Calculate average power in one cycle.

[yungman]

Homework Statement

This is part of a bigger problem, I simplify to just:

[tex]P=k sin^2 \omega t [/tex]

How do I integrate over one cycle?

The Attempt at a Solution

[tex]\left\langle P \right\rangle = \int_0^T k\; sin^2 (\omega t) d t \;\hbox { where }\; T= \frac c f = \frac {2\pi}{ \omega} [/tex]

[tex]\left\langle P \right\rangle = k \int_0^{ \frac {2\pi}{ \omega}} sin^2 (\omega t) d t = \frac 1 {\omega} \int sin^2 u du = \frac k 2 u = \frac k 2 t|_0^{\frac {2\pi}{ \omega}} = \frac {k\pi}{\omega}[/tex]

I know the answer is

[tex]\left\langle P \right\rangle = \frac k 2 [/tex]

Please show me how to get the integration over one cycle.

Thanks

Alan

 

[hikaru1221]

The average of f(x) of period T: [tex]<f(x)> = \frac{1}{T}\int ^{r+T} _r f(x)dx[/tex] with r is arbitrarily chosen

Another way to deal with your particular function is:
[tex]P = ksin^2wt = \frac{k}{2}(1-cos2wt)[/tex]

 

[stevenb]

Homework Statement

This is part of a bigger problem, I simplify to just:

[tex]P=k sin^2 \omega t [/tex]

How do I integrate over one cycle?

The Attempt at a Solution

[tex]\left\langle P \right\rangle = \int_0^T k\; sin^2 (\omega t) d t \;\hbox { where }\; T= \frac c f = \frac {2\pi}{ \omega} [/tex]

[tex]\left\langle P \right\rangle = k \int_0^{ \frac {2\pi}{ \omega}} sin^2 (\omega t) d t = \frac 1 {\omega} \int sin^2 u du = \frac k 2 u = \frac k 2 t|_0^{\frac {2\pi}{ \omega}} = \frac {k\pi}{\omega}[/tex]

I know the answer is

[tex]\left\langle P \right\rangle = \frac k 2 [/tex]

Please show me how to get the integration over one cycle.

Thanks

Alan

Your answer of kT/2 makes more sense. Wouldn't you expect the integral (area under the curve) to be proportional to T? The only issue is whether you should be integrating from 0 to T/2 rather than 0 to T, because sin^2 has double the frequency of sine, as pointed out by hikaru's trig identity. But, if you want to integrate over the period T for the unsquared sinewave, then you are correct.

EDIT: Oh wait ! You want the average power, so you have to divide by T. That's what the issue is. So k/2 is correct.

 

[yungman]

Thanks, I forgot the 1/T.

Alan

 

[paranormal]

Hello Alan,

To calculate the average power in one cycle, we can use the formula:

\left\langle P \right\rangle = \frac{1}{T} \int_0^T P(t) dt

Where T is the period of one cycle. In this case, T = \frac{2\pi}{\omega} where \omega is the angular frequency.

Substituting the given equation for P(t), we get:

\left\langle P \right\rangle = \frac{1}{\frac{2\pi}{\omega}} \int_0^{\frac{2\pi}{\omega}} k sin^2(\omega t) dt

Now, we can use the double angle formula for sine to rewrite the integral as:

\left\langle P \right\rangle = \frac{1}{\frac{2\pi}{\omega}} \int_0^{\frac{2\pi}{\omega}} \frac{k}{2} (1-cos(2\omega t)) dt

Integrating this, we get:

\left\langle P \right\rangle = \frac{k}{2} \frac{1}{\frac{2\pi}{\omega}} \left(t - \frac{sin(2\omega t)}{2\omega}\right)|_0^{\frac{2\pi}{\omega}}

Simplifying this, we get:

\left\langle P \right\rangle = \frac{k}{2} \left(\frac{2\pi}{\omega} - \frac{sin(4\pi)}{4\omega}\right)

Since sin(4\pi) = 0, this simplifies to:

\left\langle P \right\rangle = \frac{k}{2}

So, the average power in one cycle is simply half of the peak power, which is what you have already mentioned in your attempt. I hope this helps. Let me know if you have any other questions.