Aufgaben:Exercise 2.9: Coherence Time - Revision history

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

2022-02-17T11:41:20Z

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

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

2022-02-17T11:20:08Z

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

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 15:16, 16 January 2021

2021-01-16T15:16:20Z

Guenter at 15:05, 16 January 2021

2021-01-16T15:05:11Z

Guenter at 15:03, 16 January 2021

2021-01-16T15:03:08Z

Guenter at 15:02, 16 January 2021

2021-01-16T15:02:10Z

Guenter at 15:00, 16 January 2021

2021-01-16T15:00:44Z

Guenter at 14:28, 16 January 2021

2021-01-16T14:28:37Z

Javier: Text replacement - "===Sample solution===" to "===Solution==="

2020-12-16T13:16:30Z

Text replacement - "===Sample solution===" to "===Solution==="

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Mobile_Communications/The_GWSSUS_Channel_Model}}
-[[File:P_ID2180__Mob_A_2_9.png|right|frame|Doppler power spectral density and correlation function]]
+[[File:P_ID2180__Mob_A_2_9.png|right|frame|Doppler power-spectral density and correlation function]]
 In the frequency domain, the influence of Rayleigh fading is described by the&nbsp; [[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process#ACF_and_PDS_with_Rayleigh.E2.80.93Fading|Jakes spectrum]].&nbsp; If the Rayleigh parameter is&nbsp; $\sigma = \sqrt{0.5}$&nbsp; the Jakes spectrum is
 :$${\it \Phi}_{\rm D}(f_{\rm D}) =  \frac{1}{ \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot \sqrt{1 - \left (\frac{f_{\rm D}}{f_{\rm D,\hspace{0.05cm}max}} \right )^2} } $$
 in the Doppler frequency range&nbsp; $(|f_{\rm D}| &#8804; f_{\rm D, \ max})$,&nbsp; and zero otherwise.&nbsp; This function is sketched for&nbsp; $f_{\rm D, \ max} = 50 \ \rm Hz$&nbsp; (blue curve) and&nbsp; $f_{\rm D, \ max} = 100 \ \rm Hz$&nbsp; (red curve).
-The correlation function&nbsp; $\varphi_{\rm Z}(\Delta t)$&nbsp; is the inverse Fourier transform of the Doppler power spectral density&nbsp; ${\it \Phi}_{\rm D}(f)$:
+The correlation function&nbsp; $\varphi_{\rm Z}(\Delta t)$&nbsp; is the inverse Fourier transform of the Doppler power-spectral density&nbsp; ${\it \Phi}_{\rm D}(f)$:
 :$$\varphi_{\rm Z}(\Delta t ) =  {\rm J}_0(2 \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot \Delta t ) \hspace{0.05cm}.$$

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Mobile_Communications/The_GWSSUS_Channel_Model}}
-[[File:P_ID2180__Mob_A_2_9.png|right|frame|Doppler power density spectrum and correlation function]]
+[[File:P_ID2180__Mob_A_2_9.png|right|frame|Doppler power spectral density and correlation function]]
 In the frequency domain, the influence of Rayleigh fading is described by the&nbsp; [[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process#ACF_and_PDS_with_Rayleigh.E2.80.93Fading|Jakes spectrum]].&nbsp; If the Rayleigh parameter is&nbsp; $\sigma = \sqrt{0.5}$&nbsp; the Jakes spectrum is
 :$${\it \Phi}_{\rm D}(f_{\rm D}) =  \frac{1}{ \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot \sqrt{1 - \left (\frac{f_{\rm D}}{f_{\rm D,\hspace{0.05cm}max}} \right )^2} } $$
 in the Doppler frequency range&nbsp; $(|f_{\rm D}| &#8804; f_{\rm D, \ max})$,&nbsp; and zero otherwise.&nbsp; This function is sketched for&nbsp; $f_{\rm D, \ max} = 50 \ \rm Hz$&nbsp; (blue curve) and&nbsp; $f_{\rm D, \ max} = 100 \ \rm Hz$&nbsp; (red curve).
-The correlation function&nbsp; $\varphi_{\rm Z}(\Delta t)$&nbsp; is the inverse Fourier transform of the Doppler power density spectrum&nbsp; ${\it \Phi}_{\rm D}(f)$:
+The correlation function&nbsp; $\varphi_{\rm Z}(\Delta t)$&nbsp; is the inverse Fourier transform of the Doppler power spectral density&nbsp; ${\it \Phi}_{\rm D}(f)$:
 :$$\varphi_{\rm Z}(\Delta t ) =  {\rm J}_0(2 \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot \Delta t ) \hspace{0.05cm}.$$

