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Homework Help: 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

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    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

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    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

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    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

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    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

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    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.
     
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