How do I apply partial fraction expansion on $\dfrac{K}{(a+bz^{-1})(x+yz)}$?

Question

I want to apply partial fraction expansion on $\dfrac{K}{(a+bz^{-1})(x+yz)}$. I'm not able to do it in the standard way, because one term has $z^{-1}$ term and the other has $z$. What is the approach in problems like this?

Answer 1

I think this is about Z-Transformation.

In order to make the overall transfer function causal, we rewrite the given transfer function as

$$ \dfrac{K}{(a+bz^{-1})(x+yz)} = \dfrac{Kz^{-1}}{(a+bz^{-1})(y+xz^{-1})}. $$

Next, do:

$$ \dfrac{Kz^{-1}}{(a+bz^{-1})(y+xz^{-1})} = \dfrac{A}{a+bz^{-1}} + \dfrac{B}{y+xz^{-1}} $$

It is easy to see that:

$$ aA + yB = 0 \\ xA + bB = K $$

Solution of this equation set gives:

$$ A = \dfrac{Ky}{xy - ab} \\ B = \dfrac{Ka}{ab - xy} $$

Therefore, the partial fraction expression is (in the standard first order Z-Transform format):

$$ \dfrac{K}{(a+bz^{-1})(x+yz)} = \dfrac{\dfrac{Ky}{xy - ab}}{a+bz^{-1}} + \dfrac{\dfrac{Ka}{ab - xy}}{y+xz^{-1}} = \dfrac{\dfrac{Ky}{a(xy - ab)}}{1+\dfrac{b}{a}z^{-1}} + \dfrac{\dfrac{Ka}{y(ab - xy)}}{1+\dfrac{x}{y}z^{-1}} $$

And the corresponding time domain expression of the given discrete time transfer function is:

$$ f[n] = \left[\dfrac{Ky}{a(xy - ab)} e^{-\frac{b}{a}n} + \dfrac{Ka}{y(ab - xy)} e^{-\frac{x}{y}n}\right] u[n] $$