(New page: == Adam Frey's Euler Identity Summary == The identity is a special case of Euler's formula from complex analysis, which states that where j = i = <math>\sqrt{-1} </math> : <math> e^...) |
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: <math> \cos wt = \frac{e^{j w}}{2} + \frac{e^{-j w}}{2} \,\!</math> | : <math> \cos wt = \frac{e^{j w}}{2} + \frac{e^{-j w}}{2} \,\!</math> | ||
and | and | ||
− | : <math> \sin wt = \frac{e^{j w}}{ | + | : <math> \sin wt = \frac{e^{j w}}{2j} - \frac{e^{-j w}}{2j} \,\!</math> |
--[[User:Freya|Freya]] 15:04, 20 July 2009 (UTC) | --[[User:Freya|Freya]] 15:04, 20 July 2009 (UTC) | ||
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+ | = A related mystery: why is the following wrong? Can somebody please explain?= | ||
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+ | <math> e^{j 2 \pi t} = \left( e^{j 2 \pi} \right)^t= \left( cos{2 \pi } + j sin{2 \pi } \right)^t= 1^t =1 </math> | ||
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+ | * [[Answer Here]] | ||
+ | On the other hand, if n is an integer, it is true that | ||
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+ | <math> e^{j 2 \pi n} = \left( e^{j 2 \pi} \right)^n= \left( cos{2 \pi } + j sin{2 \pi } \right)^n= 1^n =1 </math> | ||
+ | |||
+ | Back to [[Exams/Quizzes]] | ||
+ | |||
+ | [https://kiwi.ecn.purdue.edu/rhea/index.php/ECE301_(HuffmalmSummer2009) Return to main] |
Latest revision as of 09:24, 23 July 2009
Adam Frey's Euler Identity Summary
The identity is a special case of Euler's formula from complex analysis, which states that where j = i = $ \sqrt{-1} $
- $ e^{jx} = \cos x + j \sin x \,\! $
for any real number x. (Note that sine and cosine should be in radians)
In particular,
- $ e^{j \pi} = \cos \pi + j \sin \pi.\,\! $
We know from trig identities that:
- $ \cos \pi = -1 \, \! $
and
- $ \sin \pi = 0,\,\! $
which results in
- $ e^{j \pi} = -1,\,\! $
which gives the identity
- $ e^{j \pi} +1 = 0.\,\! $
Also useful is the relationship in splitting sine and cosine is where
- $ \cos wt = \frac{e^{j w}}{2} + \frac{e^{-j w}}{2} \,\! $
and
- $ \sin wt = \frac{e^{j w}}{2j} - \frac{e^{-j w}}{2j} \,\! $
--Freya 15:04, 20 July 2009 (UTC)
$ e^{j 2 \pi t} = \left( e^{j 2 \pi} \right)^t= \left( cos{2 \pi } + j sin{2 \pi } \right)^t= 1^t =1 $
On the other hand, if n is an integer, it is true that
$ e^{j 2 \pi n} = \left( e^{j 2 \pi} \right)^n= \left( cos{2 \pi } + j sin{2 \pi } \right)^n= 1^n =1 $
Back to Exams/Quizzes