Q3Signals and Systems

Given $X(z)=\frac{z^3-10z^2-4z+4}{2z^2+2z-4}$ with ROC $|z|<1$.

First, simplify the denominator: $2z^2+2z-4=2(z^2+z-2)=2(z+2)(z-1)$.

Since the degree of the numerator (3) is higher than the denominator (2), we first perform long division to express it as a polynomial plus a proper rational function:

Now, apply partial fraction expansion to the rational part:

Solving for A and B yields $A=\frac{17}{3}$ and $B=\frac{13}{12}$.

Rewriting $X(z)$ in standard forms that allow easy lookup, we factor out $z^{-1}$ from the fractions so we can use the form $\frac{z}{z-p}$:

The ROC is $|z|<1$. The poles are at $z=-2$ and $z=1$. Since $|z|<1$ is strictly interior to both poles, both resulting time-domain sequences must be left-sided (anti-causal). Recall that for $|z|<|p|$, the inverse transform of $\frac{z}{z-p}$ is $-p^{n}u[-n-1]$.

Taking the inverse Z-transform term by term: 1. $\frac{1}{2}z⟹\frac{1}{2}\delta [n+1]$ 2. $-\frac{11}{4}⟹-\frac{11}{4}\delta [n]$ 3. $z^{-1}$ introduces a time delay of 1 sample ($n\to n-1$). 4. Term for $z=-2$: $-\frac{17}{3}(-2{)}^{n-1}u[-(n-1)-1]=-\frac{17}{3}(-2{)}^{n-1}u[-n]$ 5. Term for $z=1$: $-\frac{13}{12}(1{)}^{n-1}u[-(n-1)-1]=-\frac{13}{12}u[-n]$

Combining the terms, the final sequence is: