Exercise 2.9: Coherence Time

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

	The Doppler PDF is always identical in shape to the Doppler PDS.
	The Doppler PDF is identical with the Doppler PDS in this example.
	Doppler PDF and Doppler PDS differ fundamentally.

$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} \ = \ $

$\ \ \rm Hz$

$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) \ = \ $

$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} \ = \ $

$\ \ \rm ms$

	$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.

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.