relationship between poles and zeros of a transfer function and its bode plot

Discussion in 'General Electronics Chat' started by sp2821, Jun 18, 2015.

  1. sp2821

    Thread Starter New Member

    Jan 8, 2014
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    Hi all,

    I am having a hard time grasping concepts of poles/zeros of a transfer function and magnitude bode plot and relationship between them.
    From my understanding, pole and zero are any values of s that makes denominator and numerator = 0, respectively.
    This to me sounds like at zero, H(s) will be 0 and 20*log(0) will be -inf.
    Or at pole, H(s) will be +inf and 20*log(+inf) will be +inf.

    However, for example, if H(s)=K(s+2), the magnitude bode plot is flat from s=0 to s=2 with DC gain of K and have slope of +20dB/dc from s=2 to s=+inf.
    Why is this? Shouldn't it look like a notch filter?

    Thanks,
     
  2. ramancini8

    Member

    Jul 18, 2012
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    118
    First a notch filter requires more than one pole/zero to get up and down movement of the transfer function.
    H(s) = K(s + 2)/(s + 6)(s + 8) yields the upward line until (s + 6) flattens out the curve; then (s+8) causes the curve to head for -infinity. Your poor notch filter.
     
  3. tindel

    Active Member

    Sep 16, 2012
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    193
    Why do you think it should be a notch filter? I'm not following you there. A transfer function with a single zero is a differentiator.
     
  4. tindel

    Active Member

    Sep 16, 2012
    568
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    Actually - I think you mean H(s) = K*s/(s+2)^2 will give you a poor mans notch filter at 2 radians.

    s will be a differentiator out to 'DC' (-inf at s=0) and the first pole will flatten out the curve and the second pole will cause the curve to go downwards to -inf when s=+inf
     
  5. ramancini8

    Member

    Jul 18, 2012
    442
    118
    Same-same theory just said different.
     
  6. tindel

    Active Member

    Sep 16, 2012
    568
    193
    Err... yeah we're on the same page, but what you described was a low pass filter with gain in a limited band - technically it did differentiate, but just for a small part of the frequency spectrum creating a small bandpass filter. I described a traditional band-pass filter [not a notch filter... oops].

    A traditional notch filter would take the form H(s)=K*(s+1)^2/((s+0.1)(s+10))
     
  7. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Not quite!

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  8. sp2821

    Thread Starter New Member

    Jan 8, 2014
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    Well, my question is why? If H(s)=K(s+2), H(s)=0 only when s=-2. But when
    Mike could you explain why?

    At zero, H(s) has value of 0.
    At pole, H(s) has value of +inf.

    So if you take log of magnitude of H(s), shouldn't 20*log(mag(H(s)) have value of -inf at zero and +inf at pole?
     
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