← Older revision		Revision as of 12:37, 23 March 2021
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−	[[Category:~~Exercises for~~ Mobile Communications\|^2.3 The GWSSUS Channel Model^]]	+	[[Category:Mobile Communications: Exercises\|^2.3 The GWSSUS Channel Model^]]

@@ Line 72: / Line 72: @@
 '''(1)'''&nbsp; <u>Solutions 1 and 2</u> are correct:
 *Doppler PDF and Doppler PDS are generally only identical in shape.
-*But since in the example considered the integral over ${\it \Phi}_{\rm D}(f_{\rm D})$ is equal to $1$, (this can be easily seen from the correlation value $\varphi_{\rm Z}(\Delta t = 0) = 1$), the Doppler PDF and the Doppler PDS are identical in this example.
+*But since in the example considered the integral over&nbsp; ${\it \Phi}_{\rm D}(f_{\rm D})$&nbsp; is equal to&nbsp; $1$ &nbsp; &rArr; &nbsp;this can be easily seen from the correlation value&nbsp; $\varphi_{\rm Z}(\Delta t = 0) = 1$, the Doppler PDF and the Doppler PDS are identical in this example.
-*If the Rayleigh parameter $\sigma$ had been chosen differently, this would not apply.
+*If the Rayleigh parameter&nbsp; $\sigma$&nbsp; had been chosen differently, this would not apply.
 '''(2)'''&nbsp; From the axial symmetry of ${\it \Phi}_{\rm D}(f_{\rm D})$ you can see that the mean Doppler shift is $m_{\rm D} = {\rm E}\big [f_{\rm D}\big] = 0$.
-*The variance of the random variable $f_{\rm D}$ can thus be calculated directly as a mean square value:
+'''(2)'''&nbsp; From the axial symmetry of&nbsp; ${\it \Phi}_{\rm D}(f_{\rm D})$&nbsp; you can see that the mean Doppler shift is&nbsp; $m_{\rm D} = {\rm E}\big [f_{\rm D}\big] = 0$.
 :$$\sigma_{\rm D}^2 = \int_{-\infty}^{+\infty} f_{\rm D}^2 \cdot {\it \Phi}_{\rm D}(f_{\rm D}) \hspace{0.15cm}{\rm d} f_{\rm D} = \int_{-f_{\rm D,\hspace{0.05cm}max}}^{+f_{\rm D,\hspace{0.05cm}max}} \frac{f_{\rm D}^2}{ \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot \sqrt{1 - \left ({f_{\rm D}}/{f_{\rm D,\hspace{0.05cm}max}} \right )^2} } \hspace{0.15cm}{\rm d} f_{\rm D}
   \hspace{0.05cm}.$$
-*Using symmetry and the substitution $u = f_{\rm D}/f_{\rm D, \ max}$, we obtain
+*Using symmetry and the substitution&nbsp; $u = f_{\rm D}/f_{\rm D, \ max}$, we obtain
 :$$\sigma_{\rm D}^2 = \frac{2}{\pi} \cdot f_{\rm D,\hspace{0.05cm}max}^2 \cdot \int_{0}^{1} \frac{u^2}{\sqrt{1-u^2}} \hspace{0.15cm}{\rm d} u
   \hspace{0.05cm}. $$
@@ Line 89: / Line 90: @@
   \hspace{0.05cm}.$$
-*The Doppler spread is equal to the standard deviation of $f_D$, i.e., the square root of its variance:
+*The Doppler spread is equal to the standard deviation of&nbsp; $f_{\rm D}$, i.e., the square root of its variance:
 :$$B_{\rm D} = \sigma_{\rm D}  = \frac{f_{\rm D,\hspace{0.05cm}max}}{\sqrt{2}}= \left\{ \begin{array}{c} \underline{35.35\,{\rm Hz}}\\
   \underline{70.7\,{\rm Hz}}   \end{array} \right.\quad
@@ Line 95: / Line 96: @@
 \\  {\rm f\ddot{u}r} \hspace{0.15cm}f_{\rm D,\hspace{0.05cm}max} = 100\,{\rm Hz} \\ \end{array}
   \hspace{0.05cm}. $$
 '''(3)'''&nbsp; With the Bessel values given, one obtains
-* for &nbsp; $f_{\rm D, \ max} = 50 \ \rm Hz$:
+* for&nbsp; $f_{\rm D, \ max} = 50 \ \rm Hz$:
 :$$\varphi_{\rm Z}(\Delta t = 5\,{\rm ms}) =  {\rm J}_0(2 \pi \cdot 50\,{\rm Hz} \cdot 5\,{\rm ms} ) = {\rm J}_0(\pi/2) \hspace{0.1cm} \underline {= 0.472} \hspace{0.05cm},$$
-* for &nbsp; $f_{\rm D, \ max} = 100 \ \rm Hz$:
+* for&nbsp; $f_{\rm D, \ max} = 100 \ \rm Hz$:
 :$$\varphi_{\rm Z}(\Delta t = 5\,{\rm ms}) =   {\rm J}_0(\pi) \hspace{0.1cm} \underline {= -0.305} \hspace{0.05cm}.$$
-'''(4)'''&nbsp; The coherence time $T_{\rm D}$ is derived from the correlation function $\varphi_{\rm Z}(\Delta t)$. $T_{\rm D}$ is the value of $\Delta t$ where $|\varphi_{\rm Z}(\Delta t)|$ has decayed to half of its maximum value. It must hold:
+'''(4)'''&nbsp; The coherence time&nbsp; $T_{\rm D}$&nbsp; is derived from the correlation function&nbsp; $\varphi_{\rm Z}(\Delta t)$.&nbsp; $T_{\rm D}$&nbsp; is the specific value&nbsp; of $\Delta t$&nbsp; where&nbsp; $|\varphi_{\rm Z}(\Delta t)|$&nbsp; has decayed to half of its maximum value.&nbsp; It must hold:
 :$$\varphi_{\rm Z}(\Delta t = T_{\rm D}) =  {\rm J}_0(2 \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot T_{\rm D}) \stackrel {!}{=} 0.5 \hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} 2 \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot T_{\rm D} = 1.52
@@ Line 112: / Line 114: @@
 '''(5)'''&nbsp; In the subtasks '''(2)''' and '''(4)''' we obtained:
 :$$B_{\rm D} =  \frac{ f_{\rm D,\hspace{0.05cm}max}}{\sqrt{2}}\hspace{0.05cm}, \hspace{0.2cm} T_{\rm D} =  \frac{1.52}{2 \pi f_{\rm D,\hspace{0.05cm}max}}\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} B_{\rm D} \cdot T_{\rm D} =  \frac{1.52}{\sqrt{2} \cdot 2 \pi } \hspace{0.1cm}\underline {\approx 0.171}\hspace{0.05cm}.$$

