# Problems Five

Transmission line notes.

## Question 9

Define the term "isotropic radiator". A certain transmit antenna has boresight gain which is a factor 2.6 over isotropic. Express this gain as dBi. Note that antenna gain figures can be very confusing. Engineers like to work in dB, which is a logarithmic measure. Adding gains in dB is the same thing as multiplying numerical gain. There is much more insight to be had from thinking about antenna gain as a straight numerical factor. It is said that engineers cannot multiply, only add . . ["the Lord said, be fruitful and multiply, but I'm only an adder...."] so if you see a number for antenna gain quoted without any indication it is quite likely to be the dBi figure rather than the numerical factor by which the radiated power exceeds that of an isotrope. If you get this wrong, the whole caboodle fails. To compound the problem, many engineers use dBd rather than dBi. dBd refers to the gain with respect to a perfect dipole which has numerical gain over isotropic of 1.66, which is 2.2 dBi. Thus to find dBi, take the dBd figure and add 2.2

• An isotropic radiator is a hypothetical source radiating power equally in all directions. The power density incident on a large sphere centred on the source does not depend on the position on the surface of the sphere.
• A numerical gain of 2.6 is 4.15 dBi since 10 log[10](2.6) = 4.15.
• In dBd, this antenna has gain about 2dBd. So it might have been constructed from an array of two dipoles. Because, 2.2 + 2.2 = 4.4 less a little loss.

This transmit antenna is fed with a signal having power 800 Watts. Assuming that there are no scattering obstacles in the beam or the near field, and that there is no attenuation along the path, calculate the power density in watts/square metre, and the rms electric field, at a distance of 25km from the antenna along its boresight.

• At a radius of 25km = 25,000 metres, an isotropic radiator having 800 Watts total radiated power would give a power density of 800/[4 pi 25,000^2] Watts/square metre = 102 nanoWatts/square metre.
• The actual antenna has boresight gain of 2.6 so the effective isotropic radiated power for the actual antenna, on boresight, is 2.6*800 = 2080 watts.
• Therefore the received power density at 25km is 102*2.6 = 265.2 nanoWatts/sq metre.
• This power is the same as (E^2)/Zo where E is the r.m.s. electric field in Volts/metre and Zo is the impedance of free space, 120 pi = 377 Ohms.
• Thus E^2 = 120 pi 2.6 102 10^(-9) and the electric field is 10 mV/metre.

## Question 10.

A receiver is fed by an array antenna. The array consists of a broadside arrangement of 8 identical elements connected with equal weights and the same phases to the receiver. Each element has boresight gain of 6 dBi perpendicular to the plane of the broadside array. The frequency of the link is 200MHz. Calculate the array pattern gain, the total gain, and the effective area of the receive antenna array.

• The element gain of 6dBi is a factor element gain of 3.98 = 10^(0.6). The array gain for equally weighted and phased elements is equal to the number of elements, and in this case is 8.
• The combined gain is 8 * 3.98 = 31.85.
• The effective area A is obtained from the formula gain = 4 pi A/[lambda^2}. The wavelength lambda at 200MHz is 1.5 metres. The gain is 31.85 so the antenna effective area is 5.7 square metres.

If the transmitting system of Question 9 is pointing at this array from a distance of 100km, calculate the total received signal power.

• The power density at 25 km was 102 *2.6 nanoWatts/square metre. 100km is four times further away so the power has reduced by a factor 1/[4^2} = 1/16 because of the inverse square law. The area of a sphere of radius 100km is 16 times that of a sphere of radius 25km.
• The received power density at 100km is therefore 2.6*102/16 = 16.58 nanoWatts/sq metre.
• So the total received power is 5.7 * 16.58 = 94.5 nanoWatts. We multiply the power density by the effective area of the receive antenna, to find the total power received.

If the receiver noise power is due to thermal noise in 10MHz bandwidth at a source temperature of 300K, calculate the possible range of the link for the receiver signal to noise ratio to be greater than 10dB. Comment on your result.

• The noise power in a bandwidth B of 10 MHz at a temperature T of 300K is given by kTB = 1.38*10^(-23)*300*10^7 = 4.14*10^(-14) Watts. Here, k is Boltzmann's constant, 1.38*10-23 Joules/degree K. A signal of 4.14*10^(-13) Watts is 10 times larger than this.
• The received signal power to give 10dB S/N ratio is 10 times larger than this (a factor 10 = 10dB) and the received signal power at a distance R of 47,777 or about 48,000km will be at this level since [100/R]^2 * 94.5 * 10-(9) = 4.14 * 10^(-13).
• Thus with these antenna arrangements we can get a 10MHz bandwidth TV signal a distance 1.2 times the circumference of the earth. The transmit antenna has low gain so can be made omnidirectional. The power needed by the transmitter is of the order of 1 kWatt when allowing for notional antenna losses.