Phase from transferfunction

Discussion in 'Homework Help' started by Chrill3, Oct 11, 2011.

  1. Chrill3

    Thread Starter New Member

    Mar 8, 2010
    9
    0
    I try to calculate the bode-plot from this transferfunction that I derived:

    1/((1+s)^2+1)

    I got stuck on this problem and dont know how to calculate phase or amplitude:mad:. Any help is appreciated.
     
  2. jp1390

    Member

    Aug 22, 2011
    45
    2
    H(s) = \frac{1}{(s+1)^{2} + 1} = \frac{1}{s^{2} + 2s + 2}

    To take the magnitude you are going to have to bring the transfer function back in terms of 'w', by s = jw.

    H(jw) = \frac{1}{-w^{2} + 2jw + 2}

    |H(jw)| = \frac{1}{\sqrt{(2 - w^{2})^{2} + (2w)^{2}}}

    = \frac{1}{\sqrt{w^{4} + 4}}

    As for the phase:

    Phase(H(jw)) = \arctan{(\frac{0}{1})} - \arctan{(\frac{2w}{2 - w^{2}})}

    = -\arctan{(\frac{2w}{2 - w^{2}})}
     
    Chrill3 likes this.
  3. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Chrill3 likes this.
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