Calculating Inverse z-Transform for X(z) = z/(z-0.2)^2(z+0.1)

[ongxom]

Homework Statement

Find inverse z-transform of
X(z) = \frac{z}{(z-0.2)^2(z+0.1)}

Homework Equations

The Attempt at a Solution

: partial fraction
My method :\frac{X(z)}{z} = \frac{1}{(z-0.2)^2(z+0.1)}
\frac{X(z)}{z} = \frac{-100/9}{(z-0.2)} + \frac{10/3}{(z-0.2)^2} + \frac{100/9}{z+0.1}
X(z) = \frac{(-100/9)z}{z-0.2}+\frac{(10/3)z}{(z-0.2)^2}+\frac{(100/9)z}{z+0.1}
X(z) = \frac{(-100/9)}{1-0.2z^-1}+\frac{(10/3)}{(1-0.2z^-1)^-2}+\frac{(100/9)}{1+0.1z^-1}
→ x[nT]=(-100/9)0.2^n+(50/3)n.0.2^n+(100/9)(-0.1)^n

My friend :
Y(z)=\frac{4}{z-0.2} + \frac{6z}{(z-0.2)^2} + \frac{-2}{z+0.1}
Y(z)=\frac{4z^-1}{1-0.2z^-1} + \frac{6z^-1}{(1-0.2z^-1)^2} + \frac{-2z^-1}{1+0.1z^-1}
→ y[nT]=-4.(0.2)^n.u[n-1]+30n.(0.2)^n.u[n]-2.(-0.1)^nu.[n-1]

I don't know which method gives correct result.

 

[.Scott]

On these face of it, it looks like a fourth order polynomial.
I type "z=x(z+.1)(z-.2)(z-.2)" into wolframalpha.com and got something entirely different.
I suspect the course work is expecting a particular strategy.

 

[rude man]

The Attempt at a Solution

: partial fraction
My method :\frac{X(z)}{z} = \frac{1}{(z-0.2)^2(z+0.1)}
\frac{X(z)}{z} = \frac{-100/9}{(z-0.2)} + \frac{10/3}{(z-0.2)^2} + \frac{100/9}{z+0.1}
X(z) = \frac{(-100/9)z}{z-0.2}+\frac{(10/3)z}{(z-0.2)^2}+\frac{(100/9)z}{z+0.1}

Assuming your X(z) partial fraction expansion is correct to this point, the difficulty seems to be to invert the double pole at z = 0.2.

I rewrote

\frac{z}{(z-0.2)^2} = z^-1/(1 - 2az^-1 + a²z^-2) with a = 0.2.

Then I got wolfram alpha to give me x[n] = na^(n-1)u[n-1].

Maybe you can fit this in with the rest which should be conventional inverse transformations.

 

[ongxom]

This is the instruction from the book :

But it does not say if this method can be applied for the fraction which has the order of numerator is equal or larger than denominator.

Can I use the above method to find z-inverse of following fraction (order of numerator is larger and/or equal)

X(z)=\frac{4-z^{-1}}{2-2z^{-1}+z^{-2}}
X(z)=\frac{2+z^{-1}+z^{-2}}{1+0.4z^{-1}+0.4z^{-2}}
X(z)=\frac{z^{-2}+z^{-1}}{z^{-3}-z^{-2}+2z^{-1}+1}

 

[rude man]

But it does not say if this method can be applied for the fraction which has the order of numerator is equal or larger than denominator.

That is not the case with your z transform. F(z) = Az/(z-γ)² has order of denominator > order of numerator.

You can rewrire F(z) = Az^-1/(1 - γz^-1)².

I see nothing wrong with your instructions. If your table includes that expression then just follow it!
You seem to have a good z transform table!

 

[ongxom]

That is not the case with your z transform. F(z) = Az/(z-γ)² has order of denominator > order of numerator.

You can rewrire F(z) = Az^-1/(1 - γz^-1)².

I see nothing wrong with your instructions. If your table includes that expression then just follow it!
You seem to have a good z transform table!

Yes, the instruction works for the problem in my first post. But i am wondering if it works for the function like this :
X(z)=\frac{4-z^{-1}}{2-2z^{-1}+z^{-2}}

The numerator's lowest order is (-1), the denominator has (-2) as a lowest order. Can I multiply both numerator and denominator with z² and do the rest using same method ? In the instruction the order is (-6) and (-5) respectively, so I am not sure if I can follow 8 steps for the above function.

 

[rude man]

http://lpsa.swarthmore.edu/BackGrou...mial_is_not_less_than_that_of_the_denominator.

See the section under "Order of numerator polynomial is not less than that of the denominator."

Note that in this case x[0] ≠ 0. x[0] = 0 always if the order of the denominator is greater than that of the numerator.

Note also that we only care about the HIGHEST, not lowest, order of the numerator and denominator.