Aufgaben:Exercise 2.7: Coherence Bandwidth - Revision history

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

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Javier: Text replacement - "”" to """

2021-05-28T14:42:16Z

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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"

Guenter at 10:58, 16 February 2021

2021-02-16T10:58:37Z

Guenter at 15:08, 14 January 2021

2021-01-14T15:08:31Z

Guenter at 14:58, 14 January 2021

2021-01-14T14:58:10Z

Guenter at 14:52, 14 January 2021

2021-01-14T14:52:43Z

Guenter at 14:49, 14 January 2021

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@@ Line 2: / Line 2: @@
 {{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.&nbsp; With&nbsp; ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$&nbsp; we have
+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
 :$${{\it \phi}_{\rm V}(\tau)}/{{\it \phi}_{\rm 0}} = {\rm e}^{ -\tau / \tau_0 } \hspace{0.05cm}.$$
@@ Line 10: / Line 10: @@
 * 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 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 delay power spectral density&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;:
 :$$\varphi_{\rm F}(\Delta f)
   \hspace{0.2cm}  {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$
@@ Line 61: / Line 61: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Let&nbsp; ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$.&nbsp; The integral of the delay power spectral density 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 =
   {\it \Phi}_{\rm 0} \cdot \int_{0}^{+\infty} {\rm e}^{-\tau / \tau_0} \hspace{0.15cm}{\rm d} \tau =

@@ Line 2: / Line 2: @@
 {{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.&nbsp; With&nbsp; ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$&nbsp; we have
+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
 :$${{\it \phi}_{\rm V}(\tau)}/{{\it \phi}_{\rm 0}} = {\rm e}^{ -\tau / \tau_0 } \hspace{0.05cm}.$$
@@ Line 10: / Line 10: @@
 * 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 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 delay power density spectrum&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;:
 :$$\varphi_{\rm F}(\Delta f)
   \hspace{0.2cm}  {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$
@@ Line 61: / Line 61: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Let&nbsp; ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$.&nbsp; The integral of the delay power density spectrum 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 =
   {\it \Phi}_{\rm 0} \cdot \int_{0}^{+\infty} {\rm e}^{-\tau / \tau_0} \hspace{0.15cm}{\rm d} \tau =

@@ Line 22: / Line 22: @@
 ''Notes:''
 * This exercise belongs to the chapter&nbsp; [[Mobile_Communications/The_GWSSUS_Channel_Model|The GWSSUS Channel Model]].
-* This exercise requires knowledge of&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|Expected values and moments]]&nbsp; from the book &bdquo;Theory of Stochastic Signals".
+* 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:
 :$$x(t) = \left\{ \begin{array}{c} {\rm e}^{- \lambda \hspace{0.05cm}\cdot \hspace{0.05cm} t}\\

@@ Line 22: / Line 22: @@
 ''Notes:''
 * This exercise belongs to the chapter&nbsp; [[Mobile_Communications/The_GWSSUS_Channel_Model|The GWSSUS Channel Model]].
-* This exercise requires knowledge of&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|Expected values and moments]]&nbsp; from the book &bdquo;Theory of Stochastic Signals&rdquo;.
+* This exercise requires knowledge of&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|Expected values and moments]]&nbsp; from the book &bdquo;Theory of Stochastic Signals".
 * 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}\\

← Older revision		Revision as of 12:37, 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^]]

@@ Line 80: / Line 80: @@
-'''(3)'''&nbsp; 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 $\sigma^2 = m_2 \, &ndash;m_1^2$.
+'''(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 $m_2 = 2 \cdot \tau_0^2$, and therefore
+*This yields&nbsp; $m_2 = 2 \cdot \tau_0^2$, and therefore
 :$$\sigma_{\rm V}^2 = m_2 - m_1^2 = 2 \cdot \tau_0^2 - (\tau_0)^2 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
    \sigma_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm &micro; s}}
@@ Line 88: / Line 88: @@
-'''(4)'''&nbsp; ${\it \Phi}_{\rm V}(\tau)$ is identical to $x(t)$ in the given Fourier transform pair if $t$ is replaced by $\tau$ and $\lambda$ by $1/\tau_0$.
+'''(4)'''&nbsp; ${\it \Phi}_{\rm V}(\tau)$&nbsp; is identical to&nbsp; $x(t)$&nbsp; in the given Fourier transform pair if&nbsp; $t$&nbsp; is replaced by&nbsp; $\tau$&nbsp; and&nbsp; $\lambda$&nbsp; by&nbsp; $1/\tau_0$.
-*Thus, $\varphi_{\rm F}(\delta f)$ is equal to $X(f)$ with the substitution $f &#8594; \delta f$:
+*Thus,&nbsp; $\varphi_{\rm F}(\Delta f)$&nbsp; is equal to&nbsp; $X(f)$&nbsp; with the substitution&nbsp; $f &#8594; \Delta f$:
 :$$\varphi_{\rm F}(\Delta f) = \frac{1}{1/\tau_0 + {\rm j} \cdot 2\pi \Delta f}
   = \frac{\tau_0}{1 + {\rm j} \cdot 2\pi \cdot \tau_0 \cdot \Delta f}\hspace{0.05cm}.$$
@@ Line 104: / Line 104: @@
   B_{\rm K}= \frac{\sqrt{3}}{2\pi \cdot \tau_0} \approx \frac{0.276}{ \tau_0}\hspace{0.05cm}. $$
-*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}}
 [[Category:Exercises for Mobile Communications|^2.3 The GWSSUS Channel Model^]]

@@ Line 74: / Line 74: @@
-'''(2)'''&nbsp; The&nbsp; $k$-th moment of an&nbsp; [[Theory_of_Stochastic_Signals/Exponentialverteilte_Zufallsgr%C3%B6%C3%9Fen#Einseitige_Exponentialverteilung| exponential random variable]]&nbsp; is&nbsp; $m_k = k! \cdot \tau_0^k$.
+'''(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&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}}

@@ Line 36: / Line 36: @@
 <quiz display=simple>
 {What is the probability density function&nbsp; $f_{\rm V}(\tau)$&nbsp; of the delay time?
-|type="[]"}
+|type="()"}
 - $f_{\rm V}(\tau) = {\rm e}^{-\tau/\tau_0}$.
 + $f_{\rm V}(\tau) = 1/\tau_0 \cdot {\rm e}^{-\tau/\tau_0}$,