Difference between revisions of "HW1.5 Joseph Mazzei Energy and Power of sqrt(t) ECE301Fall2008mboutin" - Rhea

Latest revision as of 12:34, 5 September 2008

Energy from $t_{1}$ to $t_{2}$

$E = \int_{t_1}^{t_2}\!|x(t)|^2\ dt$

$E = \int_{t_1}^{t_2}\!|\sqrt{t}|^2\ dt$

$E = \int_{t_1}^{t_2}\!t\ dt$

$E = \frac{1}{2}t^{2}|^{t_{2}}_{t_{1}}$

$E = \frac{1}{2}(t^{2}_{2}-t^{2}_{1})$

Average power in time interval from [ $t_{1},t_{2}$ ]:

$P_{avg} = \frac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}\!|x(t)|^2\ dt$

$P_{avg} = \frac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}\!|\sqrt{t}|^2\ dt$

$P_{avg} = \frac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}\!t\ dt$

$P_{avg} = \frac{1}{{t_2}-{t_1}}(\frac{1}{2}t^{2}|^{t_{2}}_{t_{1}})$

$P_{avg} = \frac{1}{{t_2}-{t_1}}(\frac{1}{2}(t^{2}_{2}-t^{2}_{1}))$

@@ Line 1: / Line 1: @@
-== In CT ==
+== For a Continuous Time Signal==
 Energy from <math>t_{1} </math> to <math>t_{2}</math>
 <math>E = \int_{t_1}^{t_2}\!|x(t)|^2\ dt</math>
+<math>E = \int_{t_1}^{t_2}\!|\sqrt{t}|^2\ dt</math>
+<math>E = \int_{t_1}^{t_2}\!t\ dt</math>
+<math>E =  \frac{1}{2}t^{2}|^{t_{2}}_{t_{1}}</math>
+<math>E = \frac{1}{2}(t^{2}_{2}-t^{2}_{1})</math>
 Average power in time interval from [<math>t_{1},t_{2} </math>]:
@@ Line 9: / Line 17: @@
 <math>P_{avg} = \frac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}\!|x(t)|^2\ dt</math>
-== In DT ==
+<math>P_{avg} = \frac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}\!|\sqrt{t}|^2\ dt</math>
+<math>P_{avg} = \frac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}\!t\ dt</math>
+<math>P_{avg} = \frac{1}{{t_2}-{t_1}}(\frac{1}{2}t^{2}|^{t_{2}}_{t_{1}})</math>
+<math>P_{avg} = \frac{1}{{t_2}-{t_1}}(\frac{1}{2}(t^{2}_{2}-t^{2}_{1}))</math>