Difference between revisions of "Aufgaben:Exercise 2.7: Coherence Bandwidth"

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m (Text replacement - "power spectral density" to "power-spectral density")
 
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{{quiz-Header|Buchseite=Mobile Kommunikation/Das GWSSUS–Kanalmodell}}
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{{quiz-Header|Buchseite=Mobile_Communications/The_GWSSUS_Channel_Model}}
  
[[File:P_ID2172__Mob_A_2_7.png|right|frame|Verzögerungs–LDS und <br>Frequenz&ndash;Korrelationsfunktion]]
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[[File:P_ID2172__Mob_A_2_7.png|right|frame|Delay power-spectral density and <br>frequency correlation function]]
For the power spectral density of the delay, we assume an exponential behavior. With&nbsp; ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$&nbsp; we have
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For the delay power-spectral density, we assume an exponential behavior.&nbsp; With&nbsp; ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$&nbsp; we have
 
:$${{\it \phi}_{\rm V}(\tau)}/{{\it \phi}_{\rm 0}} = {\rm e}^{ -\tau / \tau_0 } \hspace{0.05cm}.$$
 
:$${{\it \phi}_{\rm V}(\tau)}/{{\it \phi}_{\rm 0}} = {\rm e}^{ -\tau / \tau_0 } \hspace{0.05cm}.$$
  
The constant&nbsp; $\tau_0$&nbsp; can be determined from the tangent in the point&nbsp; $\tau = 0$&nbsp; according to the upper graph. Note that&nbsp; ${\it \Phi}_{\rm V}(\tau)$&nbsp; has dimension&nbsp; $[1/\rm s]$&nbsp;. Furthermore,
+
The constant&nbsp; $\tau_0$&nbsp; can be determined from the tangent in the point&nbsp; $\tau = 0$&nbsp; according to the upper graph.&nbsp; Note that&nbsp; ${\it \Phi}_{\rm V}(\tau)$&nbsp; has unit&nbsp; $[1/\rm s]$&nbsp;.&nbsp; Furthermore,
* The probability density function (PDF)&nbsp; $f_{\rm V}(\tau)$&nbsp; has the same form as&nbsp; ${\it \Phi}_{\rm V}(\tau)$, but is normalized to area &nbsp;$1$&nbsp;.
+
* The probability density function&nbsp; $\rm (PDF)$&nbsp; $f_{\rm V}(\tau)$&nbsp; has the same form as&nbsp; ${\it \Phi}_{\rm V}(\tau)$, but is normalized to area &nbsp;$1$&nbsp;.
* The &nbsp;<b>average excess delay</b> or <b>mean excess delay</b>&nbsp; $m_{\rm V}$&nbsp; is equal to the linear expectation&nbsp; $E\big [\tau \big]$&nbsp; and can be determined from the PDF&nbsp; $f_{\rm V}(\tau)$&nbsp;.
+
* The &nbsp;<b>average excess delay</b> or <b>mean excess delay</b>&nbsp; $m_{\rm V}$&nbsp; is equal to the linear expectation&nbsp; $E\big [\tau \big]$&nbsp; and can be determined from the PDF $f_{\rm V}(\tau)$&nbsp;.
* The &nbsp;<b>multipath spread</b> or <b>delay spread</b>&nbsp; $\sigma_{\rm V}$&nbsp; gives the standard deviation (dispersion) of the random variable&nbsp; $\tau$&nbsp;. In the theory part we also use the term&nbsp; $T_{\rm V}$ for this.
+
* The &nbsp;<b>multipath spread</b> or <b>delay spread</b>&nbsp; $\sigma_{\rm V}$&nbsp; gives the standard deviation of the random variable&nbsp; $\tau$&nbsp;.&nbsp; In the theory part we also use the term&nbsp; $T_{\rm V}$ for this.
* The displayed frequency correlation function&nbsp; $\varphi_{\rm F}(\delta f)$&nbsp; can be calculated as the Fourier transform of the power spectral density of the delay&nbsp; ${\it \Phi}_{\rm V}(\tau)$&nbsp;:
+
* The displayed frequency correlation function&nbsp; $\varphi_{\rm F}(\Delta f)$&nbsp; can be calculated as the Fourier transform of the delay power-spectral density&nbsp; ${\it \Phi}_{\rm V}(\tau)$&nbsp;:
 
:$$\varphi_{\rm F}(\Delta f)
 
:$$\varphi_{\rm F}(\Delta f)
 
  \hspace{0.2cm}  {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$
 
  \hspace{0.2cm}  {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$
* The <b>coherence bandwidth</b>&nbsp; $B_{\rm K}$&nbsp; is the value of &nbsp; $\Delta f$ at which the frequency correlation function&nbsp; $\varphi_{\rm F}(\Delta f)$&nbsp; has dropped to half in absolute value.
+
* The <b>coherence bandwidth</b>&nbsp; $B_{\rm K}$&nbsp; is the value of $\Delta f$ at which the frequency correlation function&nbsp; $\varphi_{\rm F}(\Delta f)$&nbsp; has dropped to half in absolute value.
  
