Euler’s relation and the energy of a complex exponential signal

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Question :

Euler’s relation and the energy of a complex exponential signal

I’m having a little hard time understanding the equation:

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Why Euler’s relation’s square is equal to 1?

I know how to write Euler’s equation in terms of sin and cos:

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But when I do the math, the square of this equality is nothing but 1. I can’t get rid of from the sin and cos in the equation:
(ejwt)2=cos2(jwt)+2jcos(jwt)sin(jwt)+j2sin2(jwt)

=cos2(jwt)sin2(jwt)+2jcos(jwt)sin(jwt)

I’m probably missing something with the Euler’s relation here so I would like to hear what am I missing.

Asked By: Berk

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Answer #1:

Notice, that ?x?R:

|exp(xi)|=1

Because:

|exp(xi)|=|cos(x)+sin(x)i|=?cos2(x)+sin2(x)?= 1=?1=1

So, for your integral we get:

?T00|exp(?0tj)|2 dt=?T00|cos(?0t)+sin(?0t)j|2 dt=


?T00(?cos2(?0t)+sin2(?0t)?= 1)2 dt=?T00(?1)2?= 1 dt=


?T001 dt=[t]T00=T0?0=T0

Answered By: Jan

Answer #2:

I believe the vertical bars inside the integration denote magnitude. This exponent is a phasor whose magnitude is one and angle is (wt). The square of one is also one.

The magnitude of the phasor (or complex number) is the square root of: the real part squared plus the imaginary part (without j) squared. When you add sin^2 and cos^2, the result is one.

Hope this helps.

Answered By: Ahmad KFUPM

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