Difference between revisions of "Aufgaben:Exercise 2.9: Coherence Time"
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| − | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Mobile_Communications/The_GWSSUS_Channel_Model}} |
| − | [[File:P_ID2180__Mob_A_2_9.png|right|frame| | + | [[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 [[ | + | In the frequency domain, the influence of Rayleigh fading is described by the [[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process#ACF_and_PDS_with_Rayleigh.E2.80.93Fading|Jakes spectrum]]. If the Rayleigh parameter is $\sigma = \sqrt{0.5}$ the Jakes spectrum is |
| − | :$${\it \Phi}_{\rm D}(f_{\rm D}) = \frac{1}{ \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot \sqrt{1 - \left (\frac{f_{\rm D}}{f_{\rm D,\hspace{0.05cm}max}} \right )^2} } | + | :$${\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} } $$ |
| − | in the Doppler frequency range ( | + | in the Doppler frequency range $(|f_{\rm D}| ≤ f_{\rm D, \ max})$, and zero otherwise. This function is sketched for $f_{\rm D, \ max} = 50 \ \rm Hz$ (blue curve) and $f_{\rm D, \ max} = 100 \ \rm Hz$ (red curve). |
| − | + | The correlation function $\varphi_{\rm Z}(\Delta t)$ is the inverse Fourier transform of the Doppler power-spectral density ${\it \Phi}_{\rm D}(f)$: | |
| − | |||
| − | The correlation function $\varphi_{\rm Z}(\Delta t)$ is the inverse Fourier transform of the Doppler power density | ||
:$$\varphi_{\rm Z}(\Delta t ) = {\rm J}_0(2 \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot \Delta t ) \hspace{0.05cm}.$$ | :$$\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$ denotes the | + | ${\rm J}_0$ denotes the zeroth-order Bessel function of the first kind. The correlation function $\varphi_{\rm Z}(\Delta t)$ 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: | A characteristic value can be derived from each of these two description functions: | ||
| − | * The <b>Doppler spread</b> $B_{\rm D}$ refers to the Doppler | + | * The <b>Doppler spread</b> $B_{\rm D}$ refers to the Doppler PDS ${\it \Phi}_{\rm D}(f_{\rm D})$ and is equal to the standard deviation $\sigma_{\rm D}$ of the Doppler frequency $f_{\rm D}$. |
| − | Note that the Jakes spectrum is zero-mean, so that the variance $\sigma_{\rm D}^2$ according to Steiner's theorem is equal to the second moment ${\rm E}\big[f_{\rm D}^2\big]$ | + | ::Note that the Jakes spectrum is zero-mean, so that the variance $\sigma_{\rm D}^2$ according to Steiner's theorem is equal to the second moment ${\rm E}\big[f_{\rm D}^2\big]$. The calculation is analogous to the determination of the delay spread $T_{\rm V}$ from the delay PDS ${\it \Phi}_{\rm V}(\tau)$ ⇒ [[Aufgaben:Exercise_2.7:_Coherence_Bandwidth| Exercise 2.7]]. |
| + | |||
| + | * The <b>coherence time</b> $T_{\rm D}$ refers to the time correlation function $\varphi_{\rm Z}(\Delta t)$ . | ||
| + | :: $T_{\rm D}$ is the value of $\Delta t$ at which the magnitude $|\varphi_{\rm Z}(\Delta t)|$ first drops to half of the maximum $($at $\Delta t = 0)$ . One recognizes the analogy with the determination of the coherence bandwidth $B_{\rm K}$ from the frequency correlation function $\varphi_{\rm F}(\Delta f)$ ⇒ [[Aufgaben:Exercise_2.7:_Coherence_Bandwidth| Exercise 2.7]]. | ||
| + | |||
| + | |||
| − | |||
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''Notes:'' | ''Notes:'' | ||
| − | * This | + | *This exercise belongs to the chapter [[Mobile_Communications/The_GWSSUS_Channel_Model|The GWSSUS Channel Model]]. |
