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Feedback System Help

  1. Oct 28, 2008 #1
    Hey,

    I have the following standard feedback system:

    [​IMG]

    This shows the Z-Transform where:

    A(z) is the z transform for input a[n] and C(z) is the z-transform of the system's output c[n]

    [​IMG]
    For this question I assume I would need to get another difference equation with c[n] and b[n] terms then find c[-1] from the above question..?


    Any help would be greatly appreciated, just not sure how to get started on this one.


    Cheers,

    NW
     
  2. jcsd
  3. Oct 30, 2008 #2
    You can find the responses by doing nodal analysis. Take the left node, for instance. The sum of the quantities pointing into the node must equal that going out, so

    [tex]E(z)=A(z)+H(z)C(z)[/tex], but [tex]C(z)=E(z)G(z)[/tex] and so [tex]E(z)=A(z)+H(z)E(z)G(z)[/tex], or [tex]\frac{E(z)}{A(z)}=\frac{1}{1-H(z)G(z)}[/tex]. From here, it's obvious what C(z)/A(z) is in terms of H(z) and G(z). This quantity is the so-called impulse response. Can you do the rest?
     
  4. Oct 30, 2008 #3

    CEL

    User Avatar

    Although there is no sign indication in the input comparator, the sign for the feedback loop is probably negative, so we should have:
    [tex]E(z)=A(z)-H(z)E(z)G(z)[/tex]
    and
    [tex]\frac{E(z)}{A(z)}=\frac{1}{1+H(z)G(z)}[/tex]
     
  5. Oct 30, 2008 #4
    Yes I forgot to edit my post to say just that, you are correct.
     
  6. Oct 31, 2008 #5
    Hey,

    Thanks for the replies... I can’t believe I made that mistake... CEL is right is a negative feedback system, sorry about that, it was late when I was drawing that pic :P

    Thanks for showing the steps of how you got that jhicks :)

    For the comparator (circle) A(z) is + and the Output from the H(z) block is a -... I.e. A(z) - [Output from the H(z) block], I guess you could call it [D(z)] which equals, [tex]D(z) = H(z)E(z)G(z)[/tex]

    Which gives me [tex]\frac{C(z)}{A(z)}=\frac{G(z)}{1+H(z)G(z)}[/tex]


    Subbing in for G(z) and H(z) to get the total system transfer function I.e. solve for [tex]C(z)[/tex] I end up with:

    [tex]\frac{C(z)}{A(z)}=\frac{z(z-8)}{(z-2)(z+4)}[/tex]


    What would the unit step response be..? I think I'm starting to get the hang of it... Would I just times the above system transfer function by the standard unit step z-transform or what..?

    Also, how do I find c[n] with initial conditions that's my 3rd question... (c[-1]=-3 c[-2]=4)?


    Thanks once gain,


    NW
     
    Last edited: Oct 31, 2008
  7. Oct 31, 2008 #6

    CEL

    User Avatar

    The transfer function is
    [tex]\frac{C(z)}{A(z)}=\frac{z(z+8)}{(z+2)(z+4)}[/tex]
    For the step response you multiply the transfer function by the z-transform of the step and take the inverse z-transform of the result.
    For your last question, the inverse z-transform of the transfer function gives you the relationship of c[n] and a[n]. Substitute the initial conditions in the equation and then substitute a[n] by its values for n = 0, 1, 2, ...
     
  8. Oct 31, 2008 #7
    Hey CEL,

    Thanks very much for your fast reply :)

    Can I ask, how you got [tex]\frac{C(z)}{A(z)}=\frac{z(z+8)}{(z+2)(z+4)}[/tex]

    I had:

    [tex]\frac{z}{(z+1)} * \frac{(z+1)(z-8)}{(z+1)(z-8)+9z}[/tex]

    Simplify to
    [tex]\frac{C(z)}{A(z)}=\frac{z(z-8)}{(z-2)(z+4)}[/tex]

    Where H(z) = 9/(z-8)..?

    Also would the unit step response for the system be..?

    [tex]\frac{C(z)}{A(z)}=\frac{z^2(z+8)}{(z+2)(z+4)(z-1)}[/tex]

    Or

    [tex]\frac{C(z)}{A(z)}=\frac{z^2(z-8)}{(z-2)(z+4)(z-1)}[/tex]???

    Thanks for your help.


    Cheers,

    NW
     
  9. Oct 31, 2008 #8

    CEL

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    You are right. I read H(z) as 9/(z+8) instead of 9/(z-8)
     
  10. Oct 31, 2008 #9

    CEL could you please help me further with this?
     
    Last edited: Oct 31, 2008
  11. Oct 31, 2008 #10
    Could someone please show me how to get the difference equation of the transfer function of the system?
     
  12. Nov 1, 2008 #11

    CEL

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    Multiply the numerator and the denominator by [tex]z^{-3}[/tex].
    Then, multiply C(z) by the denominator and A(z) by the numerator.
    Remember that the inverse transform of [tex]z^{-1}F(z)[/tex] is [tex]f[n-1][/tex].
     
  13. Nov 1, 2008 #12
    What the inverse z-transform for the transfer function of this system and how do I sub in the initial conditions in this equation and then a[n]?
     
  14. Nov 1, 2008 #13

    CEL

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    When you make the calculations I suggested you will have c[n] as a function of c[n-1], c[n-2], a[n] and a[n-1].
    You have c[-2] and c[-1]. a[-1] = 0, since a[n] = 0.5nu[n].
    Now you can calculate c[0].
    The other values of c[n] follow.
     
  15. Nov 1, 2008 #14
    Thanks for that CEL :) Seem to be getting somewhere :d

    For my last question the the 1st post... With B(z) - the output from block H(z)...

    would B(z) = C(z) * H(z)

    Which equals [tex]B(z) = H(z)G(z)[R(z)-H(z)C(z)] [/tex]...>?
     
  16. Nov 1, 2008 #15
    How would I find the initial conditions terms, like would I take the Z-transform of the difference equation when i sub in the initial condition terms to the difference eqn?
     
  17. Nov 2, 2008 #16

    CEL

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    Yes, that's it.
     
  18. Nov 2, 2008 #17

    CEL

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    You use the time domain equation.
    You have
    c[n] = f{c[n-1], c[n-2], a[n], a[n-1]}
    making n = 0, you get
    c[0] = f{c[-1], c[-2], a[0], a[-1]}
    once you have c[0], you can calculate
    c[1] = f{c[0], c[-1], a[1], a[-1]}

    and so on
     
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