Aufgaben:Exercise 2.7: Coherence Bandwidth - Revision history

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:30:40Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:30, 16 March 2026
Line 11:		Line 11:
	* The  <b>multipath spread</b> or <b>delay spread</b>  $\sigma_{\rm V}$  gives the standard deviation of the random variable  $\tau$ .  In the theory part we also use the term  $T_{\rm V}$ for this.		* The  <b>multipath spread</b> or <b>delay spread</b>  $\sigma_{\rm V}$  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  ${\it \Phi}_{\rm V}(\tau)$ :		* The displayed frequency correlation function  $\varphi_{\rm F}(\Delta f)$  can be calculated as the Fourier transform of the delay power-spectral density  ${\it \Phi}_{\rm V}(\tau)$ :
	:$$\varphi_{\rm F}(\Delta f)		:$$\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>  $B_{\rm K}$  is the value of $\Delta f$ at which the frequency correlation function  $\varphi_{\rm F}(\Delta f)$  has dropped to half in absolute value.		* The <b>coherence bandwidth</b>  $B_{\rm K}$  is the value of $\Delta f$ at which the frequency correlation function  $\varphi_{\rm F}(\Delta f)$  has dropped to half in absolute value.

Line 24:		Line 23:
	* This exercise requires knowledge of  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments\|Expected values and moments]]  from the book "Theory of Stochastic Signals".		* This exercise requires knowledge of  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments\|Expected values and moments]]  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\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}.$$
	0 \end{array} \right.\quad
	\begin{array}{*{1}c} \hspace{-0.35cm} {\rm for} \hspace{0.15cm} t \ge 0
	\\ \hspace{-0.35cm} {\rm for} \hspace{0.15cm} t < 0 \\ \end{array}
	\hspace{0.4cm} {\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet} \hspace{0.4cm} X(f) = \frac{1}{\lambda + {\rm j} \cdot 2\pi f}\hspace{0.05cm}.$$


Line 62:		Line 57:
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Let  ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$.  The integral of the delay power-spectral density gives		'''(1)'''  Let  ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$.  The integral of the delay power-spectral density gives
	:$$\int_{0}^{+\infty} {\it \Phi}_{\rm V}(\tau) \hspace{0.15cm}{\rm d} \tau =		:$$\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}. $$
	{\it \Phi}_{\rm 0} \cdot \int_{0}^{+\infty} {\rm e}^{-\tau / \tau_0} \hspace{0.15cm}{\rm d} \tau =
	{\it \Phi}_{\rm 0} \cdot \tau_0
	\hspace{0.05cm}. $$

	*Then the probability density function is		*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 correct.
Line 76:		Line 67:
	'''(2)'''  The  $k$-th moment of an  [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables#One-sided_exponential_distribution\|one-sided exponential random variable]]  is  $m_k = k! \cdot \tau_0^k$.		'''(2)'''  The  $k$-th moment of an  [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables#One-sided_exponential_distribution\|one-sided exponential random variable]]  is  $m_k = k! \cdot \tau_0^k$.
	*With  $k = 1$, this results in the linear mean value  $m_1 = m_{\rm V}$:		*With  $k = 1$, this results in the linear mean value  $m_1 = m_{\rm V}$:
	:$$m_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}}		:$$m_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}}\hspace{0.05cm}. $$
	\hspace{0.05cm}. $$


	'''(3)'''  According to the  [[Theory_of_Stochastic_Signals/~~Erwartungswerte_und_Momente~~#~~Some common central moments~~\| Steiner's Theorem]], the variance of any random variable is  $\sigma^2 = m_2 \, -m_1^2$.		'''(3)'''  According to the  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments#Some_common_central_moments\| 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		*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}^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}. $$
	\sigma_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}}
	\hspace{0.05cm}. $$


Line 90:		Line 78:
	'''(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  $1/\tau_0$.		'''(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  $f → \Delta f$:		*Thus,  $\varphi_{\rm F}(\Delta f)$  is equal to  $X(f)$  with the substitution  $f → \Delta f$:
	:$$\varphi_{\rm F}(\Delta f) = \frac{1}{1/\tau_0 + {\rm j} \cdot 2\pi \Delta f}		:$$\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}.$$
	= \frac{\tau_0}{1 + {\rm j} \cdot 2\pi \cdot \tau_0 \cdot \Delta f}\hspace{0.05cm}.$$