@@ Line 46: / Line 46: @@
 - Doppler PDF and Doppler PDS differ fundamentally.
-{Determine the Doppler spread&nbsp; $B_{\rm D}$.
+{Determine the Doppler spread $B_{\rm D}$.
 |type="{}"}
 $f_{\rm D, \ max} = 50 \ {\rm Hz} \text{:} \ \hspace{0.6cm} B_{\rm D} \ = \ ${ 35.35 3% } $\ \ \rm Hz$

@@ Line 27: / Line 27: @@
 ''Notes:''
 *This exercise belongs to the chapter [[Mobile_Communications/The_GWSSUS_Channel_Model|The GWSSUS Channel Model]].
-* Reference is also made to chapter&nbsp; [[Mobile_Communications/General_Description_of_Time_Variant_Systems| General_Description_of_Time&ndash;Variant_Systems]].
+* Reference is also made to chapter&nbsp; [[Mobile_Communications/General_Description_of_Time_Variant_Systems| General Description of Time&ndash;Variant Systems]].
 * The following indefinite integral is given:
 :$$\int \frac{u^2}{\sqrt{1-u^2}} \hspace{0.15cm}{\rm d} u = -\frac{u}{2} \cdot \sqrt{1-u^2} + \frac{1}{2} \cdot {\rm arcsin}\,(u)

