(20 pts) Assuming that the sampling period, T, is .0005 sec., determine
the sampled equivalent (z-form), Gc,eq[z]. Use
the simplest algorithm for conversion.6. In the first problem, you encountered a compensator with this transfer function.
Gc(s) = (.02s + 1)/(.002s + 1)
We make a substitution for s in the compensator. The substitution is:
s <=> (z - 1)/T
If you make that substitution and simplify, the first thing you get is:
Gc[z]
= [.02(z- 1) + T]/[.002(z - 1) + T]
or
Gc[z]
= [.02(z- 1) + .0005]/[.002(z - 1) + .0005]
Gc[z]
= [40(z- 1) + 1]/[4(z - 1) + 1]
Gc[z]
= [40z- 39]/[4z - 3]
Gc[z]
= 10[z- 0.975]/[z - 0.75]
or, one last acceptable
form
Gc[z]
= [.02z- .0195/[.002z - .0015]
Next time, need to make it a requirement that the answer be in the form of a ratio of polynomials.
If the4y used s<=>(z - 1)/(Tz), then the answer is: