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 $

• The answer for the energy is correct, but the derivation is wrong. The answer for the average power is wrong. Try not to skip so many steps, it will help you to make fewer mistakes.

Answer 6

$ E_\infty=lim_{T \to \infty} \int_{-T}^T |x(t)|^2\,dt $

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

   $  E\infty= \int_{-\infty}^\infty |\sqrt{t}|^2\,dt=\int_0^\infty t\,dt $ 
   $  E\infty=(\frac{1}{2})*t^2|_{-\infty}^\infty $
   $  E\infty=(\frac{1}{2})\infty^2-0^2)=\infty $ in Joules

$ 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)}*(\frac{1}{2}t^2)|_0^T $
   $  P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(\frac{1}{2}T^2) $
   $  P\infty=lim_{T \to \infty} \ \frac{T}{4}=\infty $ in Watts

Answer 7

math> P\infty=lim_{T \to \infty} \ 1/(2T)\int_{-T}^{T} |x(t)|^2\,dt</math>

   $  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)}*(\frac{1}{2}t^2)|_0^T $
   $  P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*(\frac{1}{2}T^2) $
   $  P\infty=lim_{T \to \infty} \ \frac{T}{4}=\infty $ 

$ E_\infty=lim_{T \to \infty} \int_{-T}^T |x(t)|^2\,dt $

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

   $  E\infty= \int_{-\infty}^\infty |\sqrt{t}|^2\,dt=\int_0^\infty t\,dt $ 
   $  E\infty=(\frac{1}{2})*t^2|_{-\infty}^\infty $
   $  E\infty=(\frac{1}{2})\infty^2-0^2)=\infty $ 

Note that $ P\infty $ is allowed to be equal to $ E\infty $ which is equal to $ \infty $.

Back to CT signal energy page