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Homework Help: Stability of a linear system

  1. Jul 26, 2011 #1
    1. The problem statement, all variables and given/known data
    Determine the stability of the following linear system

    [itex]y(n) = 0.5x(n) +100x(n-2) - 20x(n-10)[/itex]

    2. Relevant equations

    [itex]x(n) = 0.5\delta(n) [/itex]

    [itex]S=\sum^{\infty}_{k=0}\left| h(k)\right|[/itex]

    3. The attempt at a solution

    [itex]Z \left\{ 0.5x(n) +100x(n-2) - 20x(n-10) \right\} [/itex]

    [itex] Z \left\{y(n) \right\} = \frac{xz}{2(z-1)^2}+100x(\frac{z}{(z-1)^2}-\frac{2z}{(z-1)})-(\frac{20x}{(z-1)^2}-\frac{10z}{(z-1)}) [/itex]

    [itex] \frac{80.5xz}{(z-1)^2}[/itex]

    Now at this point we were told that there should be a polynomial in the numerator... did I go about this all wrong? Any recommend reading would be helpful as I have exhausted Google searching for a similar problem.

    My original approach was simply to take the geometric series and use each coefficient from this equation if the formula [itex]\sum\frac{1}{1-a}[/itex]

    My result was [itex] \approx -.47 [/itex] which I though would be marginally stable as it is between -1 and 1.
     
  2. jcsd
  3. Jul 27, 2011 #2
    I don't know what you did, but typically one does not consider a specific input when determining the stability of a system; rather, one determines if the system is BIBO stable (bounded output for all bounded inputs).

    Anyway, a reasonable approach is to apply the Z-transform to the equation for y[n] to get an equation for Z{y[n]} in terms of Z{x[n]} using the "time shifting" property. Then factor out Z{x[n]} from each of the terms.

    To get the transfer function, divide by Z{x[n]}.

    You may manipulate both numerator and denominator to get polynomials for each. If you have a number of terms with different denominators added, write them in terms of a common denominator. If the numerator or denominator have negative powers of z, mutliply the numerator and denominator by a term with a corresponding positive power of z. Similiarly, if something such as 1/(1 + z) appears in the denominator, multiply both numerator and denominator by (1 + z).
     
    Last edited: Jul 27, 2011
  4. Jul 27, 2011 #3
    I had a feeling this was a strange question which doesn't help since I don't fully understand how to obtain transfer functions in the first place.

    So applying the time shift

    [itex]0.5X(n) + z^{-2}100X(z) -20z^{-10}(z)[/itex]

    Collect the [itex]X(z)[/itex]'s This transform is from the time shift right?

    [itex]X(z)= 0.5 + \frac{100}{z^{2}} - \frac{20}{z^{10}}[/itex]

    So from here I could just manipulate the numerator and denominator to get whatever it is I need, yes? Now dividing by [itex]Z\left\{x[n]\right\}[/itex] I don't understand what value that would be in this case.

    The time shift was the key for me! Thank you so much for your time.
     
    Last edited: Jul 27, 2011
  5. Jul 28, 2011 #4
    This is wrong. It should be something like:

    [itex]Y(z)= \left(0.5 + \frac{100}{z^{2}} - \frac{20}{z^{10}}\right)X(z)[/itex]

    This is a transfer function:
    [itex]\frac{Y(z)}{X(z)}= \left(0.5 + \frac{100}{z^{2}} - \frac{20}{z^{10}}\right)[/itex]
     
  6. Jul 28, 2011 #5
    Ah, so it has a transfer function of one! I didn't think that it would be that easy. This also means that is stable (poles at zero) so I think we are all set. Thanks again!
     
  7. Jul 28, 2011 #6
    No, the system's transfer function is

    [itex]\left(0.5 + \frac{100}{z^{2}} - \frac{20}{z^{10}}\right)[/itex]
     
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