Field Strength Formula — How it works

Several candidates failed to pass the Advanced licence examination recently. I asked them why they thought that they had failed. Most admitted to not doing nearly enough preparation and several admitted that their mathematic skills were very poor and had been since school days!

Yes, you can pass the examination with no maths skills at all. But why throw your chances away, and £30, for the sake of proper preparation and some maths practice. Believe me, you won’t regret it.

For instance, dealing with formulas is a very useful skill. Let’s take a typical examination question.

‘A Tx has a PEP output of 100 watts using a feeder with a 3dB loss, to a dipole with a 3dB gain. What is the field strength (FS) in volts per metre on the ground under the dipole 10 metres up’?

Well, the dipole gain and the feeder loss cancel out. This is just a red herring to make you worry! But if they’re not the same, then what happens? First, convert the 100 watts into dB. That’s 20dBW. (i.e. 20 dB above 1 watt). Let’s make the feeder very poor with a loss of 6dB. Aerial gain is +3dB, feeder loss – 6dB , + 20dB output from the Tx output and so we end up with 17db ERP = 50w. (Remember that + 3 dB doubles the power, -3dB halves it 50W is 3dB less than 100W)

Now we need the right formula to work out the Field Strength. This is… FS = __7√PEP (watts)__

Distance (metres)

So we have FS = 7__√50 __= __7 x 7.071__ = 4.95 volts per metre

10 1

(we’ve seen that 7071 number before , oh yes, the √ of ½ or 0.5 is .7071 ! Well, well!)

Ah, but supposing we had been given the distance and the field strength but not the PEP. Then what?

Same formula of course. FS = __7√PEP__ But we have to *manipulate the formula* to get

Distance (metres) the unknown PEP on one side, ON ITS OWN.

If all the people in the world had their money multiplied by 2, we would all be just as well off as we were before. Similarly, if we’d divided by two, no one is worse off. If we give everyone £5 or take £5 from everyone, we have all been treated equally. What I am getting at is that, __whatever you__ __do to one side of any formula, you__ __must do to the other, to keep the equilibrium.__

First, get rid of the distance on the RHS, by multiplying both sides (BS) by the distance………

FS x Distance = __7____√____PEP x Distance__ Distance divided by distance is 1, which gives us………

Distance

FS x Distance = __7____√____PEP x 1__ Now we must get rid of the 7 on the RHS by dividing BS by 7

1

__FS x Distance__ = __7____√____PEP__ 7 divided by 7 is 1 ! So we now have the **√**PEP on its own

7 7

Let’s now put in the numbers. __4.95 x 10 __= **√**PEP = 7.071

7

7.071 is the square root of the PEP. We must square BS to get the number, so 7.071 x 7.071 = 50w

( 50 watts…the PEP) Remember that squaring a number is just multiplying itself by itself!

Maths need practice. The more you do the easier it gets. Believe me, it works.