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THE z-TRANSFORM

141

Table 6.1 Some commonly used z-transforms

f (kT )

F(z)

δ(t)

1

z

1

z

− 1

kT

T z

(z

−z1)2

e−a kT

z

−

e−aT

aT

kT e−akT

T ze−

(z

−

e−aT )2

ak

z

z − a

aT

)

1 e−a kT

z(1 − e−

−

−

(z

−

1)(z

−

e

aT )

sin akT

z sin aT

z

2

− 2z cos aT + 1

cos akT

2

z(z − cos aT )

z

− 2z cos aT + 1

where D = ∂ D/∂ s and the xn , n = 1, 2, . . . , p, are the roots of the equation D(s) = 0. Some examples are given below.

Example 6.2

Let

1

G(s) = . s2 + 5s + 6

Determine G(z) by the methods described above.

Solution

Method 1: By finding the inverse Laplace transform. We can express G(s) as a sum of its partial fractions:

G(s) =

1

=

1

−

1

.

(6.16)

(s

+

3)(s

+

2)

s

+

2

s

+

3

The inverse Laplace transform of (6.16) is

g(t) = L−1[G(s)] = e−2t

− e−3t .

(6.17)

From the definition of the z-transforms we can write (6.17) as

∞

G(z) = (e−2nT − e−3nT )z−n

n

=

0

= (1 + e−2T z−1 + e−4T z−2 + . . .) − (1 + e−3T z−1 + e−6T z−2 + . . .)

zz

=z − e−2T − z − e−3T

G(z) =

142 SAMPLED DATA SYSTEMS AND THE Z-TRANSFORM

or

z(e−2T − e−3T )

G(z) = (z − e−2T )(z − e−3T ) .

Method 2: By using the z-transform transform tables for the partial product. From Table 6.1, the z-transform of 1/(s + a) is z/(z − e−aT ). Therefore the z-transform of (6.16) is

G(z) =

z

−

z

z

−

e

2T

z

−

e

3T

−

−

or

z(e−2T − e−3T )

(z − e−2T )(z − e−3T ) .

Method 3: By using the z-transform tables for G(s). From Table 6.1, the z-transform of

G(s)

=

(s

b − a

b)

(6.18)

+

a)(s

+

is

G(z)

= (z

z(e−aT − e−bT )

.

(6.19)

−

e

aT )(z

−

e

−

bT )

−

Comparing (6.18) with (6.16) we have, a = 2, b = 3. Thus, in (6.19) we get

G(z)

z(e−2T − e−3T )

.

= (z

−

e

−

2T )(z

−

e

−

3T )

Method 4: By using equation (6.15). Comparing our expression

G(s) =

1

s2 + 5s + 6

with (6.15), we have N (s) = 1, D(s) = s2 + 5s + 6 and D (s) = 2s + 5, and the roots of D(s) = 0 are x1 = −2 and x2 = −3. Using (6.15),

2

G(z) =

N (xn ) 1

n=1 D (xn ) 1 − exn T z−1

or, when x1 = −2,

1

1

G1(z) =

1 1

−

e

2T z

−

1

and when x1 = −3,

−

1

1

G2(z) =

.

Thus,

−1

1 − e−3T z−1

G(z) =

1

1

z

z

1

−

−

2T z

−

1

− 1

−

−

3T z

−

1

= z

−

−

2T

− z

−

−

3T

e

e

e

e

THE z-TRANSFORM

143

or

z(e−2T − e−3T )

G(z) = (z − e−2T )(z − e−3T ) .

6.2.11 Properties of z-Transforms

Most of the properties of the z-transform are analogs of those of the Laplace transforms. Important z-transform properties are discussed in this section.

1.Linearity property

Suppose that the z-transform of f (nT ) is F(z) and the z-transform of g(nT ) is G(z). Then

Z [ f (nT ) ± g(nT )] = Z [ f (nT )] ± Z [g(nT )] = F(z) ± G(z)

(6.20)

and for any scalar a

Z [a f (nT )] = a Z [ f (nT )] = a F(z)

(6.21)

2. Left-shifting property

Suppose that the z-transform of f (nT ) is F(z) and let y(nT ) = f (nT + mT ). Then

zm F(z)

m−1

−i .

Y (z)

(6.22)

=

−

f (i T )zm

=

0

i

If the initial conditions are all zero, i.e. f (i T ) = 0, i = 0,1,2, . . . , m − 1, then,

Z [ f (nT + mT )] = zm F(z).

(6.23)

3. Right-shifting property

Suppose that the z-transform of f (nT ) is F(z) and let y(nT ) = f (nT − mT ). Then

z−m F(z)

m−1

mT )z−i .

Y (z)

=

f (i T

(6.24)

+

=

0

−

i

If f (nT ) = 0 for k < 0, then the theorem simplifies to

Z [ f (nT

−

mT )]

=

z−m F(z).

(6.25)

4.Attenuation property

Suppose that the z-transform of f (nT ) is F(z). Then,

Z [e−anT f (nT )]

=

F[zeaT ].

(6.26)

This result states that if a function is multiplied by the exponential e−anT

then in the

z-transform of this function z is replaced by zeaT .

5.Initial value theorem

Suppose that the z-transform of f (nT ) is F(z). Then the initial value of the time response is given by

lim f (nT ) = lim F(z).

(6.27)

n→0

z→∞

144 SAMPLED DATA SYSTEMS AND THE Z-TRANSFORM

6.Final value theorem

Suppose that the z-transform of f (nT ) is F(z). Then the final value of the time response is given by

lim

f (nT ) = lim(1 − z−1)F(z).

(6.28)

n→∞

z→1

Note that this theorem is valid if the poles of (1 − z−1)F(z) are inside the unit circle or at z = 1.

Example 6.3

The z-transform of a unit ramp function r(nT ) is

T z

R(z) = (z − 1)2 .

Find the z-transform of the function 5r(nT ).

Solution

Using the linearity property of z-transforms,

Z [5r(nT )] = 5R(z) =

5T z

.

(z

−

1)2

Example 6.4

The z-transform of trigonometric function r(nT ) = sin nwT is

R(z) =

z sin wT

.

z2 − 2z cos wT + 1

find the z-transform of the function y(nT )

=

e−2T sin nW T .

Solution

Using property 4 of the z-transforms,

Z [y(nT )]

=

Z [e−2T r(nT )]

=

R[ze2T ].

Thus,

Z [y(nT )] =

ze2T sin wT

=

ze2T sin wT

(ze2T )2 − 2ze2T cos wT + 1

z2e4T − 2ze2T cos wT + 1

or, multiplying numerator and denominator by e−4T ,

Z [y(nT )] =

ze−2T sin wT

.

z2 − 2ze−2T + e−4T

Example 6.5

Given the function

G(z) =

0.792z

,

(z − 1)(z2 − 0.416z + 0.208)

find the final value of g(nT ).

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