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 fills in toward the origin

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

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 $

notice: ROC is type 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 $

notice: ROC is type 2

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. 3 Find the Inverse Z transform of the following signal

$ X(z)=\frac{z}{1-5z}, \text{ ROC } |z|<\frac{1}{5} $

notice: ROC is type 1

Solution

$ \begin{align} X(x) &= \frac{z}{1+5z}\\ &= \frac{z}{1-(-5z)}\\ &= \sum_{n=0}^{\infty} z(-5z^{n}) \text{ if } |z| < \frac{1}{5}\\ &= \sum_{n=-\infty}^{\infty} (-5)^{n}z^{n+1} u[n]\\ &\text{let }k = -n-1\\ &= \sum_{k=-\infty}^{\infty} (-5)^{-k-1}z^{-k} u[-k-1]\\ &= \sum_{k=-\infty}^{\infty}(-5)^{-k-1}u[-k-1] z^{-k}\\ &\text{By comparison with the Z Transform definition...}\\ x[n] &= (-5)^{-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 $

notice: ROC is type 2

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} $

Ex. 5 Find the Inverse Z transform of the following signal

$ X(z)=\frac{-1}{z^2-4z-5}, \text{ ROC } 1<|z|<5 $

notice: ROC is type 3

Solution

$ \begin{align} X(z) &= \frac{-1}{z^2-4z-5}\\ &\text{ by partial fraction expansion}\\ &= \Big( \frac{1}{6} \Big)\Big( \frac{1}{z+1}+\frac{1}{5-z} \Big)\\ &= \Big( \frac{1}{6} \Big) \Bigg( \sum_{n=0}^{\infty} (\frac{1}{z})(\frac{-1}{z})^{n} + \sum_{m=0}^{\infty} (\frac{1}{5})(\frac{z}{5})^m \Bigg) \text{ if } 1<|z|< 5\\ &=\Big( \frac{1}{6} \Big) \Bigg( \sum_{n=-\infty}^{\infty} (-1)^n(z)^{-n-1}u[n] + \sum_{m=-\infty}^{\infty} (5)^{-m-1)}(z)^{m}u[m] \Bigg)\\ & \text{let } k=n+1, l=-m\\ &= \Big( \frac{1}{6} \Big) \Bigg( \sum_{k=-\infty}^{\infty} (-1)^{k-1}(z)^{-k}u[k-1] + \sum_{l=-\infty}^{\infty} (5)^{l-1)}(z)^{-l}u[-l] \Bigg)\\ & \text{using the linearity principle of Z transforms}\\ x[n] &= \frac{1}{6}(-1)^{n-1}u[n-1] + \frac{1}{6} (5)^{n-1)}(z)^{-n}u[-n] \end{align} $