Stability of a linear system

In summary, the stability of the given linear system can be determined by applying the Z-transform to the equation for y[n] and factoring out Z{x[n]} from each term. The resulting transfer function can be manipulated to obtain a polynomial in the numerator and denominator, and the system is considered to be stable if the poles are at zero.
  • #1
lucidlobster
5
0

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.
 
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  • #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).
 
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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
lucidlobster said:
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]

lucidlobster said:
...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
lucidlobster said:
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]
 

1. What is stability of a linear system?

The stability of a linear system refers to its ability to return to its equilibrium state after experiencing a disturbance or perturbation. In other words, a stable system will remain close to its equilibrium point and not exhibit chaotic behavior.

2. How is stability determined in a linear system?

There are various methods to determine the stability of a linear system, such as using the eigenvalues of the system's matrix or analyzing the system's transfer function. Generally, a system is considered stable if all of its eigenvalues have negative real parts.

3. What is the difference between stable and unstable systems?

A stable system will eventually return to its equilibrium state after a disturbance, while an unstable system will continue to deviate from its equilibrium and exhibit chaotic behavior. This can have significant implications for the predictability and control of the system.

4. Can a stable system become unstable?

Yes, a stable system can become unstable if the parameters or inputs of the system change. This can be caused by external factors or changes within the system itself. It is important to regularly monitor and analyze a system's stability to prevent unexpected behavior.

5. How is the stability of a linear system important in real-world applications?

The stability of a linear system is crucial in many real-world applications, such as in engineering, economics, and biology. In these fields, stability ensures the proper functioning and predictability of systems, such as control systems, financial markets, and ecological systems. Understanding and maintaining stability is essential for the success and safety of these systems.

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