Javier: Javier verschob die Seite Exercises:Exercise 2.8: COST Delay Models nach Aufgaben:Exercise 2.8: COST Delay Models

#WEITERLEITUNG [[Aufgaben:Exercise 2.8: COST Delay Models]]

2020-04-22T17:12:01Z

Javier: Javier verschob die Seite Exercises:Exercise 2.7Z: Coherence Bandwidth of the LTI Two-Path Channel nach Aufgaben:Exercise 2.7Z: Coherence Bandwidth of the LTI Two-Path Channel

#WEITERLEITUNG [[Aufgaben:Exercise 2.7Z: Coherence Bandwidth of the LTI Two-Path Channel]]

Aufgaben:Exercise 2.7Z: Coherence Bandwidth of the LTI Two-Path Channel

2020-04-22T17:11:33Z

Javier:

Aufgaben:Exercise 2.7Z: Coherence Bandwidth of the LTI Two-Path Channel

2020-04-22T17:09:31Z

Javier:

{{quiz-Header|Buchseite=Mobile Kommunikation/Das GWSSUS–Kanalmodell}}

[[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 &bdquo;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  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. It is defined as the value of $\Delta f$ at which the magnitude of the frequency correlation function  $\varphi_{\rm F}(\Delta f)$  first drops to half its maximum value:
:$$|\varphi_{\rm F}(\Delta f = B_{\rm K})| \stackrel {!}{=} {1}/{2} \cdot |\varphi_{\rm F}(\Delta f = 0)| \hspace{0.05cm}.$$

The relationship between  ${\it \Phi}_{\rm V}(\tau)$  and  $\varphi_{\rm F}(\Delta f)$  is given by the Fourier transform:
:$$\varphi_{\rm F}(\Delta f)
\hspace{0.2cm} {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$

*Both definitions are only partially suitable for a time-invariant channel.
*For a time invariant two-path channel (i.e. one with constant path weights according to the above graph), the following approximation for the coherence bandwidth is often used:
:$$B_{\rm K}\hspace{0.01cm}' = \frac{1}{\tau_{\rm max} - \tau_{\rm min}} \hspace{0.05cm}.$$

In this task we want to clarify
* why there are different definitions for the coherence bandwidth 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]].

===Questionnaire===
<quiz display=simple>
{What is the approximate coherence bandwidth   $B_{\rm K}\hspace{0.01cm}'$  for channels  $\rm A$  and  $\rm B$?
|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% } $\ \ \rm kHz$

{Let $G$  be the weight of the second path. What is the PDF  $f_{\rm V}(\tau)$?
|type="()"}
- $f_{\rm V}(\tau) = \delta(\tau) + G \cdot \delta(\tau \, –\tau_0)$,
- $f_{\rm V}(\tau) = \delta(\tau) + G^2 \cdot \delta(\tau \, –\tau_0)$,
+ $f_{\rm V}(\tau) = 1/(1 + G^2) \cdot \delta(\tau) + G^2/(1 + G^2) \cdot \delta(\tau \, –\tau_0)$.

{Calculate the delay spread  $ T_{\rm V}$.
|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  $B_{\rm K}$  of channel  ${\rm A}$ ?
|type="[]"}
+   $B_{\rm K} = 333 \ \rm kHz$.
-   $B_{\rm K} = 500 \ \rm kHz$.
-   $B_{\rm K} = 1 \ \rm MHz$.
- $B_{\rm K}$  cannot be calculated according to this definition.

{What is the coherence bandwidth  $B_{\rm K}$  of channel  ${\rm B}$ ?
|type="[]"}
-   $B_{\rm K} = 333 \ \rm kHz$.
-   $B_{\rm K} = 500 \ \rm kHz$.
-   $B_{\rm K} = 1 \ \ \rm MHz$.
+ $B_{\rm K}$  cannot be calculated according to this definition.
</quiz>

===Sample solution===
{{{ML-Kopf}}
'''(1)'''  For both channels, the delay 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 graphs refer to the impulse response $h(\tau)$.
*To obtain the delay–power density spectrum, 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 of ${\it \Phi}_{\rm V}(\tau)$ is therefore $1 + G^2$.
*The probability density function (PDF), however, must have &bdquo;area 1” (i.e., the sum of the two Dirac weights must be $1$). From this follows:
:$$f_{\rm V}(\tau) = \frac{1}{1 + G^2} \cdot \delta(\tau) + \frac{G^2}{1 + G^2} \cdot \delta(\tau - \tau_0) \hspace{0.05cm}.$$

*So only solution 3 is correct.
*The first option does not describe the PDF $f_{\rm V}(\tau)$, but the impulse response $h(\tau)$.
*The second equation specifies the delay –power spectral density ${\it \Phi}_{\rm V}(\tau)$.

'''(3)'''  For channel  $\rm A$  the two impulse weights are equal.
*This means that the mean value $m_{\rm V}$ and the standard deviation $\sigma_{\rm V} = T_{\rm V}$ can be computed simply:
:$$m_{\rm V} = \frac{\tau_0}{2} \hspace{0.15cm} {= 0.5\,{\rm µ s}}\hspace{0.05cm},
\hspace{0.2cm}T_{\rm V} = \sigma_{\rm V} =\frac{\tau_0}{2} \hspace{0.15cm}\underline {= 0.5\,{\rm µ s}}
\hspace{0.05cm}.$$

For channel  $\rm B$  the Dirac weights are $1/(1+0.5^2) = 0.8$ (for $\tau = 0$) and $0.2$ (for $\tau = 1 \ \rm µ s$).
* According to the [[Stochastische_Signaltheorie/Erwartungswerte_und_Momente#Momentenberechnung_als_Scharmittelwert|basic laws]] of statistics, the noncentral first and second order moments are:
:$$m_{\rm 1} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0 + 0.2 \cdot 1\,{\rm µ s} = 0.2\,{\rm µ s} \hspace{0.05cm},\hspace{0.5cm}
m_{\rm 2} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0^2 + 0.2 \cdot (1\,{\rm µ s})^2 = 0.2\,({\rm µ s})^2 \hspace{0.05cm}.$$

*To get the result you are looking for you can use the [[Stochastische_Signaltheorie/Erwartungswerte_und_Momente#Einige_h.C3.A4ufig_benutzte_Zentralmomente| Steiner's Theorem]].
:$$\sigma_{\rm V}^2 = m_{\rm 2} - m_{\rm 1}^2 = 0.2\,({\rm µ s})^2 - (0.2\,{\rm µ s})^2 = 0.16\,({\rm µ s})^2
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}T_{\rm V} = \sigma_{\rm V} \hspace{0.15cm}\underline {= 0.4\,{\rm µ s}}\hspace{0.05cm}.$$

'''(4)'''  The frequency correlation function is the Fourier transform of ${\it \Phi}_{\rm V}(\tau) = \delta(\tau) + \delta(\tau \, – \tau_0)$:
:$$\varphi_{\rm F}(\Delta f) = 1 + {\rm exp}(-{\rm j} \cdot 2\pi \cdot \Delta f \cdot \tau_0) = 1 + {\rm cos}(2\pi \cdot \Delta f \cdot \tau_0) -{\rm j} \cdot {\rm sin}(2\pi \cdot \Delta f \cdot \tau_0) $$
[[Datei:P_ID2186__Mob_Z_2_7d.png|right|frame|Frequency correlation function and coherence bandwidth]]
:$$\Rightarrow \hspace{0.3cm} |\varphi_{\rm F}(\Delta f)| = \sqrt{2 + 2 \cdot {\rm cos}(2\pi \cdot \Delta f \cdot \tau_0) }\hspace{0.05cm}.$$

*The maximum at $\delta f = 0$ is equal to $2$.
*Therefore the equation to determine $B_{\rm K}$ is
$$|\varphi_{\rm F}(B_{\rm K})| = 1 \hspace{0.3cm} $$
$$\Rightarrow \hspace{0.3cm}|\varphi_{\rm F}(B_{\rm K})|^2 = 1
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}2 + 2 \cdot {\rm cos}(2\pi \cdot B_{\rm K} \cdot \tau_0) = 1$$
$$\Rightarrow \hspace{0.3cm}{\rm cos}(2\pi \cdot B_{\rm K} \cdot \tau_0) = -0.5 \hspace{0.3cm} $$
$$\Rightarrow \hspace{0.3cm}2\pi \cdot B_{\rm K} \cdot \tau_0 = \frac{2\pi}{3}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}B_{\rm K} = \frac{1}{3\tau_0} = 333\,{\rm kHz}\hspace{0.05cm}.$$

*Solution 1 is therefore correct. The graph (blue curve) illustrates the result.

