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Homework 4 Ben Horst:  [[HW4.1 Ben Horst _ECE301Fall2008mboutin| 4.1]]  ::  [[HW4.2 Ben Horst _ECE301Fall2008mboutin| 4.2]]  ::  [[HW4.3 Ben Horst _ECE301Fall2008mboutin| 4.3]]::  [[HW4.4 Ben Horst _ECE301Fall2008mboutin| 4.4]]::  [[HW4.5 Ben Horst _ECE301Fall2008mboutin| 4.5]]
 
Homework 4 Ben Horst:  [[HW4.1 Ben Horst _ECE301Fall2008mboutin| 4.1]]  ::  [[HW4.2 Ben Horst _ECE301Fall2008mboutin| 4.2]]  ::  [[HW4.3 Ben Horst _ECE301Fall2008mboutin| 4.3]]::  [[HW4.4 Ben Horst _ECE301Fall2008mboutin| 4.4]]::  [[HW4.5 Ben Horst _ECE301Fall2008mboutin| 4.5]]
 
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[[Homework 4_ECE301Fall2008mboutin|<< Back to Homework 4]]
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Homework 4 Ben Horst:  [[HW4.1 Ben Horst _ECE301Fall2008mboutin| 4.1]]  ::  [[HW4.2 Ben Horst _ECE301Fall2008mboutin| 4.2]]  ::  [[HW4.3 Ben Horst _ECE301Fall2008mboutin| 4.3]]::  [[HW4.4 Ben Horst _ECE301Fall2008mboutin| 4.4]]::  [[HW4.5 Ben Horst _ECE301Fall2008mboutin| 4.5]]
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==Signal==
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x(t) = 2sin(6t) + 4cos(3t)
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==Fourier Series==
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<math>
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x(t) = \sum_{k=- \infty }^ \infty a_ke^{jk\omega_0t}
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</math>
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By Euler's formula, we have:
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x(t)=2({ e^{j 6t} + -e^{-j6t} \over 2j}) + 4({ e^{j 6t} + e^{-j6t} \over 2})

Revision as of 12:45, 25 September 2008

<< Back to Homework 4

Homework 4 Ben Horst: 4.1  :: 4.2  :: 4.3:: 4.4:: 4.5


<< Back to Homework 4

Homework 4 Ben Horst: 4.1  :: 4.2  :: 4.3:: 4.4:: 4.5


Signal

x(t) = 2sin(6t) + 4cos(3t)

Fourier Series

$ x(t) = \sum_{k=- \infty }^ \infty a_ke^{jk\omega_0t} $

By Euler's formula, we have: x(t)=2({ e^{j 6t} + -e^{-j6t} \over 2j}) + 4({ e^{j 6t} + e^{-j6t} \over 2})

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood