inverse z-transform $E(z)=\frac{z}{(z-2)(z-1)^2}$

Question

The question is: Find the inverse z-transform of $E(z)=\frac{z}{(z-2)(z-1)^2}$

so far I have: using partial fractions with repeating factors... $$\frac{z}{(z-2)(z-1)^2}=\frac{Az}{(z-2)}+\frac{Bz}{(z-1)}+\frac{Cz}{(z-1)^2}\\ z=Az(z-1)^2+Bz(z-2)(z-1)^2+Cz(z-2)$$

z=1: $$1=A1(1-1)^2+B(1(1-2)(1-1)^2+C1(1-2)^2\\ 1=0+0+C(-1)^2\\ 1=-C\\ C=-1$$

setting z=2 results in: $$A=1$$

How do I go about solving B?

$$E(z)=\frac{z}{z-2}+\frac{Bz}{z-1}-\frac{z}{(z-2)^2}$$ taking the inverse z-transform then... $$e(k)=2^k+???+\frac{1}{2}k2^k$$

Please tell me this is remotely correct. If not, please show me where I am going wrong.

Try getting rid of $z$ in the nominator, i.e. $$\frac{z}{(z-2)(z-1)^2}=\frac{A}{(z-2)}+\frac{B}{(z-1)}+\frac{C}{(z-1)^2}$$ and go from there. — Daniel R, Oct 31 '13 at 07:47
How can we just get rid of the z? Do we not have to multiply it into the other side then? — Gineer, Oct 31 '13 at 13:02
See my answer below. On another note: it's not a good idea to plug in values for $z$. The equation must generally hold, not just for specific values of $z$. — Daniel R, Oct 31 '13 at 13:29

score 0 · Answer 1 · answered Oct 31 '13 at 13:28

Your assumption that $\frac{z}{(z-2)(z-1)^2}=\frac{Az}{(z-2)}+\frac{Bz}{(z-1)}+\frac{Cz}{(z-1)^2}$ is wrong. It should be: $$\frac{z}{(z-2)(z-1)^2}=\frac{A}{(z-2)}+\frac{B}{(z-1)}+\frac{C}{(z-1)^2}$$

See here for examples why.

Now, multiply by the denominator on the left hand side to get $$z=A(z-1)^2+B(z-2)(z-1)+C(z-2) \implies\\ (A+B)z^2-(1+2A+3B+C)z+A+2B-2C=0$$

For this equation to generally hold, the coefficients of each of the powers of $z$ must be equal on both sides (the coefficients on the right hand side is of course zero). This leads to the following equation system.

$$\begin{cases} A+B=0\\1+2A+3B-C=0\\A+2B-2C=0\\ \end{cases} $$ Now solve it, and plug in the resulting $A$, $B$ and $C$ above and apply the inverse z-transform on the terms.

score 0 · Answer 2 · edited Apr 18 '14 at 00:38

Instead of using partial fractions method use residue method or convolution theorem: they are very easy and you can get the answer without any confusion. I'm attaching residue method photo once refer to it!

At last for $z=1$ you'll get $-n-1$ as answer. Then add two residues i.e., $ 2^n-n-1$ is the answer! Likewise you can get anything through this method if you find difficulty in writing partial fractions.

inverse z-transform $E(z)=\frac{z}{(z-2)(z-1)^2}$

2 Answers2