'''(5)'''  For 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.
*Therefore, solution 4 is correct.

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 &bdquo;coherence bandwidths” can be very different.

{{ML-Fuß}}

[[Category:Exercises for Mobile Communications|^2.3 The GWSSUS Channel Model^]]

Aufgaben:Exercise 2.7Z: Coherence Bandwidth of the LTI Two-Path Channel

2020-04-22T17:09:15Z

Javier:

{{quiz-Header|Buchseite=Mobile Kommunikation/Das GWSSUS–Kanalmodell}}

[[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 &bdquo;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  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. It is defined as the value of $\Delta f$ at which the magnitude of the frequency correlation function  $\varphi_{\rm F}(\Delta f)$  first drops to hlf its maximum value:
:$$|\varphi_{\rm F}(\Delta f = B_{\rm K})| \stackrel {!}{=} {1}/{2} \cdot |\varphi_{\rm F}(\Delta f = 0)| \hspace{0.05cm}.$$

The relationship between  ${\it \Phi}_{\rm V}(\tau)$  and  $\varphi_{\rm F}(\Delta f)$  is given by the Fourier transform:
:$$\varphi_{\rm F}(\Delta f)
\hspace{0.2cm} {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$

*Both definitions are only partially suitable for a time-invariant channel.
*For a time invariant two-path channel (i.e. one with constant path weights according to the above graph), the following approximation for the coherence bandwidth is often used:
:$$B_{\rm K}\hspace{0.01cm}' = \frac{1}{\tau_{\rm max} - \tau_{\rm min}} \hspace{0.05cm}.$$

In this task we want to clarify
* why there are different definitions for the coherence bandwidth 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]].

===Questionnaire===
<quiz display=simple>
{What is the approximate coherence bandwidth   $B_{\rm K}\hspace{0.01cm}'$  for channels  $\rm A$  and  $\rm B$?
|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% } $\ \ \rm kHz$

{Let $G$  be the weight of the second path. What is the PDF  $f_{\rm V}(\tau)$?
|type="()"}
- $f_{\rm V}(\tau) = \delta(\tau) + G \cdot \delta(\tau \, –\tau_0)$,
- $f_{\rm V}(\tau) = \delta(\tau) + G^2 \cdot \delta(\tau \, –\tau_0)$,
+ $f_{\rm V}(\tau) = 1/(1 + G^2) \cdot \delta(\tau) + G^2/(1 + G^2) \cdot \delta(\tau \, –\tau_0)$.

{Calculate the delay spread  $ T_{\rm V}$.
|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  $B_{\rm K}$  of channel  ${\rm A}$ ?
|type="[]"}
+   $B_{\rm K} = 333 \ \rm kHz$.
-   $B_{\rm K} = 500 \ \rm kHz$.
-   $B_{\rm K} = 1 \ \rm MHz$.
- $B_{\rm K}$  cannot be calculated according to this definition.

{What is the coherence bandwidth  $B_{\rm K}$  of channel  ${\rm B}$ ?
|type="[]"}
-   $B_{\rm K} = 333 \ \rm kHz$.
-   $B_{\rm K} = 500 \ \rm kHz$.
-   $B_{\rm K} = 1 \ \ \rm MHz$.
+ $B_{\rm K}$  cannot be calculated according to this definition.
</quiz>

===Sample solution===
{{{ML-Kopf}}
'''(1)'''  For both channels, the delay 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 graphs refer to the impulse response $h(\tau)$.
*To obtain the delay–power density spectrum, 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 of ${\it \Phi}_{\rm V}(\tau)$ is therefore $1 + G^2$.
*The probability density function (PDF), however, must have &bdquo;area 1” (i.e., the sum of the two Dirac weights must be $1$). From this follows:
:$$f_{\rm V}(\tau) = \frac{1}{1 + G^2} \cdot \delta(\tau) + \frac{G^2}{1 + G^2} \cdot \delta(\tau - \tau_0) \hspace{0.05cm}.$$

*So only solution 3 is correct.
*The first option does not describe the PDF $f_{\rm V}(\tau)$, but the impulse response $h(\tau)$.
*The second equation specifies the delay –power spectral density ${\it \Phi}_{\rm V}(\tau)$.

