Inverse z transform using partial fraction

Question

Here's my attempt at an inverse z transform using partial fraction. I was going through my textbook and it stated that all the z terms need to be converted to z inverse before using partial fraction expansion, yet I have hit a roadblock. Please advise.

Answer 1

Since the Z-transform is linear, you can just use a Transform table to find the discrete time equivalent of $$ X(z) = \frac{-7/3}{z-1} + \frac{22/3}{z-4}.$$ Look separetly for an inverse of, $$\frac{-7/3}{z-1} = A(z), $$ and $$ \frac{22/3}{z-4} = B(z).$$ Then add the two inverses to get the discrete time \$x[k] = a[k] + b[k]\$.

Answer 2

For example:

\$\small Y(z)=\large\frac {a}{z-b}= \small z^{-1}\large \frac{az}{z-b}\$

hence:

\$y[k]=a\:b^{k-1}\:u(k-1)\$, where \$ u(k-1)\$ is the unit step sequence.