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

 
{{quiz-Header|Buchseite=Mobile_Communications/The_GWSSUS_Channel_Model}}

[[File:EN_Mob_A_2_7Z.png|right|frame|Two two-path channels]]
For the GWSSUS model, two parameters are given, which both statistically capture the resulting delay&nbsp; $\tau$.&nbsp; More information on the topic "Multipath Propagation" can be found in the section&nbsp; [[Mobile_Communications/The_GWSSUS_Channel_Model#Parameters_of_the_GWSSUS_model| Parameters of the GWSSUS model]]&nbsp; of the theory part.
* The &nbsp;<b>delay spread</b>&nbsp; $T_{\rm V}$&nbsp; is by definition equal to the standard deviation of the random variable &nbsp;$\tau$. <br>This can be determined from the probability density function&nbsp; $f_{\rm V}(\tau)$.&nbsp; The PDF&nbsp; $f_{\rm V}(\tau)$&nbsp; has the same shape as the delay power-spectral density&nbsp; ${\it \Phi}_{\rm V}(\tau)$. 
* The &nbsp;<b>coherence bandwidth</b>&nbsp; $B_{\rm K}$&nbsp; describes the same situation in the frequency domain. <br> It is defined as the value of $\Delta f$ at which the magnitude of the frequency correlation function&nbsp; $\varphi_{\rm F}(\Delta f)$&nbsp; 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&nbsp; ${\it \Phi}_{\rm V}(\tau)$&nbsp; and&nbsp; $\varphi_{\rm F}(\Delta f)$&nbsp; 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&nbsp; $B_{\rm K}$&nbsp; and&nbsp; $B_{\rm K}\hspace{0.01cm}'$,&nbsp; and
* which definitions make sense for which boundary conditions.






''Notes:''
*This exercise belongs to the chapter&nbsp; [[Mobile_Communications/The_GWSSUS_Channel_Model|The GWSSUS Channel Model]].
*This task also refers to some theory pages in chapter&nbsp;  [[Mobile_Communications/Multi-Path_Reception_in_Mobile_Communications| Multi-Path Reception in Mobile Communications]].
 



===Questionnaire===
<quiz display=simple>
{What is the approximate coherence bandwidth&nbsp; $B_{\rm K}\hspace{0.01cm}'$&nbsp; for channels &nbsp;$\rm A$&nbsp; and &nbsp;$\rm B$?
|type="{}"}
Channel &nbsp;${\rm A} \text{:} \hspace{0.4cm} B_{\rm K}\hspace{0.01cm}' \ = \ ${ 1000 3% } $\ \ \rm kHz$
Channel &nbsp;${\rm B} \text{:} \hspace{0.4cm} B_{\rm K}\hspace{0.01cm}' \ = \ ${ 1000 3% } $\ \ \rm kHz$

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

{Calculate the delay spread&nbsp; $ T_{\rm V}$.
|type="{}"}
Channel &nbsp;${\rm A} \text{:} \hspace{0.4cm} T_{\rm V} \ = \ ${ 0.5 3% } $\ \rm &micro; s$
Channel &nbsp;${\rm B} \text{:} \hspace{0.4cm} T_{\rm V} \ = \ ${ 0.4 3% } $\ \rm &micro; s$

{What is the coherence bandwidth&nbsp; $B_{\rm K}$&nbsp; of channel &nbsp;${\rm A}$&nbsp;?
|type="()"}
+ &nbsp; $B_{\rm K} = 333 \ \rm kHz$.
- &nbsp; $B_{\rm K} = 500 \ \rm kHz$.
- &nbsp; $B_{\rm K} = 1 \ \rm MHz$.
- $B_{\rm K}$&nbsp; cannot be calculated according to this definition.

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

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; For both channels, the delay difference is&nbsp; $\Delta \tau = \tau_{\rm max} \, - \tau_{\rm min} = 1 \ \ \rm &micro; s$.
* That's why both channels have the same value:
:$$B_{\rm K}\hspace{0.01cm}' \ \ \underline {= 1000 \ \rm kHz}.$$


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

*Only <u>solution 3</u> is correct. 
*The first option does not describe the PDF&nbsp; $f_{\rm V}(\tau)$, but the impulse response&nbsp; $h(\tau)$. 
*The second equation specifies the delay power-spectral density&nbsp; ${\it \Phi}_{\rm V}(\tau)$.