  
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''Notes:''
 
''Notes:''
* This task belongs to the topic of the chapter&nbsp; [[Mobile_Kommunikation/Das_GWSSUS%E2%80%93Kanalmodell|GWSSUS&ndash;Kanalmodell]].
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* This exercise belongs to the chapter&nbsp; [[Mobile_Communications/The_GWSSUS_Channel_Model|The GWSSUS Channel Model]].
* This task requires knowledge of &nbsp; [[Stochastische_Signaltheorie/Erwartungswerte_und_Momente#Momentenberechnung_als_Scharmittelwert| computation of moments]]&nbsp; of random variables from the book &bdquo;Stochastic Signal Theory&rdquo;.
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* This exercise requires knowledge of&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|Expected values and moments]]&nbsp; from the book "Theory of Stochastic Signals".
 
* In addition, the following Fourier transform is given:
 
* In addition, the following Fourier transform is given:
 
:$$x(t) = \left\{ \begin{array}{c} {\rm e}^{- \lambda \hspace{0.05cm}\cdot \hspace{0.05cm} t}\\
 
:$$x(t) = \left\{ \begin{array}{c} {\rm e}^{- \lambda \hspace{0.05cm}\cdot \hspace{0.05cm} t}\\
 
  0  \end{array} \right.\quad
 
  0  \end{array} \right.\quad
\begin{array}{*{1}c} \hspace{-0.35cm} {\rm f\ddot{u}r} \hspace{0.15cm} t \ge 0
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\begin{array}{*{1}c} \hspace{-0.35cm} {\rm for} \hspace{0.15cm} t \ge 0
\\ \hspace{-0.35cm} {\rm f\ddot{u}r} \hspace{0.15cm} t < 0 \\ \end{array}  
+
\\ \hspace{-0.35cm} {\rm for} \hspace{0.15cm} t < 0 \\ \end{array}  
 
  \hspace{0.4cm}  {\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet} \hspace{0.4cm} X(f) = \frac{1}{\lambda + {\rm j} \cdot 2\pi f}\hspace{0.05cm}.$$
 
  \hspace{0.4cm}  {\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet} \hspace{0.4cm} X(f) = \frac{1}{\lambda + {\rm j} \cdot 2\pi f}\hspace{0.05cm}.$$
 
   
 
   
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===Questionnaire===
 
===Questionnaire===
 
<quiz display=simple>
 
<quiz display=simple>
{What is the probability density&nbsp; $f_{\rm V}(\tau)$&nbsp; of the delay?
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{What is the probability density function&nbsp; $f_{\rm V}(\tau)$&nbsp; of the delay time?
|type="[]"}
+
|type="()"}
 
- $f_{\rm V}(\tau) = {\rm e}^{-\tau/\tau_0}$.
 
- $f_{\rm V}(\tau) = {\rm e}^{-\tau/\tau_0}$.
 
+ $f_{\rm V}(\tau) = 1/\tau_0 \cdot {\rm e}^{-\tau/\tau_0}$,
 
+ $f_{\rm V}(\tau) = 1/\tau_0 \cdot {\rm e}^{-\tau/\tau_0}$,
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$m_{\rm V} \ = \ ${ 1 3% } $\ \rm &micro; s$
 
$m_{\rm V} \ = \ ${ 1 3% } $\ \rm &micro; s$
  
{Which value results for the multipath widening with&nbsp; $\tau_0 = 1 \ \ \rm &micro; s$?
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{Which value results for the multipath spread with&nbsp; $\tau_0 = 1 \ \ \rm &micro; s$?
 
|type="{}"}
 
|type="{}"}
 
$\sigma_{\rm V} \ = \ ${ 1 3% } $\ \rm &micro; s$
 
$\sigma_{\rm V} \ = \ ${ 1 3% } $\ \rm &micro; s$
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</quiz>
 
