Exercise 2.8: COST Delay Models

From LNTwww

COST delay models

On the right, four delay power density spectra are plotted logarithmically as a function of the delay time  $\tau$ 

$$10 \cdot {\rm lg}\hspace{0.15cm} ({{\it \Phi}_{\rm V}(\tau)}/{\it \Phi}_{\rm 0}) \hspace{0.05cm}.$$

Here the abbreviation  $\phi_0 = \phi_{\rm V}(\tau = 0)$  is used.  These are the so-called COST delay models.

The upper sketch contains the two profiles  ${\rm RA}$  ("Rural Area") and  ${\rm TU}$  ("Typical Urban").  Both of these are exponential:

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

The value of the parameter  $\tau_0$  (time constant of the autocorrelation function) should be determined from the graphic in subtask  (1).  Note the specified values of   $\tau_{-30}$ for  ${\it \Phi}_{\rm V}(\tau_{-30})=-30 \ \rm dB$:

$${\rm RA}\text{:}\hspace{0.15cm}\tau_{-30} = 0.75\,{\rm µ s} \hspace{0.05cm},\hspace{0.2cm} {\rm TU}\text{:}\hspace{0.15cm}\tau_{-30} = 6.9\,{\rm µ s} \hspace{0.05cm}. $$

The lower graph applies to less favourable conditions in

  • urban areas ("Bad Urban",  ${\rm BU}$):
$${{\it \Phi}_{\rm V}(\tau)}/{{\it \Phi}_{\rm 0}} = \left\{ \begin{array}{c} {\rm e}^{ -\tau / \tau_0} \\ 0.5 \cdot {\rm e}^{ (5\,{\rm \mu s}-\tau) / \tau_0} \end{array} \right.\quad \begin{array}{*{1}c} \hspace{-0.55cm} {\rm if}\hspace{0.15cm}0 < \tau < 5\,{\rm µ s}\hspace{0.05cm},\hspace{0.15cm}\tau_0 = 1\,{\rm µ s} \hspace{0.05cm}, \\ \hspace{-0.15cm} {\,\, \,\, \rm if}\hspace{0.15cm}5\,{\rm µ s} < \tau < 10\,{\rm µ s}\hspace{0.05cm},\hspace{0.15cm}\tau_0 = 1\,{\rm µ s} \hspace{0.05cm}, \end{array}$$
  • in rural areas ("Hilly Terrain",  ${\rm HT}$):
$${{\it \Phi}_{\rm V}(\tau)}/{{\it \Phi}_{\rm 0}} = \left\{ \begin{array}{c} {\rm e}^{ -\tau / \tau_0} \\ {0.04 \cdot \rm e}^{ (15\,{\rm \mu s}-\tau) / \tau_0} \end{array} \right.\quad \begin{array}{*{1}c} \hspace{-0.55cm} {\rm if}\hspace{0.15cm}0 < \tau < 2\,{\rm µ s}\hspace{0.05cm},\hspace{0.15cm}\tau_0 = 0.286\,{\rm µ s} \hspace{0.05cm}, \\ \hspace{-0.35cm} {\rm if}\hspace{0.15cm}15\,{\rm µ s} < \tau < 20\,{\rm µ s}\hspace{0.05cm},\hspace{0.15cm}\tau_0 = 1\,{\rm µ s} \hspace{0.05cm}. \end{array}$$

For the models  ${\rm RA}$,  ${\rm TU}$  and  ${\rm BU}$  the following parameters are to be determined:

  • The  delay spread  $T_{\rm V}$  is the standard deviation of the delay  $\tau$.
    If the delay power density spectrum  ${\it \Phi}_{\rm V}(\tau)$  has an exponential course as with the profiles  ${\rm RA}$  and  ${\rm TU}$, then  $T_{\rm V} = \tau_0$, see  Exercise 2.7.
  • The coherence bandwidth  $B_{\rm K}$  is the value of  $\Delta f$ at which the magnitude of the frequency correlation function  $\varphi_{\rm F}(\Delta f)$  has dropped to half its value for the first time. With exponential  ${\it \Phi}_{\rm V}(\tau)$  as with  ${\rm RA}$  and  ${\rm TU}$  the product is  $T_{\rm V} \cdot B_{\rm K} \approx 0.276$, see  Exercise 2.7.




