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Polar notation for a complex exponential function

  1. Aug 25, 2006 #1
    I am trying to find the polar notation for

    1 + e^(j4)

    I know that e^(jx) = cos x + jsin x
    = cos(4) + jsin(4)

    I can then find the magnitude and angle.

    This is nowhere close to the answers below.

    1) cos(2) + 1
    2) e^(j2)[2cos(2)]
    3) e^(-j4)sin(2)
    4) e^(j2)(1+2cos(2))

    Can anyone point me in the right direction?
     
  2. jcsd
  3. Aug 25, 2006 #2

    shmoe

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    So you know 1+e^(j4)=(1+cos(4)) + jsin(4). What is the absolute value of this number? (some trig identities will help). This will be enough to pick the answer from those 4.

    You can go another way, if you pull out an e^(j2) you get:

    1+e^(j4)=e^(j2)*(e^(-j2)+e^(j2))

    do you see how this helps?
     
  4. Aug 26, 2006 #3

    HallsofIvy

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    This is a bit confusing. You say you want polar form: "magnitude and angle", which are both real numbers. Yet three of the "answers" you give are complex numbers and one is a single real number. What exactly do those "answers" represent?
     
  5. Aug 26, 2006 #4

    shmoe

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    [tex]r e^{j\theta}[/tex] where theta and r are real is sometimes called polar form/notation.
     
  6. Aug 26, 2006 #5

    HallsofIvy

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    Yes, but none of the answers given are of that form.
     
  7. Aug 26, 2006 #6

    shmoe

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    [tex]e^{j\theta}r[/tex] then?

    What are you seeing that I'm not?
     
  8. Aug 27, 2006 #7
    Thanks Shmoe! I was able to work through to a solution.

    I agree with HallsofIvy That this question is confusing.
     
  9. Aug 28, 2006 #8

    HallsofIvy

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    Yes, the question is confusing but my remarks were incorrect and didn't help! As Shmoe said originally, 1+e^(j4)=(1+cos(4)) + jsin(4) which has modulus (absolute value)
    [tex]\sqrt{(1+ cos(4))^2+ sin^2(4)}= \sqrt{1+ 2cos(4)+cos^2(4)+ sin^2(4)}[/tex]
    [tex]= \sqrt{2+ 2 cos(4)}= \sqrt{2+ 2cos^2(2)- 2sin^2(2)}[/tex]
    (using the double angle formula cos(2x)= cos2(x)- sin2(x))
    [tex]= \sqrt{2(1- sin^2(2))+ 2cos^2(2)}= \sqrt{4cos^2(2)}[/tex]
    [tex]= 2cos(2)[/tex]

    Does that help?
     
  10. Aug 28, 2006 #9
    That was the exact answer that I came up with!

    Thanks all for the replies!
     
  11. Aug 28, 2006 #10

    shmoe

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    You should make sure you can find the angle as well, not all question will be multiple choice where the magnitude is enough to pick out the answer!

    I prefer the route of my second suggestion:

    1+e^(j4)=e^(j2)*(e^(-j2)+e^(j2))

    And you know cos(z)=(e^(jz)+e^(-jz))/2, so you get answer 2) without having to muck about with double angle formulas that I would likely have to derive on the spot anyways.
     
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