Now, you know the minimum R, so solve for T.

NOTE: We don't care what the S/N is at the transmitter. We know the signal at this point. We are very concerned what the S/N ratio is at the receiver because that is where we are trying interpret the incoming message. When we talk about S/N, it is almost always at a receive point. For this class, it will be the receive S/N all the time.

For problem 11:
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It doesn't matter what order you do them in. Remember that the original units are multiplied and divided. The rules shown in lesson 3 discuss how the units are manipulated, but the rules are based on multiplication and division of units so you can figure out the results without memorizing the rules.

Examples:

6 dBm + 8 dBm = 14 dBm^2 !! (miliwatts times miliwatts is miliwatts squared) 33dBW - 3dBW = 30 dB !! (Watts divided by Watts cancels out, leaving no units) 10dBft + 10dB - 20dB = 0 dBft !! (the sum is zero, ft times no units divided by no units leaves ft, NOTE: this does not equal zero in absolute)

For problem 11 c, you'll notice that you have km x km / km. That is fairly straight forward.

Let's look at problem 12:
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You are asked for the transmit power. Like problem 9, you have to work backwards from the receive power. Like problem 9 you have to get the receive power from the S/N ratio. The main difference here is that you are working the problem in dB instead of absolute.

If S/N = R / N, then in dB ...

10 log S/N = 10 log [ R / N ] = 10 log(R) - 10 log(N)

I use SNR to represent the dB value of the Signal to Noise Ratio.

SNR = R_dB - N_dB

In English, The signal to noise ratio in dB is equal to the receive power in dB minus the noise power in dB. Solve this equation for R in dB.

Unlike problem 9, the Transmit power only attenuates one time. For this problem, T x some percentage gives us R. We can call the percentage, gain or G. If G is greater than one, the it increases the power, like a megaphone or an amplifier. If G is a fraction, between 0 and 1, then you get a decrease in power.

T x G = R

Whoa, we have to do this in dB, right?

10 log T + 10 log G = 10 log R

T_dB + G_dB = R_dB

We know R in dB so we need to figure out what the gain is. The problem says, that 10 dB are lost, that is -10dB. That would be 1/10 in absolute numbers. So the Gain is 1/10, or -10dB. Keep track of the minus sign, don't lose it.

Put in -10 dB for the gain in dB and plug in the value you calculated for R in dB. Solve for T in dB and give that as your answer.