Underwater propagation losses and device range

In this example, we understand the Thorp propagation model, the sources of under-water noise, the passive sonar equation and how device range can be estimated based on received SNR. Refer to section 3.1 for the underlying theory on signal level, transmission losses, and the passive sonar equation.

### Network setup#

• Create a scenario with two underwater devices

Figure 4‑4:Network Scenario

• Channel characteristics as PATHLOSS_ONLY

• Change the Following parameters, Right Click on Underwater_Device_1 Properties

Device Properties > Physical Layer
Source Level (dB//1μPa) 190.8, 187.78,183.91
Data Rate (kbps) 20
Modulation Technique QPSK, BPSK, FSK, 16QAM, 64QAM, 256QAM

Table 4‑7: Device Properties

• Create a CBR Application with Default Properties with Source ID as 1 and Destination ID as 2.
• Run the Simulation for 1000 sec.

### Analytical computations#

In the Thorp model, the $db/km$ attenuation is given by.

$$10log_{10}\alpha(f)= \begin{cases} 0.11 \times \bigg( \frac{f^2}{1+f^2}\bigg)+44\times \bigg(\frac{f^2}{4100+f^2}\bigg)+2.75\times 10^{-4}\times f^2 + 0.003 & f \geq 0.4KHz\\0.02+0.11 \times \bigg(\frac{f^2}{1+f}\bigg)+0.0011 \times f & f < 0.4 KHz \end{cases}$$

For this example, substituting $f = 20,$ we get $10\log_{10}{\alpha(f)} = 4.133\ db/km$, we see that the total pathloss is

$$10\ log\ A(d,\ f) = \ k \times \ 10\ log\ (d_{m})\ + d_{km} \times 10\ log\ \alpha(f)$$

Using input parameters $K\ (spread\ coefficient) = 2,\ f = 20\ kHz\$and distance between the source and destination, $d = 18\ km$, and we get the total transmission loss, $TL,$ as

$$TL = \ 10\ log\ A(d,\ f) = 153.51\ dB$$

Next, we turn to noise level $NL.\$ The turbulence, shipping, wind, and thermal, noise level in dB is given by

$$10\ logN_{t}(f) = \ 17\ - \ 30log(\ f)$$

$$10\ logN_{s}(f) = \ 40\ + \ 20 \times (s\ - \ 0.5) + \ 26\ log\ f\ - 60 \times log(f\ + \ 0.03)$$

$$10\ logN_{w}(f) = \ 50\ + \ 7.5 \times \sqrt{w}\ + \ 20\ log\ f - 40\ log(f\ + \ 0.4)$$

$$10\ logN_{th}(f)\ = \ - 15\ + \ 20\ log\ f$$

Substituting $f = 20\ kHz,$ shipping factor $s = 0.5,\$surface windspeed $w = 0\ m/s,\$ we get $N_{t} = - 22.03\ dB$, $N_{s} = - 4.27\ dB,$ $N_{w} = 23.63\ dB$, and $N_{th} = 11.02\ \ dB$. As explained in section 3.1.4 we see that wind noise has the most impact. After adding these noises in the linear scale and then converting back to $dB$, Total noise, $N_{Total}^{dB} = 23.87$. From the passive sonar equation

$$SNR = SL - TL - (NL - DI)$$

Substituting we get

$$SNR = 190.80 - 153.51 - (23.87 - 0) = 7.41\ dB$$

### Results: Packet Error Rate vs Distance#

For the above SNR, we plot PER vs. distance for different modulation schemes given default packet size of 14B.

Generally, range is defined as the Tx-Rx distance at which the PER is 10 %. From these plots we can determine a device's range. In summary, we see how the device range is dependent on Source Level, Noise, MCS and packet size.