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

For channel &nbsp;$\rm B$&nbsp; the Dirac weights are&nbsp; $1/(1+0.5^2) = 0.8$&nbsp; $($for&nbsp; $\tau = 0)$&nbsp; and&nbsp; $0.2$&nbsp; $($for&nbsp; $\tau = 1 \ \rm &micro; s)$.
* According to the&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|basic laws]]&nbsp; of statistics, the non-central first and second order moments are:
:$$m_{\rm 1} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0 + 0.2 \cdot 1\,{\rm &micro; s} = 0.2\,{\rm &micro; s} \hspace{0.05cm},\hspace{0.5cm}
m_{\rm 2} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0^2 + 0.2 \cdot (1\,{\rm &micro; s})^2 = 0.2\,({\rm &micro; s})^2 \hspace{0.05cm}.$$

*To get the result you are looking for you can use the&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments#Some_common_central_moments| Steiner's Theorem]]:
:$$\sigma_{\rm V}^2 = m_{\rm 2} - m_{\rm 1}^2 = 0.2\,({\rm &micro; s})^2 - (0.2\,{\rm &micro; s})^2 = 0.16\,({\rm &micro; s})^2
  \hspace{0.3cm}\Rightarrow \hspace{0.3cm}T_{\rm V} = \sigma_{\rm V}  \hspace{0.15cm}\underline {= 0.4\,{\rm &micro; s}}\hspace{0.05cm}.$$


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

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

<u>Solution 1</u> is correct.&nbsp; The graph (blue curve) illustrates the result.


'''(5)'''&nbsp; For channel &nbsp;${\rm B}$&nbsp; the corresponding equations are
:$${\it \Phi}_{\rm V}(\tau) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1^2 \cdot \delta(\tau) + (-0.5)^2 \cdot \delta(\tau - \tau_0) \hspace{0.05cm},\hspace{0.05cm}$$
:$$\varphi_{\rm F}(\Delta f) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  1 + 0.25 \cdot {\rm cos}(2\pi \cdot \Delta f \cdot \tau_0) -{\rm j} \cdot 0.25 \cdot  {\rm sin}(2\pi \cdot \Delta f \cdot \tau_0)\hspace{0.05cm},$$
:$$|\varphi_{\rm F}(\Delta f)| \hspace{-0.1cm} \ = \  \sqrt{\frac{17}{16} +  \frac{1}{2} \cdot {\rm cos}(2\pi \cdot \Delta f \cdot \tau_0) }\hspace{0.3cm}$$
:$$\Rightarrow \hspace{0.3cm}{\rm Max}\hspace{0.1cm}|\varphi_{\rm F}(\Delta f)| = 1.25\hspace{0.05cm},\hspace{0.2cm}{\rm Min}\hspace{0.1cm}|\varphi_{\rm F}(\Delta f)| = 0.75\hspace{0.05cm}.$$

You can see from this result that the&nbsp; $50\%$&ndash;coherence bandwidth cannot be specified here &nbsp; &rArr; &nbsp;  <u>solution 4</u> is correct.


This result is the reason why there are different definitions for the coherence bandwidth in the literature, for example
* the&nbsp; $90\%$&ndash;coherence bandwidth&nbsp; $($in the example&nbsp; $B_{\rm K, \hspace{0.03cm} 90\%} =184 \ \ \rm kHz$$)$,
* the very simple approximation&nbsp; $B_{\rm K}\hspace{0.01cm}'$&nbsp; given above&nbsp; $($in the example&nbsp; $B_{\rm K}\hspace{0.01cm}' =1 \ \ \rm MHz)$.


You can see from these numerical values that all the information on this topic is very vague and that the individual "coherence bandwidths" can be very different.

{{ML-Fuß}}


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