Exercise 2.7: Coherence Bandwidth

Delay power density spectrum and
frequency correlation function

For the delay power density spectrum, we assume an exponential behavior. With ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$ we have

$${{\it \phi}_{\rm V}(\tau)}/{{\it \phi}_{\rm 0}} = {\rm e}^{ -\tau / \tau_0 } \hspace{0.05cm}.$$

The constant $\tau_0$ can be determined from the tangent in the point $\tau = 0$ according to the upper graph. Note that ${\it \Phi}_{\rm V}(\tau)$ has unit $[1/\rm s]$ . Furthermore,

The probability density function $\rm (PDF)$ $f_{\rm V}(\tau)$ has the same form as ${\it \Phi}_{\rm V}(\tau)$, but is normalized to area $1$ .
The average excess delay or mean excess delay $m_{\rm V}$ is equal to the linear expectation $E\big [\tau \big]$ and can be determined from the PDF $f_{\rm V}(\tau)$ .
The multipath spread or delay spread $\sigma_{\rm V}$ gives the standard deviation of the random variable $\tau$ . In the theory part we also use the term $T_{\rm V}$ for this.
The displayed frequency correlation function $\varphi_{\rm F}(\Delta f)$ can be calculated as the Fourier transform of the delay power density spectrum ${\it \Phi}_{\rm V}(\tau)$ :

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

The coherence bandwidth $B_{\rm K}$ is the value of $\Delta f$ at which the frequency correlation function $\varphi_{\rm F}(\Delta f)$ has dropped to half in absolute value.

Notes:

This exercise belongs to the chapter The GWSSUS Channel Model.
This exercise requires knowledge of Moments calculation of random variables from the book „Theory of Stochastic Signals”.
In addition, the following Fourier transform is given:

$$x(t) = \left\{ \begin{array}{c} {\rm e}^{- \lambda \hspace{0.05cm}\cdot \hspace{0.05cm} t}\\ 0 \end{array} \right.\quad \begin{array}{*{1}c} \hspace{-0.35cm} {\rm for} \hspace{0.15cm} t \ge 0 \\ \hspace{-0.35cm} {\rm for} \hspace{0.15cm} t < 0 \\ \end{array} \hspace{0.4cm} {\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet} \hspace{0.4cm} X(f) = \frac{1}{\lambda + {\rm j} \cdot 2\pi f}\hspace{0.05cm}.$$

Questionnaire

	$f_{\rm V}(\tau) = {\rm e}^{-\tau/\tau_0}$.
	$f_{\rm V}(\tau) = 1/\tau_0 \cdot {\rm e}^{-\tau/\tau_0}$,
	$f_{\rm V}(\tau) = {\it \Phi}_0 \cdot {\rm e}^{-\tau/\tau_0}$.

$m_{\rm V} \ = \ $

$\ \rm µ s$

$\sigma_{\rm V} \ = \ $

$\ \rm µ s$

	$\varphi_{\rm F}(\Delta f) = \big[1/\tau_0 + {\rm j} \ 2 \pi \cdot \delta f \big]^{-1}$,
	$\varphi_{\rm F}(\Delta f) = {\rm e}^ {-(\tau_0 \hspace{0.05cm}\cdot \hspace{0.05cm}\Delta f)^2}$.

$B_{\rm K} \ = \ $

$\ \ \rm kHz$

Solution

(1) Let ${\it \Phi}_0 = {\it \Phi}_{\rm V}(\tau = 0)$. The integral of the delay power density spectrum gives

$$\int_{0}^{+\infty} {\it \Phi}_{\rm V}(\tau) \hspace{0.15cm}{\rm d} \tau = {\it \Phi}_{\rm 0} \cdot \int_{0}^{+\infty} {\rm e}^{-\tau / \tau_0} \hspace{0.15cm}{\rm d} \tau = {\it \Phi}_{\rm 0} \cdot \tau_0 \hspace{0.05cm}. $$

The probability density function is then

$$f_{\rm V}(\tau) = \frac{{\it \Phi}_{\rm V}(\tau) }{{\it \Phi}_{\rm 0} \cdot \tau_0}= \frac{1}{\tau_0} \cdot {\rm e}^{-\tau / \tau_0} \hspace{0.05cm}.$$

Solution 2 is therefore correct.

(2) The $k$-th moment of an exponential random variable is $m_k = k! \cdot \tau_0^k$.

With $k = 1$, this results in the linear mean value $m_1 = m_{\rm V}$:

$$m_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}} \hspace{0.05cm}. $$

(3) According to the Steiner's Theorem, the variance of any random variable is $\sigma^2 = m_2 \, –m_1^2$.

This yields $m_2 = 2 \cdot \tau_0^2$, and therefore

$$\sigma_{\rm V}^2 = m_2 - m_1^2 = 2 \cdot \tau_0^2 - (\tau_0)^2 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \sigma_{\rm V} = \tau_0 \hspace{0.1cm} \underline {= 1\,{\rm µ s}} \hspace{0.05cm}. $$

(4) ${\it \Phi}_{\rm V}(\tau)$ is identical to $x(t)$ in the given Fourier transform pair if $t$ is replaced by $\tau$ and $\lambda$ by $1/\tau_0$.

Thus, $\varphi_{\rm F}(\delta f)$ is equal to $X(f)$ with the substitution $f → \delta f$:

$$\varphi_{\rm F}(\Delta f) = \frac{1}{1/\tau_0 + {\rm j} \cdot 2\pi \Delta f} = \frac{\tau_0}{1 + {\rm j} \cdot 2\pi \cdot \tau_0 \cdot \Delta f}\hspace{0.05cm}.$$

The first expression is correct.

(5) The coherence bandwidth is implicit in the following equation:

$$|\varphi_{\rm F}(\Delta f = B_{\rm K})| \stackrel {!}{=} \frac{1}{2} \cdot |\varphi_{\rm F}(\Delta f = 0)| = \frac{\tau_0}{2}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}|\varphi_{\rm F}(\Delta f = B_{\rm K})|^2 = \frac{\tau_0^2}{1 + (2\pi \cdot \tau_0 \cdot B_{\rm K})^2} \stackrel {!}{=} \frac{\tau_0^2}{4}$$

$$\Rightarrow \hspace{0.3cm}(2\pi \cdot \tau_0 \cdot B_{\rm K})^2 = 3 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} B_{\rm K}= \frac{\sqrt{3}}{2\pi \cdot \tau_0} \approx \frac{0.276}{ \tau_0}\hspace{0.05cm}. $$

With $\tau_0 = 1 \ \ \rm µ s$, the coherence bandwidth is $B_{\rm K} \ \underline {= 276 \ \ \rm kHz}$.