306 SAMPLED-DATA CONTROL SYSTEMS 
condition. The differential equation for the system is 
a(t) = LS + Ri 
Taking the Laplace transform of both sides, there results 
E,(s) = L{sI(s) — 7(0)] + RI(s) 
where i(0) is the initial value of the current. Solving for /(s), 
E\(s) Ti(0) 
Ls -b R Ls + R 


iG = 
If the output voltage e,(¢) is desired, 
RE;\(s) zs RLi(0) 
L+kR Ls+R 
It is readily ascertained that the model shown in Fig. 11.16a has an 
over-all relationship between E,(s) and E;(s) given by this expression. 
It is seen that the only dynamical 
element is a perfect integrator, which 
can be replaced by a sampled approxi- 
mant. The initial condition 7(0) is 
inserted as an impulse applied as 
shown. 
Figure 11.166 shows the integration 
process replaced by a numerical 
equivalent using the polygonal approximation. The initial conditions 
are applied at the intermediate point, and it is noted that not only 
must the initial condition 7(0) be inserted but also 7/2 times the 
derivative of the current 2’(0). This corresponds to the quantity «(0) 
in Fig. 11.14 and Eq. (11.54), which, being the integrand in the integra- 
tion process, is also the derivative of the integral y(¢). Thus, to solve 
the problem by numerical methods with an initial condition of cur- 
rent, the initial rate change of current must also be ascertained. In 
this simple problem, this rate change is easily shown to be ez(0)/L, 
where ez,(0) is the initial value of the voltage across the inductance. 
In this problem, this initial voltage is the entire voltage e1(0). 
If, for this problem, it is assumed that the input e:(¢) is a unit step 
and the initial current 7(0) (but not the initial rate change of current) is 
zero, then the z transform of the output voltage H,(z) is 
Fed Nel ea Dred Rey oo)! 
He (3)e— es) 



Fig. 11.15. Resistance-capacitance cir- 
cuit used in example. 
2L (1 — 27})? 2h lisler 
DO 
1 OE Wea 