| − | * Reference is also made to chapter [[ | + | * Reference is also made to chapter [[Mobile_Communications/General_Description_of_Time_Variant_Systems| General Description of Time–Variant Systems]]. |
| − | * The following | + | * 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) | :$$\int \frac{u^2}{\sqrt{1-u^2}} \hspace{0.15cm}{\rm d} u = -\frac{u}{2} \cdot \sqrt{1-u^2} + \frac{1}{2} \cdot {\rm arcsin}\,(u) | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
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{Which statements apply to the probability density function (PDF) of the Doppler frequency in this example? | {Which statements apply to the probability density function (PDF) of the Doppler frequency in this example? | ||
|type="[]"} | |type="[]"} | ||
| − | + The Doppler PDF is always identical in shape to the Doppler | + | + The Doppler PDF is always identical in shape to the Doppler PDS. |
| − | + The Doppler PDF is | + | + The Doppler PDF is identical with the Doppler PDS in this example. |
| − | - Doppler | + | - Doppler PDF and Doppler PDS differ fundamentally. |
| − | {Determine the Doppler spread | + | {Determine the Doppler spread $B_{\rm D}$. |
|type="{}"} | |type="{}"} | ||
$f_{\rm D, \ max} = 50 \ {\rm Hz} \text{:} \ \hspace{0.6cm} B_{\rm D} \ = \ ${ 35.35 3% } $\ \ \rm Hz$ | $f_{\rm D, \ max} = 50 \ {\rm Hz} \text{:} \ \hspace{0.6cm} B_{\rm D} \ = \ ${ 35.35 3% } $\ \ \rm Hz$ | ||
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{What is the time correlation value for $\Delta t = 5 \ \ \rm ms$? | {What is the time correlation value for $\Delta t = 5 \ \ \rm ms$? | ||
|type="{}"} | |type="{}"} | ||
| − | $f_{\rm D, \ max} = 50 \ {\rm Hz} \text{:} \ \hspace{0.6cm} \varphi_{\rm Z}(\ | + | $f_{\rm D, \ max} = 50 \ {\rm Hz} \text{:} \ \hspace{0.6cm} \varphi_{\rm Z}(\Delta t = 5 \ \rm ms) \ = \ ${ 0.472 3% } |
| − | $f_{\rm D, \ max} = 100 \ {\rm Hz} \text{:} \ \hspace{0.4cm} \varphi_{\rm Z}(\ | + | $f_{\rm D, \ max} = 100 \ {\rm Hz} \text{:} \ \hspace{0.4cm} \varphi_{\rm Z}(\Delta t = 5 \ \rm ms) \ = \ ${ -0.31415--0.29585 } |
{What are the coherence times $T_{\rm D}$ for both parameter sets? | {What are the coherence times $T_{\rm D}$ for both parameter sets? | ||
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$f_{\rm D, \ max} = 100 \ {\rm Hz} \text{:} \ \hspace{0.4cm} T_{\rm D} \ = \ ${ 2.42 3% } $\ \ \rm ms$ | $f_{\rm D, \ max} = 100 \ {\rm Hz} \text{:} \ \hspace{0.4cm} T_{\rm D} \ = \ ${ 2.42 3% } $\ \ \rm ms$ | ||
| − | {What is the relationship between the Doppler spread $B_{\rm D}$ and the coherence time $T_{\rm D}$, when the Doppler | + | {What is the relationship between the Doppler spread $B_{\rm D}$ and the coherence time $T_{\rm D}$, when the Doppler PDS is the Jakes spectrum? |
|type="()"} | |type="()"} | ||
- $B_{\rm D} \cdot T_{\rm D} \approx $1, | - $B_{\rm D} \cdot T_{\rm D} \approx $1, | ||
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</quiz> | </quiz> | ||
| − | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
'''(1)''' <u>Solutions 1 and 2</u> are correct: | '''(1)''' <u>Solutions 1 and 2</u> are correct: | ||
| − | *Doppler PDF and Doppler | + | *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$ | + | *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. |
| − | *If the Rayleigh parameter $\sigma$ had been chosen differently, this would not apply. | + | *If the Rayleigh parameter $\sigma$ had been chosen differently, this would not apply. |
