Signals and Systems - Electrical Engineering

540 C H A P T E R 9: The Z-Transform

FIGURE 9.8

Negative-feedback system with plantG(z).

+

−

x[n] e[n] y[n]

K

G(z)

w[n]

Solution

ForG(z)= 1 /( 1 −0.5z−^1 ), the Z-transform of the error signal is

E(z)=X(z)−W(z)=X(z)−KG(z)E(z)

and forX(z)= 1 /( 1 −z−^1 ),

E(z)=

X(z)

1 +KG(z)

=

1

( 1 −z−^1 )( 1 +KG(z))

The initial value of the error signal is then

e[0]=lim

z→∞

E(z)=

1

1 +K

sinceG(∞)=1.

The steady-state or final value of the error is

lim

n→∞

e[n]=lim

z→ 1

(z− 1 )E(z)=lim

z→ 1

(z− 1 )X(z)

1 +KG(z)

=lim

z→ 1

z(z− 1 )

(z− 1 )( 1 +KG(z))

=

1

1 + 2 K

sinceG( 1 )=2. If we want the steady-state error to go to zero, thenKmust be large. In that case,

the initial error is also zero.

IfG(z)= 1 /( 1 −z−^1 )(i.e., the plant is unstable), the initial value of the error function remains the

same,e[0]= 1 /( 1 +K), but the steady state error goes to zero sinceG( 1 )→∞. n

Tables 9.1 and 9.2 provide a list of one-side Z-transforms and the basic properties of the one-sided

Z-transform.