# Two-Port Model: “What is the amplitude of the input signal?” Solution Misunderstanding

I cannot understand the solution to the following problem: The specific maths I can't understand are outlined in the following figure, along with what I thought the answer was. Is $P=\frac{V^2}{R}$ as the power dissipated over a resistor correct? So why isn't it the case in the solutions?

If someone could meticulously explain the second bubble, that would be great as well, since I don't get it based off of my understanding. Should it just be $\frac{(45.3mV)^2}{1}$? • Yeah, that formula. That was my mistake in the post, but that is what I did. I applied that formula. – Carl Sep 20 '16 at 22:42

There's a difference between the amplitude of a signal and the RMS wich is used for power calculations. The amplitude is $\hat{V}$.
$$P = \frac{V_{eff}^2}{R}$$
$$V_{eff} = \frac{\hat{V}}{\sqrt{2}} \rightarrow V_{eff}^2 = \frac{\hat{V}^2}{2}$$
• this is where the 2's in both bubbles are coming from. $V_o$ is the amplitude of the signal, so you have to divide it by $\sqrt{2}$ to get the "DC-equivalent". If you square $V_o/\sqrt{2}$ you get the two in the denominator. – Felix S Sep 21 '16 at 8:28