'''(3)'''  For channel  $\rm A$  the two impulse weights are equal.
*This means that the mean value $m_{\rm V}$ and the standard deviation $\sigma_{\rm V} = T_{\rm V}$ can be computed simply:
:$$m_{\rm V} = \frac{\tau_0}{2} \hspace{0.15cm} {= 0.5\,{\rm µ s}}\hspace{0.05cm},
\hspace{0.2cm}T_{\rm V} = \sigma_{\rm V} =\frac{\tau_0}{2} \hspace{0.15cm}\underline {= 0.5\,{\rm µ s}}
\hspace{0.05cm}.$$

For channel  $\rm B$  the Dirac weights are $1/(1+0.5^2) = 0.8$ (for $\tau = 0$) and $0.2$ (for $\tau = 1 \ \rm µ s$).
* According to the [[Stochastische_Signaltheorie/Erwartungswerte_und_Momente#Momentenberechnung_als_Scharmittelwert|basic laws]] of statistics, the noncentral first and second order moments are:
:$$m_{\rm 1} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0 + 0.2 \cdot 1\,{\rm µ s} = 0.2\,{\rm µ s} \hspace{0.05cm},\hspace{0.5cm}
m_{\rm 2} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0^2 + 0.2 \cdot (1\,{\rm µ s})^2 = 0.2\,({\rm µ s})^2 \hspace{0.05cm}.$$

*To get the result you are looking for you can use the [[Stochastische_Signaltheorie/Erwartungswerte_und_Momente#Einige_h.C3.A4ufig_benutzte_Zentralmomente| Steiner's Theorem]].
:$$\sigma_{\rm V}^2 = m_{\rm 2} - m_{\rm 1}^2 = 0.2\,({\rm µ s})^2 - (0.2\,{\rm µ s})^2 = 0.16\,({\rm µ s})^2
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}T_{\rm V} = \sigma_{\rm V} \hspace{0.15cm}\underline {= 0.4\,{\rm µ s}}\hspace{0.05cm}.$$

'''(4)'''  The frequency correlation function is the Fourier transform of ${\it \Phi}_{\rm V}(\tau) = \delta(\tau) + \delta(\tau \, – \tau_0)$:
:$$\varphi_{\rm F}(\Delta f) = 1 + {\rm exp}(-{\rm j} \cdot 2\pi \cdot \Delta f \cdot \tau_0) = 1 + {\rm cos}(2\pi \cdot \Delta f \cdot \tau_0) -{\rm j} \cdot {\rm sin}(2\pi \cdot \Delta f \cdot \tau_0) $$
[[Datei:P_ID2186__Mob_Z_2_7d.png|right|frame|Frequency correlation function and coherence bandwidth]]
:$$\Rightarrow \hspace{0.3cm} |\varphi_{\rm F}(\Delta f)| = \sqrt{2 + 2 \cdot {\rm cos}(2\pi \cdot \Delta f \cdot \tau_0) }\hspace{0.05cm}.$$

*The maximum at $\delta f = 0$ is equal to $2$.
*Therefore the equation to determine $B_{\rm K}$ is
$$|\varphi_{\rm F}(B_{\rm K})| = 1 \hspace{0.3cm} $$
$$\Rightarrow \hspace{0.3cm}|\varphi_{\rm F}(B_{\rm K})|^2 = 1
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}2 + 2 \cdot {\rm cos}(2\pi \cdot B_{\rm K} \cdot \tau_0) = 1$$
$$\Rightarrow \hspace{0.3cm}{\rm cos}(2\pi \cdot B_{\rm K} \cdot \tau_0) = -0.5 \hspace{0.3cm} $$
$$\Rightarrow \hspace{0.3cm}2\pi \cdot B_{\rm K} \cdot \tau_0 = \frac{2\pi}{3}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}B_{\rm K} = \frac{1}{3\tau_0} = 333\,{\rm kHz}\hspace{0.05cm}.$$

*Solution 1 is therefore correct. The graph (blue curve) illustrates the result.

'''(5)'''  For 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.
*Therefore, solution 4 is correct.

