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Doubt on e^jwt

  1. May 1, 2010 #1
    it is given that e^jwt=cos(wt)+jsin(wt)...
    but j=1 with angle 90 degrees..now by multiplying any function by j means multiplying the magnitude by 1 and phase change of 90 degree...
    so,
    jsin(wt)=sin(wt)*1 with angle 90 degrees
    = sin(wt+90)
    =cos(wt)
    so can we write e^jwt=cos(wt)+cos(wt)
    =2cos(wt)
     
  2. jcsd
  3. May 1, 2010 #2

    D H

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    Staff Emeritus
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    j is not 1. It is the square root of -1.
     
  4. May 1, 2010 #3
    square root of -1 is 1 with angle of 90 degrees or 1 with angle of -90 degrees
     
  5. May 1, 2010 #4
    You could see it this way, if you like, but then you should be aware that sin(x) is composed of two parts with different phases! In particular

    [tex]\sin(x)=\frac12\left(e^{i(x-\frac{\pi}{2})}+e^{i(-x+\frac{\pi}{2})}\right)[/tex]
    Note the minus sign in the second term.

    Now if you advances both phases you get
    [tex]i\sin(x)=\frac12\left(e^{ix}+e^{i(-x+\pi)}\right)[/tex]

    Which is not
    [tex]\sin(x+\frac{\pi}{2})=\frac12\left(e^{ix}+e^{-ix}\right)[/tex]

    So this rule of thumb rule you mention isn't really working.

    The explanation is: Advancing phases in the sense you have learned is not the same as multiplying with complex numbers! That only works for pure exponentials.
     
  6. May 2, 2010 #5

    uart

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    No it's the square root of (-1).

    Usually it's given the symbol "i" but in Engineering it's quite common to use the symbol "j" instead. The reason is because in Electrical Engineering it's very often used in AC circuit analysis where the symbol "i" is already used to denote electrical current.
     
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