Notes:

$$\frac{1}{\tau_0} \cdot \int_{0}^{\infty}\hspace{-0.15cm} {\rm e}^{ -\tau / \tau_0} \hspace{0.15cm}{\rm d} \tau = 1 \hspace{0.05cm},\hspace{0.6cm} \frac{1}{\tau_0} \cdot \int_{0}^{\infty}\hspace{-0.15cm} {\tau} \cdot{\rm e}^{ -\tau / \tau_0}\hspace{0.15cm}{\rm d} \tau = \tau_0 \hspace{0.05cm},\hspace{0.6cm} \frac{1}{\tau_0} \cdot \int_{0}^{\infty} \hspace{-0.15cm}{\tau^2} \cdot{\rm e}^{ -\tau / \tau_0}\hspace{0.15cm}{\rm d} \tau = 2\tau_0^2\hspace{0.05cm}.$$


Questionnaire

1

Specify the parameter  $\tau_0$  of the delay power density spectrum for the profiles  ${\rm RA}$  and  ${\rm TU}$ .

${\rm RA} \text{:} \ \hspace{0.4cm} \tau_0 \ = \ $

$\ \rm µ s$
${\rm TU} \text{:} \ \hspace{0.4cm} \tau_0 \ = \ $

$\ \rm µ s$

2

How large is the delay spread  $T_{\rm V}$  of these channels?

${\rm RA} \text{:} \ \hspace{0.4cm} T_{\rm V} \ = \ $

$\ \rm µ s$
${\rm TU} \text{:} \ \hspace{0.4cm} T_{\rm V} \ = \ $

$\ \rm µ s$

3

What is the coherence bandwidth  $B_{\rm K}$  of these channels?

${\rm RA} \text{:} \ \hspace{0.4cm} B_{\rm K} \ = \ $

$\ \ \rm kHz$
${\rm TU} \text{:} \ \hspace{0.4cm} B_{\rm K} \ = \ $

$\ \ \rm kHz$

4

For which channel does frequency selectivity play a greater role?

Rural Area  $({\rm RA})$.
Typical urban  $({\rm TU})$.

5

How large is the (normalized) power density for  "Bad Urban"  $({\rm BU})$   with   $\tau = 5.001 \ \rm µ s$  and with   $\tau = 4.999 \ \rm µ s$?

${\it \Phi}_{\rm V}(\tau = 5.001 \ \rm µ s) \ = \ $

$\ \cdot {\it \Phi}_0$
${\it \Phi}_{\rm V}(\tau = 4.999 \ \rm µ s) \ = \ $

$\ \cdot {\it \Phi}_0$

6

We consider  ${\rm BU}$ again. Let $P_1$ be the power of the signal between $0$  and  $5 \ \rm µ s$, and let $P_2$ be the remaining signal power.
What percentage of the total signal power comes from the interval $0< t < 5 \ \rm µ s$?

$P_1/(P_1 + P_2) \ = \ $

$\ \rm \%$

7

Calculate the delay spread  $T_{\rm V}$  of the profile  ${\rm BU}$.
Note:  The average delay is  $m_{\rm V} = E[\tau] = 2.667 \ \rm µ s$.

$T_{\rm V} \ = \ $

$\ \rm µ s$


Solution

(1)  The following property can be seen from the graph:

$$10 \cdot {\rm lg}\hspace{0.1cm} (\frac{{\it \Phi}_{\rm V}(\tau_{\rm -30})}{{\it \Phi}_0}) = 10 \cdot {\rm lg}\hspace{0.1cm}\left [{\rm exp}[ -\frac{\tau_{\rm -30}}{ \tau_{\rm 0}}]\right ] \stackrel {!}{=} -30\,{\rm dB}$$
$$\Rightarrow \hspace{0.3cm} {\rm lg}\hspace{0.1cm}\left [{\rm exp}[ -\frac{\tau_{\rm -30}}{ \tau_{\rm 0}}]\right ] = -3 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\rm ln}\hspace{0.1cm}\left [{\rm exp}[ -\frac{\tau_{\rm -30}}{ \tau_{\rm 0}}]\right ] = -3 \cdot {\rm ln}\hspace{0.1cm}(10)\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \tau_{\rm 0} = \frac{\tau_{\rm -30}}{ 3 \cdot {\rm ln}\hspace{0.1cm}(10)}\approx \frac{\tau_{\rm -30}}{ 6.9} \hspace{0.05cm}.$$

Here  $\tau_{-30}$  denotes the delay that leads to the logarithmic ordinate value  $-30 \ \rm dB$.  Thus one obtains

  • for "Rural Area"  $\rm (RA)$  with  $\tau_{–30} = 0.75 \ \rm µ s$:
$$\tau_{\rm 0} = \frac{0.75\,{\rm µ s}}{ 6.9} \hspace{0.1cm}\underline {\approx 0.109\,{\rm µ s}} \hspace{0.05cm},$$
  • for urban and suburban areas  ⇒   "Typical Urban" $\rm (TU)$  with  $\tau_{–30} = 6.9 \ \rm µ s$:
$$\tau_{\rm 0} = \frac{6.9\,{\rm µ s}}{ 6.9} \hspace{0.1cm}\underline {\approx 1\,{\rm µ s}} \hspace{0.05cm},$$


(2)  In  Exercise 2.7, it was shown that the delay spread is  $T_{\rm V} =\tau_0$  when the delay power density spectrum decreases exponentially according to  ${\rm e}^{-\tau/\tau_0}$.  Thus the following applies:

  • for "Rural Area":  $\hspace{0.4cm} T_{\rm V} \ \underline {= 0.109 \ \rm µ s}$,
  • for "Typical Urban":  $\hspace{0.4cm} T_{\rm V} \ \underline {= 1 \ \rm µ s}$.


