# Find inverse z-transform

1. Aug 9, 2014

### ongxom

1. The problem statement, all variables and given/known data
Find inverse z-transform of
$X(z) = \frac{z}{(z-0.2)^2(z+0.1)}$
2. Relevant equations

3. 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.

Last edited: Aug 9, 2014
2. Aug 9, 2014

### .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.

3. Aug 11, 2014

### rude man

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.

Last edited: Aug 11, 2014
4. Aug 18, 2014

### 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)

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

5. Aug 18, 2014

### rude man

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!

6. Aug 19, 2014

### ongxom

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.

7. Aug 19, 2014