	*The <u>first expression</u> is correct.		*The <u>first expression</u> is correct.
Line 98:		Line 85:

	'''(5)'''  The coherence bandwidth is implicit in the following equation:		'''(5)'''  The coherence bandwidth is implicit in the following equation:
	:$$\|\varphi_{\rm F}(\Delta f = B_{\rm K})\| \stackrel {!}{=} \frac{1}{2} \cdot \|\varphi_{\rm F}(\Delta f = 0)\| = \frac{\tau_0}{2}\hspace{0.3cm}		:$$\|\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}$$
	\Rightarrow \hspace{0.3cm}\|\varphi_{\rm F}(\Delta f = B_{\rm K})\|^2		:$$\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}. $$
	= \frac{\tau_0^2}{1 + (2\pi \cdot \tau_0 \cdot B_{\rm K})^2} \stackrel {!}{=} \frac{\tau_0^2}{4}$$
	:$$\Rightarrow \hspace{0.3cm}(2\pi \cdot \tau_0 \cdot B_{\rm K})^2 = 3 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
	B_{\rm K}= \frac{\sqrt{3}}{2\pi \cdot \tau_0} \approx \frac{0.276}{ \tau_0}\hspace{0.05cm}. $$

	*With  $\tau_0 = 1 \ \ \rm µ s$, the coherence bandwidth is  $B_{\rm K} \ \underline {= 276 \ \ \rm kHz}$.		*With  $\tau_0 = 1 \ \ \rm µ s$, the coherence bandwidth is  $B_{\rm K} \ \underline {= 276 \ \ \rm kHz}$.

Maintenance script: Add German interlanguage link

2026-03-16T12:27:07Z

Add German interlanguage link

← Older revision		Revision as of 14:27, 16 March 2026
Line 108:		Line 108:

	[[Category:Mobile Communications: Exercises\|^2.3 The GWSSUS Channel Model^]]		[[Category:Mobile Communications: Exercises\|^2.3 The GWSSUS Channel Model^]]
			[[de:Exercises:Exercise_2.7:_Coherence_Bandwidth]]

Bene: Text replacement - "power spectral density" to "power-spectral density"

2022-02-17T11:41:31Z

Text replacement - "power spectral density" to "power-spectral density"

Bene: Text replacement - "power density spectrum" to "power spectral density"

2022-02-17T11:20:27Z

Text replacement - "power density spectrum" to "power spectral density"

Javier: Text replacement - "„" to """

2021-05-28T14:49:18Z

Text replacement - "„" to """

← Older revision		Revision as of 16:49, 28 May 2021
Line 22:		Line 22:
	''Notes:''		''Notes:''
	* This exercise belongs to the chapter  [[Mobile_Communications/The_GWSSUS_Channel_Model\|The GWSSUS Channel Model]].		* This exercise belongs to the chapter  [[Mobile_Communications/The_GWSSUS_Channel_Model\|The GWSSUS Channel Model]].
	* This exercise requires knowledge of  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments\|Expected values and moments]]  from the book ~~&bdquo;~~Theory of Stochastic Signals".		* This exercise requires knowledge of  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments\|Expected values and moments]]  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}\\

Javier: Text replacement - "”" to """

2021-05-28T14:42:16Z

Text replacement - "”" to """

← Older revision		Revision as of 16:42, 28 May 2021
Line 22:		Line 22:
	''Notes:''		''Notes:''
	* This exercise belongs to the chapter  [[Mobile_Communications/The_GWSSUS_Channel_Model\|The GWSSUS Channel Model]].		* This exercise belongs to the chapter  [[Mobile_Communications/The_GWSSUS_Channel_Model\|The GWSSUS Channel Model]].
	* This exercise requires knowledge of  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments\|Expected values and moments]]  from the book &bdquo;Theory of Stochastic Signals~~”~~.		* This exercise requires knowledge of  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments\|Expected values and moments]]  from the book &bdquo;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}\\

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

2021-03-23T12:37:32Z

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

← Older revision		Revision as of 14:37, 23 March 2021
Line 107:		Line 107:
	{{ML-Fuß}}		{{ML-Fuß}}

