# Thread: Tricky Partial Fractions for Control

1. ## Tricky Partial Fractions for Control

Hi,

I am half way through a control question and there is a bit of algebra bashing I can;t get my head around. I am trying to go from:

$\displaystyle \frac{4z^{2} + z + 1 }{(z + \frac{1}{2})(z + \frac{1}{4} )}$

to

$\displaystyle 8 + \frac{12z}{z + \frac{1}{2}} + \frac{-16z}{z + \frac{1}{4}}$

It needs to be in this form for control analysis but I am unsure on the steps to get to it, I have tried normal partial fractions to no avail

Jez

2. well, let's work backwards:

8 + 12z/(z+(1/2)) + -16/(z+(1/4)) =

[8(z+(1/2))(z+(1/4)) + 12z(z+(1/4)) + -16z(z+(1/2))]/((z+(1/2))(z+1/4)) =

[8(z^2 + (3/4)z + (1/8)) + 12z^2 + 3z - 16z^2 - 8z]/((z+(1/2))(z+1/4)) =

[8z^2 + 6z + 1 + 12z^2 + 3z - 16z^2 - 8z]/((z+(1/2))(z+1/4)) =

[(8+12-16)z^2 + (6+3-8)z + 1]/((z+(1/2))(z+1/4)) =

(4z^2 + z + 1)/((z+(1/2))(z+1/4)), and voila! there it is. so to get from this to your intended destination, read bottom-to-top.

3. Originally Posted by Deveno
well, let's work backwards:

8 + 12z/(z+(1/2)) + -16/(z+(1/4)) =

[8(z+(1/2))(z+(1/4)) + 12z(z+(1/4)) + -16z(z+(1/2))]/((z+(1/2))(z+1/4)) =

[8(z^2 + (3/4)z + (1/8)) + 12z^2 + 3z - 16z^2 - 8z]/((z+(1/2))(z+1/4)) =

[8z^2 + 6z + 1 + 12z^2 + 3z - 16z^2 - 8z]/((z+(1/2))(z+1/4)) =

[(8+12-16)z^2 + (6+3-8)z + 1]/((z+(1/2))(z+1/4)) =

(4z^2 + z + 1)/((z+(1/2))(z+1/4)), and voila! there it is. so to get from this to your intended destination, read bottom-to-top.

Yes thankyou, however splitting 4 into 8+12-16 would not be a step taken unless the answer was known . . .

Is there a way of working it through forwards to get it into the correct form, $\displaystyle \frac{az}{z+b}$

4. I think partial fractions only works if the degree of the numerator is strictly less than the degree of the denominator. First step: perform long division on

$\displaystyle \frac{4z^{2}+z+1}{z^{2}+\frac{3z}{4}+\frac{1}{8}}.$

Then do partial fractions on the result. I get

$\displaystyle 4+\frac{-2z+\frac{1}{2}}{(z+\frac{1}{2})(z+\frac{1}{4})}.$

I don't know where that 8 came from, but this is what I get. Can you continue?