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 &bdquo;coherence bandwidths” can be very different.

{{ML-Fuß}}

[[Category:Exercises for Mobile Communications|^2.3 The GWSSUS Channel Model^]]

Aufgaben:Exercise 2.7Z: Coherence Bandwidth of the LTI Two-Path Channel

2020-04-22T17:08:24Z

Javier:

{{quiz-Header|Buchseite=Mobile Kommunikation/Das GWSSUS–Kanalmodell}}

[[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 &bdquo;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  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. It is defined as the value of $\delta f$ at which the magnitude of the frequency correlation function  $\varphi_{\rm F}(\delta f)$  first drops to hlf its maximum value:
:$$|\varphi_{\rm F}(\Delta f = B_{\rm K})| \stackrel {!}{=} {1}/{2} \cdot |\varphi_{\rm F}(\Delta f = 0)| \hspace{0.05cm}.$$

The relationship between  ${\it \Phi}_{\rm V}(\tau)$  and  $\varphi_{\rm F}(\Delta f)$  is given by the Fourier transform:
:$$\varphi_{\rm F}(\Delta f)
\hspace{0.2cm} {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$

*Both definitions are only partially suitable for a time-invariant channel.
*For a time invariant two-path channel (i.e. one with constant path weights according to the above graph), the following approximation for the coherence bandwidth is often used:
:$$B_{\rm K}\hspace{0.01cm}' = \frac{1}{\tau_{\rm max} - \tau_{\rm min}} \hspace{0.05cm}.$$

In this task we want to clarify
* why there are different definitions for the coherence bandwidth 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]].

===Questionnaire===
<quiz display=simple>
{What is the approximate coherence bandwidth   $B_{\rm K}\hspace{0.01cm}'$  for channels  $\rm A$  and  $\rm B$?
|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% } $\ \ \rm kHz$

{Let $G$  be the weight of the second path. What is the PDF  $f_{\rm V}(\tau)$?
|type="()"}
- $f_{\rm V}(\tau) = \delta(\tau) + G \cdot \delta(\tau \, –\tau_0)$,
- $f_{\rm V}(\tau) = \delta(\tau) + G^2 \cdot \delta(\tau \, –\tau_0)$,
+ $f_{\rm V}(\tau) = 1/(1 + G^2) \cdot \delta(\tau) + G^2/(1 + G^2) \cdot \delta(\tau \, –\tau_0)$.

{Calculate the delay spread  $ T_{\rm V}$.
|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  $B_{\rm K}$  of channel  ${\rm A}$ ?
|type="[]"}
+   $B_{\rm K} = 333 \ \rm kHz$.
-   $B_{\rm K} = 500 \ \rm kHz$.
-   $B_{\rm K} = 1 \ \rm MHz$.
- $B_{\rm K}$  cannot be calculated according to this definition.

{What is the coherence bandwidth  $B_{\rm K}$  of channel  ${\rm B}$ ?
|type="[]"}
-   $B_{\rm K} = 333 \ \rm kHz$.
-   $B_{\rm K} = 500 \ \rm kHz$.
-   $B_{\rm K} = 1 \ \ \rm MHz$.
+ $B_{\rm K}$  cannot be calculated according to this definition.
</quiz>

===Sample solution===
{{{ML-Kopf}}
'''(1)'''  For both channels, the delay 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 graphs refer to the impulse response $h(\tau)$.
*To obtain the delay–power density spectrum, 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 of ${\it \Phi}_{\rm V}(\tau)$ is therefore $1 + G^2$.
*The probability density function (PDF), however, must have &bdquo;area 1” (i.e., the sum of the two Dirac weights must be $1$). From this follows:
:$$f_{\rm V}(\tau) = \frac{1}{1 + G^2} \cdot \delta(\tau) + \frac{G^2}{1 + G^2} \cdot \delta(\tau - \tau_0) \hspace{0.05cm}.$$

*So only solution 3 is correct.
*The first option does not describe the PDF $f_{\rm V}(\tau)$, but the impulse response $h(\tau)$.
*The second equation specifies the delay –power spectral density ${\it \Phi}_{\rm V}(\tau)$.