</quiz>
  
===Sample solution===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Let ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$. The integral of the power spectral density of the delay gives
+
'''(1)'''&nbsp; Let&nbsp; ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$.&nbsp; The integral of the delay power-spectral density gives
 
:$$\int_{0}^{+\infty} {\it \Phi}_{\rm V}(\tau) \hspace{0.15cm}{\rm d} \tau =   
 
:$$\int_{0}^{+\infty} {\it \Phi}_{\rm V}(\tau) \hspace{0.15cm}{\rm d} \tau =   
 
  {\it \Phi}_{\rm 0} \cdot \int_{0}^{+\infty} {\rm e}^{-\tau / \tau_0} \hspace{0.15cm}{\rm d} \tau =  
 
  {\it \Phi}_{\rm 0} \cdot \int_{0}^{+\infty} {\rm e}^{-\tau / \tau_0} \hspace{0.15cm}{\rm d} \tau =  
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  \hspace{0.05cm}. $$
 
  \hspace{0.05cm}. $$
  
*The probability density function is then
+
*Then the probability density function is
 
:$$f_{\rm V}(\tau)  = \frac{{\it \Phi}_{\rm V}(\tau) }{{\it \Phi}_{\rm 0} \cdot \tau_0}= \frac{1}{\tau_0} \cdot  {\rm e}^{-\tau / \tau_0}  
 
:$$f_{\rm V}(\tau)  = \frac{{\it \Phi}_{\rm V}(\tau) }{{\it \Phi}_{\rm 0} \cdot \tau_0}= \frac{1}{\tau_0} \cdot  {\rm e}^{-\tau / \tau_0}  
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
 
+
*<u>Solution 2</u> is correct.
*<u>Solution 2</u> is therefore correct.
 
 
 
  
  
'''(2)'''&nbsp; The $k$-th moment of an [[Stochastische_Signaltheorie/Exponentialverteilte_Zufallsgr%C3%B6%C3%9Fen#Einseitige_Exponentialverteilung| exponential random variable]] is $m_k = k! \cdot \tau_0^k$.  
+
'''(2)'''&nbsp; The&nbsp; $k$-th moment of an&nbsp; [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables#One-sided_exponential_distribution|one-sided exponential random variable]]&nbsp; is&nbsp; $m_k = k! \cdot \tau_0^k$.  
*With $k = 1$, this results in the linear mean value $m_1 = m_{\rm V}$:
+
*With&nbsp; $k = 1$, this results in the linear mean value&nbsp; $m_1 = m_{\rm V}$:
 
:$$m_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm &micro; s}}
 
:$$m_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm &micro; s}}
 
   \hspace{0.05cm}. $$
 
   \hspace{0.05cm}. $$
  
'''(3)'''&nbsp; According to the [[Stochastische_Signaltheorie/Erwartungswerte_und_Momente#Einige_h.C3.A4ufig_benutzte_Zentralmomente| Steiner's Theorem]], the  variance of any random variable is $\sigma^2 = m_2 \, &ndash;m_1^2$.  
+
 
*This yields $m_2 = 2 \cdot \tau_0^2$, and therefore
+
'''(3)'''&nbsp; According to the&nbsp; [[Theory_of_Stochastic_Signals/Erwartungswerte_und_Momente#Some common central moments| Steiner's Theorem]], the  variance of any random variable is&nbsp; $\sigma^2 = m_2 \, -m_1^2$.  
 +
*This yields&nbsp; $m_2 = 2 \cdot \tau_0^2$, and therefore
 
:$$\sigma_{\rm V}^2 = m_2 - m_1^2 = 2 \cdot \tau_0^2 - (\tau_0)^2 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
 
:$$\sigma_{\rm V}^2 = m_2 - m_1^2 = 2 \cdot \tau_0^2 - (\tau_0)^2 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
 
   \sigma_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm &micro; s}}
 
   \sigma_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm &micro; s}}
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'''(4)'''&nbsp; ${\it \Phi}_{\rm V}(\tau)$ is identical to $x(t)$ in the given Fourier transform pair if $t$ is replaced by $\tau$ and $\lambda$ by $1/\tau_0$.  
+
'''(4)'''&nbsp; ${\it \Phi}_{\rm V}(\tau)$&nbsp; is identical to&nbsp; $x(t)$&nbsp; in the given Fourier transform pair if&nbsp; $t$&nbsp; is replaced by&nbsp; $\tau$&nbsp; and&nbsp; $\lambda$&nbsp; by&nbsp; $1/\tau_0$.  
*Thus, $\varphi_{\rm F}(\delta f)$ is equal to $X(f)$ with the substitution $f &#8594; \delta f$:
+
*Thus,&nbsp; $\varphi_{\rm F}(\Delta f)$&nbsp; is equal to&nbsp; $X(f)$&nbsp; with the substitution&nbsp; $f &#8594; \Delta f$:
 