(3)  In Exercise 2.7 it was also shown that for the coherence bandwidth  $B_{\rm K} \approx 0.276/\tau_0$  applies.  It follows:

  • $B_{\rm K} \ \underline {\approx 2500 \ \rm kHz}$ ("Rural Area"),
  • $B_{\rm K} \ \underline {\approx 276 \ \ \rm kHz}$ ("Typical Urban")



(4) The second solution is correct:

  • Frequency selectivity of the mobile radio channel is present if the signal bandwidth  $B_{\rm S}$  is larger than the coherence bandwidth  $B_{\rm K}$  (or at least of the same order of magnitude).
  • The smaller  $B_{\rm K}$  is, the more often this happens.


(5)  According to the given equation, we have  ${\it \Phi}_{\rm V}(\tau = 5.001 \ \rm µ s)/{\it \Phi}_0 \hspace{0.15cm}\underline{\approx0.5}$.

  • On the other hand, for slightly smaller  $\tau$  $($for example $\tau = 4.999 \ \rm µ s)$  we have approximately
$${{\it \Phi}_{\rm V}(\tau = 4.999\,{\rm \mu s})}/{{\it \Phi}_{\rm 0}} = {\rm e}^{ -{4.999\,{\rm µ s}}/{ 1\,{\rm \mu s}}} \approx {\rm e}^{-5} \hspace{0.1cm}\underline {= 0.00674 }\hspace{0.05cm}.$$


(6)  The power  $P_1$  of all signal components with delays between  $0$  and  $5 \ µ\rm   s$  is:

$$P_1 = {\it \Phi}_{\rm 0} \cdot \int_{0}^{5\,{\rm \mu s}} {\rm exp}[ -{\tau}/{ \tau_0}] \hspace{0.15cm}{\rm d} \tau \hspace{0.15cm} \approx \hspace{0.15cm} {\it \Phi}_{\rm 0} \cdot \int_{0}^{\infty} {\rm e}^{ -{\tau}/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau = {\it \Phi}_{\rm 0} \cdot \tau_0 \hspace{0.05cm}.$$
  • The power outside  $[0\;µ \mathrm{s}, 5\;µ \mathrm{s}]$  is
$$P_2 = \frac{{\it \Phi}_{\rm 0}}{2} \cdot \int_{5\,{\rm µ s}}^{\infty} {\rm exp}[ \frac{5\,{\rm µ s} -\tau}{ \tau_0}] \hspace{0.15cm}{\rm d} \tau \hspace{0.15cm} \approx \hspace{0.15cm} \frac{{\it \Phi}_{\rm 0}}{2} \cdot \int_{0}^{\infty} {\rm exp}[ -{\tau}/{ \tau_0}] \hspace{0.15cm}{\rm d} \tau = \frac{{\it \Phi}_{\rm 0} \cdot \tau_0}{2} \hspace{0.05cm}. $$
Delay power density of the COST profiles  ${\rm BU}$  and  ${\rm HT}$
  • Correspondingly, the percentage of power between  $0$  and  $5 \ µ\rm   s$  is
$$\frac{P_1}{P_1+ P_2} = \frac{2}{3} \hspace{0.15cm}\underline {\approx 66.7\%}\hspace{0.05cm}.$$

The figure shows  ${\it \Phi}_{\rm V}(\tau)$  in linear scale.  The areas  $P_1$  and  $P_2$  are labeled.

  • The left graph is for  ${\rm BU}$, the right graph is for  ${\rm HT}$.
  • For the latter, the power percentage of all later echoes  $($later than  $15 \ \rm µ s)$  is only about  $12\%$.


(7)  The area of the entire power density spectrum gives  $P = 1.5 \cdot \phi_0 \cdot \tau_0$.

