1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Feedback System Help

  1. Oct 28, 2008 #1

    I have the following standard feedback system:

    http://img112.imageshack.us/img112/7152/feedbacksystemuq5.jpg [Broken]

    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]

    http://img504.imageshack.us/img504/9418/feedbacksystemscanpm3.jpg [Broken]
    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.


    Last edited by a moderator: May 3, 2017
  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


    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:
  5. Oct 30, 2008 #4
    Yes I forgot to edit my post to say just that, you are correct.
  6. Oct 31, 2008 #5

    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:


    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,

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


    User Avatar

    The transfer function is
    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

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

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




    Thanks for your help.


  9. Oct 31, 2008 #8


    User Avatar

    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


    User Avatar

    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


    User Avatar

    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


    User Avatar

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


    User Avatar

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook