm K} = 1 \ \ \rm MHz$. |
− | + $B_{\rm K}$ | + | + $B_{\rm K}$ cannot be specified according to this definition. |
</quiz> | </quiz> | ||
− | === | + | ===Sample solution=== |
− | {{ML-Kopf}} | + | {{{ML-Kopf}} |
− | '''(1)''' | + | '''(1)''' For both channels the runtime difference is $\Delta \tau = \tau_{\rm max} \, - \tau_{\rm min} = 1 \ \ \rm µ s$. |
− | * | + | * That's why both channels have the same value: |
− | + | $$B_{\rm K}\hspace{0.01cm}' \ \ \underline {= 1000 \ \rm kHz}.$$ | |
+ | '''(2)''' The graphics refer to the impulse response $h(\tau)$. | ||
+ | *To obtain the delay–LDS, the weights must be squared: | ||
+ | $${\it \Phi}_{\rm V}(\tau) = 1^2 \cdot \delta(\tau) + G^2 \cdot \delta(\tau - \tau_0) \hspace{0.05cm}.$ | ||
− | + | *The integral over ${\it \Phi}_{\rm V}(\tau)$ is therefore $1 + G^2$. | |
− | * | + | *The probability density function (WDF), however, must give the „area 1” (sum of the two Dirac weights equals $1$). From this follows: |
− | + | $$f_{\rm V}(\tau) = \frac{1}{1}{1 + G^2} \cdot \delta(\tau) + \frac{G^2}{1 + G^2} \cdot \delta(\tau - \tau_0) \hspace{0.05cm}.$ | |
− | * | + | *So only the <u>solution 3</u> is correct. |
− | * | + | *The first proposal does not describe the WDF $f_{\rm V}(\tau)$, but the impulse response $h(\tau)$. |
− | + | *The second equation specifies the delay –LDS ${\it \Phi}_{\rm V}(\tau)$. | |
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− | + | '''(3)''' For channel $\rm A$ the two impulse weights are equal. | |
− | '''(3)''' | + | *This means that for the mean value $m_{\rm V}$ and the standard deviation $\sigma_{\rm V} = T_{\rm V}$ can be written without a big calculation: |
− | * | + | $$m_{\rm V} = \frac{\frost_0}{2} {\hspace{0.15cm} {= 0.5\,{\rm µ s}}\hspace{0.05cm} |
− | + | \hspace{0.2cm}T_{\rm V} = \sigma_{\rm V} =\frac{\tau_0}{2} \hspace{0.15cm}\underline {= 0.5\,{\rm µ s} | |
− | \hspace{0.2cm}T_{\rm V} = \sigma_{\rm V} =\frac{\tau_0}{2} \hspace{0.15cm}\underline {= 0.5\,{\rm µ s | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
− | + | For channel $\rm B$ the pulse weights are $1/(1+0.5^2) = 0.8$ (for $\tau = 0$) and $0.2$ (for $\tau = 1 \ \rm µ s$). | |
− | * | + | * This gives for the linear and the quadratic mean value according to the [[Stochastic_Signal Theory/Expected Values_and_Moments#Moment Calculation_as_Sharsh Mean Value|basic Laws]] of statistics: |
:$$m_{\rm 1} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0 + 0.2 \cdot 1\,{\rm µ s} = 0.2\,{\rm µ s} \hspace{0.05cm},\hspace{0.5cm} | :$$m_{\rm 1} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0 + 0.2 \cdot 1\,{\rm µ s} = 0.2\,{\rm µ s} \hspace{0.05cm},\hspace{0.5cm} | ||
m_{\rm 2} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0^2 + 0.2 \cdot (1\,{\rm µ s})^2 = 0.2\,({\rm µ s})^2 \hspace{0.05cm}.$$ | m_{\rm 2} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0^2 + 0.2 \cdot (1\,{\rm µ s})^2 = 0.2\,({\rm µ s})^2 \hspace{0.05cm}.$$ | ||
− | * | + | *To get the result you are looking for you can use the [[Stochastic_Signaltheorie/Expected_values_and_Moments#Some_h.C3.A4Used_central_moments| Theorem of Steiner]]. |
