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==Problem== | ==Problem== | ||
− | + | Calculate the energy <math>E_\infty</math> and the average power <math>P_\infty</math> for the CT signal | |
<math>x(t)=tu(t)</math> | <math>x(t)=tu(t)</math> | ||
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---- | ---- | ||
==Solution 1== | ==Solution 1== | ||
+ | <math>E_\infty = \int_{-\infty}^\infty |tu(t)|^2\,dt = \int_{0}^\infty t^2\,dt=\infty</math> | ||
+ | |||
+ | <math>P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \int_{-T}^T |tu(t)|^2\,dt = lim_{T \to \infty} \ \frac{1}{2T} \int_{0}^T t^2\,dt =\frac{\infty}{\infty}=1</math> | ||
+ | |||
+ | <span style="color:red"> Your energy is correct, but you distributed the limit too early when you computed the average power, so your answer came out wrong. </span> | ||
+ | ---- | ||
+ | ==Solution 2== | ||
<math>E_\infty = \int_{-\infty}^\infty |tu(t)|^2\,dt)</math> | <math>E_\infty = \int_{-\infty}^\infty |tu(t)|^2\,dt)</math> | ||
Line 34: | Line 41: | ||
<math>P_\infty = \infty</math> | <math>P_\infty = \infty</math> | ||
+ | |||
+ | <span style="color:green"> Looks pretty good! </span> | ||
---- | ---- | ||
[[Signal_energy_CT|Back to CT signal energy page]] | [[Signal_energy_CT|Back to CT signal energy page]] |
Latest revision as of 17:06, 25 February 2015
Problem
Calculate the energy $ E_\infty $ and the average power $ P_\infty $ for the CT signal
$ x(t)=tu(t) $
Solution 1
$ E_\infty = \int_{-\infty}^\infty |tu(t)|^2\,dt = \int_{0}^\infty t^2\,dt=\infty $
$ P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \int_{-T}^T |tu(t)|^2\,dt = lim_{T \to \infty} \ \frac{1}{2T} \int_{0}^T t^2\,dt =\frac{\infty}{\infty}=1 $
Your energy is correct, but you distributed the limit too early when you computed the average power, so your answer came out wrong.
Solution 2
$ E_\infty = \int_{-\infty}^\infty |tu(t)|^2\,dt) $
$ E_\infty = \int_{0}^\infty t^2\,dt) $
$ E_\infty =\frac{t^3}{3}\bigg]_0^\infty) $
$ E_\infty =\infty-0 = \infty $
Calculating $ P_\infty $
$ P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \int_{-T}^T |tu(t)|^2\,dt $
$ P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \int_{0}^T t^2\,dt $
$ P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \frac{t^3}{3}\bigg]_0^T $
$ P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \frac{T^3}{3} $
$ P_\infty = lim_{T \to \infty} \ \frac{T^2}{6} $
$ P_\infty = \infty $
Looks pretty good!