Power of a signal

[EngIntoHW]

Say that the average power of a signal i(t) is G.

What is the average power of i(t)cos(ωt)?

I think it might be 0.5G, but I don't understand why.

Can you help please explain it?

Thank you.

 

[Ratchit]

EngIntoHW,

What is the average power of i(t)cos(ωt)?

The average of any sinusoidal wave is zero. So according to your question, the average power of i(t)cos(ωt) is zero.

Ratch

 

[EngIntoHW]

The average power of i(t) is:
1/T * ∫i^2 = G

The average power of i(t)cos(ωt) is
1/T * ∫i^2 cos^2 = ?

For some reason, it equals 1/2 * G, but I dont understand why.
Keep in mind that i(t) is not constant, so you can't just take it outside of the integral.

 

[Ratchit]

PMEngIntoHW,

The average power of i(t) is:
1/T * ∫i^2 = G

As it is given.

The average power of i(t)cos(ωt) is
1/T * ∫i^2 cos^2 = ?

So will it be solved.

For some reason, it equals 1/2 * G, but I dont understand why.

Behold below.

Keep in mind that i(t) is not constant, so you can't just take it outside of the integral.

Yes, you can. The average power G is constant for every period of length 2∏.

Therefore G*(1/(2∏))*∫(cos(wt)^2)dt for interval 0 to 2∏ = G/2 .

Ratch

 

[MrAl]

Say that the average power of a signal i(t) is G.

What is the average power of i(t)cos(ωt)?

I think it might be 0.5G, but I don't understand why.

Can you help please explain it?

Thank you.

Hi,


This question is a little ill posed. On the one hand you are asking about something that sounds common but on the other hand you are asking a question that is very uncommon. Let me try to explain a few points here to help clear this up...

The average power of a current all by itself is zero. That's because a current with no voltage produces no power. This makes us assume that you are probably asking about the one ohm energy content of the signal. That would mean that the resistance is assumed to be 1 ohm and since we know the current we can calculate the instantaneous power and so we can find the average power.

The second point is that it is a sinusoidal wave which has an average of zero, but when we deal with a sinusoid we dont usually average over the entire period but over 1/2 the total period. Thus if the frequency is f, we dont average over the time period 1/f (which would give us zero which doesnt do us any good) but instead we average over 1/2f, which is half the period. This gives us the average because power from a sinusoid is generated in both halves of the sine wave. Doing it this way we would find that the average is i^2/2 for some peak value represented by i, which would be a constant. But you stated that i is not a constant and is really an unspecified function of time.

The third point, and probably the most critical, is that you are stating that i(t) is not constant. That means that i(t) is a function of time. This really throws the problem out the window because there's no way to know what the average is of a wave represented by a function that is not known or not specified.
However, since you have specified the average of i(t) as G, what we need to prove then is that the integral of the square of a function i(t) times the integral of the square of the function cos(wt) is equal to the integral of the square of the two functions multiplied together over the period T/2 (and each integral scaled by 2/T or 2/Ta):
(2/Ta)*Integral i(t)^2 dt * (2/T)*Integral cos^2(wt) dt = (2/T)*Integral ((i(t)*cos(wt))^2) dt (all over the period 0 to T/2 or Ta/2)
or in Latex:

[LATEX]\frac{2}{Ta}\int^{Ta/2}_{0} i(t)^2\,\, dt * \frac{2}{T}\int^{T/2}_{0} cos^2(wt) \,\,dt = \frac{2}{T}\int^{T/2}_{0} (i(t)*cos(wt))^2 \,\,dt[/LATEX]

I dont think this is true in general unless some constraint is imposed on the function i(t) such as maybe it has to be the same frequency as cos(wt). If there is some constraint on i(t) you should let us know now. See if you can find at least one function for i(t) that doesnt work in the above.
This assumes i understand your question correctly.

Interesting, it looks like it works for sinusoidal signals so far when the signals are harmonically related.

 

[EngIntoHW]

Hi Mr Al,

I thank you so much for this post!
I came to understanding many additional things after reading your post.

I understand the problem with i(t) not being constant.

I'll investigate it a bit more to see if there're additional constraints.

Thanks!

 

[MrAl]

Hi,

Oh sure you're welcome. Just to note though in the first part i had assumed that we would have i(t) symmetrical about zero amps, but that may not be true either so the more general form would be to integrate that first part from 0 to Ta instead of 0 to Ta/2 and multiply by 1/Ta instead of 2/Ta. This may or may not need splitting into two or more pieces to get the correct average value.

Later...
Ok, what i tried was this:
i(t)=2+sin(w*t)

which is simply a pseudo randomly chosen function that makes the proof a little simpler because it has the same period as cos(w*t). This means we can try to prove this equation:
[LATEX]
\[\int_{0}^{T}{\mathrm{cos}\left( t\,w\right) }^{2}dt\,*\,\int_{0}^{T}{\left( \mathrm{sin}\left( t\,w\right) +2\right) }^{2}dt=T*\,\int_{0}^{T}{\mathrm{cos}\left( t\,w\right) }^{2}\,{\left( \mathrm{sin}\left( t\,w\right) +2\right) }^{2}dt\][/LATEX]

where the previous Ta=T as we only have one period to worry about now, the period from 0 to T.
Also, since we can choose a frequency of 1Hz that means T=1 also, which really simplifies this a lot.

Doing the math, that equation evaluates to two different numerical values on each side:
[LATEX]
\frac{9}{4}=\frac{17}{8}
[/LATEX]

so we've proven that there is at least one function i(t) where G/2 will not work. This means there has to be some constraints placed on the function i(t), and that doesnt necessarily mean this concept will be rendered useless because those constraints may just happen to lead to some very common functions that we would encounter often that could help with certain specific problems quite a bit.