# Inverse z transform using partial fraction

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.

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]\$$.

• But the books says the answer is: delta [n] /2 - 7[1^n]/3 + 11 [4^n]/6. Where did the third term come from? Commented Jan 19, 2020 at 19:49
• Looking at what Chu wrote, that delta term is there to make $x[0] = 0$ isn't it?
– jDAQ
Commented Jan 19, 2020 at 20:03

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.