Wimax transmit power calculation

Do we need to consider return loss of the device along with insertion loss when we calculate the output power of the particular device?

For example Balun has return loss of 12dB and IL of 2.5 dB, when i give input to the balun as 0dBm what will be the out put of the balun power available?
output power= Input power-IL or
output power= Input power-(IL+RL)?

In such case if the input to the BALUN is 0dBm then the output of the Balun will be -15dBm because of the return loss 12 dB and insertion loss 2.6dB, Is this calculation correct?

If there is really a 12 dB return loss in the balun then the calculation is correct.

Iam just wondering why the RL of the Balun is 12dB. I think it is quite large. We use balun to match impedance and minimize RL.

Can you tell how the value of RL obtained?

Do not consider the return loss when making your link budget or EIRP calculations. The insersion loss is to be considered (only).

The return loss is the indicator of the health of your cable and antenna together. The antenna is an impedence matching device from the cable (50 ohms) to free space (377 ohms) and is frequency dependant.

Since there is no perfect impedence match there will be some reflected power. A return loss of 13 db means that 1/20 th of the power was reflected. If you were transmitting 10 watts than .5 watts was refected. This number is too little to worry about.

However, if you see reflected power start to rise you must ceck your cable and antenna and jumpers.

i over looked return loss as loss due to return instead of loss of the return. thats why i think its value is high for a loss.

to consider it theoretically in your calculation. just subtract the linear value of RL form 1 then convert it back to dB

for 12dB RL means 1/16 of incident is reflected hence 15/16 is for the load (1-1/16)

So your loss due to return is 10log(15/16)= - 0.28dB

[get this widget]