Transform of fraction

[Cantafford]

I'm trying to figure out an exercise from a book which does not explain anything good, skips steps, etc.
I have a fraction let's call it f = [kp*(1-a)*z] / [(z-1)*(z-a)] . (kp is a constant)
Then f is given as:

f = (kp*z) / (z-1) - (kp*z) / (z-a)

I need to remember which method was used to perform this transformation on the f fraction. Thank you for reading!

 

[Cantafford]

Yes, I have already solved with that method and it gave me same result as yours.
But still my book gives f = (kp*z) / (z-1) - (kp*z) / (z-a) which is still correct because the result is the same as the initial f. Unfortunatlly this book doesn't explain anything at all. Useless teachers

PS: How about if I have something like: f = kz / [(z-1)*(z^2-2*n*a*z+a^2)] , would I need to start with: A/(z-1) + B/(z^2-2*n*a*z+a^2)? Because I remember it beiing a bit diffrent.

 

[Cantafford]

ok what they've done is factor (kp z) out of f(z) and then solve the partial fraction expansion of

\(\displaystyle \dfrac{(1-a)}{(z-a)(z-1)}=\dfrac{1}{z-1}-\dfrac{1}{z-a}\)

and then multiply that result by the (kp z) that was factored out to obtain

\(\displaystyle f(z)=kp\cdot z \left(\dfrac{1}{z-1}-\dfrac{1}{z-a}\right)\)

Thank you very much! Can you please tell me that if I have something like:
f = kz / [(z-1)*(z^2-2*n*a*z+a^2)] , would I need to start with: A/(z-1) + B/(z^2-2*n*a*z+a^2)? Because I remember it beiing a bit diffrent.

 

[Cantafford]

Thank you!