− | + | $$\sigma_{\rm V}^2 = m_{\rm 2} - m_{\rm 1}^2 = 0.2\,({\rm µ s})^2 - (0.2\,{\rm µ s})^2 = 0.16\,({\rm µ s})^2 | |
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}T_{\rm V} = \sigma_{\rm V} \hspace{0.15cm}\underline {= 0.4\,{\rm µ s}}\hspace{0.05cm}.$$ | \hspace{0.3cm}\Rightarrow \hspace{0.3cm}T_{\rm V} = \sigma_{\rm V} \hspace{0.15cm}\underline {= 0.4\,{\rm µ s}}\hspace{0.05cm}.$$ | ||
− | '''(4)''' | + | '''(4)''' The frequency–correlation function is the Fourier transform of ${\it \Phi}_{\rm V}(\tau) = \delta(\tau) + \delta(\tau \, – \tau_0)$: |
− | + | $$$\varphi_{\rm F}(\delta f) = 1 + {\rm exp}(-{\rm j} \cdot 2\pi \cdot \delta f \cdot \cdot \tau_0) = 1 + {\rm cos}(2\pi \cdot \delta f \cdot \tau_0) -{\rm j} \cdot {\rm sin}(2\pi \cdot \delta f \cdot \tau_0) $$ | |
− | [[ | + | [[File:P_ID2186__Mob_Z_2_7d.png|right|frame|Frequency correlation function and coherence bandwidth]] |
− | + | $$\Rightarrow \hspace{0.3cm} |\varphi_{\rm F}(\delta f)| = \sqrt{2 + 2 \cdot {\rm cos}(2\pi \cdot \delta f \cdot \cdot \tau_0) }\hspace{0.05cm}.$ | |
− | * | + | *The function maximum at $\delta f = 0$ is equal to $2$. |
− | * | + | *Therefore the equation of determination for $B_{\rm K}$ is |
− | + | $$|\varphi_{\rm F}(B_{\rm K})| = 1 \hspace{0.3cm} $$ | |
− | + | $$\Rightarrow \hspace{0.3cm}|\varphi_{\rm F}(B_{\rm K})|^2 = 1 | |
− | \hspace{0.3cm} \Rightarrow \hspace{0.3cm}2 + | + | \hspace{0.3cm} \Rightarrow \hspace{0.3cm}2 + 2 \cdot {\rm cos}(2\pi \cdot B_{\rm K} \cdot \tau_0) = 1$$ |
− | + | $$\Rightarrow \hspace{0.3cm}{\rm cos}(2\pi \cdot B_{\rm K} \cdot \tau_0) = -0.5 \hspace{0.3cm} $$ | |
− | + | $$\Rightarrow \hspace{0.3cm}2\pi \cdot B_{\rm K} \cdot \tau_0 = \frac{2\pi}{3}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}B_{\rm K} = \frac{1}{3\tau_0} = 333\,{\rm kHz}\hspace{0.05cm}.$$ | |
− | * | + | *Correct is therefore the <u>solution 1</u>. The graphic (blue curve) illustrates the result. |
+ | '''(5)''' For the channel ${\rm B}$ the corresponding equations are | ||
+ | $${\it \Phi}_{\rm V}(\tau) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1^2 \cdot \delta(\tau) + (-0.5)^2 \cdot \delta(\tau - \tau_0) \hspace{0.05cm},\hspace{0.05cm} | ||
+ | \varphi_{\rm F}(\delta f) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1 + 0.25 \cdot {\rm cos}(2\pi \cdot \delta f \cdot \tau_0) -{\rm j} \cdot 0.25 \cdot {\rm sin}(2\pi \cdot \delta f \cdot \tau_0)\hspace{0.05cm},$$ | ||
+ | $$$\varphi_{\rm F}(\delta f)| \hspace{-0.1cm} \ = \ \hspace{-0.1cm}= \sqrt{\frac{17}{16} + \frac{1}{2} \cdot {\rm cos}(2\pi \cdot \delta f \cdot \tau_0) }\hspace{0.3cm} | ||
+ | \Rightarrow \hspace{0.3cm}{\rm Max}\hspace{0.1cm}|\varphi_{\rm F}(\delta f)| = 1.25\hspace{0.05cm},\hspace{0.2cm}{\rm Min}\hspace{0.1cm}|\varphi_{\rm F}(\delta f)| = 0.75\hspace{0.05cm}.$$ | ||
+ | *You can see from this result that the $50\%$–coherence bandwidth cannot be specified here. | ||
+ | *The correct solution is therefore the <u>solution proposal 4</u>. | ||
