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Magnitude & Phase

  1. Jan 31, 2013 #1
    1. The problem statement, all variables and given/known data

    H(e^jw) = (1-1.25e^(-jw))/(1-0.8e^(-jw))

    Prove |H(e^jw)|^2 = G^2, and what is G
    Find Magnitude & Phase

    2. Relevant equations

    H(e^jw) = (e^(jw)-1.25)/(e^(jw)-0.8)

    H(e^jw) = 1 - (0.45e^(-jw))/(1-0.8e^(-jw))

    H(e^jw) = 1 - 0.45/(e^(jw)-0.8)

    3. The attempt at a solution

    I don't know how to approch this question. I can rewrite H(e^jw) in 4 different formats, but none of which make me understand how to attempt to get the magnitude in the first place.
  2. jcsd
  3. Jan 31, 2013 #2

    Simon Bridge

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    H(e^jw) = (1-1.25e^(-jw))/(1-0.8e^(-jw))
    this would be:$$H(e^{j\omega})=\frac{1-(1.25)e^{-j\omega}}{1-(0.8)e^{-j\omega}}$$

    What is the definition of the magnitude of a complex exponential/number?
  4. Jan 31, 2013 #3

    $$Magnitude = \sqrt{Re(z)^2 + Im(z)^2}$$
  5. Jan 31, 2013 #4

    Simon Bridge

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    Great - so all you need to do is identify the real and imaginary parts of ##H(e^{j\omega})## ... how do you do that?
    Hint: the exponential describes a phasor.
  6. Jan 31, 2013 #5
    Because |Z|^2 = z * zbar

    we have



    $$|H(e^{j\omega})|^2=(\frac{1-(1.25)x^-1-(1.25)x + 1.25^2}{1-(0.8)x^-1-(0.8)x+0.8^2})(\frac{x}{x})$$

    $$|H(e^{j\omega})|^2=\frac{x-(1.25)-(1.25)x^2 + 1.25^2x}{x-(0.8)-(0.8)x^2+0.8^2x}$$

    $$|H(e^{j\omega})|^2=\frac{-1.25}{-0.80}\frac{x^2 + 2.05x - 1}{x^2 + 2.05x -1}$$

    $$|H(e^{j\omega})|^2=\frac{-1.25}{-0.80} = 1.5625$$

    $$|H(e^{j\omega})|= G = 1.25 = Magnatude$$

    I got the Phase by computing the two angles, one on top with arctan (Im/Re), and the one on bottom, then phase total = top - bottom.
  7. Jan 31, 2013 #6

    Simon Bridge

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    See - you didn't need me :)
    Hmmm ... I notice that 1.25 is the magnitude of one of the phasors in the combination which is a little startling. The second to last line says that the magnitude is the square-root of the ratio of the magnitudes of the phasors. You could explore to see if this is a general result or just a judicious choice of amplitudes.


    Some latex notes (just sayin'):
    you can use \left ( <some stuff> \right ) to fit brackets around the bigger stuff so
    \left ( \frac{q}{p-1} \right ) gives you $$\left ( \frac{q}{p-1} \right )$$

    the complex conjugate is usually better represented by a star notation as in ##z^*## or ##z^\star## - the bar gets tricky to typeset after a bit and you risk confusing ##z_{ave}=\bar{z}## ...
  8. Feb 1, 2013 #7
    I've never used latex before I just quoted yours and edited it as needed but I'll keep it in mind for the future. Thx
  9. Feb 1, 2013 #8

    Simon Bridge

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    You did well at that too :) Most people just go "meh".
    Them I don't usually bother to give pointers ;)
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