Exercise 2.7Z: Coherence Bandwidth of the LTI Two-Path Channel

Zwei Zweiwegekanäle

For the GWSSUS–model, two parameters are given, which both statistically capture the resulting delay $\tau$ . More information on the topic „multipath propagation” can be found in section Simulation gemäß dem GWSSUS–Modell of the theory part.

The delay spread $T_{\rm V}$ is by definition equal to the standard deviation of the random variable $\tau$.
This can be determined from the probability density $f_{\rm V}(\tau)$ . The PDF $f_{\rm V}(\tau)$ has the same shape as the delay power density spectrum ${\it \Phi}_{\rm V}(\tau)$.
The coherence bandwidth $B_{\rm K}$ describes the same situation in the frequency domain.
It is defined as the value of $\Delta f$ at which the magnitude of the frequency correlation function $\varphi_{\rm F}(\Delta f)$ first drops to half its maximum value:

$$|\varphi_{\rm F}(\Delta f = B_{\rm K})| \stackrel {!}{=} {1}/{2} \cdot |\varphi_{\rm F}(\Delta f = 0)| \hspace{0.05cm}.$$

The relationship between ${\it \Phi}_{\rm V}(\tau)$ and $\varphi_{\rm F}(\Delta f)$ is given by the Fourier transform:

$$\varphi_{\rm F}(\Delta f) \hspace{0.2cm} {\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ} \hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$$

Both definitions are only partially suitable for a time-invariant channel.
For a time invariant two-path channel (i.e. one with constant path weights according to the above graph), the following approximation for the coherence bandwidth is often used:

$$B_{\rm K}\hspace{0.01cm}' = \frac{1}{\tau_{\rm max} - \tau_{\rm min}} \hspace{0.05cm}.$$

In this task we want to clarify

why there are different definitions for the coherence bandwidth in the literature,
which connection exists between $B_{\rm K}$ and $B_{\rm K}\hspace{0.01cm}'$ and
which definitions make sense for which boundary conditions.

Notes:

This task belongs to the chapter Das GWSSUS–Kanalmodell.
This task also refers to some theory pages in chapter Mehrwegeempfang beim Mobilfunk.

Questionnaire

Sample solution

Musterlösung

(1) For both channels, the delay difference is $\Delta \tau = \tau_{\rm max} \, - \tau_{\rm min} = 1 \ \ \rm µ s$.

That's why both channels have the same value:

$$B_{\rm K}\hspace{0.01cm}' \ \ \underline {= 1000 \ \rm kHz}.$$

(2) The graphs refer to the impulse response $h(\tau)$.

To obtain the delay–power density spectrum, the weights must be squared:

$${\it \Phi}_{\rm V}(\tau) = 1^2 \cdot \delta(\tau) + G^2 \cdot \delta(\tau - \tau_0) \hspace{0.05cm}.$$

The integral of ${\it \Phi}_{\rm V}(\tau)$ is therefore $1 + G^2$.
The probability density function (PDF), however, must have „area 1” (i.e., the sum of the two Dirac weights must be $1$). From this follows:

$$f_{\rm V}(\tau) = \frac{1}{1 + G^2} \cdot \delta(\tau) + \frac{G^2}{1 + G^2} \cdot \delta(\tau - \tau_0) \hspace{0.05cm}.$$

So only solution 3 is correct.
The first option does not describe the PDF $f_{\rm V}(\tau)$, but the impulse response $h(\tau)$.
The second equation specifies the delay –power spectral density ${\it \Phi}_{\rm V}(\tau)$.

(3) For channel $\rm A$ the two impulse weights are equal.

This means that the mean value $m_{\rm V}$ and the standard deviation $\sigma_{\rm V} = T_{\rm V}$ can be computed simply:

$$m_{\rm V} = \frac{\tau_0}{2} \hspace{0.15cm} {= 0.5\,{\rm µ s}}\hspace{0.05cm}, \hspace{0.2cm}T_{\rm V} = \sigma_{\rm V} =\frac{\tau_0}{2} \hspace{0.15cm}\underline {= 0.5\,{\rm µ s}} \hspace{0.05cm}.$$

For channel $\rm B$ the Dirac weights are $1/(1+0.5^2) = 0.8$ (for $\tau = 0$) and $0.2$ (for $\tau = 1 \ \rm µ s$).

