# Energy and power of time-continous signal $x(t)=t$?

Preparing for my exams, I stumbled upon this particular problem.

I tried to calculate it using following formulae: $$E = \int_{-\infty}^{\infty}|t|^2dt$$

Dividing it into two cases because of the absolute value of $$t$$, I got: $$E = -\int_{-\infty}^{0}t²dt + \int_{0}^{\infty}t²dt$$

As result for the previous integral I get $$-\infty + \infty$$ which in an indeterminate form.

Then I tried using the formula:

$$E = \lim_{T\to\infty} (-\int_{-\frac{T}{2}}^{0}t²dt + \int_{0}^{\frac{T}{2}}t²dt)$$

which yields $$\lim_{T\to\infty}(-\frac{T³}{24}+\frac{T³}{24})=0$$

Using that, I tried to calculate the power: $$P = \lim_{T\to\infty} \frac{1}{T} (-\int_{-\frac{T}{2}}^{0}t²dt + \int_{0}^{\frac{T}{2}}t²dt)$$

Solving the integral once again it simplifies to: $$\lim_{T\to\infty} \frac{1}{T}\cdot 0 = 0$$

I'm having doubts whether my solution is correct, could you point me into the right direction?

• What about $t^2$ for $t<0$? Commented Nov 24, 2023 at 22:29
• @robertbristow-johnson oh so it would be $(-t)²$? Commented Nov 24, 2023 at 22:42
• Which equals $t^2$
The first step of your analysis is incorrect, there is no need to introduce a negative sign for $$t < 0$$. Note that $$|t|^2 = t^2$$ for all real numbers $$t$$, so we can reduce the expression for the energy to $$E = \int_{-\infty}^{\infty} t^2 \, dt.$$ This is easily integrable and diverges to $$+ \infty$$.
Your calculation for the power can be simplified in the same way, $$P = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} t^2 \, dt.$$ Again, the integration is straightforward from here.