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Homework Help: Poles with Increasing Phase

  1. Feb 2, 2014 #1
    1. The problem statement, all variables and given/known data

    Understanding the bode plot for,

    [tex]G(s) = \frac{1}{2s-1}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    Usually when I am drawing bode plots for either real poles or zeros, I simply solve for the break frequency knowing that a zero/pole will give me a +/-20dB/decade asymptote after the break frequency on my magnitude plot, and a +/-90° phase shift across two decades, with the center (i.e. the +/-45° point) at my break frequency on my phase plot.

    This has always worked when I have,

    [tex](s+a) \quad a>0[/tex]

    but it seems to change when,

    [tex](s - a) \quad a>0 \quad \text{or, } \quad (-s+a) \quad a>0[/tex]

    I can't wrap my head around why a pole has an increasing phase that starts at -180°. I was expecting a phase decrease from 0° to -90°.

    Can someone explain what I am misunderstanding?

    Attached Files:

    Last edited: Feb 2, 2014
  2. jcsd
  3. Feb 3, 2014 #2


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    You did solve the bode plot correctly.

    When a pole is in the right half plane (greater than zero) the phase will increase during the pole.
    It is important to note that this means the system is unstable.
  4. Feb 4, 2014 #3

    rude man

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    There is no such thoing as a Bode plot for poles in the right-hand plane.

    Zeros, OK.
  5. Feb 4, 2014 #4
    A Bode plot is usually just defined as a magnitude and phase plot of [itex]H(s)\rvert_{s = j\omega}[/itex] versus [itex]\omega[/itex], where [itex]H(s)[/itex] is the transfer function of the system. So strictly speaking, regardless of the implications of having poles in the right-half plane, the Bode plot for the system is still well-defined.

    I'm not sure why you would expect that. When in doubt, go back to basics:
    \left. H(s) \right|_{s = j\omega} = \left. \frac{1}{2s - 1} \right|_{s = j\omega} = \frac{1}{2j\omega - 1} = \frac{1}{-1 + j2\omega} \Rightarrow\\
    \arg(H(j\omega)) = \arg\left(\frac{1}{-1 + j2\omega}\right) = \arg(1) - \arg(-1 + j2\omega) = 0 - (\arctan(-2\omega) + \pi) = -\arctan(-2\omega) - \pi
    Try evaluating that for [itex]\omega \ll 1[/itex] and [itex]\omega \gg 1[/itex].
  6. Feb 4, 2014 #5

    rude man

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    No. The gain is undefined. The output approaches infinity for any finite input. So a "magnitude" of gain does not exist, neither does a "phase".
  7. Feb 4, 2014 #6
    It's true that the Bode plot for an unstable system does not give you information about its steady-state gain and phase shift (since it has no steady state to speak of), but that doesn't mean the Bode plot itself is undefined, which, if you read my post carefully, was what I was commenting on.

    [itex]H(s)\rvert_{s = j\omega}[/itex] evaluates fine for [itex]0 < \omega < \infty[/itex], be it for a stable or unstable [itex]H(s)[/itex].
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