Properties of the z-Transform

4.2 Properties of the z-Transform

In this section we discuss the properties of the z-transform that are useful in obtain- ing z-transform pairs and in applying the transform to the analysis of discrete-time linear time-invariant systems. These properties are very similar to those of the Laplace transform and Fourier transform.

Most of the properties are commonly applicable to the bilateral and unilateral z -transforms. There are, however, a few exceptions such as the time shifting prop- erty, time reversal property, and initial/final value theorems. In particular, it is the time shifting property that makes the unilateral z-transform very useful for solving difference equations with initial conditions.

4.2.1 Linearity

Let the z-transforms of two time sequences x[n] and y[n] be Z{x[n]} = X[z] with ROC R x and Z{y[n]} = Y[z] with ROC R y . Then we can express the z-transform of their linear combination in terms of X[z] and Y[z] as

1 ∩R 2 ) (4.2.1) where the ROC is generally the intersection of the two ROCs R x and R y , but it can

ax z [n] + βy[n] ↔ αX[z] + βY[z] with ROC R ⊃ (R

be larger when any pole of X[z] and Y[z] is cancelled by the zeros resulting from the addition.

214 4 The z-Transform

4.2.2 Time Shifting – Real Translation

Let the bilateral z-transform of x[n] be Z{x[n]} = X[z] with ROC R x . Then the bilateral z-transform of a time-shifted version x[n − n 1 ] is x z [n − n

1 ] ↔z −n 1 X[z], ROC R x (possibly with z = 0 or ∞ added or removed) (4.2.2)

The addition/removal of z = ∞ to/from the ROC occurs when a causal/non-causal sequence is shifted to become a non-causal/causal sequence. Similarly, the addi-

tion/removal of the origin z = 0 to/from the ROC is owing to the possible transition between anticausal and non-anticausal sequences resulting from the shift operation.

On the other hand, the time-shifting property of the unilateral z-transform is as follows:

< Case 1> x[n − n 1 ], n 1 > 0 (delayed sequence)

Z{x[n − n ∞

n−n 1 =m,n=m+n 1 −1

x [m]z −(m+n 1 )

x [m]z −(m+n 1 ) m=−n 1 +

m=0

=z −n 1 ∞

x [m]z −m +

x [m]z −m ;

m=0

m=−n 1

x [n − n 1 ] ↔z −n 1 X [z] +

x [m]z −m (4.2.3a)

m=−n 1

where the second term in the RHS will disappear when x[n] is causal. < Case 2> x[n + n 1 ], n 1 > 0 (advanced sequence)

−n n+n 1 1 1 ∞ x 1 ]z =m,n=m−n x [m]z −(m−n 1 Z{x[n + n ) ]} = =

n 1 =z −1 1 x [m]z −m

x [m]z −m ;

x [m]z −m (4.2.3b)

m=0

Example 4.5 Applying Linearity and Time Shifting Properties of the z-Transform For a rectangular pulse sequence of duration N given by

n=···−1

0 1 ···N −1 N ···

x [n] = u s [n] − u s [n − N ] = {· · · 0 1 1 · · · 1 0···} (E4.5.1)

4.2 Properties of the z-Transform 215 we can use Eq. (E4.1.1) (with a = 1) together with the linearity (4.2.1) and time-

shifting property (4.2.2) to get its z-transform as

X[z] −1

(E4.1.1) with a=1

(E4.5.2) (z − 1)

Note that X[z] has multiple pole of order N − 1 at z = 0 and N − 1 zeros at j z=e 2mπ/N (for m = 1, 2, · · · , N − 1) on the unit circle where the zero at z = 1

(resulting from the addition) cancels the pole (of Z{u s [n]} = z/(z −1)) at that point. Due to this pole-zero cancellation, the ROC of X[z] becomes |z| > 0 (the entire

z -plane except the origin z = 0), which is larger than that of Z{u s [n]} = z/(z − 1), i.e., |z| > 1.

4.2.3 Frequency Shifting – Complex Translation

Let the bilateral z-transform of x[n] be Z{x[n]} = X[z] with ROC R x = {z : r x − < |z| < r x + }. Then we have the z-transform of a frequency-shifted or modulated

version z n 1 x [n] as

e z jΩ 1 n x [n] ↔ X[e − jΩ 1 z ] with ROC : R x = {z : r x − < |z| < r + x } (4.2.4a) z n

1 x [n] ↔ X[z/z 1 ] with ROC : |z 1 |R x = {z : |z 1 |r − x < |z| < |z 1 |r x + } (4.2.4b)

4.2.4 Time Reversal

Let the bilateral z-transform of x[n] be Z{x[n]} = X[z] with ROC R x = {z : r x − < |z| < r x + }. Then we have the bilateral z-transform of a time-reversed version x [−n] as

x z [−n] ↔ X[z −1 ] with ROC : 1/ R x = {z : 1/r + x < |z| < 1/r x − } (4.2.5) where if z ∈ R x , then 1/z ∈ 1/R x .