| − | '''(2)''' 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)''' 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: | ||
:$$\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} | :$$\sigma_{\rm D}^2 = \int_{-\infty}^{+\infty} f_{\rm D}^2 \cdot {\it \Phi}_{\rm D}(f_{\rm D}) \hspace{0.15cm}{\rm d} f_{\rm D} = \int_{-f_{\rm D,\hspace{0.05cm}max}}^{+f_{\rm D,\hspace{0.05cm}max}} \frac{f_{\rm D}^2}{ \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot \sqrt{1 - \left ({f_{\rm D}}/{f_{\rm D,\hspace{0.05cm}max}} \right )^2} } \hspace{0.15cm}{\rm d} f_{\rm D} | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
| − | *Using symmetry and the substitution $u = f_{\rm D}/f_{\rm D, \ max}$, we obtain | + | *Using symmetry and the substitution $u = f_{\rm D}/f_{\rm D, \ max}$, we obtain |
:$$\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 | :$$\sigma_{\rm D}^2 = \frac{2}{\pi} \cdot f_{\rm D,\hspace{0.05cm}max}^2 \cdot \int_{0}^{1} \frac{u^2}{\sqrt{1-u^2}} \hspace{0.15cm}{\rm d} u | ||
\hspace{0.05cm}. $$ | \hspace{0.05cm}. $$ | ||
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\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
| − | *The Doppler spread is equal to the standard deviation of $ | + | *The Doppler spread is equal to the standard deviation of $f_{\rm D}$, i.e., the square root of its variance: |
:$$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}}\\ | :$$B_{\rm D} = \sigma_{\rm D} = \frac{f_{\rm D,\hspace{0.05cm}max}}{\sqrt{2}}= \left\{ \begin{array}{c} \underline{35.35\,{\rm Hz}}\\ | ||
\underline{70.7\,{\rm Hz}} \end{array} \right.\quad | \underline{70.7\,{\rm Hz}} \end{array} \right.\quad | ||
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\\ {\rm f\ddot{u}r} \hspace{0.15cm}f_{\rm D,\hspace{0.05cm}max} = 100\,{\rm Hz} \\ \end{array} | \\ {\rm f\ddot{u}r} \hspace{0.15cm}f_{\rm D,\hspace{0.05cm}max} = 100\,{\rm Hz} \\ \end{array} | ||
\hspace{0.05cm}. $$ | \hspace{0.05cm}. $$ | ||
| + | |||
'''(3)''' With the Bessel values given, one obtains | '''(3)''' With the Bessel values given, one obtains | ||
| − | * for $f_{\rm D, \ max} = 50 \ \rm Hz$: | + | * for $f_{\rm D, \ max} = 50 \ \rm Hz$: |
:$$\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},$$ | :$$\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},$$ | ||
| − | * for $f_{\rm D, \ max} = 100 \ \rm Hz$: | + | * for $f_{\rm D, \ max} = 100 \ \rm Hz$: |
:$$\varphi_{\rm Z}(\Delta t = 5\,{\rm ms}) = {\rm J}_0(\pi) \hspace{0.1cm} \underline {= -0.305} \hspace{0.05cm}.$$ | :$$\varphi_{\rm Z}(\Delta t = 5\,{\rm ms}) = {\rm J}_0(\pi) \hspace{0.1cm} \underline {= -0.305} \hspace{0.05cm}.$$ | ||
| − | '''(4)''' The coherence time $T_{\rm D}$ is derived from the correlation function $\varphi_{\rm Z}(\Delta t)$. $T_{\rm D}$ is the value of $\Delta t$ where $|\varphi_{\rm Z}(\ | + | '''(4)''' The coherence time $T_{\rm D}$ is derived from the correlation function $\varphi_{\rm Z}(\Delta t)$. $T_{\rm D}$ is the specific value of $\Delta t$ where $|\varphi_{\rm Z}(\Delta t)|$ has decayed to half of its maximum value. It must hold: |
:$$\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} | :$$\varphi_{\rm Z}(\Delta t = T_{\rm D}) = {\rm J}_0(2 \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot T_{\rm D}) \stackrel {!}{=} 0.5 \hspace{0.3cm} | ||
\Rightarrow \hspace{0.3cm} 2 \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot T_{\rm D} = 1.52 | \Rightarrow \hspace{0.3cm} 2 \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot T_{\rm D} = 1.52 | ||
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| − | '''(5)''' In the subtasks '''(2)''' and '''(4)''' we obtained: | + | |
| + | '''(5)''' In the subtasks '''(2)''' and '''(4)''' we obtained: | ||
:$$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} | :$$B_{\rm D} = \frac{ f_{\rm D,\hspace{0.05cm}max}}{\sqrt{2}}\hspace{0.05cm}, \hspace{0.2cm} T_{\rm D} = \frac{1.52}{2 \pi f_{\rm D,\hspace{0.05cm}max}}\hspace{0.3cm} | ||
\Rightarrow \hspace{0.3cm} B_{\rm D} \cdot T_{\rm D} = \frac{1.52}{\sqrt{2} \cdot 2 \pi } \hspace{0.1cm}\underline {\approx 0.171}\hspace{0.05cm}.$$ | \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}.$$ | ||
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| − | [[Category: | + | [[Category:Mobile Communications: Exercises|^2.3 The GWSSUS Channel Model^]] |
Latest revision as of 12:41, 17 February 2022
Doppler power-spectral density and correlation function
In the frequency domain, the influence of Rayleigh fading is described by the Jakes spectrum. If the Rayleigh parameter is $\sigma = \sqrt{0.5}$ the Jakes spectrum is
- $${\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} } $$
in the Doppler frequency range $(|f_{\rm D}| ≤ f_{\rm D, \ max})$, and zero otherwise. This function is sketched for $f_{\rm D, \ max} = 50 \ \rm Hz$ (blue curve) and $f_{\rm D, \ max} = 100 \ \rm Hz$ (red curve).
The correlation function $\varphi_{\rm Z}(\Delta t)$ is the inverse Fourier transform of the Doppler power-spectral density ${\it \Phi}_{\rm D}(f)$:
- $$\varphi_{\rm Z}(\Delta t ) = {\rm J}_0(2 \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot \Delta t ) \hspace{0.05cm}.$$
${\rm J}_0$ denotes the zeroth-order Bessel function of the first kind. The correlation function $\varphi_{\rm Z}(\Delta t)$ which is also symmetrical, is drawn below, but for space reasons only the right half.
A characteristic value can be derived from each of these two description functions:
- The Doppler spread $B_{\rm D}$ refers to the Doppler PDS ${\it \Phi}_{\rm D}(f_{\rm D})$ and is equal to the standard deviation $\sigma_{\rm D}$ of the Doppler frequency $f_{\rm D}$.
- Note that the Jakes spectrum is zero-mean, so that the variance $\sigma_{\rm D}^2$ according to Steiner's theorem is equal to the second moment ${\rm E}\big[f_{\rm D}^2\big]$. The calculation is analogous to the determination of the delay spread $T_{\rm V}$ from the delay PDS ${\it \Phi}_{\rm V}(\tau)$ ⇒ Exercise 2.7.
- The coherence time $T_{\rm D}$ refers to the time correlation function $\varphi_{\rm Z}(\Delta t)$ .
- $T_{\rm D}$ is the value of $\Delta t$ at which the magnitude $|\varphi_{\rm Z}(\Delta t)|$ first drops to half of the maximum $($at $\Delta t = 0)$ . One recognizes the analogy with the determination of the coherence bandwidth $B_{\rm K}$ from the frequency correlation function $\varphi_{\rm F}(\Delta f)$ ⇒ Exercise 2.7.
Notes:
- This exercise belongs to the chapter The GWSSUS Channel Model.
- Reference is also made to chapter General Description of Time–Variant Systems.
- The following indefinite integral is given:
- $$\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) \hspace{0.05cm}.$$
- We also provide the following few values of the zeroth-order Bessel function of the first kind $({\rm J}_0)$:
- $${\rm J}_0(\pi/2) = 0.472\hspace{0.05cm},\hspace{0.4cm}{\rm J}_0(1.52) = 0.500\hspace{0.05cm},\hspace{0.4cm}{\rm J}_0(\pi) = -0.305\hspace{0.05cm},\hspace{0.4cm} {\rm J}_0(2\pi) = 0.221 \hspace{0.05cm}.$$
Questionnaire
Solution
(1) Solutions 1 and 2 are correct:
- Doppler PDF and Doppler PDS are generally only identical in shape.
