91Ó°ÊÓ

Consider an 18 GHz point-to-point communication system. Parabolic antennas are mounted on masts and the LOS between the antennas is just above the tree line. As a result, power falls off as \(1 / d^{3},\) where \(d\) is the distance between the antennas. The gain of the transmit antenna is \(20 \mathrm{dBi}\) and the gain of the receive antenna is \(15 \mathrm{dBi}\). The antennas are aligned so that they are in each other's main beam. The distance between the antennas is \(1 \mathrm{~km}\). The transmit antenna is driven by a power amplifier with an output power of \(100 \mathrm{~W}\). The amplifier drives a coaxial cable that is connected between the amplifier and the transmit antenna. The cable loses \(75 \%\) of its power due to resistive losses. On the receive side, the receive antenna is directly connected to a masthead amplifier with a gain of \(10 \mathrm{~dB}\) and then a short cable with a loss of \(3 \mathrm{~dB}\) before entering the receive base station. (a) Draw the signal path. (b) What is the loss and gain of the transmitter coaxial cable in decibels? (c) What percentage of the power input to the receive coaxial cable is lost in the receive cable? (d) Express the power of the transmit amplifier in \(\mathrm{dBW}\) and \(\mathrm{dBm}\). (e) What is the propagation loss in decibels? (f) Determine the total power in watts delivered to the receive base station.

Step by step solution

01

Draw the Signal Path

Illustrate the path the signal takes from the transmitter to the receiver, including all the components mentioned: the power amplifier, coaxial cable, transmit antenna, propagation path, receive antenna, masthead amplifier, and the receive base station.

02

Calculate Loss and Gain of the Transmitter Coaxial Cable

The coaxial cable loses 75% of its power due to resistive losses. Therefore, the loss in power is \[ L_{loss} = 10 \times \text{log}_{10}(0.25) \]which solves to approximately 6 dB loss.

03

Calculate Power Loss in the Receive Coaxial Cable

The coaxial cable has a loss of 3 dB, which means a loss of \[ P_{loss, \text{receive}} = 1 - 10^{-3/10} = 1 - 0.501 = 49.9\text{~%} \] of the power input.

04

Convert Transmit Amplifier Power to dBW and dBm

The power of the transmit amplifier is 100 W. In decibels, this translates to:\[ P_{\text{dBW}} = 10 \times \text{log}_{10}(100 \text{~W}) = 20 \text{~dBW} \]and \[ P_{\text{dBm}} = 10 \times \text{log}_{10}(100 \times 10^3 \text{~mW}) = 10 \times \text{log}_{10}(100,000) \text{~mW} = 50 \text{~dBm} \]

05

Determine the Propagation Loss in Decibels

The propagation loss follows the rule \[ L_{\text{propagation}} = 10 \times \text{log}_{10} \bigg( \frac{1}{d^3} \bigg) \] where \( d = 1 \text{~km} = 1,000 \text{~m} \).Thus, \[ L_{\text{propagation}} = 10 \times \text{log}_{10} \bigg( \frac{1}{(1,000)^3} \bigg) = -90 \text{~dB} \]

06

Calculate Total Power Delivered to the Receive Base Station

We combine the power components to find the total received power. The transmit power after coaxial loss (-6 dB) is \[ P_{\text{transmit}} = 100 \text{~W} \times 10^{-6/10} = 25 \text{~W} \]The received power from the propagation loss (-90 dB) is \[ P_{\text{received}} = 25 \text{~W} \times 10^{-90/10} = 2.5 \times 10^{-7} \text{~W} \]Adding gains of antennas and masthead (20 dB, 15 dB, 10 dB respectively) and cable loss (-3 dB), the total gain is\[ G = 20+15+10-3 = 42 \text{~dB} \]Thus, the final received power is\[ P_{\text{final}} = 2.5 \times 10^{-7} \text{~W} \times 10^{42/10} = 0.099 \text{~W} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

signal path diagram

Understanding the signal path in a point-to-point communication system is crucial. It shows how the signal travels from the transmitter to the receiver, including all key components. For our 18 GHz system, here is the path:
The signal starts at the power amplifier and travels through the transmitter coaxial cable. It then reaches the transmit antenna.
Next, it propagates through the air until it reaches the receive antenna. From there, it goes through the masthead amplifier and finally enters the receive base station via a short coaxial cable.
Visualizing the signal path ensures that we account for all gains and losses in the system. This helps us calculate the total received power accurately.

attenuation and gain calculations

Attenuation and gain calculations tell us how much signal is lost or amplified at different stages.
In this system, the transmitter coaxial cable has a loss of 75%. This loss can be converted to decibels (dB) using the formula:

• \( L_{loss} = 10 \times \log_{10}(0.25) \approx -6 \text{ dB} \)

The gains of the antennas and the masthead amplifier also affect the signal. The transmit antenna has a gain of 20 dBi, and the receive antenna has a gain of 15 dBi. The masthead amplifier adds another 10 dB gain.
Combining these values helps us determine the total gain or loss in the entire path.

power conversion (Watts to dBW/dBm)

Converting power measurements from Watts to decibels (dBW or dBm) helps with easier calculations and comparisons.
The transmit amplifier's output is 100 W. Converting this to dBW involves using the formula:

• \( P_{\text{dBW}} = 10 \times \log_{10}(100) \approx 20 \text{ dBW} \)

Similarly, converting it to dBm (decibels referenced to 1 milliwatt) gives:

• \( P_{\text{dBm}} = 10 \times \log_{10}(100 \times 10^3) \approx 50 \text{ dBm} \)

These conversions help in more straightforward signal calculations, especially when dealing with large distances and various gain/loss elements.

propagation loss

Propagation loss occurs as the signal travels through the air. For this example, the signal has an inverse cubic relation with distance, i.e., it falls off as \( \frac{1}{d^3} \).
Using the distance of 1 km (or 1,000 meters), the propagation loss in decibels is calculated as:

• \( L_{\text{propagation}} = 10 \times \log_{10}(\frac{1}{(1000)^3}) \approx -90 \text{ dB} \)

This significant loss compels us to ensure sufficient gain from amplifiers and antennas to maintain signal integrity.

total received power calculation

Calculating the total received power involves accounting for all gains and losses along the signal path.
Starting with 100 W (or 20 dBW) at the transmitter, the signal after -6 dB loss in the coaxial cable is:

• \( P_{\text{transmit}} = 100 W \times 10^{-6/10} \approx 25 W \)

After -90 dB propagation loss, the received power is:

• \( P_{\text{received}} = 25 W \times 10^{-90/10} = 2.5 \times 10^{-7} W \)

Adding the gains of the antennas (20 dB, 15 dB) and the masthead amplifier (10 dB), minus the receive coaxial cable loss (3 dB), the total gain is:

• \( G = 20 + 15 + 10 - 3 = 42 dB \)

Thus, the final received power is:

• \( P_{\text{final}} = 2.5 \times 10^{-7} W \times 10^{42/10} = 0.099 W \)

This calculation ensures we can determine if the received signal strength is adequate for reliable communication.