Feedback With Disturbances

The questions below are due on Friday September 15, 2023; 09:59:00 AM.
 
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Consider a feedback sytem in which v is being controlled so that it tracks w.

w[n] = \begin{cases} 0 & \text{if } n \leq 5, \\ 1 & \text{if } n > 5 \end{cases}

We will add a disturbance term x.

A model for the feedback system with the disturbance is given by the difference equation:

v[n] = v[n-1] + \Delta T\left(K_p e[n-1] + x[n-1] \right)

where where \Delta T = 0.01, e[n] = w[n] - v[n], and x[n] = 2 for all n>5, or written another way:

x[n] = \begin{cases} 0 & \text{if } n \leq 5, \\ 2 & \text{if } n > 5 \end{cases}

What is the smallest K_p > 0 such that the \lim_{ n\to\infty} |e[n]| \le 0.1?

Smallest K_p > 0

Suppose K_p is greater than your answer to the previous question. What is the largest K_{bound} > 0 so that if K_p < K_{bound}, \lim_{ n\to\infty} |e[n]|\le 0.1 for any initial value of v.

K_{bound}=

What is the smallest e_{bound} > 0 such that the \lim_{ n\to\infty} |e[n]| < e_{bound}.?

e_{bound} =

Notice the careful use of less-than rather than less-than-or-equal-to in the above questions. They are used to avoid a boundary case, when a natural frequency of a difference equation has a magnitude of exactly one (|\lambda| = 1). In that case, errors typically accumulate rather than decaying away, but do not grow exponentially fast.

Suppose we want the difference equation to achieve steady-state as quickly as possible? In that case, we might try to pick K_p so that \lambda = 0. For such a K_p, what is \lim_{ n\to\infty} |e[n]|?

\lambda = 0 case steady state error:

Notice that picking K_p to achieve the fastest response does NOT minimize the steady-state error.