7.4 The inverse z-transform

The z-transform for a signal $x[n]$ has been defined as:$X(z)=\sum\limits_{n=-\infty}^{\infty}x[n]z^{-n}$As with other transforms such as the DFT and DTFT, we would like to recover $x[n]$ from its transform. The technical definition for this inverse transform is:$x[n] = \oint_{\mathcal{C}} X(z) z^n \frac{dz}{j 2\pi z}$Evaluation of this contour integral requires knowledge of complex analysis, but there is another way to find the inverse transform, which we will now consider.

Inverse transform by inspection

Key to this process is the fact that if a z-transform is rational (which is the case for virtually all of our cases of interest), then its numerator and denominator can be factored according to its poles and zeros. This means that an $X(z)$ of the form: $X(z)=\frac{b_0+b_1z^{-1}+b_2z^{-2}+\cdots+b_Mz^{-M}}{1+a_1z^{-1}+a_2z^{-2}+\cdots+a_Nz^{-N}}$can equivalently be expressed as: $X(z)=z^{N-M} \frac{(1-\zeta_1 z^{-1})(1-\zeta_2 z^{-1}) \cdots (1-\zeta_M z^{-1})}{(1-p_1 z^{-1})(1-p_2 z^{-1}) \cdots (1-p_N z^{-1})}$ Note how this expression is starting to look similar to $\frac{1}{1-\alpha z^{-1}}$. We can do a little more work on $X(z)$ to help with things even more. A rational function can be expressed in terms of products of $\frac{1}{1-\alpha z^{-1}}$ terms, but even better, it can also be represented as a sum of such terms. Therefore, any $X(z)$ of the form$X(z)=z^{N-M} \frac{(1-\zeta_1 z^{-1})(1-\zeta_2 z^{-1}) \cdots (1-\zeta_M z^{-1})} {(1-p_1 z^{-1})(1-p_2 z^{-1}) \cdots (1-p_N z^{-1})}$can be re-arranged to be expressed as $X(z)=C_0+\frac{C_1}{1-p_1 z^{-1}}+\frac{C_2}{1-p_2 z^{-1}}+\cdots+\frac{C_N}{1-p_N z^{-1}}$Now, this is something we can work with. Using the linearity property of the z-transform, we can take the inverse transform of that whole expression to yield (assuming the ROC of $X(z)$ extends outwards from the outermost pole) $x[n]=C_0\delta[n]+C_1 p_1^n u[n]+C_2 p_2^n u[n]+\cdots+C_N p_N^n u[n]$

Partial fractions

The first step of the partial fraction decomposition is to make sure the order of the denominator polynomial is greater than that of the numerator. If not, then we first need to carry out a polynomial long division and then continue with the remainder. For example, suppose $X(z)=\frac{5 +2z^{-1} -z^{-2}}{1 -\frac{1}{6} z^{-1} -\frac{1}{6}z^{-2}}~~,~|z|\gt\frac{1}{2}$The long division is a single step: \begin{align*}&~~~~~~~~~~~~~~~~~6\\ 1 -\frac{1}{6} z^{-1}-\frac{1}{6}z^{-2}~~&\overline{\smash{)}5 +2z^{-1} -z^{-2}}\\&\underline{~6-1z^{-1}-z^{-2}}\\&-1+3z^{-1} \end{align*}$So $X(z)=\frac{5 +2z^{-1} -z^{-2}}{1 -\frac{1}{6} z^{-1} -\frac{1}{6}z^{-2}}=6+\frac{-1+3z^{-1}}{1 -\frac{1}{6} z^{-1} -\frac{1}{6}z^{-2}}$ We can now work on that second term, as the denominator's order is greater than the numerator. The next step, then, will be to factor the denominator:$\frac{-1+3z^{-1}}{1 -\frac{1}{6} z^{-1} -\frac{1}{6}z^{-2}}=\frac{-1+3z^{-1}}{(1-\frac{1}{2}z^{-1})(1+\frac{1}{3}z^{-1})}$ From here, the next step is to express that in terms of a sum:$\frac{-1+3z^{-1}}{(1-\frac{1}{2}z^{-1})(1+\frac{1}{3}z^{-1})}=\frac{C_1}{1-\frac{1}{2}z^{-1}}+\frac{C_2}{1+\frac{1}{3}z^{-1}}$ The above form assumes that each of the poles is unique; if a pole is repeated, e.g. $\frac{1}{(1-az^{-1})^2}$, then that single term in the product would correspond to two in the sum, $\frac{C_1}{1-az^{-1}}+\frac{C_{1'}}{(1-az^{-1})^2}.At this point, we simply solve for the $C$ coefficients: $\begin{align*}\frac{-1+3z^{-1}}{(1-\frac{1}{2}z^{-1})(1+\frac{1}{3}z^{-1})}&=\frac{C_1}{1-\frac{1}{2}z^{-1}}+\frac{C_2}{1+\frac{1}{3}z^{-1}}\\ -1+3z^{-1}&=C_1(1+\frac{1}{3}z^{-1})+C_2(1-\frac{1}{2}z^{-1})\\ -1+3z^{-1}&=(C_1+C_2)+z^{-1}(\frac{1}{3}C_1-\frac{1}{2}C_2)\\ C_1+C_2=-1,&~\frac{1}{3}C_1-\frac{1}{2}C_2=3\rightarrow C_1=3~,~C_2=-4 \end{align*}$Given these coefficients, we now have that $X(z)=6+\frac{3}{1-\frac{1}{2}z^{-1}}-\frac{4}{1+\frac{1}{3}z^{-1}}~~,~|z|\gt\frac{1}{2}$

Having split the transform into a sum through partial fractions, the inverse transform is now straightforward because (due to the linearity of the z-transform) we can take the inverse of each piece of the sum. So for: $X(z)=6+\frac{3}{1-\frac{1}{2}z^{-1}}-\frac{4}{1+\frac{1}{3}z^{-1}}~~,~|z|\gt\frac{1}{2}$The inverse transform of $1$ is $\delta[n]$, the inverse of $\frac{1}{1-\frac{1}{2}z^{-1}}$ is $(\frac{1}{2})^n u[n]$, and the inverse of $\frac{1}{1+\frac{1}{3}z^{-1}}$ is $(\frac{-1}{3})^n u[n]$. For the latter two terms the ROC of the transform is must-know information; if the ROC extended inwards instead of outwards, the inverses would have been left-sided instead of right-sided. Putting all of this together through linearity, we have that the inverse transform is: $x[n]=6\delta[n]+3(\frac{1}{2})^n u[n]-4(\frac{-1}{3})^n u[n]$.