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− | * | + | This result is the reason why there are different definitions for the coherence range in the literature, for example |
− | * | + | * the $90\%$–coherence bandwidth (in the example $B_{\rm K, \hspace{0.03cm} 90\%} =184 \ \ \rm kHz$), |
+ | * the very simple approximation $B_{\rm K}\hspace{0.01cm}'$ given above (in the example $B_{\rm K}\hspace{0.01cm}' =1 \ \ \rm MHz$) | ||
− | + | You can see from these numerical values that all the information on this is very vague and that the individual „coherence bandwidths” can differ by factors. | |
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{{ML-Fuß}} | {{ML-Fuß}} | ||
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[[Category:Exercises for Mobile Communications|^2.3 The GWSSUS Channel Model^]] | [[Category:Exercises for Mobile Communications|^2.3 The GWSSUS Channel Model^]] |
Version vom 22. April 2020, 18:46 Uhr
For the GWSSUS–model, two parameters are given, which both statistically capture the resulting delay $\tau$ . More information on the topic „multipath propagation” can be found in section Simulation gemäß dem GWSSUS–Modell of the theory part.
- The delay spread $T_{\rm V}$ is by definition equal to the standard deviation of the random variable $\tau$.
This can be determined from the probability density $f_{\rm V}(\tau)$ . The PDF $f_{\rm V}(\tau)$ has the same shape as the delay power density spectrum ${\it \Phi}_{\rm V}(\tau)$. - The coherence bandwidth $B_{\rm K}$ describes the same situation in the frequency domain.
This is implicitly defined by the frequency–correlation function $\varphi_{\rm F}(\delta f)$ defined as the $\delta f$–value at which its amount first dropped to half:
$$$\varphi_{\rm F}(\Delta f = B_{\rm K})| \stackrel {!}{=} {1}/{2} \cdot |\varphi_{\rm F}(\delta f = 0)| \hspace{0.05cm}.$$
The connection between ${\it \Phi}_{\rm V}(\tau)$ and $\varphi_{\rm F}(\delta f)$ is given by the Fourier transform: $$\varphi_{\rm F}(\delta f) \{\hspace{0.2cm} {\bullet\!} {\hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$ *Both definitions are only partially suitable for a time invariant channel. *Often one uses for a time invariant two-way channel (i.e. with constant path weights according to the above graphic) as an approximation for the coherence bandwidth: :$$B_{\rm K}\hspace{0.01cm}' = \frac{1}{\frac_{\rm max} - \frac_{\rm min}} \hspace{0.05cm}.$$ In this task we want to clarify * why there are different definitions for the coherence band in the literature, * which connection exists between $B_{\rm K}$ and $B_{\rm K}\hspace{0.01cm}'$ and * which definitions make sense for which boundary conditions. ''Notes:'' *This task belongs to the chapter [[Mobile_Kommunikation/Das_GWSSUS%E2%80%93Kanalmodell| Das GWSSUS–Kanalmodell]]. *This task also refers to some theory pages in chapter [[Mobile_Kommunikation/Mehrwegeempfang_beim_Mobilfunk| Mehrwegeempfang beim Mobilfunk]]. ==='"`UNIQ--h-0--QINU`"'Questionnaire=== '"`UNIQ--quiz-00000002-QINU`"' ==='"`UNIQ--h-1--QINU`"'Sample solution=== {'"`UNIQ--html-00000003-QINU`"' '''(1)''' For both channels the runtime difference is $\Delta \tau = \tau_{\rm max} \, - \tau_{\rm min} = 1 \ \ \rm µ s$. * That's why both channels have the same value: $$B_{\rm K}\hspace{0.01cm}' \ \ \underline {= 1000 \ \rm kHz}.