Inverse ZTransform
The forward Ztransform helped us express samples in time as an analytic function on which we can use our algebra tools. Eventually, we have to return to the time domain using the Inverse Ztransform.\(\)
The inverse Ztransform can be derived using Cauchy’s integral theorem.
Start with the definition of the Ztransform
$$ \def\lfz#1{\overset{\Large#1}{\,\circ\kern6mu\kern7mu\kern7mu\kern6mu\bullet\,}} \def\ztransform{\lfz{\mathcal{Z}}} \begin{align} f[m]\,\ztransform\,&F(z)=\sum_{m=0}^\infty z^{m}\ f[m]\nonumber \end{align} \nonumber $$
Multiply both sides by \(z^{n1}\)
Integrate with a counterclockwise contour integral for which the contour encloses the origin and lies entirely within the region of convergence of \(F(z)\)
A special case of Caughy integral theorem states
$$ \frac{1}{2\pi j}\,\oint_Cz^{l}\,\mathrm{d}z\,=\, \begin{cases} 1 & l = 1 \\ 0 & l\neq1 \end{cases} \nonumber $$
That implies we can replace the integral with a \(\delta\) function.
The Inverse Ztransform follows as
Cauchy’s Residue Theorem
When the transfer functions is rational, a ratio of polynomials, we may use the method described below to calculate the Inverse Ztransform.
Denote the unique poles of \(F(z)\) as \(p_{1\ldots{\small N}}\) and their algebraic multiplicities as \(m_{1\ldots{\small N}}\). As long as \(N\) is finite, which is the case if \(F(z)\) is rational, we can evaluate the inverse ZTransform via Cauchy’s residue theorem that states
$$ f[n] = \frac{1}{2\pi j}\oint_C F(z)\,z^{n1}\,\mathrm{d}z=\sum_{p_k\text{ inside }C}\mathrm{Res}\large{(}\,F(z)\,z^{n1},\,p_k,\,m_k\,{\large{)}} \nonumber $$ where $$ \mathrm{Res}{\Large(}\,F(z)\,z^{n1},\,p_k,\,m_k\,{\Large)}=\frac{1}{(m_k1)!}\left[\frac{\text{d}^{(m_k1)}}{\text{d}z^{(m_k1)}}\,{\Large[}\,(zp_k)^{m_k}\,F(z)\,z^{n1}\,{\Large]}\right]_{z=p_k} \nonumber $$ for a single pole $$ \mathrm{Res}{\Large(}\,F(z)\,z^{n1},\,p_k,\,1\,{\Large)}=\left.\,(zp_k)\,F(z)\,z^{n1}\,\right_{z=p_k} \nonumber $$
Cauchy’s residue theorem allows us to compute the contour integral by computing derivatives, however tedious.
Other techniques
Given that the \(z\)transform is a particular type of Laurent series, and the Laurent series in a given annulus of convergence is unique, any technique can be used to generate a power series for \(F(z)\) that converges in the outermost annulus of convergence to obtain the inverse \(z\)transform.
Inversion techniques available are
 using the binomial theorem
 using the convolution theorem
 performing long division
 using the initialvalue theorem
 expanding \(F(z)\) in partial fractions
 power series expansion (for nonrational ztransforms).
Inverse Unilateral ZTransform
The inverse Ztransform, can be evaluated using Cauchy’s integral. Which is an integral taken over a counterclockwise closed contour \(C\) in the region of converge of \(Y(z)\). When the ROC is causal, this means the path \(C\) must encircle all the poles of \(Y(z)\).
Let’s try some simplifications:

When all poles of \(Y(z)\) poles are inside the unit circle, \(Y(z)\) is stable and \(C\) can be the unit circle. Thus the contour integral simplifies to the inverse discretetime Fourier transform (DTFT) of the periodic values of the Ztransform around the unit circle . To proof we take the unit circle \(z=1\), and parameterize contour \(C\) by \(z(\omega)=\mathrm{e}^{j\omega}\), with \(\pi\leq \omega\leq\pi\) so \(\frac{\text{d}z}{\text{d}\omega}=j\mathrm{e}^{j\omega}\)
$$ \begin{align} y[n] &= \frac{1}{2\pi j}\oint_C Y(z)\,z^{n1}\,\mathrm{d}z \nonumber \\ &= \frac{1}{2\pi \bcancel{j}}\int_{\pi}^{\pi} Y(\mathrm{e}^{j\omega})\,(\mathrm{e}^{j\omega})^{n\cancel{1}}\bcancel{j}\cancel{{\mathrm{e}^{j\omega}}}\,\,\mathrm{d}\omega \nonumber \\ &= \frac{1}{2\pi}\int_{\pi}^{\pi} Y(\mathrm{e}^{j\omega })\,\mathrm{e}^{j\omega n}\,\mathrm{d}\omega \nonumber \end{align} $$

If a system is represented by a linear constantcoefficient difference equations (LCCDE), it is said to be rational. The output is in the form \((N\gt M)\)
$$ \sum_{k=0}^N a_k y[nk]=\sum_{k=0}^M b_k x[nk] \label{eq:rational} $$this allows us to find the impulse response \(h[n]\) and frequency response \(H(\mathrm{e}^{j\omega})\) of this LTI system similarly to the methods to solve a continuous LCCDE problems.
For rational systems captured by equation \(\eqref{eq:rational}\) the output in the Zdomain output can be expressed as
We will examine solution methods for rational systems in the following sections.
Long Division
Longdivision of the polynomials directly is a simple but not so practical method for obtaining a power series expansion for \(Y(z)\). Using the definition of the Ztransform, the terms of the sequence can then be identified one at a time. Problem with this method is that it is labor intensive, and does not produce a closedform expression for \(y[n]\).
Direct Computation
When \(x[n]=\delta[n]\), \(y[n]=h[n]\). For \(n=0\), we obtain the initial condition:
For \(n>0\), we plug the general solution \(h[n]=Az^n\) into the DE and get
From which we get \(z=a\) and \(h[n]=Aa^n\). But as \(h[0]=1\), we have \(A=1\) and
The Fourier spectrum of \(h[n]\) is the corresponding frequency response
src: http://fourier.eng.hmc.edu/e101/lectures/handout3/node9.html
Partial Fraction Expansion
2BD: see “discrete transfer functions
Eigenequation method ??
Consider a linear time invariant system \(H\) with impulse response \(h\) operating on some space of infinite length continuous time signals. Recall that the output \(H\big(x(t)\big)\) of the system for a given input \(x(t)\) is given by the continuous time convolution of the impulse response with the input
Consider the input \(x(t)=\mathrm{e}^{st}\) where \(s\in \mathbb{C}\), the output
Define
The eigenvalue follows as
This makes it particularly easy to calculate the output of a system when an eigenfunction is the input because the output is simply the eigenfunction scaled by the associated eigenvalue.
src: http://pilot.cnxproject.org/content/collection/col10064/latest/module/m34639/latest
Use the eigenequation of the LTI system.
——– If the input is a complex exponential
Substitute \(z=e^{j\omega}\)
Substituting \(x[n]\) and \(y[n]\) into the given DE, we can obtain \(H(e^{j\omega})\).
Fourier transform ??
Take Fourier transform on both sides of the given DE, and use the linearity and timeshifting properties:
Due to the linearity property, this becomes
and due to the time shifting property, we get
From which we find
2BD …
src: http://fourier.eng.hmc.edu/e101/lectures/handout3/node9.html
Similar to Laplace: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Signals_and_Systems/node2.html
… page 111 in http://web.stanford.edu/~kairouzp/teaching/ece310/secure/Chapter5.pdf