Further z z z -transform properties

4 Further $z$ -transform properties

We showed earlier that the results

$&integers; {v_{n} + w_{n}} = V (z) + W (z)$ and similarly $&integers; {v_{n} - w_{n}} = V (z) - W (z)$

follow from the linearity property.

You should be clear that there is no comparable result for the product of two sequences.

$&integers; {v_{n} w_{n}}$ is not equal to $V (z) W (z)$

For two specific products of sequences however we can derive useful results.

4.1 Multiplication of a sequence by $a^{n}$

Suppose $f_{n}$ is an arbitrary sequence with z-transform $F (z)$ .

Consider the sequence ${v_{n}}$ where

$v_{n} = a^{n} f_{n}$ i.e. ${v_{0}, v_{1}, v_{2}, \dots} = {f_{0}, a f_{1}, a^{2} f_{2}, \dots}$

By the z-transform definition

\begin{array}{rcl} &integers; {v_{n}} & = & v_{0} + v_{1} z^{- 1} + v_{2} z^{- 2} + \dots \\ = & f_{0} + a f_{1} z^{- 1} + a^{2} f_{2} z^{- 2} + \dots \\ = & \sum_{n = 0}^{\infty} a^{n} f_{n} z^{- n} \\ = & \sum_{n = 0}^{\infty} f_{n} {(\frac{z}{a})}^{- n} \\ But F (z) & = & \sum_{n = 0}^{\infty} f_{n} z^{- n} \end{array}

Thus we have shown that $&integers; {a^{n} f_{n}} = F (\frac{z}{a})$

Key Point 8

$&integers; {a^{n} f_{n}} = F (\frac{z}{a})$

That is, multiplying a sequence ${f_{n}}$ by the sequence ${a^{n}}$ does not change the form of the z-transform $F (z)$ . We merely replace $z$ by $\frac{z}{a}$ in that transform.

$&integers; {cos n} = \frac{z^{2} - z cos 1}{z^{2} - 2 z cos 1 + 1}$

So, replacing $z$ by $\frac{z}{(\frac{1}{2})} = 2 z,$

$&integers; \{{(\frac{1}{2})}^{n} cos n\} = \frac{{(2 z)}^{2} - (2 z) cos 1}{{(2 z)}^{2} - 4 z cos 1 + 1}$

Task!

Using Key Point 8, write down the z-transform of the sequence ${v_{n}}$ where

$v_{n} = e^{- 2 n} sin 3 n$

We have, $&integers; {sin 3 n} = \frac{z sin 3}{z^{2} - 2 z cos 3 + 1}$

so with $a = e^{- 2}$ we replace $z$ by $z e^{+ 2}$ to obtain

\begin{array}{rcl} &integers; {v_{n}} = &integers; {e^{- 2 n} sin 3 n} & = & \frac{z e^{2} sin 3}{{(z e^{2})}^{2} - 2 z e^{2} cos 3 + 1} \\ = & \frac{z e^{- 2} sin 3}{z^{2} - 2 z e^{- 2} cos 3 + e^{- 4}} \end{array}

Task!

Using the property just discussed write down the z-transform of the sequence ${w_{n}}$ where

$w_{n} = e^{- α n} cos ω n$

We have, $&integers; {cos ω n} = \frac{z^{2} - z cos ω}{z^{2} - 2 z cos ω + 1}$

So replacing $z$ by $z e^{α}$ we obtain

\begin{array}{rcl} &integers; {w_{n}} = &integers; {e^{- α n} cos ω n} & = & \frac{{(z e^{α})}^{2} - z e^{α} cos ω}{{(z e^{α})}^{2} - 2 z e^{α} cos ω + 1} \\ = & \frac{z^{2} - z e^{- α} cos ω}{z^{2} - 2 z e^{- α} cos ω + e^{- 2 α}} \end{array}

Key Point 9

\begin{array}{rcl} &integers; {e^{- α n} cos ω n} & = & \frac{z^{2} - z e^{- α} cos ω}{z^{2} - 2 z e^{- α} cos ω + e^{- 2 α}} \\ &integers; {e^{- α n} sin ω n} & = & \frac{z e^{- α} sin ω}{z^{2} - 2 z e^{- α} cos ω + e^{- 2 α}} \end{array}

Note the same denominator in each case.

4.2 Multiplication of a sequence by $n$

An important sequence whose z-transform we have not yet obtained is the unit ramp sequence ${r_{n}}$ :

$r_{n} = \{\begin{matrix} 0 & n = - 1, - 2, - 3, \dots \\ n & n = 0, 1, 2, \dots \end{matrix}$

Figure 5

No alt text was set. Please request alt text from the person who provided you with this resource.

Figure 5 clearly suggests the nomenclature ‘ramp’.

