Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.
Hi,
I think u've to correct u'r question.Its a bit ambiguous.However, my answer is Average signal power is calculated based on the amplitude of the signal irrespective of sine or cosine.
For ex., if the signal is represented as Acos2Πfmt, then its average signal power would be A²/2.
Regards
Bhanumurthy.
Well. Let’s check it, my friends. Of course, for circular functions Sin(wt) and Cos(wt) the areas under positive and negative halves of period are equal and of the opposite signs (if area can be negative), hence total value is zero. Therefore, mathematically the average (mean) value for the whole period is zero and here you are right. But electrical engineers are not mathematicians, their horizon must be wider and more close to the real life. What does it mean for this particular case? Let’s see. First of all, power is not a vector it is scalar. It has no sign or phase. Moreover, it is not real physical phenomenon, it is just artificially created tool for describing some real electrical processes. The initial question was about the amount of the energy that exchanges for one Cosine period. Definitely, it cannot be zero or it will be in contrary with the first law of thermodynamics. Because power is not more than artificial tool, we must use the absolute value in order to get the actual rate of the speed of energy exchange into the circuit. Electrical engineers usually call this rate as power (Watt=Joule/second).
It is not difficult to show that mean (or active as it more often called) power will be equal to U*I*Cos(phi). Here U and I are the RMS values for voltage and current and Cos(phi) we should consider equal to unit (or phi=0 degrees) because for our discussion we are talking about pure Cosine signal without any phase considerations. I think that now it is clear that this simple formula will provide resulting coefficient of 0.5.
This is simple material that usually studied on the first or second semester of electrical theory course and it is easy to check with any available textbook.
The initial question was about the average power. RMS is probably easier because it is used every day for electrical engineering tasks. Mean power is not so popular as RMS, but used a lot.
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.