EEL6507sp09L07

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EEL6507 Spring 2009, Lecture 07, Friay 2008/01/23 (Notes created by Xiaoyuan Li)

Introduce Z-Transform for the solution of transient state probabilities

An example of Z-Transform

Find the expression of ai given that a_{i+1}\ =3a_i\ +4,\ a_0=0,\ a_1=4,\ a_2=16,\ a_3=52.

We introduce Z-Transform

a\left(z\right)\ =\ \sum_{i=0}^{\infty} a_iz^i
\sum_{i=0}^{\infty} a_{i+1}z^i\ = \sum_{i=0}^{\infty} \left(3a_i+4\right)z^i =\ 3\sum_{i=0}^{\infty} a_iz^i\ +\ 4\sum_{i=0}^{\infty} z^i\ =3a\left(z\right)+4\left(\frac{1}{1-z}\right)
\sum_{i=0}^{\infty} a_{i+1}z^i\ = \left ( \frac{1}{z} \right )\sum_{i=1}^{\infty} a_{i}z^i\ = \left ( \frac{1}{z} \right)\left(a\left(z\right)-a_0\right)\ = 3a\left(z\right)+4\left(\frac{1}{1-z}\right)

Then we get

a\left(z\right)\ = \frac{a_0}{1-3z}+\frac{4z}{\left(1-3z\right)\left(1-z\right)}

Using Partial Fraction

M\ = \frac{1}{\left(1-3z\right)\left(1-z\right)}\ = \frac{A}{1-3z}+\frac{B}{1-z}

Let z\ =\frac{1}{3}, A\ = M\left(1-3z\right)\ =\frac{3}{2}

Let z\ =\ 1, B\ = M\left(1-z\right)\ = -\frac{1}{2}

Therefore

\begin{align} \sum_{i=0}^{\infty} a_{i}z^i\ &= \sum_{i=0}^{\infty} a_{0}3^{i}z^{i}\ +4z\left(\frac{\frac{3}{2}}{1-3z}\ -\frac{\frac{1}{2}}{1-z}\right)\\ &= \sum_{i=0}^{\infty} a_{0}3^{i}z^{i}\ +4z\left(\frac{2}{3}\sum_{i=0}^{\infty}3^{i}z^{i}\ -\frac{1}{2}\sum_{i=0}^{\infty}z^{i}\right)\\ &=\sum_{i=0}^{\infty} a_{0}3^{i}z^{i}\ +\sum_{i=0}^{\infty}\left(6*3^i-2\right)z^{i+1}\\ &=\sum_{i=0}^{\infty} a_{0}3^{i}z^{i}\ +\sum_{i=1}^{\infty}\left(6*3^{i-1}-2\right)z^{i} \end{align}
a_0\ +\sum_{i=1}^{\infty} a_{i}3^{i}z^{i}\ = a_0\ +\sum_{i=1}^{\infty}\left(a_{0}3^{i}\ +6*3^{i-1}\ -2\right)z^i

Obviously

a_i\ =a_{0}3^i\ +6*3^{i-1}\ -2

Given the initial condition \ a_0=0,\ a_1=4,\ a_2=16,\ a_3=52,

a_i\ =6*3^{i-1}\ -2

Transient State Probabilities

Recall that, \vec \pi(z) = \vec \pi{(0)}\left(I-zP\right)^{-1}

We want to expand \left(I-zP\right)^{-1} as powers of z.

Example

We have P\ = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ 1 & 0 \end{bmatrix}

A\left(z\right)\ = I\ - zP\ = \begin{bmatrix} 1-\frac{1}{2}z & -\frac{1}{2}z \\ -z & 1 \end{bmatrix}

Recall that if X\ = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, then X^{-1}\ = \frac{1}{det\left(A\right)}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

where det\left(A\right)\ = ad\ -bc

Based on the expression of X − 1 and det\left(A\right)\ = 1-\frac{1}{2}z-\frac{1}{2}z^2=\frac{1}{2}\left(z+2\right)\left(1-z\right)

we can get

\begin{align} A^{-1}&= \frac{1}{\frac{1}{2}\left(z+2\right)\left(1-z\right)} \begin{bmatrix} 1 & \frac{1}{2}z \\ z & 1-\frac{1}{2}z \end{bmatrix}\\ &=\frac{2}{\left(z+2\right)\left(1-z\right)} \begin{bmatrix} 1 & \frac{1}{2}z \\ z & 1-\frac{1}{2}z \end{bmatrix}\\ &=\frac{2}{3}\left(\frac{1}{1-z}+\frac{\frac{1}{2}}{1+\frac{z}{2}}\right) \begin{bmatrix} 1 & \frac{1}{2}z \\ z & 1-\frac{1}{2}z \end{bmatrix}\\ &=\frac{2}{3}\left(\sum_{i=0}^{\infty} z^i+\frac{1}{2}\sum_{i=0}^{\infty} \left(-\frac{1}{2}\right)^{i}z^i\right) \begin{bmatrix} 1 & \frac{1}{2}z \\ z & 1-\frac{1}{2}z \end{bmatrix}\\ &=\frac{2}{3}\left(\sum_{i=0}^{\infty} \left(1-\left(-\frac{1}{2}\right)^{i+1}\right)z^i\right)\left(I+Mz\right) \end{align}

where I=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} and M=\begin{bmatrix} 0 & \frac{1}{2} \\ 1 & -\frac{1}{2} \end{bmatrix}

\begin{align} A^{-1}&=\frac{2}{3}\left(\sum_{i=0}^{\infty} \left(1-\left(-\frac{1}{2}\right)^{i+1}\right)z^i*I+\sum_{i=0}^{\infty} \left(1-\left(\frac{1}{2}\right)^{i+1}\right)z^{i+1}*M\right)\\ &=\frac{2}{3}\left(\sum_{i=0}^{\infty} \left(\left(I+M\right)-\left(-\frac{1}{2}\right)^{i}\left(-\frac{1}{2}I+M\right)\right)z^i\right)\\ &=\sum_{i=0}^{\infty} P^i z^i \end{align}
\begin{align} P^i &= \frac{2}{3}\left(I+M\right)-\frac{2}{3}\left(M-\frac{7}{2}\right)\left(-\frac{1}{2}\right)^i\\ P^i &= \frac{2}{3}\begin{bmatrix} 1 & \frac{1}{2} \\ 1 & \frac{1}{2} \end{bmatrix}-\frac{2}{3}\begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ 1 & 1 \end{bmatrix}\left(-\frac{1}{2}\right)^i \end{align}

Therefore,

\lim_{i \to \infty}P^i = \begin{bmatrix} \frac{2}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{1}{3} \end{bmatrix}
\begin{bmatrix} P & 1-P \end{bmatrix} \begin{bmatrix} \frac{2}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{1}{3} \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & \frac{1}{3} \end{bmatrix}\left[P+1-P\right]

Therefore

\vec \pi = \left(\begin{bmatrix} \frac{2}{3} & \frac{1}{3} \end{bmatrix}\right)

Lecture Recording

lecture_recording

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