Phase shift

Discussion in 'General Electronics Chat' started by lkgan, Dec 22, 2009.

  1. lkgan

    Thread Starter Member

    Dec 18, 2009
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    Hi everyone,

    From the attached picture, when we reduce the amount of feedback (decreasing the β value), the magnitude of the bode plot will decrease but the phase shift will remain the same without affected by the feedback β value, why? What is the relationship between feedback, magnitude and phase shift?

    Appreciate for answering my questions....
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    The transfer function of this is:

    Y(s)/X(s)=G(s)=H(s)/[1+β*H(s)]

    As you increase β the gain is reduced and vice versa.

    If Y(s) has an angle φ and X(s) has an angle ω. Then H(s), without feedback will have an angle φ-ω.

    With feedback, G(s)=H(s)/[1+β*H(s)]=H(s)<φ-ω/[1+β*H(s)<φ-ω]

    As it is seen the angle of H(s) is not affected by β because β is a real number. If β was a complexi number then the phase would change too.
     
  3. lkgan

    Thread Starter Member

    Dec 18, 2009
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    Why when without feedback, H(s) will have an angle φ-ω and not φ / ω as β = 0 in the transfer function? And can you please explain more about the equation with feedback? Thanks...:)
     
  4. mik3

    Senior Member

    Feb 4, 2008
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    When you divide two complex numbers, the magnitude of the first is divided by the magnitude of the second. However, their angles are subtracted and not divided.

    Which equation?
     
  5. lkgan

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    Dec 18, 2009
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    Alright, I have understood what's your explanation, thanks and appreciate it. :)
     
  6. Ron H

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    But the phase of G(s) is affected by β.
     
  7. lkgan

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    Dec 18, 2009
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    Why G(s) is affected by β since β is just a real number?
     
  8. Ron H

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    Take the case where H(s) is a single pole. The closed loop pole of G(s) moves as a function of β, and the phase shift will follow the shift in the pole frequency.
     
  9. lkgan

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    Dec 18, 2009
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    Sorry, I don't quite understand what you mean, can you please explain by proving formula or any other easier way? Thanks....
     
  10. Ron H

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    From mik3's post,

    G(s)=\frac{H(s)}{1+\beta*H(s)}

    Let H(s) have the transfer function of a single pole op amp:

    H(s)=\frac{K}{s*\tau+1}

    where K is the DC open loop gain of the op amp.

    G(s)=\frac{\frac{K}{\tau}}{s+\frac{1+K*\beta}{\tau}}

    The pole of G(s) is at

    s=-\frac{1+K*\beta}{\tau}

    which is obviously a function of β. The phase shift will be -45° at the pole frequency.
     
  11. mik3

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    The closed loop pole is affected by β but the phase shift of the output relative to the input is the same in both the open and closed loop case.
     
  12. Ron H

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    Am I misunderstanding you?

    This is only true when the input frequency is substantially lower than the frequency of the closed loop pole. For example, when the input frequency is at the closed loop pole frequency, F(-3dB), the phase shift will be -45°.

    Here are the results of a simulation, with a single pole op amp having open loop gain (Aol) = 100,000 and Gain-bandwidth product (GBW) = 1Meg. This results in an open loop pole at GBW/Aol=10Hz.
    Note the positions of the closed loop poles as β is changed from 0 to 1. Note the phase shift, which is relative to the input.

    The curves correspond to the equations I posted.
     
  13. mik3

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    Feb 4, 2008
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    RonH,

    You are right, the phase shift would be the same if H(s) was just a real number with no phase shift. With a single pole case, the phase shift depends on β and equals:

    φ=-atan(ω/(1+βK))
     
  14. Ron H

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    Shouldn't that be

    \phi=-atan(\frac{\frac{\omega}{\omega_0}}{1+\beta*K}),

    where \omega_0 is the corner frequency (1/time constant) of H(s)?
     
  15. mik3

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    Yes, this is the general case for a single pole system. In the example you gave intially, ωo was 1.
     
  16. Ron H

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    I believe that in my initial example, ωo was 1/\tau.
    We are nitpicking this to death, but I don't like to leave threads hanging with errors that may trip up a reader.
     
  17. lkgan

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    Dec 18, 2009
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    Hi gentlemen,

    Thanks for replying and having a good discussion. Refering back to my initial diagram, the phase shift, <[βH(ω)] would not be affected by the value of β right? And hence the phase diagram wouldn't change by manipulating the value of β right?
     
  18. Ron H

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    You are correct (assuming β is a real number), but it seems to me that you would be interested in the angle of Y(ω), which is affected by β. Why are you interested in the angle of βH(ω)?
     
  19. lkgan

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    Dec 18, 2009
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    From the bode plot attachement, and refering to my initial system diagram, when we reduce the amount of feedback, we decrease β, obtaining the gray shaded magnitude plot. Reducing the β only make the gain crossover point moves toward the origin while the phase crossover point remains constant, resulting in a more stable system. If you look carefully, the phase plot is using the phase βH(ω) and not Y(s)/X(s). And from this, it leads to another question, why did the phase plot did not use Y(s)/X(s) to determine the stability of the control system but instead using βH(ω)? Appreciate if anyone got the answer......
     
  20. t_n_k

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    Since the relationship 1+βH(ω)=0 is the important condition for instability, one plots the bode diagram of βH(ω). This immediately gives the observer an idea of how closely one is approaching the stability limit condition of βH(ω)=-1.
     
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