Several questions about power of periodic signal

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.

• i presume your waveform is a sawtooth. this is correct: $$P = \frac{1}{T}\int_{0}^{T}|s(t)|^20dt$$ but this is not: $$P \ne \frac{1}{2}\int_{0}^{2}t^2dt$$ Commented Oct 29, 2019 at 21:19
• and $$x(t) = t \qquad \qquad \forall -\infty < t < \infty$$ is not a periodic function. Commented Oct 29, 2019 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. Commented Oct 30, 2019 at 5:43
• 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. Commented Oct 30, 2019 at 6:39

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...

• i think you want to say it like this: $$s_1(t) = \begin{cases}{ t - 2 ~~~, 1 < t < 2 \\ t ~~~~~~~~~~ , ~ 0< t < 1}\end{cases}$$ Commented Oct 30, 2019 at 6:41
• 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 $$s_1(t) = \begin{cases} {t+2 ~~~,~-1<t<0 \\ t ~~~~~~~~~~,~~~~~~ 0< t < 1 }\end{cases}$$ Commented Oct 30, 2019 at 9:42
• but that's not how the OP got it defined. Commented Oct 30, 2019 at 19:50