5 Shifting properties of the z-transform

In this subsection we consider perhaps the most important properties of the z-transform. These properties relate the z-transform $Y\left(z\right)$ of a sequence $\left\{y_n\right\}$ to the z-transforms of

1. right shifted or delayed sequences $\left\{y_{n-1}\right\}\left\{y_{n-2}\right\}$ etc.
2. left shifted or advanced sequences $\left\{y_{n+1}\right\},\left\{y_{n+2}\right\}$ etc.

The results obtained, formally called shift theorems, are vital in enabling us to solve certain types of difference equation and are also invaluable in the analysis of digital systems of various types.

5.1 Right shift theorems

Let $\left\{v_n\right\}=\left\{y_{n-1}\right\}$ i.e. the terms of the sequence $\left\{v_n\right\}$ are the same as those of $\left\{y_n\right\}$ but shifted one place to the right. The z-transforms are, by definition,

i.e.

$V\left(z\right)=\mathbb{Z}\left\{y_{n-1}\right\}=y_{-1}+z^{-1}Y\left(z\right)$

Task!

Obtain the z-transform of the sequence $\left\{w_n\right\}=\left\{y_{n-2}\right\}$ using the method illustrated above.

Answer

Clearly, we could proceed in a similar way to obtains a general result for $\mathbb{Z}\left\{y_{n-m}\right\}$ where $m$ is any positive integer. The result is

$\mathbb{Z}\left\{y_{n-m}\right\}=y_{-m}+y_{-m+1}{z}^{-1}+\dots +y_{-1}{z}^{-m+1}+z^{-m}Y\left(z\right)$

For the particular case of causal sequences (where $y_{-1}=y_{-2}=\dots =0$ ) these results are particularly simple:

$\left.\begin{array}{c}\hfill \mathbb{Z}\left\{y_{n-1}\right\}=z^{-1}Y\left(z\right)\\ \hfill \mathbb{Z}\left\{y_{n-2}\right\}=z^{-2}Y\left(z\right)\\ \hfill \\ \hfill \mathbb{Z}\left\{y_{n-m}\right\}=z^{-m}Y\left(z\right)\end{array}\right\}\text{(causal systems only)}$

You may recall from earlier in this Workbook that in a digital system we represented the right shift operation symbolically in the following way:

Figure 6

The significance of the $z^{-1}$ factor inside the rectangles should now be clearer. If we replace the ‘input’ and ‘output’ sequences by their z-transforms:

$\mathbb{Z}\left\{y_n\right\}=Y\left(z\right)\mathbb{Z}\left\{y_{n-1}\right\}=z^{-1}Y\left(z\right)$

it is evident that in the z-transform ‘domain’ the shift becomes a multiplication by the factor $z^{-1}$ .

N.B. This discussion applies strictly only to causal sequences.

Notational point:

A causal sequence is sometimes written as $y_n{u}_n$ where $u_n$ is the unit step sequence

$u_n=\left\{\begin{array}{ccc}\hfill 0\hfill & \hfill \hfill & n=-1,-2,\dots \hfill \\ \hfill 1\hfill & \hfill \hfill & n=0,1,2,\dots \hfill \end{array}\right.$

The right shift theorem is then written, for a causal sequence,

$\mathbb{Z}\left\{y_{n-m}{u}_{n-m}\right\}=z^{-m}Y\left(z\right)$

Examples

Recall that the z-transform of the causal sequence $\left\{a^n\right\}$ is $\frac{z}{z-a}$ . It follows, from the right shift theorems that

1. $\mathbb{Z}\left\{a^{n-1}\right\}=\mathbb{Z}\left\{0,1,a,a^2,\dots \right\}=\frac{z{z}^{-1}}{z-a}=\frac{1}{z-a}$

   $\uparrow$

2. $\mathbb{Z}\left\{a^{n-2}\right\}=\mathbb{Z}\left\{0,0,1,a,a^2,\dots \right\}=\frac{z^{-1}}{z-a}=\frac{1}{z\left(z-a\right)}$

   $\uparrow$

Task!

Write the following sequence $f_n$ as a difference of two unit step sequences. Hence obtain its z-transform.

Answer

5.2 Left shift theorems

Recall that the sequences $\left\{y_{n+1}\right\},\left\{y_{n+2}\right\}\dots$ denote the sequences obtained by shifting the sequence $\left\{y_n\right\}$ by $1,2,\dots$ units to the left respectively. Thus, since $Y\left(z\right)=\mathbb{Z}\left\{y_n\right\}=y_0+y_1{z}^{-1}+y_2{z}^{-2}+\dots$ then

The term in brackets is the z-transform of the unshifted sequence $\left\{y_n\right\}$ apart from its first two terms:

thus

$\mathbb{Z}\left\{y_{n+1}\right\}=y_1+z\left(Y\left(z\right)-y_0-y_1{z}^{-1}\right)$

$\therefore Z\left\{y_{n+1}\right\}=zY\left(z\right)-z{y}_0$

Task!

Obtain the z-transform of the sequence $\left\{y_{n+2}\right\}$ using the method illustrated above.

Answer

if $y_0=0\mathbb{Z}\left\{y_{n+1}\right\}=zY\left(z\right)$

if $y_0=y_1=0\mathbb{Z}\left\{y_{n+2}\right\}=z^{2}Y\left(z\right)$

if $y_0=y_1=\dots y_{m-1}=0\mathbb{Z}\left\{y_{n+m}\right\}=z^{m}Y\left(z\right)$

Note carefully the occurrence of positive powers of $z$ in the left shift theorems and of negative powers of $z$ in the right shift theorems. Table 1: z-transforms

This table has been copied to the back of this Workbook (page 96) for convenience.