:$$\varphi_{\rm F}(\Delta f) = \frac{1}{1/\tau_0 + {\rm j} \cdot 2\pi \Delta f}
 
:$$\varphi_{\rm F}(\Delta f) = \frac{1}{1/\tau_0 + {\rm j} \cdot 2\pi \Delta f}
 
  = \frac{\tau_0}{1 + {\rm j} \cdot 2\pi \cdot \tau_0 \cdot \Delta f}\hspace{0.05cm}.$$
 
  = \frac{\tau_0}{1 + {\rm j} \cdot 2\pi \cdot \tau_0 \cdot \Delta f}\hspace{0.05cm}.$$
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  B_{\rm K}= \frac{\sqrt{3}}{2\pi \cdot \tau_0} \approx \frac{0.276}{ \tau_0}\hspace{0.05cm}. $$
 
  B_{\rm K}= \frac{\sqrt{3}}{2\pi \cdot \tau_0} \approx \frac{0.276}{ \tau_0}\hspace{0.05cm}. $$
  
*With $\tau_0 = 1 \ \ \rm &micro; s$, the coherence bandwidth is $B_{\rm K} \ \underline {= 276 \ \ \rm kHz}$.
+
*With&nbsp; $\tau_0 = 1 \ \ \rm &micro; s$, the coherence bandwidth is&nbsp; $B_{\rm K} \ \underline {= 276 \ \ \rm kHz}$.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
[[Category:Exercises for Mobile Communications|^2.3 The GWSSUS Channel Model^]]
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[[Category:Mobile Communications: Exercises|^2.3 The GWSSUS Channel Model^]]

Latest revision as of 12:41, 17 February 2022

Delay power-spectral density and
frequency correlation function

For the delay power-spectral density, we assume an exponential behavior.  With  ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$  we have

$${{\it \phi}_{\rm V}(\tau)}/{{\it \phi}_{\rm 0}} = {\rm e}^{ -\tau / \tau_0 } \hspace{0.05cm}.$$

The constant  $\tau_0$  can be determined from the tangent in the point  $\tau = 0$  according to the upper graph.  Note that  ${\it \Phi}_{\rm V}(\tau)$  has unit  $[1/\rm s]$ .  Furthermore,

  • The probability density function  $\rm (PDF)$  $f_{\rm V}(\tau)$  has the same form as  ${\it \Phi}_{\rm V}(\tau)$, but is normalized to area  $1$ .
  • The  average excess delay or mean excess delay  $m_{\rm V}$  is equal to the linear expectation  $E\big [\tau \big]$  and can be determined from the PDF $f_{\rm V}(\tau)$ .
  • The  multipath spread or delay spread  $\sigma_{\rm V}$  gives the standard deviation of the random variable  $\tau$ .  In the theory part we also use the term  $T_{\rm V}$ for this.
  • The displayed frequency correlation function  $\varphi_{\rm F}(\Delta f)$  can be calculated as the Fourier transform of the delay power-spectral density  ${\it \Phi}_{\rm V}(\tau)$ :
$$\varphi_{\rm F}(\Delta f) \hspace{0.2cm} {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$
  • The coherence bandwidth  $B_{\rm K}$  is the value of $\Delta f$ at which the frequency correlation function  $\varphi_{\rm F}(\Delta f)$  has dropped to half in absolute value.




Notes:

$$x(t) = \left\{ \begin{array}{c} {\rm e}^{- \lambda \hspace{0.05cm}\cdot \hspace{0.05cm} t}\\ 0 \end{array} \right.\quad \begin{array}{*{1}c} \hspace{-0.35cm} {\rm for} \hspace{0.15cm} t \ge 0 \\ \hspace{-0.35cm} {\rm for} \hspace{0.15cm} t < 0 \\ \end{array} \hspace{0.4cm} {\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet} \hspace{0.4cm} X(f) = \frac{1}{\lambda + {\rm j} \cdot 2\pi f}\hspace{0.05cm}.$$



Questionnaire

1

What is the probability density function  $f_{\rm V}(\tau)$  of the delay time?