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

Guenter at 10:58, 16 February 2021

2021-02-16T10:58:37Z

← Older revision		Revision as of 12:58, 16 February 2021
Line 22:		Line 22:
	''Notes:''		''Notes:''
	* This exercise belongs to the chapter  [[Mobile_Communications/The_GWSSUS_Channel_Model\|The GWSSUS Channel Model]].		* This exercise belongs to the chapter  [[Mobile_Communications/The_GWSSUS_Channel_Model\|The GWSSUS Channel Model]].
	* This exercise requires knowledge of  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments\|~~Moments calculation of random variables~~]]  from the book &bdquo;Theory of Stochastic Signals”.		* This exercise requires knowledge of  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments\|Expected values and moments]]  from the book &bdquo;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}\\

Guenter at 15:08, 14 January 2021

2021-01-14T15:08:31Z

← Older revision		Revision as of 17:08, 14 January 2021
Line 80:		Line 80:


	'''(3)'''  According to the [[Theory_of_Stochastic_Signals/Erwartungswerte_und_Momente#~~Einige_h.C3.A4ufig_benutzte_Zentralmomente~~\| Steiner's Theorem]], the variance of any random variable is $\sigma^2 = m_2 \, ~~–~~m_1^2$.		'''(3)'''  According to the  [[Theory_of_Stochastic_Signals/Erwartungswerte_und_Momente#Some common central moments\| 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		*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}^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}}		\sigma_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}}
Line 88:		Line 88:


	'''(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 $1/\tau_0$.		'''(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 $f → \~~delta~~ f$:		*Thus,  $\varphi_{\rm F}(\Delta f)$  is equal to  $X(f)$  with the substitution  $f → \Delta f$:
	:$$\varphi_{\rm F}(\Delta f) = \frac{1}{1/\tau_0 + {\rm j} \cdot 2\pi \Delta f}		:$$\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}.$$		= \frac{\tau_0}{1 + {\rm j} \cdot 2\pi \cdot \tau_0 \cdot \Delta f}\hspace{0.05cm}.$$
Line 104:		Line 104:
	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 $\tau_0 = 1 \ \ \rm µ s$, the coherence bandwidth is $B_{\rm K} \ \underline {= 276 \ \ \rm kHz}$.		*With  $\tau_0 = 1 \ \ \rm µ s$, the coherence bandwidth is  $B_{\rm K} \ \underline {= 276 \ \ \rm kHz}$.
	{{ML-Fuß}}		{{ML-Fuß}}

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

Guenter at 14:58, 14 January 2021

2021-01-14T14:58:10Z

← Older revision		Revision as of 16:58, 14 January 2021
Line 74:		Line 74:


	'''(2)'''  The  $k$-th moment of an  [[Theory_of_Stochastic_Signals/~~Exponentialverteilte_Zufallsgr%C3%B6%C3%9Fen~~#~~Einseitige_Exponentialverteilung~~\| exponential random variable]]  is  $m_k = k! \cdot \tau_0^k$.		'''(2)'''  The  $k$-th moment of an  [[Theory_of_Stochastic_Signals/Exponentially_Distributed_Random_Variables#One-sided_exponential_distribution\|one-sided exponential random variable]]  is  $m_k = k! \cdot \tau_0^k$.
	*With  $k = 1$, this results in the linear mean value  $m_1 = m_{\rm V}$:		*With  $k = 1$, this results in the linear mean value  $m_1 = m_{\rm V}$:
	:$$m_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}}		:$$m_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}}

← Older revision		Revision as of 13:41, 17 February 2022
Line 2:		Line 2:
	{{quiz-Header\|Buchseite=Mobile_Communications/The_GWSSUS_Channel_Model}}		{{quiz-Header\|Buchseite=Mobile_Communications/The_GWSSUS_Channel_Model}}

	[[File:P_ID2172__Mob_A_2_7.png\|right\|frame\|Delay power spectral density and <br>frequency correlation function]]		[[File:P_ID2172__Mob_A_2_7.png\|right\|frame\|Delay power-spectral density and <br>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		For the delay power-spectral density, we assume an exponential behavior.  With  ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$  we have
	:$${{\it \phi}_{\rm V}(\tau)}/{{\it \phi}_{\rm 0}} = {\rm e}^{ -\tau / \tau_0 } \hspace{0.05cm}.$$		:$${{\it \phi}_{\rm V}(\tau)}/{{\it \phi}_{\rm 0}} = {\rm e}^{ -\tau / \tau_0 } \hspace{0.05cm}.$$

