Inverse Z-transform

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 powerseries 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 $$ \begin{align} H(e^{j\omega})&: {\cal F}[h[n]]=\sum_{n=-\infty}^\infty h[n]e^{-jn\omega}\\ &:\sum_{n=0}^\infty a^n e^{-jn\omega}=\frac{1}{1-ae^{-j\omega}} \end{align} $$ from:

Partial Fraction Expansion

.. 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. src: 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 $$H(e^{j\omega})=\frac{Y(e^{j\omega})}{X(e^{j\omega})}= \frac{\sum_{k=0}^M b_k e^{-jk\omega}}{\sum_{k=0}^N a_k e^{-jk\omega}}$$ … src: Similar to Laplace: … page 111 in
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