According to the basic laws of statistics, the noncentral first and second order moments are:

$$m_{\rm 1} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0 + 0.2 \cdot 1\,{\rm µ s} = 0.2\,{\rm µ s} \hspace{0.05cm},\hspace{0.5cm} m_{\rm 2} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \cdot 0^2 + 0.2 \cdot (1\,{\rm µ s})^2 = 0.2\,({\rm µ s})^2 \hspace{0.05cm}.$$

To get the result you are looking for you can use the Steiner's Theorem.

$$\sigma_{\rm V}^2 = m_{\rm 2} - m_{\rm 1}^2 = 0.2\,({\rm µ s})^2 - (0.2\,{\rm µ s})^2 = 0.16\,({\rm µ s})^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}T_{\rm V} = \sigma_{\rm V} \hspace{0.15cm}\underline {= 0.4\,{\rm µ s}}\hspace{0.05cm}.$$

(4) The frequency correlation function is the Fourier transform of ${\it \Phi}_{\rm V}(\tau) = \delta(\tau) + \delta(\tau \, – \tau_0)$:

$$\varphi_{\rm F}(\Delta f) = 1 + {\rm exp}(-{\rm j} \cdot 2\pi \cdot \Delta f \cdot \tau_0) = 1 + {\rm cos}(2\pi \cdot \Delta f \cdot \tau_0) -{\rm j} \cdot {\rm sin}(2\pi \cdot \Delta f \cdot \tau_0) $$

Frequency correlation function and coherence bandwidth

$$\Rightarrow \hspace{0.3cm} |\varphi_{\rm F}(\Delta f)| = \sqrt{2 + 2 \cdot {\rm cos}(2\pi \cdot \Delta f \cdot \tau_0) }\hspace{0.05cm}.$$

The maximum at $\delta f = 0$ is equal to $2$.
Therefore the equation to determine $B_{\rm K}$ is

$$|\varphi_{\rm F}(B_{\rm K})| = 1 \hspace{0.3cm} $$ $$\Rightarrow \hspace{0.3cm}|\varphi_{\rm F}(B_{\rm K})|^2 = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}2 + 2 \cdot {\rm cos}(2\pi \cdot B_{\rm K} \cdot \tau_0) = 1$$ $$\Rightarrow \hspace{0.3cm}{\rm cos}(2\pi \cdot B_{\rm K} \cdot \tau_0) = -0.5 \hspace{0.3cm} $$ $$\Rightarrow \hspace{0.3cm}2\pi \cdot B_{\rm K} \cdot \tau_0 = \frac{2\pi}{3}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}B_{\rm K} = \frac{1}{3\tau_0} = 333\,{\rm kHz}\hspace{0.05cm}.$$

Solution 1 is therefore correct. The graph (blue curve) illustrates the result.

(5) For channel ${\rm B}$ the corresponding equations are

$${\it \Phi}_{\rm V}(\tau) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1^2 \cdot \delta(\tau) + (-0.5)^2 \cdot \delta(\tau - \tau_0) \hspace{0.05cm},\hspace{0.05cm} \varphi_{\rm F}(\Delta f) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1 + 0.25 \cdot {\rm cos}(2\pi \cdot \Delta f \cdot \tau_0) -{\rm j} \cdot 0.25 \cdot {\rm sin}(2\pi \cdot \Delta f \cdot \tau_0)\hspace{0.05cm},$$

$$|\varphi_{\rm F}(\Delta f)| \hspace{-0.1cm} \ = \ \hspace{-0.1cm}= \sqrt{\frac{17}{16} + \frac{1}{2} \cdot {\rm cos}(2\pi \cdot \Delta f \cdot \tau_0) }\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm Max}\hspace{0.1cm}|\varphi_{\rm F}(\Delta f)| = 1.25\hspace{0.05cm},\hspace{0.2cm}{\rm Min}\hspace{0.1cm}|\varphi_{\rm F}(\Delta f)| = 0.75\hspace{0.05cm}.$$

You can see from this result that the $50\%$–coherence bandwidth cannot be specified here.
Therefore, solution 4 is correct.

This result is the reason why there are different definitions for the coherence range in the literature, for example

the $90\%$–coherence bandwidth (in the example $B_{\rm K, \hspace{0.03cm} 90\%} =184 \ \ \rm kHz$),
the very simple approximation $B_{\rm K}\hspace{0.01cm}'$ given above (in the example $B_{\rm K}\hspace{0.01cm}' =1 \ \ \rm MHz$)

You can see from these numerical values that all the information on this is very vague and that the individual „coherence bandwidths” can be very different.

	$B_{\rm K} = 333 \ \rm kHz$.
	$B_{\rm K} = 500 \ \rm kHz$.
	$B_{\rm K} = 1 \ \rm MHz$.
	$B_{\rm K}$ cannot be calculated according to this definition.

	$B_{\rm K} = 333 \ \rm kHz$.
	$B_{\rm K} = 500 \ \rm kHz$.
	$B_{\rm K} = 1 \ \ \rm MHz$.
	$B_{\rm K}$ cannot be calculated according to this definition.

	$f_{\rm V}(\tau) = \delta(\tau) + G \cdot \delta(\tau \, –\tau_0)$,
	$f_{\rm V}(\tau) = \delta(\tau) + G^2 \cdot \delta(\tau \, –\tau_0)$,
	$f_{\rm V}(\tau) = 1/(1 + G^2) \cdot \delta(\tau) + G^2/(1 + G^2) \cdot \delta(\tau \, –\tau_0)$.