# Questions about a problem in AM

1. Jan 29, 2014

Hi.
I may post a few questions here and there in the upcoming days in this thread (all related to AM). I begin with the first one:

and its solution:

http://www.israup.net/images/633efdf02562124d723708218182a185.png

Relevant equations

I have a two questions, both related to understanding the core of the AM process here:

1. What exactly is the peak envelope power? If it equals 32 kW, then why do I need to square the whole (2Ac) term? If Am=Ac (equal amplitudes of both message signal and carrier signal), then why do I need that square operation? I feel like that's a purely basic understanding I lack to completely understand what's going on here.

2. Sx=1/2
Sx is supposed to be the average power of the message signal. Assuming only the amplitude of the carrier signal is known now (i.e Am=Ac=8kW) - How it's possible to calculate that 1/2 value?

Thanks.

2. Jan 29, 2014

### rude man

Power goes as the square of amplitude.
You can't write " ... i.e Am=Ac=8kW ... "

Amplitude is V/m, power is E*H = poynting vector magnitude = (V^2)/Z with Z = 377 ohms for a plane wave in vacuo.