Exercise 2.2Z: Real Two-Path Channel

Two-path scenario

The sketched scenario is considered in which the transmitted signal $s(t)$ reaches the antenna of the receiver via two paths:

$$r(t) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} r_1(t) + r_2(t) =k_1 \cdot s( t - \tau_1) + k_2 \cdot s( t - \tau_2) \hspace{0.05cm}.$$

Note the following:

The delays $\tau_1$ and $\tau_2$ of the main and the secondary paths can be calculated from the path lengths $d_1$ and $d_2$ using the speed of light $c = 3 \cdot 10^8 \ \rm m/s$ .
The amplitude factors $k_1$ and $k_2$ are obtained according to the path loss model with path loss exponent $\gamma = 2$ (free–space attenuation).
The height of the transmit antenna is $h_{\rm S} = 500 \ \rm m$. The height of the receiving antenna is $h_{\rm E} = 30 \ \rm m$. The antennas are separated by a distance of $d = 10 \ \ \rm km$.
The reflection on the secondary path causes a phase change of $\pi$, so that the partial signals must be subtracted. This is taken into account by a negative $k_2$ value.

Note:

This task belongs to the chapter Multi–Path Reception in Mobile Communications.

Questionnaire

$d_1 \ = \ $

$\ \ \rm m$

$d_2 \ = \ $

$\ \ \rm m$

$\Delta d \ = \ $

$\ \ \rm m$

$\Delta \tau \ = \ $

$\ \ \rm ns$

	$\Delta \tau = (h_{\rm S} \ - h_{\rm E})/d$,
	$\Delta \tau = (h_{\rm S} \ - h_{\rm E})/(c \cdot d)$,
	$\Delta \tau = 2 \cdot h_{\rm S} \cdot h_{\rm E}/(c \cdot d)$.

	The coefficients $k_1$ and $k_2$ are almost equal in magnitude.
	The magnitudes $\|k_1\|$ and $\|k_2\|$ differ significantly.
	The coefficients $\|k_1\|$ and $\|k_2\|$ differ in sign.

Solution

(1) According to Pythagoras:

$$d_1 = \sqrt{d^2 + (h_{\rm S}- h_{\rm E})^2} = \sqrt{10^2 + (0.5- 0.03)^2} \,\,{\rm km} \hspace{0.1cm} \underline {=10011.039\,{\rm m}} \hspace{0.05cm}.$$

Actually, specifying such a length with an accuracy of one millimeter is not very useful and contradicts the mentality of an engineer.
We have done this anyway to be able to check the accuracy of the approximation in subtask (4).

(2) If you fold the reflected beam on the right side of $x_{\rm R}$ downwards (reflection on the ground), you get again a right triangle. From this follows:

$$d_2 = \sqrt{d^2 + (h_{\rm S}+ h_{\rm E})^2} = \sqrt{10^2 + (0.5+ 0.03)^2} \,\,{\rm km} \hspace{0.1cm} \underline {=10014.035\,{\rm m}} \hspace{0.05cm}.$$

(3) Using the results from (1) and (2), the length and delay differences are:

$$\Delta d = d_2 - d_1 = \hspace{0.1cm} \underline {=2.996\,{\rm m}} \hspace{0.05cm},\hspace{1cm} \Delta \tau = \frac{\Delta d}{c} = \frac{2.996\,{\rm m}}{3 \cdot 10^8 \,{\rm m/s}} \hspace{0.1cm} \underline {=9.987\,{\rm ns}} \hspace{0.05cm}.$$

(4) With $h_{\rm S} + h_{\rm E} \ll d$ the above equation can be expressed as follows:

$$d_1 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} d \cdot \sqrt{1 + \frac{(h_{\rm S}- h_{\rm E})^2}{d^2}} \approx d \cdot \left [ 1 + \frac{(h_{\rm S}- h_{\rm E})^2}{2d^2} \right ] \hspace{0.05cm},\hspace{1cm} d_2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} d \cdot \sqrt{1 + \frac{(h_{\rm S}+ h_{\rm E})^2}{d^2}} \approx d \cdot \left [ 1 + \frac{(h_{\rm S}+ h_{\rm E})^2}{2d^2} \right ] $$

$$\Rightarrow \hspace{0.3cm} \Delta d = d_2 - d_1 \approx \frac {1}{2d} \cdot \left [ (h_{\rm S}+ h_{\rm E})^2 - (h_{\rm S}- h_{\rm E})^2 \right ] = \frac {2 \cdot h_{\rm S}\cdot h_{\rm E}}{d}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \Delta \tau = \frac{\Delta d}{c} \approx \frac {2 \cdot h_{\rm S}\cdot h_{\rm E}}{c \cdot d} \hspace{0.05cm}.$$

So the correct solution is the solution 3. With the given numerical values, we have

$$\Delta \tau \approx \frac {2 \cdot 500\,{\rm m}\cdot 30\,{\rm m}}{3 \cdot 10^8 \,{\rm m/s} \cdot 10000\,{\rm m}} = 10^{-8}\,{\rm s} = 10\,{\rm ns} \hspace{0.05cm}.$$

The relative error with respect to the actual value according to the subtask (3) is only $0.13\%$.
In solutions 1 and 2, the dimensions are wrong.
In solution 2, there would be no propagation delay if both antennas were the same height. This is clearly not true.

(5) The path loss exponent $\gamma = 2$ implies that the reception power $P_{\rm E}$ decreases quadratically with distance.

The signal amplitude thus decreases with $1/d$, so for some constant $K$ we have

$$k_1 = \frac {K}{d_1} \hspace{0.05cm},\hspace{0.2cm}|k_2| = \frac {K}{d_2} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \frac {|k_2|}{k_1} = \frac {d_1}{d_2}= \frac {10011,039\,{\rm m}}{10014,035\,{\rm m}} \approx 0.99 \hspace{0.05cm}.$$

The two path weights thus only differ in magnitude by about $1\%$.
In addition, the coefficients $k_1$ and $k_2$ have different signs ⇒ Answers 1 and 3 are correct.