Revision as of 13:01, 24 February 2015

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

$ E\infty= \int_{-\infty}^{\infty}|x(t)|^2dt= \int_{-\infty}^{\infty}|t| dt = \int_{-\infty}^{0}-t dt+\int_{0}^{\infty} t dt=\infty+\infty=\infty. $

$ P\infty= limit_{T\rightarrow \infty} \frac{1}{2T}\int_{-T}^{T}|x(t)|^2dt= limit_{T\rightarrow \infty}\frac{1}{2T} \int_{-T}^{T} |t| dt = limit_{T\rightarrow \infty} \frac{1}{2T}\left( \int_{-T}^{0} -t dt+\int_{0}^{T} t dt\right) =limit_{T\rightarrow \infty} \frac{1}{2T}\left( \frac{T^2}{2}+\frac{T^2}{2}\right)=limit_{T\rightarrow \infty} \frac{T}{2}=\infty. $

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