$$ '''(2)''' The graphics refer to the impulse response $h(\tau)$. *To obtain the delay–LDS, the weights must be squared: $${\it \Phi}_{\rm V}(\tau) = 1^2 \cdot \delta(\tau) + G^2 \cdot \delta(\tau - \tau_0) \hspace{0.05cm}.$ *The integral over ${\it \Phi}_{\rm V}(\tau)$ is therefore $1 + G^2$. *The probability density function (WDF), however, must give the „area 1” (sum of the two Dirac weights equals $1$). From this follows: '"`UNIQ-MathJax34-QINU`"'m_{\rm V} = \frac{\frost_0}{2} {\hspace{0.15cm} {= 0.5\,{\rm µ s}}\hspace{0.05cm} \hspace{0.2cm}T_{\rm V} = \sigma_{\rm V} =\frac{\tau_0}{2} \hspace{0.15cm}\underline {= 0.5\,{\rm µ s} \hspace{0.05cm}.'"`UNIQ-MathJax35-QINU`"'m_{\rm 1} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0 + 0.2 \cdot 1\,{\rm µ s} = 0.2\,{\rm µ s} \hspace{0.05cm},\hspace{0.5cm} m_{\rm 2} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0^2 + 0.2 \cdot (1\,{\rm µ s})^2 = 0.2\,({\rm µ s})^2 \hspace{0.05cm}.'"`UNIQ-MathJax36-QINU`"'\sigma_{\rm V}^2 = m_{\rm 2} - m_{\rm 1}^2 = 0.2\,({\rm µ s})^2 - (0.2\,{\rm µ s})^2 = 0.16\,({\rm µ s})^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}T_{\rm V} = \sigma_{\rm V} \hspace{0.15cm}\underline {= 0.4\,{\rm µ s}}\hspace{0.05cm}.'"`UNIQ-MathJax37-QINU`"'$\varphi_{\rm F}(\delta f) = 1 + {\rm exp}(-{\rm j} \cdot 2\pi \cdot \delta f \cdot \cdot \tau_0) = 1 + {\rm cos}(2\pi \cdot \delta f \cdot \tau_0) -{\rm j} \cdot {\rm sin}(2\pi \cdot \delta f \cdot \tau_0) $$ [[File:P_ID2186__Mob_Z_2_7d.png|right|frame|Frequency correlation function and coherence bandwidth]] $$\Rightarrow \hspace{0.3cm} |\varphi_{\rm F}(\delta f)| = \sqrt{2 + 2 \cdot {\rm cos}(2\pi \cdot \delta f \cdot \cdot \tau_0) }\hspace{0.05cm}.$ *The function maximum at $\delta f = 0$ is equal to $2$. *Therefore the equation of determination for $B_{\rm K}$ is '"`UNIQ-MathJax39-QINU`"' '"`UNIQ-MathJax40-QINU`"' '"`UNIQ-MathJax41-QINU`"' '"`UNIQ-MathJax42-QINU`"' *Correct is therefore the <u>solution 1</u>. The graphic (blue curve) illustrates the result. '''(5)''' For the channel ${\rm B}$ the corresponding equations are '"`UNIQ-MathJax43-QINU`"' '"`UNIQ-MathJax44-QINU`"' *You can see from this result that the $50\%$–coherence bandwidth cannot be specified here. *The correct solution is therefore the <u>solution proposal 4</u>. This result is the reason why there are different definitions for the coherence range in the literature, for example * the $90\%$–coherence bandwidth (in the example $B_{\rm K, \hspace{0.03cm} 90\%} =184 \ \ \rm kHz$), * the very simple approximation $B_{\rm K}\hspace{0.01cm}'$ given above (in the example $B_{\rm K}\hspace{0.01cm}' =1 \ \ \rm MHz$)
You can see from these numerical values that all the information on this is very vague and that the individual „coherence bandwidths” can differ by factors.