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Poles and Zeroes of a Transfer Function

Discussion in 'Electronics Homework Help' started by LordSputnik, Oct 20, 2012.

  1. LordSputnik

    LordSputnik

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    0
    Aug 11, 2011
    So, after foolishly thinking I knew all about poles and zeroes of transfer functions after a few lectures on it, we're given this problem:

    [​IMG]

    Firstly, I didn't know that a transfer function could be between a current quantity and a voltage quantity - I thought it was a ratio between two relative quantities, such as input voltage and output voltage. This just looks like an impedance, relating voltage to current.

    Still, I made an attempt an attempt at the problem - I felt that it was just a simple case of applying Ohm's law to the total impedance, so V = iZ:

    [​IMG]

    However, as you can see, doing this I get a real zero, and two complex poles. While the question doesn't ask us to explain them, I can't seem to work out how these values are related to the input frequency. It just seems wrong, somehow.

    We were told in the lecture that, if a zero has a value s=X, then at frequency X Hz, there will be a +20db/decade slope. Likewise, a pole with a value s=X will have a -20db/decade slope at X Hz. These combine together to give the bode plot of the transfer function.

    However, to me, this seems to imply that all the values of s will be real - what happens if a pole or zero value has an imaginary part?

    Or have I gone about this in completely the wrong way? I also had a go at writing differential equations, but I couldn't really find one that applied.

    If anyone has the time, please post how you would do it, or let me know if/where I'm going wrong. Please remember that I'm a simple student and can only do it using impedance functions or differential equations and the Laplace transform!

    Thanks very much for anything you can do to help me fully understand this!
    Ben
     

    Attached Files:

    Last edited by a moderator: Oct 21, 2012
  2. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    The following reference should answer your questions -
    http://www.ee.up.ac.za/main/_media/en/undergrad/subjects/eli220/polezero.pdf

    When you have a pair of complex conjugate poles in the transfer function, there are two poles at the same frequency that together will contribute -40dB per decade past the break frequency in the Bode plot. Note that the conjugate poles are located on a circle in the complex frequency s-plane where the radius of the circle is the break frequency.
     
  3. LordSputnik

    LordSputnik

    45
    0
    Aug 11, 2011
    Thanks for your help! I think I had a look at that pdf before, but I only skim-read the first few pages.

    After reading it properly, it does indeed answer all my questions! I think I understand now...

    So, taking that into account:
    Zeroes:
    s = -2

    Poles:
    p1 = 1+2j
    p2 = 1-2j

    So, the gain will be 0db until 2 rad/s, then will increase at 20db/dec due to the zero on the left, then at 2.24 rad/s it will decrease at 20db/dec for all other frequencies, due to there being (2-1) poles on the left?
     
  4. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    I never did fully appreciate why poles & zeroes are considered important, except that they allow the Bode magnitude plot to be estimated and sketched. However, with a computer algebra program it becomes easier to numerically generate an accurate Bode plot than to make a rough sketch from the poles and zeroes.

    See the attached MathCAD sheet for this transfer function, including the Bode plot. Note that the zero frequency gain is 12 dB. Any idea why?
     

    Attached Files:

  5. LordSputnik

    LordSputnik

    45
    0
    Aug 11, 2011
    Well, when you substitute 0 rad/s into s, you get:

    (2(0)+4)/(0.2(0)+0.4(0)+1) = 4/1 = 4

    So the "gain" is 4 - which is the 12db on the graph?
     
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