$f_{\rm V}(\tau) = {\rm e}^{-\tau/\tau_0}$.
$f_{\rm V}(\tau) = 1/\tau_0 \cdot {\rm e}^{-\tau/\tau_0}$,
$f_{\rm V}(\tau) = {\it \Phi}_0 \cdot {\rm e}^{-\tau/\tau_0}$.

2

Determine the average delay time for  $\tau_0 = 1 \ \ \rm µ s$.

$m_{\rm V} \ = \ $

$\ \rm µ s$

3

Which value results for the multipath spread with  $\tau_0 = 1 \ \ \rm µ s$?

$\sigma_{\rm V} \ = \ $

$\ \rm µ s$

4

What is the frequency–correlation function  $\varphi_{\rm F}(\Delta f)$?

$\varphi_{\rm F}(\Delta f) = \big[1/\tau_0 + {\rm j} \ 2 \pi \cdot \delta f \big]^{-1}$,
$\varphi_{\rm F}(\Delta f) = {\rm e}^ {-(\tau_0 \hspace{0.05cm}\cdot \hspace{0.05cm}\Delta f)^2}$.

5

Determine the coherence bandwidth  $B_{\rm K}$.

$B_{\rm K} \ = \ $

$\ \ \rm kHz$


Solution

(1)  Let  ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$.  The integral of the delay power-spectral density gives

$$\int_{0}^{+\infty} {\it \Phi}_{\rm V}(\tau) \hspace{0.15cm}{\rm d} \tau = {\it \Phi}_{\rm 0} \cdot \int_{0}^{+\infty} {\rm e}^{-\tau / \tau_0} \hspace{0.15cm}{\rm d} \tau = {\it \Phi}_{\rm 0} \cdot \tau_0 \hspace{0.05cm}. $$
  • Then the probability density function is
$$f_{\rm V}(\tau) = \frac{{\it \Phi}_{\rm V}(\tau) }{{\it \Phi}_{\rm 0} \cdot \tau_0}= \frac{1}{\tau_0} \cdot {\rm e}^{-\tau / \tau_0} \hspace{0.05cm}.$$
  • Solution 2 is correct.


(2)  The  $k$-th moment of an  one-sided exponential random variable  is  $m_k = k! \cdot \tau_0^k$.

  • With  $k = 1$, this results in the linear mean value  $m_1 = m_{\rm V}$:
$$m_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}} \hspace{0.05cm}. $$


(3)  According to the  Steiner's Theorem, the variance of any random variable is  $\sigma^2 = m_2 \, -m_1^2$.

  • This yields  $m_2 = 2 \cdot \tau_0^2$, and therefore
$$\sigma_{\rm V}^2 = m_2 - m_1^2 = 2 \cdot \tau_0^2 - (\tau_0)^2 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \sigma_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}} \hspace{0.05cm}. $$


(4)  ${\it \Phi}_{\rm V}(\tau)$  is identical to  $x(t)$  in the given Fourier transform pair if  $t$  is replaced by  $\tau$  and  $\lambda$  by  $1/\tau_0$.

  • Thus,  $\varphi_{\rm F}(\Delta f)$  is equal to  $X(f)$  with the substitution  $f → \Delta f$:
$$\varphi_{\rm F}(\Delta f) = \frac{1}{1/\tau_0 + {\rm j} \cdot 2\pi \Delta f} = \frac{\tau_0}{1 + {\rm j} \cdot 2\pi \cdot \tau_0 \cdot \Delta f}\hspace{0.05cm}.$$
  • The first expression is correct.


(5)  The coherence bandwidth is implicit in the following equation:

$$|\varphi_{\rm F}(\Delta f = B_{\rm K})| \stackrel {!}{=} \frac{1}{2} \cdot |\varphi_{\rm F}(\Delta f = 0)| = \frac{\tau_0}{2}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}|\varphi_{\rm F}(\Delta f = B_{\rm K})|^2 = \frac{\tau_0^2}{1 + (2\pi \cdot \tau_0 \cdot B_{\rm K})^2} \stackrel {!}{=} \frac{\tau_0^2}{4}$$
$$\Rightarrow \hspace{0.3cm}(2\pi \cdot \tau_0 \cdot B_{\rm K})^2 = 3 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} B_{\rm K}= \frac{\sqrt{3}}{2\pi \cdot \tau_0} \approx \frac{0.276}{ \tau_0}\hspace{0.05cm}. $$
  • With  $\tau_0 = 1 \ \ \rm µ s$, the coherence bandwidth is  $B_{\rm K} \ \underline {= 276 \ \ \rm kHz}$.