4.2.5 Real Convolution

Let the z-transforms of two time sequences g[n] and x[n] be Z{g[n]} = G[z] with ROC R g and Z{x[n]} = X[z] with ROC R x . Then we can express the z-transform of their convolution in terms of G[z] and X[z] as

y z [n] = g[n] ∗ x[n] ↔ Y[z] = G[z] X[z] with ROC R ⊃ (R

g ∩R x ) (4.2.6)

216 4 The z-Transform (proof)

Z{g[n] ∗ x[n]} ∞ =

(4.1.1a)

n=−∞ (g[n] ∗ x[n])z −n

[m]x[n − m]

∞ = . g [m] x −(n−m) z −m

∞ . = (4.1.1a) g [m]z −m x [n]z −n

= G[z]X[z] This convolution property holds for causal sequences, which implies that the uni-

n−m→n, n→n+m

m=−∞

n=−∞

lateral z-transform also has the same property. It is very useful for describing the input-output relationship of a discrete-time LTI system with the input x[n], out- put y[n], and impulse response g[n] where the transform of the impulse response, Z{g[n]} = G[z], is called the system or transfer function (see Sects. 1.2.3 and 1.2.4).

4.2.6 Complex Convolution

Let the z-transforms of x[n] and y[n] be Z{x[n]} = X[z] with ROC R x = {z : r x − < |z| < r x + } and Z{y[n]} = Y[z] with ROC R y = {z : r y − < |z| < r + y }, respectively. Then the z-transform of the product of the two sequences can be expressed in terms

of X[z] and Y[z] as Z / 1

x [n]y[n] ↔ X[z/v]Y[v]v −1 dv with ROC − r − < + r + 2π j

R = {z : r x y |z| < r x y }

(4.2.7a) Z / 1

x [n]y[n] ↔ X[v]Y[z/v]v −1 dv with ROC 2π j

R = {z : r − x r − y < |z| < r x + r + y }

(4.2.7b) where 0

C i means the complex integral along a closed contour C i within the intersec- tion of the ROCs of X[z/v] and Y[v] or X[v] and Y[z/v] (see [O-2], Sect. 2.3.9 for

its proof).

4.2.7 Complex Differentiation

Let the z-transform of x[n] be Z{x[n]} = X[z] with ROC R x . Then the z-transform of nx[n] can be expressed in terms of X[z] as

nx [n] ↔ −z X[z] with ROC R x

dz

4.2 Properties of the z-Transform 217 This can be proved by differentiating the definition of the z-transform (4.1.1)

w.r.t. z.

4.2.8 Partial Differentiation

If a z-transform pair is given in the form of Z{x(nT, a)} = X(z, a) with a parameter

a , we can differentiate it w.r.t. a to obtain another z-transform pair as

Example 4.6 Complex Differentiation and Partial Differentiation

2 a For y[n] = n n u s [n], we can apply the complex differentiation property (4.2.8) twice for Eq. (E4.1.1) to write

− 2az(z − a) ↔ −z dz

d az

4 = (z − a) 3 (z − a) (z − a) (E4.6.2)

2 = −z

Alternatively, we can apply the partial differentiation property for Eq. (E4.1.1) to get the same results:

(z − a) + 2az(z − a) (z + a)

Multiplying both sides of these equations by a yields Eqs. (E4.6.1) and (E4.6.2), which are listed in Table B.9(10) and (12).

4.2.9 Initial Value Theorem

For a causal sequence x[n] such that x[n] = 0 ∀ n < 0, we can get its initial value x [0] from its z-transform as

This can easily be shown by substituting z = ∞ into the z-transform definition (4.1.1b).

218 4 The z-Transform

4.2.10 Final Value Theorem

For a causal sequence x[n] such that x[n] = 0 ∀ n < 0, we can get its final value x [∞] from its z-transform as

x [∞] = lim (z − 1)X[z] = lim (1 − z −1 )X [z]

This requires that x[n] should converge or equivalently, all the poles of its z- transform X [z] should lie inside the unit circle possibly with the exception of a simple pole at z = 1. (proof)

We can use the z-transform definition (4.1.1b) to write

Z{x[n + 1] − x[n]} = lim . x

n=0 [n + 1]z −n −

x [n]z −n

x [n]z −(n−1)

− x[0] − . x [n]z −n

n=1

On the other hand, from the time shifting property (4.2.3b), we can write Z{x[n + 1] − x[n]} (4.2.1) = Z{x[n + 1]} − Z{x[n]}

(4.2.3) = z(X[z] − x[0]) − X[z] = (z − 1)X[z] − zx[0] Noting that these two equations are commonly the z-transform of (x[n + 1] − x[n]),

we can equate their RHSs and substitute z = 1 to get the desired result.