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Homework Help: How do I find the magnitude of this function?

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm asked to find the magnitude of the following complex function:

    [tex]R(j \omega) = 1 + \exp{(-j \omega)} + \exp{(-j2 \omega)} + \exp{(-j3 \omega)} + \exp{(-j4 \omega)}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    I did the following but got stuck immediately as I'm not sure on how to proceed:

    [tex]|R(j \omega)| = |1 + \exp{(-j \omega)} + \exp{(-j2 \omega)} + \exp{(-j3 \omega)} + \exp{(-j4 \omega)}|[/tex]
  2. jcsd
  3. Oct 4, 2011 #2


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    Staff: Mentor

    I'm not sure what you are doing. Maybe finding the peak value of a signal and its harmonics?

    I think the idea would be to use the identity from maths: ejW=cosW + j sinW
  4. Oct 4, 2011 #3
    I'm trying to find the magnitude. If I use euler's identity, then I would get multiple values of sin's and cosine's:
    [tex]R(j \omega) = 1 + \cos{(\omega)} - j\sin{(\omega)} + \cos{(2\omega)} -j\sin{(2\omega)} ... [/tex]

    How do you find the magnitude of that?
  5. Oct 4, 2011 #4


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    Staff Emeritus
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    Gold Member

    Initially, I would stick with exponential notation and recall that [itex]|z|^2=zz^*[/itex]. Once you have computed [itex]|z|^2=zz^*[/itex] and simplified, you should find that you are left with a rather nice expression, which you should be able to easily convert into a real, non-negative, trigonometric expression.
  6. Oct 4, 2011 #5
    Does that mean I take the conjugates like this:

    [tex]R^{*} = 1 - \exp{(-j \omega)} - \exp{(-j2 \omega)} - \exp{(-j3 \omega)} - \exp{(-j4 \omega)} [/tex]


    [tex]R^{*} = 1 + \exp{(j \omega)} + \exp{(j2 \omega)} + \exp{(j3 \omega)} + \exp{(j4 \omega)} [/tex]
  7. Oct 4, 2011 #6


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    No. Like this:
    Remember that if [itex]z = Re^{i\theta}[/itex], then [itex]z^* = (e^{i\theta})^*R^*[/itex], but since R is real: [itex]z^* = R(e^{i\theta})^* = Re^{-i\theta}[/itex] (check this using Euler's formula if you are unsure).
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