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We have this signal:

$$\operatorname{rect}\left(\cos\left(\frac{\pi t} {2}\right)\right) $$

I must find the average power , how can i get there ?

My solution:

I have seen that
$$-\frac 12 < \operatorname{rect}\left(\cos\left(\frac{\pi t}{2}\right)\right) < 1$$ is for
$$\frac 23+k \pi <t < \frac 43 + k\pi$$ so in a period it should be rectangular pulse , but how do I calculate the average ?

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    $\begingroup$ Hi! Homework ? Online-Quiz ? Where have you been stuck ? What's that rect() function ? $\endgroup$ – Fat32 Nov 1 '20 at 21:13
  • $\begingroup$ Hi , rect is the function that is 1 between -1/2 <t < 1/2 $\endgroup$ – Giovanni Cerciello Nov 1 '20 at 21:18
  • $\begingroup$ Good! So what's the problem ? $\endgroup$ – Fat32 Nov 1 '20 at 21:19
  • $\begingroup$ and i was doing this exercise and i got 1/3 as average , but i have written that it should be 1/2 , so which one is correct ? $\endgroup$ – Giovanni Cerciello Nov 1 '20 at 21:19
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    $\begingroup$ You need to show how you arrived at your result. Then we can tell where you went wrong. $\endgroup$ – Matt L. Nov 2 '20 at 6:46
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The given signal is clearly periodic. So just follow these steps and find the solution:

  1. Find the period $T$.
  2. Figure out the interval within a period for which the signal equals $1$.
  3. Compute the average: $$\overline{x(t)}=\frac{1}{T}\int_0^Tx(t)dt$$

If you're convinced that you did everything right, don't worry about a given solution which is different.

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  • $\begingroup$ Doing this i find that the period is 4 , and the average is 1/3 is it right ? $\endgroup$ – Giovanni Cerciello Nov 3 '20 at 8:22
  • $\begingroup$ @GiovanniCerciello: Sounds like a reasonable result. $\endgroup$ – Matt L. Nov 3 '20 at 12:16

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