2

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.

MBaz
  • 15,314
  • 9
  • 30
  • 44
Michal
  • 21
  • 1
  • i presume your waveform is a sawtooth.

    this is correct:

    $$ P = \frac{1}{T}\int_{0}^{T}|s(t)|^20dt $$

    but this is not:

    $$ P \ne \frac{1}{2}\int_{0}^{2}t^2dt $$

     – robert bristow-johnson Oct 29 '19 at 21:19
  • 1
    and

    $$ x(t) = t \qquad \qquad \forall -\infty < t < \infty $$

    is not a periodic function.

     – robert bristow-johnson Oct 29 '19 at 21:20
  • 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. – Michal Oct 30 '19 at 05:43
  • 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. – robert bristow-johnson Oct 30 '19 at 06:39

1 Answers1

2

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...

Fat32
  • 28,152
  • 3
  • 24
  • 50