Stability of a linear system

  • #1

Homework Statement


Determine the stability of the following linear system

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

Homework Equations



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

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

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.
 

Answers and Replies

  • #2
760
69
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).
 
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  • #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.
 
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  • #4
760
69
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]
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]

...I don't fully understand how to obtain transfer functions in the first place...Now dividing by [itex]Z\left\{x[n]\right\}[/itex] I don't understand what value that would be in this case.
This is a transfer function:
[itex]\frac{Y(z)}{X(z)}= \left(0.5 + \frac{100}{z^{2}} - \frac{20}{z^{10}}\right)[/itex]
 
  • #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!
 
  • #6
760
69
Ah, so it has a transfer function of one!
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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