# 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. 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:

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

$= -\arctan{(\frac{2w}{2 - w^{2}})}$

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3. ### Georacer Moderator

Nov 25, 2009
5,154
1,280
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