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

$${\it \Phi}_{\rm D}(f_{\rm D}) = \frac{1}{ \pi \cdot f_{\rm D,\hspace{0.05cm}max} \cdot \sqrt{1 - \left (\frac{f_{\rm D}}{f_{\rm D,\hspace{0.05cm}max}} \right )^2} } \hspace{0.05cm}.$$

in the Doppler frequency range  ($|f_{\rm D}| ≤ f_{\rm D, \ max}$), and $0$ 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 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 (PDF) of the Doppler frequency in this example?

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

2

Determine the Doppler spread  $B_{\rm D}$.

$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 coherence times  $T_{\rm D}$  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 spread  $B_{\rm D}$  and the coherence time  $T_{\rm D}$, when the Doppler PSD is 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 $1$, (this can be easily seen from the correlation value $\varphi_{\rm Z}(\Delta t = 0) = 1$), the Doppler PDF and the Doppler PSD 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 $f_D$, 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 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.