DIGITAL COMPENSATION OF SAMPLED-DATA SYSTEMS 165 
Since the system is to respond to a unit ramp with zero system error, 
the second condition must hold, namely, 
1 — 2-1)2(1 + b2-! Salty 
1=K@=S=* Se 

As in the example in the previous section, only a1, a2, and b; are required 
to satisfy the two preceding equations. Solving for a; and az, 
a, = 0.56 
ag = —0.405 
This yields an over-all pulse transfer function K(z) given by 
0.562—1 + 0.8952-? — 0.9482- 
Ke) = r= 08 @e3h) 

The response of this system to a ramp input and to a step input is found 
by multiplying K(z) by the respective R(z), resulting in an output 
pulse transform 
C(z) = 
for the ramp input and 
Oley = 0.562-* + 0.89527? — 0.94827% 
1 = 1.524 + 0.5272 
for the step input. 
Inversion of these z transforms gives the pulse sequences, which are 
plotted in Figs. 7.12 and 7.13, respectively. It is seen that the system 
no longer has a finite settling time at sampling instants as did the 
minimum finite-settling-time prototype. Onthe other hand, the system 
approaches a tolerable error in a reasonable time, and the ripple com- 
ponent is not severe. The response of the system to a step input 
shows that the peak overshoot is less than that for the finite-settling- 
time prototype, being approximately 175 per cent, as opposed to 225 
per cent in the minimum prototype system. The system is seen to 
approach steady state gradually, with some ripple component which is 
not as severe as that for the minimum prototype. 
Substituting in (7.5) gives the pulse transfer function of the digital 
controller, 
0.562-2 + 0.8952-3 — 0.9482-4 
1 — 2.5271 + 22-2 — 0.5273 

0.56 — 0.8182—! + 0.3742? — 0.0527’ 
1 + 0.1127! — 0.942? — 0.1512-3 (7.32) 
D(z) = 
This pulse transfer function should be compared to that of the minimum 
- prototype. The comparison shows that the various coefficients and 
signs are different but that the required number of storages in the 
digital controller are the same. This means that a system with stale- 
