From Continuous to Discrete

In a discrete system, we can transform the previous equation to a discrete form ($t$ is a natural number):

$$k(t)=K_{p}e(t)+K_{d}(e(t)-e(t-n))+K_{i}\sum{e(t)}$$ (9.2)

Note that the division of $n$ for the differential term is absorbed by $K_{d}$.

Why choose $n$ instead of 1? To get a differential, we just need to compare the error of time $t$ to the error at time $t-1$. This issue will be addressed later.

It should be noted that the sum term should be reset every time the reference is changed. This is because the integration of errors should be particular to a particular reference.