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

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.
This is implicitly defined by the frequency–correlation function $\varphi_{\rm F}(\delta f)$ defined as the $\delta f$–value at which its amount first dropped to half:

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

The connection 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\!} {\hspace{0.2cm} {\it \Phi}_{\rm V}(\tau)\hspace{0.05cm}.$

Both definitions are only partially suitable for a time invariant channel.
Often one uses for a time invariant two-way channel (i.e. with constant path weights according to the above graphic) as an approximation for the coherence bandwidth:

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

In this task we want to clarify

why there are different definitions for the coherence band 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

{

Solution

(1) For both channels the runtime 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 graphics refer to the impulse response $h(\tau)$.

To obtain the delay–LDS, 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 over ${\it \Phi}_{\rm V}(\tau)$ is therefore $1 + G^2$.
The probability density function (WDF), however, must give the „area 1” (sum of the two Dirac weights equals $1$). From this follows:

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

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

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

This means that for the mean value $m_{\rm V}$ and the standard deviation $\sigma_{\rm V} = T_{\rm V}$ can be written without a big calculation:

$$m_{\rm V} = \frac{\frost_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 pulse weights are $1/(1+0.5^2) = 0.8$ (for $\tau = 0$) and $0.2$ (for $\tau = 1 \ \rm µ s$).

This gives for the linear and the quadratic mean value according to the basic Laws of statistics:

$$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 Theorem of Steiner.

$$\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 ${\it \Phi}_{\rm V}(\tau) = \delta(\tau) + \delta(\tau \, – \tau_0)$: $$$\varphi_{\rm F}(\delta f) = 1 + {\rm exp}(-{\rm j} \cdot 2\pi \cdot \delta f \cdot \cdot \tau_0) = 1 + {\rm cos}(2\pi \cdot \delta f \cdot \tau_0) -{\rm j} \cdot {\rm sin}(2\pi \cdot \delta f \cdot \tau_0) $$

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 \cdot \tau_0) }\hspace{0.05cm}.$

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

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

Correct is therefore the solution 1. The graphic (blue curve) illustrates the result.

(5) For the 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.
The correct solution is therefore the solution proposal 4.

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 differ by factors.

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

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

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