@@ Line 3: / Line 3: @@
 [[File:P_ID2180__Mob_A_2_9.png|right|frame|Doppler power density spectrum and correlation function]]
-In the frequency domain, the influence of Rayleigh fading is described by the&nbsp; [[Mobile_Communications/Statistische_Bindungen_innerhalb_des_Rayleigh-Prozesses#AKF_und_LDS_bei_Rayleigh.E2.80.93Fading| Jakes spectrum]]&nbsp;. If the Rayleigh parameter is&nbsp; $\sigma = \sqrt{0.5}$&nbsp; the Jakes spectrum is
+In the frequency domain, the influence of Rayleigh fading is described by the&nbsp; [[Mobile_Communications/Statistische_Bindungen_innerhalb_des_Rayleigh-Prozesses#AKF_und_LDS_bei_Rayleigh.E2.80.93Fading| Jakes spectrum]].&nbsp; If the Rayleigh parameter is&nbsp; $\sigma = \sqrt{0.5}$&nbsp; the Jakes spectrum is
-:$${\it \Phi}_{\rm D}(f_{\rm D}) =  \frac{1}{ \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot \sqrt{1 - \left (\frac{f_{\rm D}}{f_{\rm D,\hspace{0.05cm}max}} \right )^2} } \hspace{0.05cm}.$$
+:$${\it \Phi}_{\rm D}(f_{\rm D}) =  \frac{1}{ \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot \sqrt{1 - \left (\frac{f_{\rm D}}{f_{\rm D,\hspace{0.05cm}max}} \right )^2} } $$
-in the Doppler frequency range&nbsp; ($|f_{\rm D}| &#8804; f_{\rm D, \ max}$), and $0$ otherwise.
+in the Doppler frequency range&nbsp; $(|f_{\rm D}| &#8804; f_{\rm D, \ max})$,&nbsp; and zero otherwise.&nbsp; This function is sketched for&nbsp; $f_{\rm D, \ max} = 50 \ \rm Hz$&nbsp; (blue curve) and&nbsp; $f_{\rm D, \ max} = 100 \ \rm Hz$&nbsp; (red curve).
 The correlation function&nbsp; $\varphi_{\rm Z}(\Delta t)$&nbsp; is the inverse Fourier transform of the Doppler power density spectrum&nbsp; ${\it \Phi}_{\rm D}(f)$:
 :$$\varphi_{\rm Z}(\Delta t ) =  {\rm J}_0(2 \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot \Delta t ) \hspace{0.05cm}.$$
-${\rm J}_0$&nbsp; denotes the <i>zeroth-order Bessel function of the first kind</i>. The correlation function&nbsp; $\varphi_{\rm Z}(\Delta t)$&nbsp; which is also symmetrical, is drawn below, but for space reasons only the right half.
+${\rm J}_0$&nbsp; denotes the zeroth-order Bessel function of the first kind.&nbsp; The correlation function&nbsp; $\varphi_{\rm Z}(\Delta t)$&nbsp; which is also symmetrical, is drawn below, but for space reasons only the right half.
 A characteristic value can be derived from each of these two description functions:
-* The &nbsp;<b>Doppler spread</b>&nbsp; $B_{\rm D}$ refers to the Doppler PDS&nbsp; ${\it \Phi}_{\rm D}(f_{\rm D})$&nbsp; and is equal to the standard deviation $\sigma_{\rm D}$ of the Doppler frequency $f_D$.
+* The &nbsp;<b>Doppler spread</b>&nbsp; $B_{\rm D}$&nbsp; refers to the Doppler PDS&nbsp; ${\it \Phi}_{\rm D}(f_{\rm D})$&nbsp; and is equal to the standard deviation&nbsp; $\sigma_{\rm D}$&nbsp; of the Doppler frequency&nbsp; $f_{\rm D}$.&nbsp;
-Note that the Jakes spectrum is zero-mean, so that the variance&nbsp; $\sigma_{\rm D}^2$&nbsp; according to Steiner's theorem is equal to the second moment&nbsp; ${\rm E}\big[f_{\rm D}^2\big]$&nbsp;. The calculation is analogous to the determination of the delay spread&nbsp; $T_{\rm V}$&nbsp; from the delay PDS&nbsp; ${\it \Phi}_{\rm V}(\tau)$ &nbsp; &#8658; &nbsp; [[Aufgaben:Exercise_2.7:_Coherence_Bandwidth| Exercise 2.7]].
+::Note that the Jakes spectrum is zero-mean, so that the variance&nbsp; $\sigma_{\rm D}^2$&nbsp; according to Steiner's theorem is equal to the second moment&nbsp; ${\rm E}\big[f_{\rm D}^2\big]$.&nbsp; The calculation is analogous to the determination of the delay spread&nbsp; $T_{\rm V}$&nbsp; from the delay PDS&nbsp; ${\it \Phi}_{\rm V}(\tau)$ &nbsp; &#8658; &nbsp; [[Aufgaben:Exercise_2.7:_Coherence_Bandwidth| Exercise 2.7]].
@@ Line 27: / Line 28: @@
 * This task belongs to chapter&nbsp; [[Mobile_Communications/Das_GWSSUS%E2%80%93Kanalmodell| GWSSUS&ndash;Kanalmodell]].
 * Reference is also made to chapter&nbsp; [[Mobile_Communications/Allgemeine_Beschreibung_zeitvarianter_Systeme| Allgemeine Beschreibung zeitvarianter Systeme]].
-* The following undefined integral is given:
+* The following indefinite integral is given:
 :$$\int \frac{u^2}{\sqrt{1-u^2}} \hspace{0.15cm}{\rm d} u = -\frac{u}{2} \cdot \sqrt{1-u^2} + \frac{1}{2} \cdot {\rm arcsin}\,(u)
   \hspace{0.05cm}.$$

@@ Line 67: / Line 67: @@
 </quiz>
-===Sample solution===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; <u>Solutions 1 and 2</u> are correct: