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I am currently learning Fourier series (periodic signals) and there are few things I am not sure about.

(1)

$$ P = \frac{1}{T}\int_{-T/2}^{T/2}|s(t)|^2dt = \frac{1}{T}\int_{0}^{T}|s(t)|^2dt $$

Let's say, that period is $T = 2$ and my signal $s(t) = t$:

$$ P = \frac{1}{T}\int_{-T/2}^{T/2}|s(t)|^2dt = \frac{1}{2}\int_{-1}^{1}t^2dt = \frac{1}{3} $$

but

$$ P = \frac{1}{2}\int_{0}^{T}|s(t)|^2dt = \frac{1}{2}\int_{0}^{2}t^2dt = \frac{4}{3} $$

What am I missing here? Are these equations wrong? I've checked several sources and all of them state it this way. What kind of power do I calculate this way? Is it total power of periodic signal?

(2) I also have a problem, I don't understand how to calculate. I am given:

\begin{align} x(t) &= t\\T &= \frac{1}{3} \end{align} what is the power on interval $$ \left(\frac{1}{3};\frac{2}{3}\right) $$ How do I calculate this?

(3) Am I right if I say, that I calculate power of first $N$ harmonics with Parseval's theorem?

$$ P_n = a_0^2 + \frac{1}{2}(a_n^2 + b_n^2) $$

I am really trying to understand it, but the more I read, the more confused I am. So I really need to point me in the right direction.

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  • $\begingroup$ i presume your waveform is a sawtooth. this is correct: $$ P = \frac{1}{T}\int_{0}^{T}|s(t)|^20dt $$ but this is not: $$ P \ne \frac{1}{2}\int_{0}^{2}t^2dt $$ $\endgroup$
    – robert bristow-johnson
    Commented Oct 29, 2019 at 21:19
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    $\begingroup$ and $$ x(t) = t \qquad \qquad \forall -\infty < t < \infty $$ is not a periodic function. $\endgroup$
    – robert bristow-johnson
    Commented Oct 29, 2019 at 21:20
  • $\begingroup$ x(t) = t was just an example, but if I make it periodic with base period T = 2, it is sawtooth wave. How do I calculate power of such signal correctly? I might understand it better on solved example. Thank you. $\endgroup$
    – Michal
    Commented Oct 30, 2019 at 5:43
  • $\begingroup$ when $1<t<2$, then $x(t) \ne t$ you need to express that integral from 0 to 2 differently. you need to split the integral into 2 integrals. $\endgroup$
    – robert bristow-johnson
    Commented Oct 30, 2019 at 6:39

1 Answer 1

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You are missing the effect of the integration interval on the integrand function...

If you choose a period; T = 2 , and a periodic signal whose base period is $s_0(t) = t$ in the interval [0,2]... Then the integrand in the shifted interval [-1,1] will be different as given by

$$ s_1(t) = \begin{cases}{ t + 2 ~~~, -1 < t < 0 \\ t ~~~~~~~~~~ , ~ 0< t < 1}\end{cases} $$

Then you should write the integrals as follows: : $$ P = \frac{1}{2}\int_{-1}^{1}|s_1(t)|^2dt = \frac{1}{2}\int_{0}^{2}|s_0(t)|^2dt $$

You should get the same result now...

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  • $\begingroup$ i think you want to say it like this: $$ s_1(t) = \begin{cases}{ t - 2 ~~~, 1 < t < 2 \\ t ~~~~~~~~~~ , ~ 0< t < 1}\end{cases} $$ $\endgroup$
    – robert bristow-johnson
    Commented Oct 30, 2019 at 6:41
  • $\begingroup$ Not actually. Given $s_0(t) = t $ for the interval $ 0 < t < 2 $, as the base period, then I define $s_1(t)$ for the interval [-1,1] as $$ s_1(t) = \begin{cases} {t+2 ~~~,~-1<t<0 \\ t ~~~~~~~~~~,~~~~~~ 0< t < 1 }\end{cases}$$ $\endgroup$
    – Fat32
    Commented Oct 30, 2019 at 9:42
  • $\begingroup$ but that's not how the OP got it defined. $\endgroup$
    – robert bristow-johnson
    Commented Oct 30, 2019 at 19:50

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