# Path Loss - Solved Numerical Problems from Wireless Communications

Digital ElectronicsElectronics & ElectricalElectron

Let us understand the significance of path loss by solving some numericals.

## Example 1 − Problem Solution

For a microwave terrestrial-based line-of-sight communication operating at 10 GHz, what is the maximum faithful coverage distance that the signal could make before requiring a repeater? The following details are provided −

• Signal transmission power = 27.78 dBW

• Transmit antenna gain = 18 dBi

• Receive antenna gain = 20 dBi

• Signal transmission bandwidth = 4 MHz

• 2- sided Noise power spectral density = 10-10 W/Hz

Solution − We are provided with the following data −

ParameterValue
Pt30 dBW = 1000 W
Gt35 dBi = 3162.22
Gr35 dBi = 3162.22
f10 GHz
B4 MHz
N0/210-10 W/Hz

We know that the channel noise power is given by

$$N=N_{0}B$$

Here, N0 = 2 x 10-10 W/Hz

We want the signal power to be always greater than the noise power at the receiver side or at any intermediate repeater station. We use Friis transmission equation for modeling the signal power.

We have the signal power, at a distance‘d’ from the transmitter, expressed as

$$P_{r}=P_{t}G_{t}G_{r}(\frac{\lambda}{4\pi\:d})^{2}$$

Now, the condition is that the signal power must always be greater than the noise power.

$$P_{t}G_{t}G_{r}(\frac{\lambda}{4\pi\:d})^{2}>N_{0}B\:---(1a)$$

Upon simplifying this equation, we arrive at the relation (1 b)

Faithful distance covered by the signal.

$$d<\frac{\lambda}{4\pi}\sqrt{\frac{P_{t}G_{t}G_{r}}{N_{0}B}}\:---(1b)$$

Equation (1 b) represents the distance signal can faithfully cover before getting buried in noise. For distances exceeding ‘d’, the signal fades out completely due to noise. Thus, a repeater is required to continue the communication from here.

The wavelength λ can be found using the standard formula.

$$\lambda\frac{c}{f}=30mm$$

Plugging in the values from the table into (1 b), we get

$$d<8.44km$$

Thus, for distances upto 8.44 km, there is no need fora repeater.

We can also reaffirm this. Let us calculate the received signal power at a distance d = 8.44 km (as obtained in this problem as the cut-off distance).

P_{r}(d=8.44km)=P_{t}G_{t}G_{r}(\frac{\lambda}{4\pi\:d})^{2}\approx\:800\mu\:W

How to calculate noise power level?

Let us calculate the noise power level.

$$N=N_{0}B=(2\times\:10^{-10}W/Hz)(4MHz)=8\times\:10^{-4}W$$

Thus, we can observe that at a distance of 8.44 km from the transmitter, the signal power almost equals the noise power.

Thus, the SNR is nearly 0 dB. This is undesirable as the signal and noise become indistinguishable at this point.

We can further increase the coverage range by increasing the transmission power and/or using high gain antennas. However, we can’t keep increasing the transmission power as it might cause interference to adjacent channels. There are regulations that limit the operable power levels.

## Example 2 − Problem Solution

We are required to establish a wireless communication link between a transmitter and receiver that are 50 km apart from each other. Power available at the transmitter is 5 kW. The link operates at 12 GHz. The transmission power available at each repeater is 3 dB lower than that available at the source. The signal transmission bandwidth is 5 MHz. If the antenna gains are 25 dBi each (both at the source, repeaters and receiver), how many repeaters are required to complete the communication link?

Solution − Table. Given data

ParameterValue
Pt (source)5000 W
Pt (each repeater)2500 W (3 dB less than Pt source)
Gt25 dBi = 316.22
Gr25 dBi = 316.22
f12 GHz
B5 MHz
N0/210-10 W/Hz

With the given data, let us first compute the faithful distance covered by the signal from the source. Using (1 b), we find that after about 14 km, the signal gets buried in noise. Thus, we don’t need a repeater till 14 km. After this cut-off distance, we are in need of a repeater. The repeaters operate at power levels 3 dB lower than that available at the source. Hence, 5 kW reduced by 3 dB gives us 2.5 kW.

We proceed to find the faithful distance covered by the signal from the first repeater. From (1 b), we can find that this cut-off distance is 9.95 km. Thus, the distance covered so far is 23.95 km. After this, the signal fades out and hence we are in need of the next repeater.

Since the parameters related to all the repeaters are the same, we can complete the cycle.

Table − Operating power level and operable range of each repeater

StationOperating powerDistance covered
Transmitter (source)5 kW14 km
1st repeater2.5 kW9.95 km
2nd repeater2.5 kW9.95 km
3rd repeater2.5 kW9.95 km
4th repeater2.5 kW9.95 km

0 to 14 km – No repeater required

After 14 km till 23.95 km – covered by 1st repeater

After 23.95 km till 33.9 km – covered by 2nd repeater

After 33.9 km till 43.85 km – covered by 3rd repeater

After 43.85 km till destination – covered by 4th repeater

Thus, we require 4 repeaters each operating at 2.5 kW and displaced by 9.95 km from each other to complete the communication link between the transmitter and the receiver. The transmitter takes care of the first 14 km.

We can reduce the number of repeaters required by operating them at high power levels but again, there might be constraints on the permissible levels of operation.

## Path Loss and Received Power Level

From (8), we can observe that whenever the received power level is high, the path loss must have been less. By deploying high gain antennas we can reduce the path loss. It is important to note that the received power levels vary dynamically over several orders of magnitude.

We can compute how much path loss can affect the received signal level. But we know that apart from path loss, channel effects play a major role in influencing the magnitude of the received signal levels. Reflection, diffraction, scattering, absorption and fading contribute to signal distortion (and attenuation).

Published on 23-Jun-2021 14:42:02