# Working out EIRP

There is a debate about the proper way of how to calculate the EIRP, so I need your help on which method is wrong and why please!

Let's suppose we have the following:

Output power: 165 W

Reflected power: 1 W

Amp's gain: 62 dBi

Losses: 1.3 dB

Method 1:

Carried out by converting all powers from W to dBW

EIRP = 10log(165) - 10log(1) + 62 - 1.3 = 82.8748 dBW

Method 2:

Carried out by converting losses units (dB to W)

So, 1.3 dB = 1.3489 W

then, 1W + 1.348963W = 2.3489 W

EIRP = 10log(162.651) + 10log(2.3489) + 62 = 80.46608 dBW

The EIRP should be > 82.5 dBW in order to pass the specs

How can we proof which one is passed and which is failed? and WHY please?

Regards

• Where is the antenna gain being introduced? Is it lumped into "output power"? – Envidia Jun 15 at 15:44

10log(165) - 10log(1)

is wrong; 1W is reflected; remember, subtraction of logarithm yields the same as division of numbers the logarithm is taken from.

By the way, log(1) is always 0, so this should be a giveaway that this operation can't be right.

So, 1.3 dB = 1.3489 W

no. Definitily not. 1.3 dB is a unitless factor.

You need to refresh what a decibel is, and what dBW is!

• Thank you Marcus Actually, I know the what is wrong in method 2, but I haven't noticed subtracting the logs was wrong! – W.Dabbas Jun 16 at 8:20

Method 1 is close and Method 2 is even more off. There are a few mistakes:

The dB is a relative value, it is always calculated from a ratio of something. It is up to you to know what the reference is. You cannot convert direct values to and from dB and apply them correctly without keeping track of how those values were calculated in the first place.

## Method 1

As Marcus status, the fact that $$10\space log_{10}(1) = 0 \space dB$$ should clue you in that something is wrong. In this case, you must recognize that, ignoring other losses, the output power that makes it to the output port is 164 W since 1 W was lost to reflections. In this case, you do not calculate the loss by $$10\space log_{10}(1)$$ but by computing

$$10\space log_{10}\left(\frac{164}{165}\right) = -0.0264 \space dB$$

A more intuitive example might be if your reflected power was 82.5 W. By your logic that amounts to a loss of $$10\space log_{10}(82.5) = 19.165 \space dB$$ (that's a lot!), but it's easy to see that your loss is half of the 165 W of power, so -3 dB is the value to use. Note that these values are negative, so don't accidentally make them positive by subtracting.

## Method 2

Here we freely convert between the 1 W loss to dB blindly. Similarly, we convert the loss value from dB to W and then add them together. Again, this is wrong because we lost the reference from which we are converting.

Method 1 was close, and making the change described above should give you the correct value given the numbers you provided.

• Thank you for your explanation Envidia Correction: (I have changed the reflected power to 10 W to give more sense) Output power: 165 W = 22.17483 dBW Reflected power: 10 W = -0.2715 dBW Amp's gain: 62 dBi Losses: 1.3 dB So, EIRP = 22.17483 - 0.2715 + 62 - 1.3 = 82.60333 dBW – W.Dabbas Jun 16 at 8:27
• @W.Dabbas Good to see that you calculated the loss due to reflections correctly. However, be careful to write it as "10 W = -0.2175 dB". You should explicitly think of it as "155 W output given 165 W of input is a loss of -0.2175 dB". – Envidia Jun 16 at 17:37