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PULSE TRANSFER FUNCTION AND MANIPULATION OF BLOCK DIAGRAMS

159

Solution

The transfer function of the zero-order hold is

G

1

(s)

=

1 − e−T s

s

and that of the RC system is

G(s) =

1

.

s

+

1

For this system we can write

y(s) = u*(s)G1G2(s)

and

y*(s) = u*(s)[G1G2]*(s)

or, taking z-transforms,

y(z) = u(z)G1G2(z).

Now, T = 1 s and

1 − e−s

1

G

G

2

(s)

=

,

s

1

1

s

+

and by partial fraction expansion we can write

1

1

G1G2(s) = (1 − e−s )

−

.

s

s

1

From the z-transform tables we then find that

+

G1G2(z) = (1 − z−1)

z

−

z

=

0.

63

.

z

−

1

z

e

−

1

z

−

0.37

−

(i) For a unit step input,

u(z) =

z

z

−

1

and the system output response is given by

y(z) =

0.63z

.

(z

−

1)(z

−

0.37)

Using the partial fractions method, we can write

y(z)

=

A

+

B

,

z

z

−

1

z

−

0.37

where A = 1 and B = −1; thus,

z

z

y(z) =

−

.

z

−

1

z

−

0.37

160 SAMPLED DATA SYSTEMS AND THE Z-TRANSFORM

y(nT)

0.98 0.95

0.86

0.63

T

0

T

2T

3T

4T

Figure 6.24 Step input time response of Example 6.19

From the inverse z-transform tables we find that the time response is given by

y(nT ) = a − (0.37)n ,

where a is the unit step function; thus

y(nT ) = 0.63δ(t − 1) + 0.86δ(t − 2) + 0.95δ(t − 3) + 0.98δ(t − 4) + . . . .

The time response in this case is shown in Figure 6.24.

(ii) For a unit ramp input,

u(z) =

T z

(z − 1)2

and the system output response (with T = 1) is given by

y(z) =

0.63z

=

0.63z

.

(z − 1)2(z − 0.37)

z3 − 2.37z2 + 1.74z − 0.37

Using the long division method, we obtain the first few output samples as

y(z) = 0.63z−2 + 1.5z−3 + 2.45z−4 + 3.43z−5 + . . .

and the output response is given as

y(nT ) = 0.63δ(t − 2) + 1.5δ(t − 3) + 2.45δ(t − 4) + 3.43δ(t − 5) + . . . ,

as shown in Figure 6.25.

Example 6.20

The open-loop block diagram of a system with a zero-order hold is shown in Figure 6.26. Calculate and plot the system response when a step input is applied to the system, assuming that T = 1 s.

PULSE TRANSFER FUNCTION AND MANIPULATION OF BLOCK DIAGRAMS

161

y(nT)

2.45

1.5

0.63

T

2T

3T

4T

Figure 6.25 Ramp input time response of Example 6.19

G1(s)

G2(s)

u(s)

u*(s)

Z.O.H

1

y(s)

s(s + 1)

Figure 6.26 Open-loop system with zero-order hold

Solution

The transfer function of the zero-order hold is

G1(s) =

1 − e−T s

s

and that of the plant is

1

G(s) = . s(s + 1)

For this system we can write

y(s) = u*(s)G1G2(s)

and

y*(s) = u*(s)[G1G2]*(s)

or, taking z-transforms,

y(z) = u(z)G1G2(z).

Now, T = 1 s and

1 − e−s

G

G

(s)

1

2

= s2(s

+

1)

or, by partial fraction expansion,

G1G2(s) = (1 − e−s )

1

1

1

−

+

s2

s

s

+

1

162 SAMPLED DATA SYSTEMS AND THE Z-TRANSFORM

y(nT)

0.9145

0.7675

0.3678

T

0

T

2T

3T

Figure 6.27 Output response

and the z-transform is given by

− s

+ s 1

1 .

G1G2(z) = (1 − z−1)Z s2

1

1

From the z-transform tables we obtain

+

G

G

(z)

(1

z−1)

z

z

z

ze−1 + 1 − 2e−1

=

−

1)2 − z

1

+ z

e

1

=

1

2

(z

−

−

−

−

(z

−

1)(z

−

e

−

1)

0.3678z + 0.2644

.

= z2 − 1.3678z + 0.3678

After long division we obtain the time response

y(nT ) = 0.3678δ(t − 1) + 0.7675δ(t − 2) + 0.9145δ(t − 3) + . . . ,

shown in Figure 6.27.

6.3.3 Closed-Loop Systems

Some examples of manipulating the closed-loop system block diagrams are given in this section.

Example 6.21

The block diagram of a closed-loop sampled data system is shown in Figure 6.28. Derive an expression for the transfer function of the system.

Solution

For the system in Figure 6.28 we can write

e(s) = r(s) − H (s)y(s)

(6.38)

and

y(s) = e*(s)G(s).

(6.39)

PULSE TRANSFER FUNCTION AND MANIPULATION OF BLOCK DIAGRAMS

163

r(s)

e(s)

e*(s)

y(s)

G(s)

+ −

H(s)

Figure 6.28 Closed-loop sampled data system

Substituting (6.39) into (6.38),

e(s) = r(s) − G(s)H (s)e*(s)

or

e*(s) = r*(s) − GH*(s)e*(s) and, solving for e*(s), we obtain

e*(s) =

r*(s)

1 + GH*(s)

and, from (6.39),

y(s) = G(s)

r*(s)

.

1 + GH*(s)

The sampled output is then

y*(s) =

r*(s)G*(s)

1 + GH*(s)

Writing (6.43) in z-transform format,

y(z) =

r(z)G(z)

1

+

GH(z)

and the transfer function is given by

y(z)

G(z)

=

.

r(z)

1

+

GH(z)

Example 6.22

(6.40)

(6.41)

(6.42)

(6.43)

(6.44)

(6.45)

The block diagram of a closed-loop sampled data system is shown in Figure 6.29. Derive an expression for the output function of the system.

Solution

For the system in Figure 6.29 we can write

y(s) = e(s)G(s)

(6.46)

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