If you attended the free Webinar on Wednesday I told everyone that I would answer the question posed in my last post, and I will. But, based on some of the comments I think a little more explanation is needed. Forgive me for my typo’s, but here it is:
The question asks;
*
"Suppose I have an Access-Point that is capable of 50-mW of output power. This AP is connected to a cable that has 3 dB of loss, then to an antenna with of 6 dB of gain. How would you figure out the actual power radiated?"*
First off notice that the AP is capable of 50-mW of output power. In the last post I began by discussing the Watt. The way is kinda too big of a number when talking about wireless transmissions. A transmitter of 4 Watts could send a wireless signal for miles, and you are probably aware that out Cisco APs and even the ones we get for our homes dont work outside of a few hundred feet. Thats because they transmit at a lower power level. In this case the AP transmits at 50-mW of output power.
Now as the signal transmits it gets weaker. This happens pretty fast. So if we were to measure how much loss we have had once the signal is received we might say the signal is 99.9 milliwatts weaker. This, from a human standpoint, is hard to measure. But, Fractions on the other hand, we like those. So we could say the signal is 1/10th weaker and its a little easier to work with. In fact, make the numbers bigger and you’ll see what I mean. For example; 99.999999 vs 1/10,000th. Get it?
So with that said, the mW is not the best value to use when we calculate wireless outputs. mW is a linear value. The decibel on the other hand measures relative strength between signals. It is specifically designed to measure power loss and gain.
We are going to use decibels to measure the gain and losses from the time the AP transmits at 50-mW until it leaves the antenna. So here comes the Powers of 3’s and 10’s that I mentioned in the last post.

Here’s how it works; with -3dB we just divide by 2, with +3dB we multiply by 2. with -10dB we divide by 10 and with +10dB we multiply by 10. Sounds easy right?
Now lets apply that to the problem. Remember *the AP is transmitting at 50-mW. This AP is connected to a cable that has 3 dB of loss*, in other words, 50-mW / 2 = 25-mW.
The problem continues, *then to an antenna with of 6 dB of gain.* So we take the 25-mW we had after the loss from the cable and multiply it by 2, then 2 again. Since its 6-dB of gain we solve like this: +3db +3dB. This leaves us with 100-mW of power radiated.
Well, that wasn’t so bad right? Here are a few more to spin around:
**Question 1:**
An AP is transmitting at 100-mW. The AP connects to a cable with 6 dB of loss then to an amplifier with 9-dB of gain. The amplifier connects to a cable with 3-dB of loss and then to an antenna with 3-dB of gain. Whats the power radiated from the antenna in milliwatts?
**Question 2:**
An AP is transmitting at 30-mW and connects to an antenna with 6-dB of gain. What’s the power radiated in milliwatts?
**Question 3:**
An AP is transmitting at 30-mW and is connected to a cable with 3-dB of loss, then to an amplifier adding 6-dB of gain, then another cable with 3 dB of loss and finally out of an antenna with 7-dB of gain. (this is a tough one!) What’s the radiated power in milliwatts?
Well that should keep you busy for a bit. But there is still more. Still ahead are some other values to discuss when it comes to RF-math such as dBi and dBm, but thats going to have to wait.