(New page: Let us find the energy and average power of a signal x(t) = ==Energy== ==Average Power==)
 
 
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Let us find the energy and average power of a signal x(t) =  
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Let us find the energy and average power of a signal <math>x(t) = 5e^{5t}</math> for the time interval [0,5]
  
 
==Energy==
 
==Energy==
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Energy expanded from [0,5]
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<math>E = \int_0^{5}{(5e^{5t})^2dt}</math>
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<math>E =\int_0^{5}{25e^{10t}dt}</math>
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<math>E =25\int_0^{5}{e^{10t}dt}</math>
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<math>E =25|\frac{e^{10t}}{10}|_{t=0}^{t=5}</math>
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<math>E =\frac{25}{10}|e^{50}-e^0|</math>
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<math>E =2.5(e^{50}-1)</math>
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==Average Power==
 
==Average Power==
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<math>P =\frac{1}{5-0}\int_0^{5}{(5e^{5t})^2dt}</math>
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<math>P =\frac{1}{5}\int_0^{5}{25e^{10t}dt}</math>
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<math>P =5\int_0^{5}{e^{10t}dt}</math>
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<math>P =5\int_0^{5}{e^{10t}dt}</math>
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<math>P =5|\frac{e^{10t}}{10}|_{t=0}^{t=5}</math>
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<math>P =0.5[e^{50} - e^0]</math>
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<math>P = 0.5(e^{50} - 1)</math>

Latest revision as of 10:12, 4 September 2008

Let us find the energy and average power of a signal $ x(t) = 5e^{5t} $ for the time interval [0,5]

Energy

Energy expanded from [0,5]

$ E = \int_0^{5}{(5e^{5t})^2dt} $

$ E =\int_0^{5}{25e^{10t}dt} $

$ E =25\int_0^{5}{e^{10t}dt} $

$ E =25|\frac{e^{10t}}{10}|_{t=0}^{t=5} $

$ E =\frac{25}{10}|e^{50}-e^0| $

$ E =2.5(e^{50}-1) $


Average Power

$ P =\frac{1}{5-0}\int_0^{5}{(5e^{5t})^2dt} $

$ P =\frac{1}{5}\int_0^{5}{25e^{10t}dt} $

$ P =5\int_0^{5}{e^{10t}dt} $

$ P =5\int_0^{5}{e^{10t}dt} $

$ P =5|\frac{e^{10t}}{10}|_{t=0}^{t=5} $

$ P =0.5[e^{50} - e^0] $

$ P = 0.5(e^{50} - 1) $

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