A link is described by a 2-ray ground reflection model, where r >> ht, hr and Pt = 50 W, fc = 1800 MHz, Gt = 20 dBi and Gt = 3 dBi.
a. If the received LOS signal power was measured to be Pr = 1μW at a range of r = 1 km, determine the path loss and the corresponding path-loss exponent.
b. Consider the 2-ray ground reflection model with the following system values: ht = 40 m, hr = 3 m and Gr = 3 dBi. Calculate the received power level at r = 2 km.
c. Consider the 2-ray ground reflection model of (b). Suppose the receiver antenna height hr is adjustable (but all other system values are fixed). Findthe value of hr which maximizes the received power.
The correct answer and explanation is :
To solve this, we use principles from wireless communication theory, specifically the 2-ray ground reflection model. Here’s the detailed solution:
Given:
- Pt = 50 W
- fc = 1800 MHz → λ = c / f = (3×10⁸) / (1.8×10⁹) = 0.1667 m
- Gt = 20 dBi → Gt_lin = 10^(20/10) = 100
- Gr = 3 dBi → Gr_lin = 10^(3/10) = 2
a. Path Loss and Path-Loss Exponent
Given:
- Pr = 1 μW = 1×10⁻⁶ W at r = 1 km
i. Path Loss (PL) in dB
[
PL = 10 \log_{10} \left( \frac{P_t \cdot G_t \cdot G_r}{P_r} \right)
]
Substitute:
[
PL = 10 \log_{10} \left( \frac{50 \cdot 100 \cdot 2}{1 \times 10^{-6}} \right) = 10 \log_{10} (10^4 \cdot 10^6) = 10 \log_{10}(10^{10}) = 100\ \text{dB}
]
ii. Path-Loss Exponent (n)
In free-space and many environments:
[
P_r \propto \frac{1}{r^n} \Rightarrow 10 \log_{10} P_r = -10n \log_{10} r + C
]
At r = 1 km (base), we define this as reference. To get n, we need another measurement at a different r. But since this is from a 2-ray model, at large r:
[
P_r = \frac{P_t G_t G_r h_t^2 h_r^2}{r^4}
\Rightarrow n = 4
]
✔️ Path loss = 100 dB
✔️ Path-loss exponent = 4
b. Received Power at r = 2 km using 2-Ray Model
[
P_r = \frac{P_t G_t G_r h_t^2 h_r^2}{r^4}
]
Substitute:
- ht = 40 m
- hr = 3 m
- r = 2000 m
- Pt = 50 W
- Gt = 100
- Gr = 2
[
P_r = \frac{50 \cdot 100 \cdot 2 \cdot (40)^2 \cdot (3)^2}{(2000)^4}
= \frac{50 \cdot 100 \cdot 2 \cdot 1600 \cdot 9}{(16 \times 10^{12})}
= \frac{144 \times 10^6}{1.6 \times 10^{13}} = 9 \times 10^{-6} = \boxed{9\ \text{μW}}
]
c. Maximize Received Power by Optimizing hr
From 2-ray model:
[
P_r \propto h_t^2 h_r^2 \Rightarrow \text{Maximize } h_r
]
But 2-ray model has destructive interference at some ranges, especially when:
[
r_c = \frac{4 h_t h_r}{\lambda}
]
At large distances, interference diminishes and power becomes proportional to ( h_r^2 ). Hence, maximum Pr occurs when hr is as large as possible.
However, exact maximum occurs when:
[
h_r = \frac{r \lambda}{4 h_t}
= \frac{2000 \cdot 0.1667}{4 \cdot 40}
= \frac{333.4}{160} ≈ \boxed{2.08\ \text{m}}
]
✅ Final Answers:
- (a) Path loss = 100 dB, Path-loss exponent = 4
- (b) Received power at 2 km = 9 μW
- (c) Optimum hr to maximize received power = 2.08 m