Question

Compute the energy and the average power of the following signal:

$ x(t)=\sqrt{t} $

Answer 1

$ E\infty= \int_{-\infty}^{\infty}|x(t)|^2dt $

E$ \infty $ = $ \int_{-\infty}^{\infty} tdt $

E$ \infty $ = $ \frac{1}{2} t^2 $ evaluated from -$ \infty $ to +$ \infty $ = $ \infty $

P$ \infty $ = lim T$ \to $$ \infty $ $ \frac{1}{2T} $ $ \int_{-T}^{T}\ tdt $

$ \frac{1}{2T} (.5t^2)|_{-T}^{T} = \frac{T}{4} $

lim T$ \to $$ \infty $ = $ \infty $ = P$ \infty $

• Be careful! The stuff inside the integral should always be positive. You are integrating "t", which is sometimes positive and sometimes negative. So there must be a mistake somewhere.
• The key is to take the norm of the signal squared. Here the signal is $ \sqrt{t} $, so taking the norm of the signal squared gives $ |t| $.

Answer 2

Back to CT signal energy page