Aufgaben:Exercise 2.9: Coherence Time

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Doppler–Leistungsdichtespektrum und Zeit–Korrelationsfunktion

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 $${\it \Phi}_{\rm D}(f_{\rm D}) = \frac{1}{ \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot \sqrt{1 - \left (\frac{f_{\rm D}}}{f_{\rm D,\hspace{0.05cm}max}}} \right )^2} } \hspace{0.05cm}.$$ for the Doppler frequency range  $|f_{\rm D}| ≤ f_{\rm D, \ max}$: 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 function  $\varphi_{\rm Z}(\delta t)$  is the inverse Fourier transform of the Doppler power density spectrum  ${\it \Phi}_{\rm D}(f)$:

$$\varphi_{\rm Z}(\Delta t ) = {\rm J}_0(2 \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot \Delta t ) \hspace{0.05cm}.$$

${\rm J}_0$  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 PSD  ${\it \Phi}_{\rm D}(f_{\rm D})$  and is equal to the standard deviation $\sigma_{\rm D}$ of the Doppler frequency $f_D$.

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 PSD  ${\it \Phi}_{\rm V}(\tau)$   ⇒   Exercise 2.7.

• The  coherence time $T_{\rm D}$  on the other hand, 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 task belongs to chapter  GWSSUS–Kanalmodell.
• Reference is also made to chapter  Allgemeine Beschreibung zeitvarianter Systeme.
• The following undefined integral is given:

$$\int \frac{u^2}{\sqrt{1-u^2}} \hspace{0.15cm}{\rm d} u = -\frac{u}{2} \cdot \sqrt{1-u^2} + \frac{1}{2} \cdot {\rm arcsin}\,(u) \hspace{0.05cm}.$$

• Abschließend noch einige Werte für die Besselfunktion nullter Ordnung  $({\rm J}_0)$:

$${\rm J}_0(\pi/2) = 0.472\hspace{0.05cm},\hspace{0.4cm}{\rm J}_0(1.52) = 0.500\hspace{0.05cm},\hspace{0.4cm}{\rm J}_0(\pi) = -0.305\hspace{0.05cm},\hspace{0.4cm} {\rm J}_0(2\pi) = 0.221 \hspace{0.05cm}.$$

Questionnaire

1

Which statements apply to the probability density function (WDF) of the Doppler frequency in the present example?

	The Doppler–WDF is always identical in shape to the Doppler–LDS.
	The Doppler–WDF is here identical with the Doppler–LDS.
	Doppler–WDF and Doppler–LDS differ fundamentally.

2

Determine the doppler broadenings  $B_{\rm D}$.

$f_{\rm D, \ max} = 50 \ {\rm Hz} \text \ \hspace{0.6cm} B_{\rm D} \ = \ $ $\ \ \rm Hz$

$f_{\rm D, \ max} = 100 \ {\rm Hz} \text \ \hspace{0.4cm} B_{\rm D} \ = \ $ $\ \ \rm Hz$

3

What is the time correlation value for  $\Delta t = 5 \ \ \rm ms$?

$f_{\rm D, \ max} = 50 \ {\rm Hz} \text \ \hspace{0.6cm} \varphi_{\rm Z}(\delta t = 5 \ \rm ms) \ = \ $

$f_{\rm D, \ max} = 100 \ {\rm Hz} \text \ \hspace{0.4cm} \varphi_{\rm Z}(\delta t = 5 \ \rm ms) \ = \ $

4

What are the correlation durations  $T_{\rm D}$  for both parameter sets?

$f_{\rm D, \ max} = 50 \ {\rm Hz} \text \ \hspace{0.6cm} T_{\rm D} \ = \ $ $\ \ \rm ms$

$f_{\rm D, \ max} = 100 \ {\rm Hz} \text \ \hspace{0.4cm} T_{\rm D} \ = \ $ $\ \ \rm ms$

5

What is the relationship between the Doppler broadening  $B_{\rm D}$  and the correlation duration  $T_{\rm D}$, based on the Jakes–spectrum?

	$B_{\rm D} \cdot T_{\rm D} \approx $1,
	$B_{\rm D} \cdot T_{\rm D} \approx 0.5$,
	$B_{\rm D} \cdot T_{\rm D} \approx $0.171.

