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Exercise 2.8: COST Delay Models

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COST delay models

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

10lg(ΦV(τ)/Φ0),

Here the abbreviation  ϕ0=ϕV(τ=0)  is used. These are the so-called COST–delay models.

The upper sketch contains the two profiles  RA  (Rural Area) and  TU  (Typical Urban). Both of these are exponential:

ΦV(τ)/Φ0=eτ/τ0.

The value of the parameter  τ0  (time constant of the ACF) should be determined from the graphic in subtask (1). Note the specified values of   τ30 for  ΦV(τ30)=30 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 delaypower 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  T_{\rm V} is \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 is 0  and  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


Sample 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 (RA) with \tau_{–30} = 0.75 \ \rm µ s:
\tau_{\rm 0} = \frac{0.75\,{\rm \mu s}}{ 6.9} \hspace{0.1cm}\underline {\approx 0.109\,{\rm µ s}} \hspace{0.05cm},
  • for urban and suburban areas (Typical Urban, TU) with \tau_{–30} = 6.9 \ \rm µ s:
\tau_{\rm 0} = \frac{6.9\,{\rm \mu 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 \ \mu\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\;\mu \mathrm{s}, 5\;\mu \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}.
  • Correspondingly, the percentage of power between 0 and 5 \ \mu\rm   s is
Delay power density of the COST profiles  {\rm BU}  and  {\rm HT}
\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.

  • Normalizing {\it \Phi}_{\rm V}(\tau) to this value yields the probability density function f_{\rm V}(\tau), as shown in the next graph (left diagram).
Delay PDF of profile  {\rm BU}
  • 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}.