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

AI Thread Summary
The discussion focuses on finding the inverse z-transform of X(z) = z/((z-0.2)^2(z+0.1)). The method involves using partial fraction decomposition, leading to a series of terms that can be transformed back to the time domain. Participants express uncertainty about the correctness of their methods, particularly regarding the treatment of poles and the order of the numerator and denominator. There is a debate on whether certain transformation techniques apply when the numerator's order is equal to or greater than that of the denominator. The conversation highlights the importance of following established z-transform tables and methods for accurate results.
ongxom
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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.
 
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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.
 
ongxom said:

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.
 
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This is the instruction from the book :
bfadca3eb2740d775078aa479cba59c4.png


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}
 
ongxom said:
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!
 
rude man said:
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.
 
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