DIGITAL COMPENSATION OF SAMPLED-DATA SYSTEMS 185 
mand pulse and the second over-all transfer function coefficient, ke. 
Substituting numerical values, 
ke + 0.552 = (1.5)(0.767) + (e1)(0.368) 
Simplifying, 
Also, from the expression for K(z), ki is (b1 — 1), which has already 
been found to be equal to 0.552. From this same relationship, ky is 
seen to be equal to bi, which means that 
= (= y= 01552 
and bi — ke 
Solving, 
ke = 0.448 
This value of k, is required, where the first term of K(z) is equal to 
0.552271, if the system is to respond to a unit step input without error. 
Thus, from the expression relating k. and ei, 
ay = — (0.407 
Since this value is less than the maximum allowable value of 1.5, it is a 
feasible term and the value of k. is acceptable. Thus, K(z) is given by 
K(z) = 0.5522! + 0.4482-? 
With this value of K(z), the first command input to the plant is a pulse 
whose magnitude is 1.5 and the second a pulse of magnitude 0.407. 
This assumes, of course, that the input is a unit step function. If the 
input step is higher, the command signal to the plant will be higher in 
proportion and the design is not valid. In other words, the design is 
based on the largest input expected. 
The command sequence applied to the plant when a unit step is 
applied to the input can be found by applying (7.65), using the K(z) 
found in the example. Thus, 
K (@) 
ie) a— G(2) R(z) 

which becomes 
=i fr, pat 
ae) = (0.552 + 0.4482-1)(1 0.3682—!) 
0.368 + 0.26427 

Simplifying this expression, 
(d + 0.8121) — 0.3682) 
Reins 1 + 0.7162 

Inverting this pulse transform by long division, the command seauence 
