3 Initial and final value theorems of z-transforms

These results are important in, for example, Digital Control Theory where we are sometimes particularly interested in the initial and ultimate behaviour of systems.

3.1 Initial value theorem

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If $f_n$ is a sequence with z-transform $F\left(z\right)$ then the ‘initial value’ $f_0$ is given by

$f_0=\underset{z\to \infty }{lim}F\left(z\right)$ (provided, of course, that this limit exists).

This result follows, at least informally, from the definition of the z-transform:

$F\left(z\right)=f_0+f_1{z}^{-1}+f_2{z}^{-2}+\dots$

from which, taking limits as $z\to \infty$ the required result is obtained.

Task!

Obtain the z-transform of

$f\left(n\right)=1-a^n,0<a<1$

Verify the initial value theorem for the z-transform pair you obtain.

Answer

Suppose again that $\left\{f_n\right\}$ is a sequence with z-transform $F\left(z\right)$ . We further assume that all the poles of $F\left(z\right)$ lie inside the unit circle in the $z-$ plane (i.e. have magnitude less than 1) apart possibly from a first order pole at $z=1$ .

The ‘final value’ of $f_n$ i.e. $\underset{n\to \infty }{lim}{f}_n$ is then given by $\underset{n\to \infty }{lim}{f}_n=\underset{z\to 1}{lim}\left(1-z^{-1}\right)F\left(z\right)$

Proof: Recalling the left shift property

$Z\left\{f_{n+1}\right\}=zF\left(z\right)-z{f}_0$

we have

$Z\left\{f_{n+1}-f_n\right\}=\underset{k\to \infty }{lim}{\sum }_{n=0}^k\left(f_{n+1}-f_n\right)z^{-n}=zF\left(z\right)-z{f}_0-F\left(z\right)$

or, alternatively, dividing through by $z$ on both sides:

$\left(1-z^{-1}\right)F\left(z\right)-f_0=\underset{k\to \infty }{lim}{\sum }_{n=0}^k\left(f_{n+1}-f_n\right)z^{-\left(n+1\right)}$

Hence $\left(1-z^{-1}\right)F\left(z\right)=f_0+\left(f_1-f_0\right)z^{-1}+\left(f_2-f_1\right)z^{-2}+\dots$

or as $z\to 1$

Example

Again consider the sequence $f_n=1-a^{n}0<a<1$ and its z-transform

$F\left(z\right)=\frac{z}{z-1}-\frac{z}{z-a}=\frac{1}{1-z^{-1}}-\frac{1}{1-a{z}^{-1}}$

Clearly as $n\to \infty$ then $f_n\to 1$ .

Considering the right-hand side

$\left(1-z^{-1}\right)F\left(z\right)=1-\frac{\left(1-z^{-1}\right)}{1-a{z}^{-1}}\to 1-0=1$ as $z\to 1.$

Note carefully that

$F\left(z\right)=\frac{z}{z-1}-\frac{z}{z-a}$

has a pole at $a$ $\left(0<a<1\right)$ and a simple pole at $z=1$ .

The final value theorem does not hold for z-transform poles outside the unit circle

e.g. $f_n=2^{n}F\left(z\right)=\frac{z}{z-2}$

Clearly $f_n\to \infty$ as $n\to \infty$

whereas

$\left(1-z^{-1}\right)F\left(z\right)=\left(\frac{z-1}{z}\right)\frac{z}{\left(z-2\right)}$ $\to 0$ as $z\to 1$

Exercises

1. A low pass digital filter is characterised by

   $y_n=0.1x_n+0.9y_{n-1}$

   Two such filters are connected in series. Deduce the transfer function and governing difference equation for the overall system. Obtain the response of the series system to (i) a unit step and (ii) a unit alternating input. Discuss your results.

2. The two systems

   $y_n=x_n-0.7x_{n-1}+0.4y_{n-1}$

   $y_n=0.9x_{n-1}-0.7y_{n-1}$

   are connected in series. Find the difference equation governing the overall system.

3. A system $S_1$ is governed by the difference equation

   $y_n=6x_{n-1}+5y_{n-1}$

   It is desired to stabilise $S_1$ by using a feedback configuration. The system $S_2$ in the feedback loop is characterised by

   $y_n=\alpha x_{n-1}+\beta y_{n-1}$

   Show that the feedback system $S_3$ has an overall transfer function

   $H_3\left(z\right)=\frac{H_1\left(z\right)}{1+H_1\left(z\right)H_2\left(z\right)}$

   and determine values for the parameters $\alpha$ and $\beta$ if $H_3\left(z\right)$ is to have a second order pole at $z=0.5$ . Show briefly why the feedback systems $S_3$ stabilizes the original system.

4. Use z-transforms to find the sum of squares of all integers from 1 to $n$ :

   $y_n={\sum }_{k=1}^n{k}^2$

   [Hint: $y_n-y_{n-1}=n^2$ ]

5. Evaluate each of the following convolution summations (i) directly (ii) using z-transforms:
   1. $a^n\ast b^{n}a\ne b$
   2. $a^n\ast a^n$
   3. ${\delta }_{n-3}\ast {\delta }_{n-5}$
   4. $x_n\ast x_n$ where $x_n=\left\{\begin{array}{cc}\hfill 1\hfill & \hfill n=0,1,2,3\hfill \\ \hfill 0\hfill & \hfill n=4,5,6,7\dots \hfill \end{array}\right.$

Answer