Prove integral of complex exp

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Homework Statement


Prove that the integral of a complex exponential over an integer number of periods is zero.


Homework Equations



[itex] \int_{0}^{T_{0}}e^{j (2\pi /T_{0}) kt} dt = 0 , k = integer[/itex]

The Attempt at a Solution



I am never sure how to work a proof. In this case, i can see that it would be true but not sure how you go about "proving" it. That the area from 0 to 1/2 T0 would zero out the area from 1/2 T0 to T0. Can someone point me to a good example on how to work this type of proof? or help me through this one?

[itex]\frac{1}{e^{j(2\pi /T_{0})k}}e^{j(2\pi /T_{0})kt} \mid ^{0}_{T_{0}}[/itex]
 

Answers and Replies

  • #2
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I would start with substitution to get rid of the variable T0, and consider the case k=1 first.
That the area from 0 to 1/2 T0 would zero out the area from 1/2 T0 to T0.
You can split the integral in two parts and show that they are equal apart from their sign.
 
  • #3
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So would you just say consider T0 = 1, but i would worry that would be like saying that since 2+2 = 4 and 2*2 = 4 therefor addition and multiplication are the same...
 
  • #4
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[itex]\frac{e^{j(2\pi /T_{0})kT_{0}}}{j(2\pi/T_{0})k}- \frac{e^0}{j(2\pi/T_{0})k}[/itex]

[itex]e^j\theta = cos\theta +j sin\theta [/itex]

[itex]\frac{cos(2\pi k)+j sin(2\pi k)}{j(2\pi /T_{0})k} - \frac{1}{j(2\pi /T_{0})k}[/itex]


[itex]cos(2\pi k) = 1[/itex]

[itex]j sin(2\pi k) = 0[/itex]

[itex] \frac{1}{j(2\pi /T_{0})k} - \frac{1}{j(2\pi /T_{0})k} = 0[/itex]

Is this sufficient?
 
  • #5
Dick
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[itex]\frac{e^{j(2\pi /T_{0})kT_{0}}}{j(2\pi/T_{0})k}- \frac{e^0}{j(2\pi/T_{0})k}[/itex]

[itex]e^j\theta = cos\theta +j sin\theta [/itex]

[itex]\frac{cos(2\pi k)+j sin(2\pi k)}{j(2\pi /T_{0})k} - \frac{1}{j(2\pi /T_{0})k}[/itex]


[itex]cos(2\pi k) = 1[/itex]

[itex]j sin(2\pi k) = 0[/itex]

[itex] \frac{1}{j(2\pi /T_{0})k} - \frac{1}{j(2\pi /T_{0})k} = 0[/itex]

Is this sufficient?

Yes, that's exactly what you want. Though it is true that ##e^{j 2 \pi k}=1## for k an integer, right? Probably no need to go through the sines and cosines.
 
Last edited:
  • #6
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Hmm right, my idea was more complicated than necessary. You can simply compute the integral.

So would you just say consider T0 = 1, but i would worry that would be like saying that since 2+2 = 4 and 2*2 = 4 therefor addition and multiplication are the same...
No, it is like saying T0/T0=1 for all real T0 (apart from 0). A sound mathematical proof that T0 does not influence the result.
 

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