We shall attempt to use the z-transform of ${r_{n}}$ from the definition:

$&integers; {r_{n}} = 0 + 1 z^{- 1} + 2 z^{- 2} + 3 z^{- 3} + \dots$

This is not a geometric series but we can write

\begin{array}{rcl} z^{- 1} + 2 z^{- 2} + 3 z^{- 3} & = & z^{- 1} (1 + 2 z^{- 1} + 3 z^{- 2} + \dots) \\ = & z^{- 1} {(1 - z^{- 1})}^{- 2} |z^{- 1}| < 1 \end{array}

where we have used the binomial theorem ( HELM booklet 16.3) .

Hence

\begin{array}{rcl} &integers; {r_{n}} = &integers; {n} & = & \frac{1}{z {(1 - \frac{1}{z})}^{2}} \\ = & \frac{z}{{(z - 1)}^{2}} |z| > 1 \end{array}

Key Point 10

The z-transform of the unit ramp sequence is

$&integers; {r_{n}} = \frac{z}{{(z - 1)}^{2}} = R (z)$ (say)

Recall now that the unit step sequence has z-transform $&integers; {u_{n}} = \frac{z}{(z - 1)} = U (z)$ (say) which is the subject of the next Task.

Task!

Obtain the derivative of $U (z) = \frac{z}{(z - 1)}$ with respect to $z$ .

We have, using the quotient rule of differentiation:

\begin{array}{rcl} \frac{d U}{d z} = \frac{d}{d z} (\frac{z}{z - 1}) & = & \frac{(z - 1) 1 - (z) (1)}{{(z - 1)}^{2}} \\ = & \frac{- 1}{{(z - 1)}^{2}} \end{array}

We also know that

$R (z) = \frac{z}{{(z - 1)}^{2}} = (- z) (- \frac{1}{{(z - 1)}^{2}}) = - z \frac{d U}{d z}$ (3)

Also, if we compare the sequences

$u_{n} = {0, 0, 1, 1, 1, 1, \dots}$

$↑$

$r_{n} = {0, 0, 0, 1, 2, 3, \dots}$

$↑$

we see that $r_{n} = n u_{n}$ , (4)

so from (3) and (4) we conclude that $&integers; {n u_{n}} = - z \frac{d U}{d z}$

Now let us consider the problem more generally.

Let $f_{n}$ be an arbitrary sequence with z-transform $F (z)$ :

$F (z) = f_{0} + f_{1} z^{- 1} + f_{2} z^{- 2} + f_{3} z^{- 3} + \dots = \sum_{n = 0}^{\infty} f_{n} z^{- n}$

We differentiate both sides with respect to the variable $z$ , doing this term-by-term on the right-hand side. Thus

\begin{array}{rcl} \frac{d F}{d z} & = & - f_{1} z^{- 2} - 2 f_{2} z^{- 3} - 3 f_{3} z^{- 4} - \dots = \sum_{n = 1}^{\infty} (- n) f_{n} z^{- n - 1} \\ = & - z^{- 1} (f_{1} z^{- 1} + 2 f_{2} z^{- 2} + 3 f_{3} z^{- 3} + \dots) = - z^{- 1} \sum_{n = 1}^{\infty} n f_{n} z^{- n} \end{array}

But the bracketed term is the z-transform of the sequence

${n f_{n}} = {0, f_{1}, 2 f_{2}, 3 f_{3}, \dots}$

Thus if $F (z) = &integers; {f_{n}}$ we have shown that

$\frac{d F}{d z} = - z^{- 1} &integers; {n f_{n}}$ or $&integers; {n f_{n}} = - z \frac{d F}{d z}$

We have already (equations (3) and (4) above) demonstrated this result for the case $f_{n} = u_{n}$ .

Key Point 11

If $&integers; {f_{n}} = F (z)$ then $&integers; {n f_{n}} = - z \frac{d F}{d z}$

Task!

By differentiating the z-transform $R (z)$ of the unit ramp sequence obtain the z-transform of the causal sequence ${n^{2}}$ .

We have

$&integers; {n} = \frac{z}{{(z - 1)}^{2}}$

$&integers; {n^{2}} = &integers; {n . n} = - z \frac{d}{d z} (\frac{z}{{(z - 1)}^{2}})$

By the quotient rule

\begin{array}{rcl} \frac{d}{d z} (\frac{z}{{(z - 1)}^{2}}) & = & \frac{{(z - 1)}^{2} - (z) (2) (z - 1)}{{(z - 1)}^{4}} \\ = & \frac{z - 1 - 2 z}{{(z - 1)}^{3}} = \frac{- 1 - z}{{(z - 1)}^{3}} \end{array}

Multiplying by $- z$ we obtain

$&integers; {n^{2}} = \frac{z + z^{2}}{{(z - 1)}^{3}} = \frac{z (1 + z)}{{(z - 1)}^{3}}$

Clearly this process can be continued to obtain the transforms of ${n^{3}}, {n^{4}}, \dots$ etc.

4 Further z -transform properties

4.1 Multiplication of a sequence by a n

Answer

Answer

4.2 Multiplication of a sequence by n

Answer

Answer

4 Further $z$ -transform properties

4.1 Multiplication of a sequence by $a^{n}$

4.2 Multiplication of a sequence by $n$