Delay PDF of profile  ${\rm BU}$
  • Normalizing  ${\it \Phi}_{\rm V}(\tau)$  to this value yields the probability density function  $f_{\rm V}(\tau)$, as shown in the graph on the right (left diagram).
  • With  $\tau_0 = 1 \ \ \rm µ s$  and  $\tau_5 = 5 \ \ \rm µ s$, the mean is:
$$m_{\rm V}= \int_{0}^{\infty} f_{\rm V}(\tau) \hspace{0.15cm}{\rm d} \tau$$
$$\Rightarrow \hspace{0.3cm}m_{\rm V}= \frac{2}{3\tau_0} \cdot \int_{0}^{\tau_5} \tau \cdot {\rm e}^{ - {\tau}/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau \ + $$
$$ \hspace{1.7cm}+\ \frac{1}{3\tau_0} \cdot \int_{\tau_5}^{\infty} \tau \cdot {\rm e}^{ (\tau_5 -\tau)/\tau_0}\hspace{0.15cm}{\rm d} \tau \hspace{0.05cm}. $$
  • The first integral is equal to  $2\tau_0/3$  according to the provided expression.
  • With the substitution  $\tau' = \tau \, -\tau_5$  you finally obtain using the integral solutions given above:
$$m_{\rm V} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{2\tau_0}{3} + \frac{1}{3\tau_0} \cdot \int_{0}^{\infty} (\tau_5 + \tau') \cdot{\rm e}^{ - {\tau}'/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau ' = \frac{2\tau_0}{3} + \frac{\tau_5}{3\tau_0} \cdot \int_{0}^{\infty} \cdot{\rm e}^{ - {\tau}'/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau ' + \frac{1}{3\tau_0} \cdot \int_{0}^{\infty} \tau' \cdot \cdot{\rm e}^{ - {\tau}'/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau ' $$
$$\Rightarrow \hspace{0.3cm}m_{\rm V}= \frac{2\tau_0}{3} + \frac{\tau_5}{3}+ \frac{\tau_0}{3} = \tau_0 + \frac{\tau_5}{3} \hspace{0.15cm}\underline {\approx 2.667\,{\rm µ s}} \hspace{0.05cm}. $$
  • The variance  $\sigma_{\rm V}^2$  is equal to the second moment (mean of the square) of the zero-mean random variable  $\theta = \tau \, –m_{\rm V}$, whose PDF is shown in the right graph
  • From this  $T_{\rm V} = \sigma_{\rm V}$  can be specified.


A second possibility is to first calculate the mean square value of the random variable  $\tau$  and from this the variance  $\sigma_{\rm V}^2$  using Steiner's theorem.

  • With the substitutions and approximations already described above, one obtains
$$m_{\rm V2} \hspace{-0.1cm} \ \approx \ \hspace{-0.1cm} \frac{2}{3\tau_0} \cdot \int_{0}^{\infty} \tau^2 \cdot {\rm e}^{ - {\tau}/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau + \frac{1}{3\tau_0} \cdot \int_{0}^{\infty} (\tau_5 + \tau')^2 \cdot {\rm e}^{ - {\tau}'/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau ' $$
$$\Rightarrow \hspace{0.3cm}m_{\rm V2} = \frac{2}{3} \cdot \int_{0}^{\infty} \frac{\tau^2}{\tau_0} \cdot {\rm e}^{ - {\tau}/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau + \frac{\tau_5^2}{3} \cdot \int_{0}^{\infty} \frac{1}{\tau_0} \cdot {\rm e}^{ - {\tau}'/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau ' +\frac{2\tau_5}{3} \cdot \int_{0}^{\infty} \frac{\tau '}{\tau_0} \cdot {\rm e}^{ - {\tau}'/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau ' + \frac{1}{3} \cdot \int_{0}^{\infty} \frac{{\tau '}^2}{\tau_0} \cdot {\rm e}^{ - {\tau}'/{ \tau_0}} \hspace{0.15cm}{\rm d} \tau ' \hspace{0.05cm}. $$
  • With the integrals given above, we have
$$m_{\rm V2} \approx \frac{2}{3} \cdot 2 \tau_0^2 + \frac{\tau_5^2}{3} \cdot 1 + \frac{2\tau_5}{3} \cdot \tau_0 + \frac{1}{3} \cdot 2 \tau_0^2 = 2 \tau_0^2 + \frac{\tau_5^2}{3} + \frac{2 \cdot \tau_0 \cdot \tau_5}{3} $$
$$\Rightarrow \hspace{0.3cm} \sigma_{\rm V}^2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} m_{\rm V2} - m_{\rm V}^2 = 2 \tau_0^2 + \frac{\tau_5^2}{3} + \frac{2 \cdot \tau_0 \cdot \tau_5}{3} - (\tau_0 + \frac{\tau_5}{3})^2 =\tau_0^2 + \frac{2\tau_5^2}{9} = (1\,{\rm µ s})^2 + \frac{2\cdot (5\,{\rm µ s})^2}{9} = 6.55\,({\rm µ s})^2$$
$$\Rightarrow \hspace{0.3cm} T_{\rm V} = \sigma_{\rm V} \hspace{0.15cm}\underline {\approx 2.56\,{\rm µ s}}\hspace{0.05cm}.$$

The above graph shows the parameters  $T_{\rm V}$ and $\sigma_{\rm V}$.