InverseZtransform - Rhea

Introduction The Z Transform is the generalized version of the DTFT. You can obtain the Z Transform from the DTFT by replacing $ e^{j\omega} $ with $ re^{j\omega} $ which is equivalent to z. The The DTFT is equal to the Z Transform when $ |z| =1 $

$ \begin{align} \text{DTFT: } X(w) &= \sum_{n=-\infty}^\infty x[n]e^{-j\omega n}\\ \text{Z-Transform: } X(z) &= \sum_{n=-\infty}^\infty x[n]z^{-n}\\ \text{Inv. Z-Transform: } x[n] &= \frac{1}{2\pi j}\oint_{c}X(z)z^{n-1}dz \end{align} $

Region of Convergence (ROC) The ROC determines the region on the Z Plane where the Z Transform converges. The ROC depends solely on the 'r' value that is contained in 'z'. The ROC is always one of three cases;

1. The ROC starts from a circle centered at the origin and extends outward to infinity

2. The ROC starts from a circle centered at the origin and fills in toward the origin

3. The ROC is the space in-between two circles centered at the origin.

If the ROC includes the unit circle then the DTFT converges for that function if it is not included, then it does not.

$ \begin{align} \text{Remember: } z &=re^{j\omega} \end{align} $

The ROC is determined when preforming Z transforms and is given when preforming inverse Z transforms.

Solving an inverse Z Transform To find the Inverse Z transform of signals use manipulation then direct Inversion. Do not use formula directly!

The Infinite Geometric Series formula is used in most problems involving Inv. Z transform.

$ \begin{align} \text{Infinite Geometric Series: } X(z) &= \frac{a}{1-r}\\ &= \sum_{n=0}^{\infty} (a)r^{n} , \text{if } |z| < 1\\ &= \sum_{n=-\infty}^{\infty} (a)r^{n} u[n]\\ \end{align} $

it can be seen that this general form is already starting to look like that of the Z Transform, with some change of variables we can manipulate this equation to be that of a Z transform and then by comparison find the inverse z transform.

Examples

Ex. 1 Find the Inverse Z transform of the following signal

$ X(z)=\frac{1}{1-z}, \text{ ROC } |z|<1 $

Solution

$ \begin{align} X(x) &= \frac{1}{1-z}\\ &= \sum_{n=0}^{\infty} 1(z^{n}) \text{ if } |z| < 1\\ &= \sum_{n=-\infty}^{\infty} z^{n} u[n]\\ &\text{let }k = -n\\ &= \sum_{k=-\infty}^{\infty} z^{-k} u[-k]\\ &= \sum_{k=-\infty}^{\infty}u[-k] z^{-k}\\ &\text{By comparison with the Z Transform definition..}\\ x[n] &= u[-n]\\ \end{align} $

Ex. 2 Find the Inverse Z transform of the following signal

$ X(z)=\frac{1}{1-z}, \text{ ROC } |z|>1 $

Solution

$ \begin{align} X(z) &= \frac{1}{1-z} \\ &= \frac{1}{z} \frac{1}{1-(-\frac{1}{z})}\\ & \text{Using a infinite Geometric series...}\\ &= \sum_{n=0}^{\infty} (\frac{-1}{z})^{n}\frac{1}{z} \text{ if } |-\frac{1}{z}| < 1\\ &= \sum_{n=0}^{\infty}(-1)^{n} z^{-n-1} \\ &= \sum_{n=-\infty}^{\infty} (-1)^{n} z^{-n-1} u[n]\\ &\text{ let } k=n+1 \\ &= \sum_{k=-\infty}^{\infty}(-1)^{k-1} z^{-k} u[k-1] \\ &= \sum_{k=-\infty}^{\infty} (-1)^{k-1} u[k-1] z^{-k}\\ &\text{By comparison with the Z Transform definition...}\\ x[n] &=(-1)^{n-1} u[n-1]\\ \end{align} $

Ex. 4 Find the Inverse Z transform of the following signal

$ X(z)=\frac{1}{1-2z}, \text{ ROC } |2z|>1 $

Solution

$ \begin{align} X(z) &= \frac{1}{1-2z}\\ &= \frac{1}{2z} \frac{1}{1-(-\frac{1}{2z})}\\ &= \sum_{n=0}^{\infty} (\frac{-1}{2z})^{n}\frac{1}{2z} \text{ if } |-\frac{1}{2z}| < 1\\ &= \sum_{n=0}^{\infty} (-2z)^{-n}(2z)^{-1}\\ &= \sum_{n=-\infty}^{\infty} \frac{1}{2}(-2)^{-n}z^{-n-1} u[n]\\ &\text{ let } k=n+1\\ &= \sum_{k=-\infty}^{\infty} \frac{1}{2}(-2)^{-k+1}u[k-1] z^{-k}\\ &\text{By comparison with the Z Transform definition...}\\ x[n] &= \frac{1}{2}(-2)^{-k+1}u[n-1]\\ \end{align} $