The free space model predicts that received power decays as a function of the square of the T-R separation distance (i.e. a power law function). This model is used when there is a LOS path between Tx and Rx.The free space power received by a receiver antenna which is separated from a radiating transmitter antenna by a distance d, is given by the Friis free space equation,

$P_{r}(d)=\frac{P_{t} G_{t} G r \lambda^{2}}{\left(4 \pi^{2}\right) d^{2} L}$

Where,

$P_t$ - Transmitted power

$P_{r(d)}$ - Power received at a distance d from the Tx

$G_t,G_r$ – Gain of Tx, Rx antenna

L is the system loss and $\lambda$ is the wavelength of signal given by, $\lambda=\frac{c}{f}=\frac{c}{w / 2 \pi}=\frac{2 \pi c}{w}$

Gain of any antenna

$G=\frac{4 \pi A^{2}}{\lambda}$

Free space propagation path loss in $\mathrm{dB}$

$P_{L}(d B)=10 \log \left(\frac{P t}{P r}\right)$

$\quad \quad \quad=10 \log \left(\frac{P_t(4 \pi)^{2} d^{2} L}{P_{t} G_{t} G_{r} \lambda^{2}}\right)$

$\quad \quad \quad=10 \log \left[\frac{(4 \pi)^{2} d^{2} L}{G_{t} G_{r} \lambda^{2}}\right]$

$\quad \quad \quad=-10 \log \left[\frac{G_{t} G_{r} \lambda^{2}}{(4 \pi)^{2} d^{2} L^{2}}\right]$

Assuming lossless system with $\mathrm{L}=1$ and having unity gain

$P_{L}(d B)=-10 \log \left(\frac{\lambda^{2}}{(4 \pi)^{2} d^{2}}\right)$

Thus, we see that path loss is inversely proportional to the square of the distance. This model is applicable only if the mobile lies in the far field region or fraunhoffer region i.e. outside the fraunhoffer distance ($d_f$). This is the distance from a transmitting antenna beyond which the radiative circles are prominent.

Limitations of this model are:

1. LOS path should exist between Tx and Rx.

2. The distance of Rx should be greater than fraunhoffer distance. i.e. $d\gt d_{f}$ ,where

$d_{f} : \frac{2 D^{2}}{\lambda}$

Where, D is the largest dimension of antenna

Usually large scale propagation models use a close in reference distance $d_{o}$ and the received power $P_{o}$ at this point. The power at any other point is given by

$P_{r}(d)=P_o\left(\frac{d_{o}}{d}\right)^{2}$

Expressing this equation in dBm,

$\operatorname{P_r}(d) d Bm=10 \log \left(\frac{P_{o}}{0.001}\right)+20 \log \left(\frac{d_{o}}{d}\right)$

Examples:

1) Find the far field distance for an antenna with a max dimension of 2.5 m and operating at a frequency of 900 MHz.

$\rightarrow$ $d_f = \frac{2D^2}{ \lambda}$

$d_f = 37.5 \ m$

$dBw + 30 = dBm$

(Hint: to convert dBW to dBm, just add 30 to the dBW value)

2) If a Tx produces 65W of power express the transmitted power in units of $dBm$ and $dBw$. If 50W is applied to a unity gain antenna with a 900 MHz carrier frequency find the received power in $dBm$ at a free space distance of 100 m from the antenna. What is $P_r$ at 10 km for the same case. Assume unity gain for the receiving antenna also.

$\rightarrow$ $P_t = 65 w$

$P_t \ (dBw) = \ 10 \ log \ 65 = \ 18.13 \ dBw$

$P_t \ (dBm) = \ 10 \ log \ (65 \times 10^3) = 48.13 dBm$

$P_r \ (100m) \ = \frac{P_t \ G_t \ G_r \ \lambda^2}{(4 \pi)^2 \ d^2 \ L}$

$= \ \frac{50 \times 1 \times (3*10^8)^2}{(900 MHz)^2 \times (4 \pi)^2 \times 100^2 \times 1}$

$P_r (d) = 3.51 mw$

$P_r (d) = -24.56 dBm$

$P_r(d) = P_o (\frac{d_o}{d})^2$

$P_r(10 \ km) = 3.5 \times 10^{-6} \ (\frac{100}{10 \times 10^3})^2$

=$0.351 nw$

3) An isotropic antenna is radiating at a frequency of 9 MHz. What is the free space path loss at a distance of 3 kms from the transmitter antenna?

