Inverse Z-transform (unfinished)

\( \require{AMSsymbols} \def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}} \def\lfzraised#1{\raise{10mu}{#1}} \def\laplace{\lfz{\mathscr{L}}} \def\fourier{\lfz{\mathcal{F}}} \def\ztransform{\lfz{\mathcal{Z}}} \require{cancel} \newcommand\ccancel[2][black] {\color{#1}{\cancel{\color{black}{#2}}}} \newcommand\ccancelto[3][black] {\color{#1}{\cancelto{#2}{\color{black}{#3}}}} \)

Inverse Unilateral Z-Transform

The inverse Z-transform, can be evaluated using Cauchy’s integral. Which is an integral taken over a counter-clockwise 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 \(Y(z)\). $$ y[n] = \frac{1}{2\pi j}\oint_C Y(z)\,z^{n-1}\,\mathrm{d}z $$

Let’s try some simplifications:

  1. 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 discrete-time Fourier transform (DTFT) of the periodic values of the Z-transform 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 \(\frac{\text{d}z}{\text{d}\omega}=j\mathrm{e}^{j\omega}\) $$ \require{cancel} \begin{align} y[n] &= \frac{1}{2\pi j}\oint_C Y(z)\,z^{n-1}\,\mathrm{d}z \nonumber \\ &= \frac{1}{2\pi \bcancel{j}}\int_{-\pi}^{\pi} Y(\mathrm{e}^{j\omega})\,(\mathrm{e}^{j\omega})^{n\cancel{-1}}\bcancel{j}\cancel{{\mathrm{e}^{j\omega}}}\,\,\mathrm{d}\omega \nonumber \\ &= \frac{1}{2\pi}\int_{-\pi}^{\pi} Y(\mathrm{e}^{j\omega })\,\mathrm{e}^{j\omega n}\,\mathrm{d}\omega \nonumber \end{align} $$
  2. If a system is represented by a linear constant-coefficient difference equations (LCCDE), it is said to be rational. The output is in the form \((N\gt M)\) $$ \sum_{k=0}^N a_k y[n-k]=\sum_{k=0}^M b_k x[n-k] \label{eq:rational} $$ 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 Z-domain output can be expressed as $$ Y(z)=\frac{b_0+b_1z+b_2z^2+\ldots+b_Mz^M}{a_0+a_1z+a_2z^2+\ldots+a_Nz^N} $$

We will examine solution methods for rational systems in the following sections.

Long Division

Long-division 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 Z-transform, 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 closed-form expression for \(y[n]\).

Direct Computation

When \(x[n]=\delta[n]\), \(y[n]=h[n]\). For \(n=0\), we obtain the initial condition: $$ h[0]-ah[-1]=h[0]=\delta[0]=1 $$

For \(n>0\), we plug the general solution \(h[n]=Az^n\) into the DE and get $$ Az^n-aAz^{n-1}=\delta[n]=0,\ \ n\gt 0 $$

From which we get \(z=a\) and \(h[n]=Aa^n\). But as \(h[0]=1\), we have \(A=1\) and $$ h[n]=a^n \gamma[n] $$

The Fourier spectrum of \(h[n]\) is the corresponding frequency response $$ \begin{align} H(e^{j\omega})&: {\cal F}[h[n]]=\sum_{n=-\infty}^\infty h[n]e^{-jn\omega}\\ &:\sum_{n=0}^\infty a^n e^{-jn\omega}=\frac{1}{1-ae^{-j\omega}} \end{align} $$


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 $$ H\big(x(t)\big)=\int_{-\infty}^{\infty}h(\tau)\,x(t−\tau)\,d\tau $$

Consider the input \(x(t)=\mathrm{e}^{st}\) where \(s\in \mathbb{C}\), the output $$ \begin{align} H\big(\mathrm{e}^{st}\big) &= \int_{-\infty}^{\infty}h(\tau)\,\mathrm{e}^{s(t-\tau)}\,d\tau \nonumber \\ &= \int_{-\infty}^{\infty}h(\tau)\,\mathrm{e}^{st}\mathrm{e}^{-s\tau}\,d\tau \nonumber \\ &= \mathrm{e}^{st}\int_{-\infty}^{\infty}h(\tau)\,\mathrm{e}^{-s\tau}\,d\tau \end{align} $$

Define $$ \lambda_s=\int_{-\infty}^{\infty}h(\tau)\,\mathrm{e}^{-s\tau}\,d\tau $$

The eigenvalue follows as $$ H\big(\mathrm{e}^{st}\big)=\lambda_s\mathrm{e}^{st} $$ corresponding with eigenvector \(\mathrm{e}^{st}\).

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.


Use the eigenequation of the LTI system.

——– If the input is a complex exponential $$ x[n]\ztransform z^n\\ $$ an eigenfunction of the LTI system, then $$ y[n]=\ztransform z^n\,H(z) $$

Substitute \(z=e^{j\omega}\) $$ y[\omega]=e^{j\omega n}\,H(e^{j\omega}) $$

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 time-shifting properties: $$ {\cal F}[\sum_{k=0}^N a_k y[n-k]]={\cal F}[\sum_{k=0}^M b_k x[n-k]] $$

Due to the linearity property, this becomes $$ \sum_{k=0}^N a_k {\cal F}[y[n-k]]=\sum_{k=0}^M b_k {\cal F}[x[n-k]] $$

and due to the time shifting property, we get $$ Y(e^{j\omega})[\sum_{k=0}^N a_k e^{-jk\omega}] = X(e^{j\omega})[\sum_{k=0}^M b_k e^{-jk\omega}] $$

From which we find $$ H(e^{j\omega})=\frac{Y(e^{j\omega})}{X(e^{j\omega})} = \frac{\sum_{k=0}^M b_k e^{-jk\omega}}{\sum_{k=0}^N a_k e^{-jk\omega}} $$

2BD …


Similar to Laplace:

… page 111 in

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