How to compute the energy efficiency in bits/s/watt

Question

The energy efficiency (EE) in wireless communications is usually calculated as the ratio of data rate to power consumption: how many data (in bits per second) are delivered per consumed power (in watts)? This is usually calculated at some fixed time instant.

I have multiple time instants. So, I am confused on how to compute the EE. Let us say there are two time instants, each of duration \$\Delta\$t. In the first period, the data rate is \$R_1\$ bits/s and the consumed power is \$P_1\$ watts. In the second period the data rate is \$R_2\$ bits/s and the consumed power is \$P_2\$ watts. Which one is correct $$\text{EE}=\frac{R_1+R_2}{P_1+P_2},$$ or $$\text{EE}=\frac{(R_1+R_2)}{2(P_1+P_2)}$$?

In the first formula, I just summed the data rate over the two time instants and divided by the sum of the consumed power over the two time instants.

In the second formula, I computed how many bits are delivered during the two time instants: \$\Delta R_1 + \Delta R_2\$. Then I divided the number of bits delivered by the energy consumed (in watt-seconds), which is I computed as: \$2\Delta(P_1+P_2)\$. I think this is wrong.

Answer 1

The total data delivered is $$bits = \Delta t_1 R_1 + \Delta t_2 R_2$$ The total energy consumed is $$joules = \Delta t_1 P_1 + \Delta t_2 P_2$$ So the overall efficiency is $$\frac{bits}{joules} = \frac{\Delta t_1 R_1 + \Delta t_2 R_2} {\Delta t_1 P_1 + \Delta t_2 P_2}$$ If all of the \$\Delta t\$s are the same, then they can be factored out and cancelled, and you just have $$\frac{bits}{joules} = \frac{R_1 + R_2}{ P_1 + P_2}$$

Answer 2

I believe you should use "average rate" divided by "average power" to produce anything meaningful so, your first equation will be correct.