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. $
- Looks pretty good!
Answer 3
$ E\infty=\int_{-\infty}^\infty |x(t)|^2\,dt $
$ E\infty=\int_{-\infty}^\infty |\sqrt{t}|^2\,dt=\int_0^\infty t\,dt $ (due to sqrt limiting to positive Real numbers) (*) $ E\infty=.5*t^2|_{-\infty}^\infty $ $ E\infty=.5(\infty^2-0^2)=\infty $
$ P\infty=lim_{T \to \infty} \ 1/(2T)\int_{-T}^{T} |x(t)|^2\,dt $
$ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T |\sqrt{t}|^2\,dt=\int_0^T t\,dt $ $ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*.5t^2|_0^T $ $ P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(.5T^2) $ $ P\infty=lim_{T \to \infty} \ (.25T)=\infty $
- * It is actually possible to define $ \sqrt{t} $ for negative values of t. For example, $ \sqrt{-1}=j $, the imaginary number. Remember, signals can be complex valued. So unless the signal is explicitly restricted to the positive values of t, you cannot make that assumption.
Answer 4
$ x(t) = \sqrt{t} $
$ E_{\infty} = \int_{-\infty}^{\infty}|x(t)|^{2}dt $
$ E_{\infty} = \int_{-\infty}^{\infty}|\sqrt{t}|^{2}dt $
$ E_{\infty} = \int_{-\infty}^{\infty}t dt $
$ E_{\infty} = \frac{1}{2}t^{2}|_{-\infty}^{0}+\frac{1}{2}t^{2}|_{0}^{\infty} $
$ E_{\infty} = \infty $
$ P_{\infty} = lim_{T\rightarrow\infty}\frac{1}{2T}\int_{-T}^{T}|x(t)|^{2}dt $
$ P_{\infty} = lim_{T\rightarrow\infty}\frac{1}{2T}(.5T^{2}|_{-\infty}^{0}+.5T^{2}|_{0}^{\infty}) $
$ P_{\infty} = lim_{T\rightarrow\infty}\frac{1}{4}(T|_{-\infty}^{0}+T|_{0}^{\infty}) $
$ P_{\infty} = \infty $
- I see the same mistake are in Answer 1 above.
Answer 5
E$ \infty $ = $ \int $|$ \sqrt{t} $|^2dt = $ \int $tdt =(t^2)/2|-$ \infty $,$ \infty $ = $ \infty $
P$ \infty $ = lim((1/(2*T))*$ \int $|$ \sqrt{t} $|^2dt) = lim(T-(-T)) = $ \infty $
