1. Not finding help here? Sign up for a free 30min 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!

System transfer function

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data

    x[n] = anu[n]
    A discrete system
    y[n] = −1/2y[n − 1] + x[n];
    where x[n] and y[n] in- and output of the system, respectively.
    Find the system transfer function H(z), and sketch its zeros and poles
    in the z-plane
    2. Relevant equations
    u[n] is the unit step function


    3. The attempt at a solution
    I transformed x[n] to X(z)
    but what confuses is the-1/2y[n-1],
    which basically means the previous -half of the previous output plus the input is equal to out put. but I don't know how to represent the previous output.
     
  2. jcsd
  3. Oct 8, 2012 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Do you know the time-shift theorem? And don't forget the initial condition.

    And BTW I suggest rewriting the equation so as to put it in standard form:
    y(n+1) + by(n) = x(n+1), n = 0, 1, ... in your case.
     
    Last edited: Oct 8, 2012
  4. Oct 15, 2012 #3
    I'm not so sure about the time shifting theorem, but I used the Transfer function H(z)=Y(z)/X(z). my X(z) is z/(z-a) and I tried Y(z) to be z+1/2 but I'm unsure. after that I supposed to sketch it's Zeros and poles. Am I on the right track?
     
  5. Oct 15, 2012 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    You got X(z) right but I don't know where you're going with the rest. Certainly not with z+ 1/2.

    Question to you: if Z{y(n)} = Y(z), what is Z{y(n+1)}?

    Other than that I refer you to my previous post. I can't really say more without violating the rules of this forum - until you get past this point.
     
  6. Oct 15, 2012 #5
    well in the text book I'm using it's written

    Z [x (n − n0)] = z-n0X(z);

    and if Z(Y[n])= Y(z) then Z(Y[n+1])= z1Y(z)

    that's gives me y[n]+1/2y[n-1] = z0Y(z) + 1/2z-1Y(z)

    that's where I go Y(z) = 1+1/2z-1. I don't understand where I am going wrong
     
  7. Oct 15, 2012 #6

    rude man

    User Avatar
    Homework Helper
    Gold Member

    You book should say
    If Z{y(n)} = Y(z) then
    Z{y(n+1)} = z{Y(z) - y(0)}
    which is actually quite close to your second equation above, except for the initial condition y(0).

    Distinguish between y and Y. Y is the z transformed version of y. I notice you sometimes use them interchangeably. Not good.

    Again - I urge you to rewrite your original equation in the conventional way, with n, n+1 and n+2 terms instead of n-1, n and n+1. Reason is I don't want to get confused myself & come up with the wrong answer. It's very straight-forward.
     
  8. Oct 16, 2012 #7
    Z [x (n − n0)] = z-n0X(z);
    y[n]+1/2y[n-1]=x[n]
    →y[n+1]0+1/2y[n]=x[n+1]
    Z-transform and get
    → z1Y(z) + 1/2z0Y(z)=z1X(z)
    → (z+1/2])Y(z)=zX(z)

    given that H(z) = X(z)/Y(z)

    z/(z+1/2)=X(z)/Y(z)=H(z)

    I hope this is correct now?
    though I'm still not getting the initial point part.
     
  9. Oct 16, 2012 #8

    rude man

    User Avatar
    Homework Helper
    Gold Member

    This is now correct (I assume the 0 after the first term is a typo?)
    Also, the right-hand side is X(z), not zX(z). In you case x[n] = an+1u[n+1] so the Z transform of that term is X(z) = zZ{anu(n)}. But here you're dealing with any x[n] since you're after the transfer function H = Y/X so you ignore the particular form of x[n]. See below.

    given that H(z) = X(z)/Y(z)
    [/quote]
    H= Y/X, not X/Y.
    Never mind about the initial condition. In a transfer function it is assumed that all initial conditions are zero. I should have noticed at the start that you're looking for a transfer function rather than a general solution of a finite-difference equation.

    Including x[n] = anu[n] is a red herring in this problem. If you're looking for H = Y/X then the particular form of X is irrelevant. X is the input so Y = HX no matter what X happens to be. So your final equation is correct if you transpose X and Y and get rid of the numerator z.

    If you did have to go on and solve for y[n] given the particular input x[n] then you'd have to worry about Z{x[n]}.
     
  10. Oct 16, 2012 #9

    rude man

    User Avatar
    Homework Helper
    Gold Member

    EDIT:

    This is now correct (I assume the 0 after the first term is a typo?)
    Correct.

    given that H(z) = X(z)/Y(z)
    [/quote]
    H= Y/X, not X/Y.
    Almost. If you change X(z)/Y(z) to Y(z)/X(z).

    though I'm still not getting the initial point part.[/QUOTE]

    Never mind about the initial condition. In a transfer function it is assumed that all initial conditions are zero. I should have noticed at the start that you're looking for a transfer function rather than the general solution of the full finite-difference equation.

    Including x[n] = anu[n] is a red herring in this problem. If you're looking for H = Y/X then the particular form of X is irrelevant. X is the input so Y = HX no matter what X happens to be. If you did have to go on and solve for y[n] given the particular input x[n] then you'd have to worry about Z{x[n+1]}.

    Also: I think I confused you unencessarily by having you rewrite the equation so that n=0 is the lowest n.

    You could have left it as y[n] + (1/2)y[n-1] = x[n]
    then use Z{y[n-1]} = z-1Z{y[n]}. I apologize for this unnecessary detour.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook