- #1

WhiteWolf98

- 86

- 5

- Homework Statement
- How do you decompose the following:

$$\frac{-3z^{-1}+2z^{-2}}{1-z^{-1}+z^{-2}}$$

- Relevant Equations
- None.

So the original question is from Control Theory, and the topic is the inverse z-transform. This is a part from the solution I just can't understand. The reason it has to be in this form (##z^{-1}##) is because that's the form used in the z-transform table. The question essentially is, how do you get from the left side to the right side:

$$\frac{-3z^{-1}+2z^{-2}}{1-z^{-1}+z^{-2}}=-3z^{-1}\frac{1-0.5z^{-1}}{1-z^{-1}+z^{-2}}+z^{-1}\frac{0.5z^{-1}}{1-z^{-1}+z^{-2}}$$

In the solution, it states: 'Noting that the poles in the quadratic term are complex conjugates'. I have no idea what that means. In the normal (##z##) form, it is:

$$\frac{-3z+2}{z^2-z+1}$$

The solutions of this, I know, are complex conjugates:

$$\frac{1}{2}+\frac{\sqrt{3}}{2}i,\,\frac{1}{2}-\frac{\sqrt{3}}{2}i$$

I don't see how that helps me though. Any help would be appreciated. Thank you.

$$\frac{-3z^{-1}+2z^{-2}}{1-z^{-1}+z^{-2}}=-3z^{-1}\frac{1-0.5z^{-1}}{1-z^{-1}+z^{-2}}+z^{-1}\frac{0.5z^{-1}}{1-z^{-1}+z^{-2}}$$

In the solution, it states: 'Noting that the poles in the quadratic term are complex conjugates'. I have no idea what that means. In the normal (##z##) form, it is:

$$\frac{-3z+2}{z^2-z+1}$$

The solutions of this, I know, are complex conjugates:

$$\frac{1}{2}+\frac{\sqrt{3}}{2}i,\,\frac{1}{2}-\frac{\sqrt{3}}{2}i$$

I don't see how that helps me though. Any help would be appreciated. Thank you.