The second blog about RF Math is about calculation dBm to mW and mW to dBm. The rules from the 10s and the 3s still apply, but you need to change the mindset.

A new basic rule is that 0 dBm is equal to 1 mW. With the rules of 10 and rules of 3 you can make the below table.

Those were easy. So, what if we pick some other numbers. How and where do you start?

31 dBm: how many mW is that? (Starting with the 10s.)

31 dBm = 10 dBm + 10 dBm + 10 dBm + 1 dBm.

If we use the table from the first RF math blog we know that 1 dBm is the same as + 10 – 3 – 3 – 3.

Let’s combine this all together and see the end result.

31 dBm is *10 *10 *10 *10 /2 /2 /2.

1 mW, 10 mW, 100 mW, 1000 mW, 10000 mW, 5000 mW, 2500 mW, 1250 mW.

31 dBm equals 1,250 mW (1.3 W)

This was still the easy part. Now we will calculate mW to dBm.

800 mW: how many dBm is that? We need to rewrite the basic rules.

Multiplying or dividing by 2 is a gain or loss by 3 dB.

Multiplying or dividing by 10 is a gain or loss by 10 dB.

Let’s start with the big number 10 first and see how many times that number fits in 800.

1 mW, 10 mW, 100 mW, 1000 mW, 10000 mW, 5000 mW, 2500 mW, 1250 mW.

31 dBm equals 1,250 mW (1.3 W)

This was still the easy part. Now we will calculate mW to dBm.

800 mW: how many dBm is that? We need to rewrite the basic rules.

Multiplying or dividing by 2 is a gain or loss by 3 dB.

Multiplying or dividing by 10 is a gain or loss by 10 dB.

Let’s start with the big number 10 first and see how many times that number fits in 800.

800 / 10 = 80

80 / 10 = 8

80 / 10 = 8

At this point 10 doesn’t fit in 8 anymore, so we need to use the number 2.

8 / 2 = 4

4 / 2 = 2

2 / 2 = 1

4 / 2 = 2

2 / 2 = 1

At this point 2 doesn’t fit in 1 anymore so we are done and need to see how many 10s and 2s we have. Don’t forget the 2 is a 3. We have 2 times 10 dB and 3 times 2 (2 mW is a 3 dBm).

10 + 10 + 3 + 3 + 3 is 29. So, 800 mW equals 29 dBm.

This was a number that was easy to divide by the numbers 10 and 2. Let’s pick another number: 5731 mW.

This was a number that was easy to divide by the numbers 10 and 2. Let’s pick another number: 5731 mW.

5731 / 10 = 573.1

573.1 / 10 = 57.31

57.31 / 10 = 5.731

573.1 / 10 = 57.31

57.31 / 10 = 5.731

At this point 5 is smaller than 10, so we are switching over to the number 2.

5.731 / 2 = 2.8655

2.8655 / 2 = 1.43275

2.8655 / 2 = 1.43275

At this point 1.43275 is smaller than 2, so we are done with the calculation and need to check the final result. We have 3 times 10 and 2 times 2.

The total is 10 + 10 + 10 + 3 + 3 = 36 dBm. Since the end result was 1.43275 and not 1, the number 36 dBm is approximate. If we do the online calculation on 5731 mW, the result is 37.58 dBm. So, not the same, but pretty close.

This is a good approximate answer, but was it possible to get closer to the 37.58 dBm? Yes, of course, with the 1.43275 we are under the 2 mW. We know that 2 mW is equals to 3 dBm, so we can calculate 1 dBm and 2 dBm, so let’s do that.

The total is 10 + 10 + 10 + 3 + 3 = 36 dBm. Since the end result was 1.43275 and not 1, the number 36 dBm is approximate. If we do the online calculation on 5731 mW, the result is 37.58 dBm. So, not the same, but pretty close.

This is a good approximate answer, but was it possible to get closer to the 37.58 dBm? Yes, of course, with the 1.43275 we are under the 2 mW. We know that 2 mW is equals to 3 dBm, so we can calculate 1 dBm and 2 dBm, so let’s do that.

2 dBm = +10 + 10 – 3 – 3 – 3 – 3 – 3 – 3

2 dBm = 10, 100, 50, 25, 12.5, 6.25, 3.125, 1.5625

1.5625 is more than our rest value of 1.43275.

Let’s calculate 1 dBm.

2 dBm = 10, 100, 50, 25, 12.5, 6.25, 3.125, 1.5625

1.5625 is more than our rest value of 1.43275.

Let’s calculate 1 dBm.

1 dBm = +10 – 3 – 3 – 3

1 dBm = 10, 5, 2.5, 1.25

1 dBm = 10, 5, 2.5, 1.25

1.25 is less than our rest value of 1.43275. If we subtract those values we will have a rest value of 0.18275. So, with this simple calculation we can change our answer from 36 dBm to 37 dBm. This answer is closer to 37.58.

I hope you all understand the RF math better after these two blogs.

I hope you all understand the RF math better after these two blogs.

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