# 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: http://fourier.eng.hmc.edu/e101/lectures/handout3/node9.html

### 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: 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 $$x[n]\ztransform z^n\\$$ 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: 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
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