Sample solution

Solution

(1)'  Solutions 1 and 2 are correct:

• Doppler PDF and Doppler PSD 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 $1$, recognizable by the correlation value $\varphi_{\rm Z}(\Delta t = 0) = 1$, even the identity is correct here.
• 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 value $m_{\rm D} = {\rm E}\big [f_{\rm D}\big] = 0$.

• The variance of the random variable $f_{\rm D}$ can thus be calculated directly as a square mean 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 with the substitution $u = f_{\rm D}/f_{\rm D, \ max}$ results in

$$\sigma_{\rm D}^2 = \frac{2}{\pi} \cdot f_{\rm D,\hspace{0.05cm}max}^2 \cdot \int_{0}^{1} \frac{u^2}{\sqrt{1-u^2} \hspace{0.15cm}{\rm d} u \hspace{0.05cm}. $$

• With the integral given on the data page 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 widening is equal to the scatter, i.e. the square root of the variance:

$$B_{\rm D} = \sigma_{\rm D} = \frac{f_{\rm D,\hspace{0.05cm}max}}}{\sqrt{2}}= \left\{ \begin{array}{c} \underline {35.35\,{\rm Hz}}\\\ \underline {70.7\,{\rm Hz}} \end{array} \right.\quad \begin{array}{*{*{1}c} {\rm f\ddot{u} \hspace{0.15cm}f_{\rm D,\hspace{0.05cm}max} = 50\,{\rm Hz} \\ {\rm f\ddot{u} \hspace{0.15cm}f_{\rm D,\hspace{0.05cm}max} = 100\,{\rm Hz} \\ \end{array} \hspace{0.05cm}. $$

(3)  With the Bessel values given, one obtains

• for the Doppler frequency  $f_{\rm D, \ max} = 50 \ \rm Hz$:

$$\varphi_{\rm Z}(\delta t = 5\,{\rm ms}) = {\rm J}_0(2 \pi \cdot 50\,{\rm Hz} \cdot 5\,{\rm ms} ) = {\rm J}_0(\pi/2) \hspace{0.1cm} \underline {= 0.472} \hspace{0.05cm},$$

• for the Doppler frequency  $f_{\rm D, \ max} = 100 \ \rm Hz$:

$$\varphi_{\rm Z}(\delta t = 5\,{\rm ms}) = {\rm J}_0(\pi) \hspace{0.1cm} \underline {= -0.305} \hspace{0.05cm}.$$

(4)  The correlation duration $T_{\rm D}$ is derived from the time correlation function $\varphi_{\rm Z}(\delta t)$. $T_{\rm D}$ is the $\delta t$–value where $|\varphi_{\rm Z}(\delta t)|$ has decayed to half of its maximum value. It must hold: $$\varphi_{\rm Z}(\delta t = T_{\rm D}) = {\rm J}_0(2 \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot T_{\rm D}) \stackrel {!}{=} 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 2 \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot T_{\rm D} = 1.52 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} T_{\rm D} = \frac{1.52}{2 \pi f_{\rm D,\hspace{0.05cm}max}}} = \frac{0.242}{ f_{\rm D,\hspace{0.05cm}max}}$$ $$\Rightarrow \hspace{0.3cm} f_{\rm D,\hspace{0.05cm}max} = 50\,{\rm Hz}\text{:} \hspace{-0.1cm}\hspace{0.2cm}\hspace{0.2cm} T_{\rm D} \hspace{0.1cm} \underline {\approx 4.84\,{\rm ms} \hspace{0.05cm},\hspace{0.8cm} f_{\rm D,\hspace{0.05cm}max} = 100\,{\rm Hz}\text{:} \hspace{-0.1cm}\hspace{0.2cm}\hspace{0.2cm} T_{\rm D} \hspace{0.1cm} \underline {\approx 2.42\,{\rm ms} \hspace{0.05cm}. $$

(5)  In the subtasks (2) and (4) we received $$B_{\rm D} = \frac{ f_{\rm D,\hspace{0.05cm}max}}}}{\sqrt{2}}\hspace{0.05cm}, \hspace{0.2cm} T_{\rm D} = \frac{1.52}{2 \pi f_{\rm D,\hspace{0.05cm}max}}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} B_{\rm D} \cdot T_{\rm D} = \frac{1.52}{\sqrt{2} \cdot 2 \pi } \hspace{0.1cm}\underline {\approx 0.171}\hspace{0.05cm}.$$

Correct is therefore the last proposed solution.