- But since in the example considered the integral over ${\it \Phi}_{\rm D}(f_{\rm D})$ is equal to $1$ ⇒ this can be easily seen from the correlation value $\varphi_{\rm Z}(\Delta t = 0) = 1$, the Doppler PDF and the Doppler PDS are identical in this example.
- If the Rayleigh parameter $\sigma$ had been chosen differently, this would not apply.
(2) From the axial symmetry of ${\it \Phi}_{\rm D}(f_{\rm D})$ you can see that the mean Doppler shift is $m_{\rm D} = {\rm E}\big [f_{\rm D}\big] = 0$.
- The variance of the random variable $f_{\rm D}$ can thus be calculated directly as a mean square value:
- $$\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}.$$
- Using symmetry and the substitution $u = f_{\rm D}/f_{\rm D, \ max}$, we obtain
- $$\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}. $$
- With the integral provided in the task description, you get further:
- $$\sigma_{\rm D}^2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{2}{\pi} \cdot f_{\rm D,\hspace{0.05cm}max}^2 \cdot \left [ -\frac{u}{2} \cdot \sqrt{1-u^2} + \frac{1}{2} \cdot {\rm arcsin}\,(u) \right ]_0^1 = \frac{2}{\pi} \cdot f_{\rm D,\hspace{0.05cm}max}^2 \cdot \frac{2}{2}\cdot \frac{\pi}{2} = \frac{f_{\rm D,\hspace{0.05cm}max}^2}{2} \hspace{0.05cm}.$$
- The Doppler spread is equal to the standard deviation of $f_{\rm D}$, i.e., the square root of its variance:
- $$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 \begin{array}{*{1}c} {\rm f\ddot{u}r} \hspace{0.15cm}f_{\rm D,\hspace{0.05cm}max} = 50\,{\rm Hz} \\ {\rm f\ddot{u}r} \hspace{0.15cm}f_{\rm D,\hspace{0.05cm}max} = 100\,{\rm Hz} \\ \end{array} \hspace{0.05cm}. $$
(3) With the Bessel values given, one obtains
- for $f_{\rm D, \ max} = 50 \ \rm Hz$:
- $$\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},$$
- for $f_{\rm D, \ max} = 100 \ \rm Hz$:
- $$\varphi_{\rm Z}(\Delta t = 5\,{\rm ms}) = {\rm J}_0(\pi) \hspace{0.1cm} \underline {= -0.305} \hspace{0.05cm}.$$
(4) The coherence time $T_{\rm D}$ is derived from the correlation function $\varphi_{\rm Z}(\Delta t)$. $T_{\rm D}$ is the specific value of $\Delta t$ where $|\varphi_{\rm Z}(\Delta t)|$ has decayed to half of its maximum value. It must hold:
- $$\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 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} T_{\rm D} = \frac{1.52}{2 \pi f_{\rm D,\hspace{0.05cm}max}} = \frac{0.242}{ f_{\rm D,\hspace{0.05cm}max}}$$
- $$\Rightarrow \hspace{0.3cm} f_{\rm D,\hspace{0.05cm}max} = 50\,{\rm Hz}\text{:} \ \hspace{-0.1cm}\hspace{0.2cm} T_{\rm D} \hspace{0.1cm} \underline {\approx 4.84\,{\rm ms}} \hspace{0.05cm},\hspace{0.8cm} f_{\rm D,\hspace{0.05cm}max} = 100\,{\rm Hz}\text{:} \ \hspace{-0.1cm}\hspace{0.2cm} T_{\rm D} \hspace{0.1cm} \underline {\approx 2.42\,{\rm ms}} \hspace{0.05cm}. $$
(5) In the subtasks (2) and (4) we obtained:
- $$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}.$$
Therefore, the last option is the correct one.
