Phasor help

Discussion in 'Homework Help' started by testing12, Sep 13, 2012.

  1. testing12

    Thread Starter Member

    Jan 30, 2011
    80
    2
    [​IMG]

    Hello,
    I am familiar with Euler identity, however I am not sure how the left side translates into the right side. Could someone please provide some extra detail.
    Kinds Regards
     
  2. blah2222

    Well-Known Member

    May 3, 2010
    554
    33
    cos(w_{0}t) = \frac{1}{2}[e^{jw_{0}t} + e^{-jw_{0}t}]

    Multiply in using law of exponentials with:

    e^{j\alpha t}
     
    testing12 likes this.
  3. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,795
    To see where the identity that blah2222 used comes from, consider the following:

    <br />
e^{j\theta} \,=\, \cos(\theta)\,+\,j\,\sin(\theta)<br />
\;<br />
e^{-j\theta} \,=\, \cos(-\theta)\,+\,j\,\sin(-\theta)<br />
\;<br />
\cos(-x) \,=\, \cos(x)<br />
\sin(-x) \,=\, -sin(x)<br />
\;<br />
e^{-j\theta} \,=\, \cos(\theta)\,-\,j\,\sin(\theta)<br />
\;<br />
e^{j\theta} \,+\, e^{-j\theta} \,=\, 2\cos(\theta)<br />
\;<br />
\cos(\theta) \,=\, \frac{1}{2} \left( e^{j\theta} \,+\,e^{-j\theta} \right)<br />
     
    testing12 likes this.
  4. testing12

    Thread Starter Member

    Jan 30, 2011
    80
    2
    Thank you, I will study these in greater detail this evening.
     
  5. Whyregister

    New Member

    Sep 12, 2012
    20
    0
    sin(wt) will be similar, but over 2j
     
  6. testing12

    Thread Starter Member

    Jan 30, 2011
    80
    2
    I know this is an old post, but i Would like to add to it...
    I have a new problem, see below. How is the real part of E1=2 Eo sin (B) z sin (wt) ? in the previous line there was a j in front of all of this, making it all imaginary.
    [​IMG]
     
  7. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,795
    No, e^(jwt) has both a real and an imaginary part. Remember:

    e^(jwt) = cos(wt) + jsin(wt)
     
  8. testing12

    Thread Starter Member

    Jan 30, 2011
    80
    2
    got it, thank you sir.
     
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