$\rightarrow$ Frequnecy = 9 MHz

$\lambda=\frac{C}{f}=\frac{3 \times 10^{8}}{9 \times 10^{6}} \mathrm{m}$

Free space pathloss $=\left\{\frac{4 \times \pi \times d}{\lambda}\right\}^{2}$

Path loss$\operatorname(\mathrm{dB})=20 \log \left\{\frac{4 \times \pi \times d}{\lambda}\right\}=20 \log \left\{\frac{4 \times 3.14 \times 9 \times 10^{6} \times 3 \times 10^{3}}{3*10^8}\right\}=61 \mathrm{dBs}$

Path loss= 61 dB

4) Assuming Free Space propagation model, the transmit power is 10 mW and the received power is 10-7 mW. What is the carrier frequency if the distance between the transmitter and the receiver antennas is 3 km?

$\rightarrow$ Transmit power $=10 \mathrm{mw}=10 \mathrm{d} \mathrm{Bm}$

Received power $=10^{-7} \mathrm{mw}=-70 \mathrm{dBm}$

Path loss $\operatorname(\mathrm{dB})=80 \mathrm{dB} \quad(=10 \ \mathrm{dBm}-(-70 \mathrm{dBm}))$

Free Space path loss $(\mathrm{d} \mathrm{B})=20 \log \left(\frac{4 \times \pi \times d}{\lambda}\right)=80 \mathrm{dB}$

$\frac{4 \times 3.14 \times f \times 9 \times 10^{3}}{3 \times 10^{8}}=10^{4}$

$f=79.62 \mathrm{MHz}$

5) The received power at the receiver is -95 dBm. The distance between the receiver and transmitter is 4 kms and the wavelength of propagation is 0.5 meters. Assuming the free space propagation model, what is the transmitted signal power?

$\rightarrow$ Received power = -95 dBm

Path loss $(\mathrm{dB})=20 \log \left(\frac{4 \times \pi \times d}{\lambda}\right)=20 \log \left (\frac{4 \times 3.14 \times 10^{3}}{0.5}\right)=100 \mathrm{d} \mathrm{B}$

Transmitted Signal power =Received Power +Pathloss

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$=-95 dBm+100 dB=5 dBm

Transmitted signal power = 5 dBm

6) Power of a particular signal is 1 W. Calculate the power level in dBW and dBm.

Given, Power level(P) = 1w

Power level in $\mathrm{dBW}=10 \log _{10} \mathrm{P}=10 \log _{10} 1=0 \mathrm{d} \mathrm{BW}$

Power indBm = power in $\mathrm{dBW}+30$

Therefore, Power level in $\mathrm{d} \mathrm{Bm}=0+30=30 \mathrm{dBm}$

Therefore, Answer: 0 $\mathrm{dBW}$ and 30 $\mathrm{dBm}$.

8) Consider a transmitter antenna. The output power of the transmitter amplifier is 30 W and the transmit antenna gain is 15 dB. The feeder attenuation is 5 dB. What is the EIRP (Equivalent Isotropic Radiated Power)?

( Explanation: An isotropic radiator is an ideal antenna which radiates power with unit gain uniformly in all directions, and is often used to reference antenna gains in wireless systems. The effective isotropic radiated power (EIRP) is defined as $EIRP = P_t.G_t$ and represents the maximum radiated power available from a transmitter in the direction of maximum antenna gain, as compared to an isotropic radiator.In practice, effective radiated power (ERP) is used instead of EIRP to denote the maximum radiated power as compared to a half-wave dipole antenna (instead of an isotropic antenna), Since a dipole antenna has a gain of 1.64 (2.15dB above an isotrope), the ERP will be 2.15dB smaller than the EIRP for the same transmission system. In practice, antenna gains are given in units of dBi (dB gain with respect to an isotropic source) or dBd (dB gain with respect to a half-wave dipole).

$\rightarrow$ EIRP = Transmitter output power+ Antenna Gain-Attenutation

=15+15-5=25

Answer: 25 dBw