286 SAMPLED-DATA CONTROL SYSTEMS 
The z transform of the output C(z) is thus 
Ca(z) = G(z) R(z) 
ty 0.63227} 
~ 1 — 1,3682-1 + 0.3682-2 
This transform is inverted numerically by use of long division to yield 
the sequence 
C.(z) = 0.63221 + 0.8652-2 + 0.95023 + - - - 
A check will show that the values of the samples obtained from this 
sequence are exact, as expected. 
For purposes of comparison, the response of the system to a unit 
ramp will be obtained, using the same sampled model and interval. It 
is immediately evident that the reconstructed input will be a “‘stair- 
case’’ which differs considerably from the actual input. It is expected 
that the output from the sampled model will be in error. For this 
input, R(z) is 
gol 
DNS ey 
The z transform of the output C(z), using G(z) obtained previously, is 
Ee 0.6322-1 
PES. S72 Vey de aes 
Inverting this transform by the method of long division, 
C.(z) = 0.6322-2 + 1.52-3 + 2.442-4 + 3.422-5 4+ 4.522% + - += 
The exact solution obtained by inversion of C(s) gives the following 
values at the same sampling instants: 
C(z) = 0.3682-! + 1.1352-2 + 2.052-* + 3.002-4 + +: - 
Comparison of the approximate and exact solutions shows that the former 
is always smaller than the latter at comparable sampling instants. This 
is due to the “staircase” approximation of the ramp produced by the 
zero-order hold which, aside from ripple, produces a time delay equal to 
T/2. Advancing the approximate solution by 14 sec in this example 
improves the accuracy of the resultant points. 
The illustrative example points up the fact that the data hold which is 
used to obtain a numerical solution must be chosen with care and with 
some regard as to the character of the input function r(¢) and the transfer 
function G(s). In view of the fact that the problem at hand is to compute 
and not to implement physically, there is no reason why a desirable data 
hold which is not physically realizable should not be used. In this cate- 