Line 10:		Line 10:
	* The  <b>average excess delay</b> or <b>mean excess delay</b>  $m_{\rm V}$  is equal to the linear expectation  $E\big [\tau \big]$  and can be determined from the PDF $f_{\rm V}(\tau)$ .		* The  <b>average excess delay</b> or <b>mean excess delay</b>  $m_{\rm V}$  is equal to the linear expectation  $E\big [\tau \big]$  and can be determined from the PDF $f_{\rm V}(\tau)$ .
	* The  <b>multipath spread</b> or <b>delay spread</b>  $\sigma_{\rm V}$  gives the standard deviation of the random variable  $\tau$ .  In the theory part we also use the term  $T_{\rm V}$ for this.		* The  <b>multipath spread</b> or <b>delay spread</b>  $\sigma_{\rm V}$  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  ${\it \Phi}_{\rm V}(\tau)$ :		* The displayed frequency correlation function  $\varphi_{\rm F}(\Delta f)$  can be calculated as the Fourier transform of the delay power-spectral density  ${\it \Phi}_{\rm V}(\tau)$ :
	:$$\varphi_{\rm F}(\Delta f)		:$$\varphi_{\rm F}(\Delta f)
	\hspace{0.2cm} {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$		\hspace{0.2cm} {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$
Line 61:		Line 61:
	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Let  ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$.  The integral of the delay power spectral density gives		'''(1)'''  Let  ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$.  The integral of the delay power-spectral density gives
	:$$\int_{0}^{+\infty} {\it \Phi}_{\rm V}(\tau) \hspace{0.15cm}{\rm d} \tau =		:$$\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 =

← Older revision		Revision as of 13:20, 17 February 2022
Line 2:		Line 2:
	{{quiz-Header\|Buchseite=Mobile_Communications/The_GWSSUS_Channel_Model}}		{{quiz-Header\|Buchseite=Mobile_Communications/The_GWSSUS_Channel_Model}}

	[[File:P_ID2172__Mob_A_2_7.png\|right\|frame\|Delay power density ~~spectrum~~ and <br>frequency correlation function]]		[[File:P_ID2172__Mob_A_2_7.png\|right\|frame\|Delay power spectral density and <br>frequency correlation function]]
	For the delay power density ~~spectrum~~, we assume an exponential behavior.  With  ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$  we have		For the delay power spectral density, we assume an exponential behavior.  With  ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$  we have
	:$${{\it \phi}_{\rm V}(\tau)}/{{\it \phi}_{\rm 0}} = {\rm e}^{ -\tau / \tau_0 } \hspace{0.05cm}.$$		:$${{\it \phi}_{\rm V}(\tau)}/{{\it \phi}_{\rm 0}} = {\rm e}^{ -\tau / \tau_0 } \hspace{0.05cm}.$$

Line 10:		Line 10:
	* The  <b>average excess delay</b> or <b>mean excess delay</b>  $m_{\rm V}$  is equal to the linear expectation  $E\big [\tau \big]$  and can be determined from the PDF $f_{\rm V}(\tau)$ .		* The  <b>average excess delay</b> or <b>mean excess delay</b>  $m_{\rm V}$  is equal to the linear expectation  $E\big [\tau \big]$  and can be determined from the PDF $f_{\rm V}(\tau)$ .
	* The  <b>multipath spread</b> or <b>delay spread</b>  $\sigma_{\rm V}$  gives the standard deviation of the random variable  $\tau$ .  In the theory part we also use the term  $T_{\rm V}$ for this.		* The  <b>multipath spread</b> or <b>delay spread</b>  $\sigma_{\rm V}$  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 density ~~spectrum~~  ${\it \Phi}_{\rm V}(\tau)$ :		* The displayed frequency correlation function  $\varphi_{\rm F}(\Delta f)$  can be calculated as the Fourier transform of the delay power spectral density  ${\it \Phi}_{\rm V}(\tau)$ :
	:$$\varphi_{\rm F}(\Delta f)		:$$\varphi_{\rm F}(\Delta f)
	\hspace{0.2cm} {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$		\hspace{0.2cm} {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$
Line 61:		Line 61:
	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Let  ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$.  The integral of the delay power density ~~spectrum~~ gives		'''(1)'''  Let  ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$.  The integral of the delay power spectral density gives
	:$$\int_{0}^{+\infty} {\it \Phi}_{\rm V}(\tau) \hspace{0.15cm}{\rm d} \tau =		:$$\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 =