Can This Linear System Be Considered Stable?

[lucidlobster]

Homework Statement

Determine the stability of the following linear system

y(n) = 0.5x(n) +100x(n-2) - 20x(n-10)

Homework Equations

x(n) = 0.5\delta(n)

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

The Attempt at a Solution

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

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)})

\frac{80.5xz}{(z-1)^2}

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 \sum\frac{1}{1-a}

My result was \approx -.47 which I though would be marginally stable as it is between -1 and 1.

 

[MisterX]

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).

 

[lucidlobster]

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

0.5X(n) + z^{-2}100X(z) -20z^{-10}(z)

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

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

So from here I could just manipulate the numerator and denominator to get whatever it is I need, yes? Now dividing by Z\left\{x[n]\right\} 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.

 

[MisterX]

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

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

This is wrong. It should be something like:

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

...I don't fully understand how to obtain transfer functions in the first place...Now dividing by Z\left\{x[n]\right\} I don't understand what value that would be in this case.

This is a transfer function:
\frac{Y(z)}{X(z)}= \left(0.5 + \frac{100}{z^{2}} - \frac{20}{z^{10}}\right)

 

[lucidlobster]

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!

 

[MisterX]

Ah, so it has a transfer function of one!

No, the system's transfer function is

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