Processing math: 100%

Exercise 2.2Z: Real Two-Path Channel

Aus LNTwww
Wechseln zu:Navigation, Suche

Zweiwege–Szenario

The sketched scenario is considered in which the transmitted signal  s(t)  reaches the antenna of the receiver via two paths: r(t) = r1(t)+r2(t)=k1s(tτ1)+k2s(tτ2).

Note the following:

  • The delays  τ1  and  τ2  of the main and secondary paths can be calculated from the path lengths  d1  and  d2  using the speed of light  c=3108 m/s .
  • The amplitude factors  k1  and  k2  are obtained according to the path loss model with path loss exponent  γ=2  (free-space attenuation).
  • The height of the transmit antenna is  hS=500 m. The height of the receiving antenna is  hE=30 m. The antennas are separated by a distance of  d=10  km.
  • The reflection on the secondary path causes a phase change of  π, so that the partial signals must be subtracted. This is taken into account by a negative  k2 value.



Note:



Questionnaire

1

Calculate the length  d1  of the direct path

d1 = 

  m

2

Calculate the length  d2  of the reflected path

d2 = 

  m

3

Which differences  Δd=d2 d1  and  Δτ=τ2τ1  (term) result from exact calculation?

Δd = 

  m
Δτ = 

  ns

4

What equation results for the path delay difference  δτ  with the approximation (1+ε)1+ε/2 valid for small  ε ?

Δτ=(hS hE)/d,
Δτ=(hS hE)/(cd),
Δτ=2hShE/(cd).

5

Which statements apply for the amplitude coefficients  k1  and  k2 ?

The coefficients  k1  and  k2  are almost equal in magnitude.
The magnitudes  |k1|  and  |k2|  differ significantly.
The coefficients  |k1|  and  |k2|  differ in sign.


Sample solution

(1)  According to Pythagoras: d1=d2+(hShE)2=102+(0.50.03)2km=10011.039m_.

  • 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 (4).


(2)  If you fold the reflected beam on the right side of xR downwards (reflection on the ground), you get again a right triangle. From this follows: d2=d2+(hS+hE)2=102+(0.5+0.03)2km=10014.035m_.


(3)  With the results from (1) and (2) you get for the lengths– and the runtime difference:

$$\Delta d = d_2 - d_1 = \hspace{0.1cm} \underline {=2,996\,{\rm m} \hspace{0.05cm},\hspace{1cm} \delta \tau = \frac{\delta d}{c} = \frac{2,996\,{\rm m}}}{3 \cdot 10^8 \,{\rm m/s}} \hspace{0.1cm} \underline {=9,987\,{\rm ns} \hspace{0.05cm}.$$

(4)  With hS+hEd 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 you get for this:

Δτ2500m30m3108m/s10000m=108s=10ns.

  • The relative falsification to the actual value according to the subtask '(3) is only 0.13%.
  • In solution 1 the unit is already wrong.
  • In solution 2, there would be no propagation delay if both antennas were the same height. This is certainly not true.


(5)  The path loss exponent γ=2 says that the reception power PE decreases quadratically with distance.

  • The signal amplitude thus decreases with 1/d, and with a constant K applies:
k1=Kd1,|k2|=Kd2|k2|k1=d1d2=10011,039m10014,035m0.99.
  • The two path weights thus only differ in amount by about 1%.
  • However, the coefficients k1 and k2 have different signs   ⇒   Correct are the answers 1 and 3.