'''(3)'''  For channel  $\rm A$  the two impulse weights are equal.
*This means that the mean value $m_{\rm V}$ and the standard deviation $\sigma_{\rm V} = T_{\rm V}$ can be computed simply:
:$$m_{\rm V} = \frac{\tau_0}{2} \hspace{0.15cm} {= 0.5\,{\rm µ s}}\hspace{0.05cm},
\hspace{0.2cm}T_{\rm V} = \sigma_{\rm V} =\frac{\tau_0}{2} \hspace{0.15cm}\underline {= 0.5\,{\rm µ s}}
\hspace{0.05cm}.$$

For channel  $\rm B$  the Dirac weights are $1/(1+0.5^2) = 0.8$ (for $\tau = 0$) and $0.2$ (for $\tau = 1 \ \rm µ s$).
* According to the [[Stochastische_Signaltheorie/Erwartungswerte_und_Momente#Momentenberechnung_als_Scharmittelwert|basic laws]] of statistics, the noncentral first and second order moments are:
:$$m_{\rm 1} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0 + 0.2 \cdot 1\,{\rm µ s} = 0.2\,{\rm µ s} \hspace{0.05cm},\hspace{0.5cm}
m_{\rm 2} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0^2 + 0.2 \cdot (1\,{\rm µ s})^2 = 0.2\,({\rm µ s})^2 \hspace{0.05cm}.$$

*To get the result you are looking for you can use the [[Stochastische_Signaltheorie/Erwartungswerte_und_Momente#Einige_h.C3.A4ufig_benutzte_Zentralmomente| Steiner's Theorem]].
:$$\sigma_{\rm V}^2 = m_{\rm 2} - m_{\rm 1}^2 = 0.2\,({\rm µ s})^2 - (0.2\,{\rm µ s})^2 = 0.16\,({\rm µ s})^2
\hspace{0.3cm}\Rightarrow \hspace{0.3cm}T_{\rm V} = \sigma_{\rm V} \hspace{0.15cm}\underline {= 0.4\,{\rm µ s}}\hspace{0.05cm}.$$

'''(4)'''  The frequency correlation function is the Fourier transform of ${\it \Phi}_{\rm V}(\tau) = \delta(\tau) + \delta(\tau \, – \tau_0)$:
:$$\varphi_{\rm F}(\Delta f) = 1 + {\rm exp}(-{\rm j} \cdot 2\pi \cdot \Delta f \cdot \tau_0) = 1 + {\rm cos}(2\pi \cdot \Delta f \cdot \tau_0) -{\rm j} \cdot {\rm sin}(2\pi \cdot \Delta f \cdot \tau_0) $$
[[Datei:P_ID2186__Mob_Z_2_7d.png|right|frame|Frequency correlation function and coherence bandwidth]]
:$$\Rightarrow \hspace{0.3cm} |\varphi_{\rm F}(\Delta f)| = \sqrt{2 + 2 \cdot {\rm cos}(2\pi \cdot \Delta f \cdot \tau_0) }\hspace{0.05cm}.$$

*The maximum at $\delta f = 0$ is equal to $2$.
*Therefore the equation to determine $B_{\rm K}$ is
$$|\varphi_{\rm F}(B_{\rm K})| = 1 \hspace{0.3cm} $$
$$\Rightarrow \hspace{0.3cm}|\varphi_{\rm F}(B_{\rm K})|^2 = 1
\hspace{0.3cm} \Rightarrow \hspace{0.3cm}2 + 2 \cdot {\rm cos}(2\pi \cdot B_{\rm K} \cdot \tau_0) = 1$$
$$\Rightarrow \hspace{0.3cm}{\rm cos}(2\pi \cdot B_{\rm K} \cdot \tau_0) = -0.5 \hspace{0.3cm} $$
$$\Rightarrow \hspace{0.3cm}2\pi \cdot B_{\rm K} \cdot \tau_0 = \frac{2\pi}{3}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}B_{\rm K} = \frac{1}{3\tau_0} = 333\,{\rm kHz}\hspace{0.05cm}.$$

*Solution 1 is therefore correct. The graph (blue curve) illustrates the result.

'''(5)'''  For 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.
*Therefore, solution 4 is correct.

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 &bdquo;coherence bandwidths” can be very different.

{{ML-Fuß}}

[[Category:Exercises for Mobile Communications|^2.3 The GWSSUS Channel Model^]]