0.11 Lecture 12:interconnected systems and feedback  (Page 4/5)

Two-minute miniquiz problem

Problem 9-1

For K = 1000 determine the unit step response y(t) of the position control system.

$[H\text{int:}\begin{array}{cc}& \end{array}\text{the}\begin{array}{cc}& \end{array}\text{polynomial}\begin{array}{cc}& \end{array}{s}^2+\text{101}s+\text{1100}\approx (s+\text{88}\text{.}5)(s+\text{12}\text{.}5)]$

Solution

Substituting K = 1000 into the system function yields

$Y(s)=X(s)H(s)=\left[\frac{1}{s}\right]\left[\frac{\text{1000}}{(s+1)(s+\text{100})+\text{1000}}\right]$

The denominator polynomial can be factored and expanded in a partial fraction expansion as follows

$\begin{array}{}Y(s)=\frac{\text{1000}}{s(s+\text{88}\text{.}5)(s+\text{12}\text{.}5)}\\ \begin{array}{cc}& \end{array}=\frac{\text{1000}/(\text{88}\text{.}5)(\text{12}\text{.}5)}{s}+\frac{\text{100}/(-\text{88}\text{.}5)(-\text{76})}{s+\text{88}\text{.}5}+\frac{\text{1000}/(-\text{12}\text{.}5)(\text{76})}{s+\text{12}\text{.}5}\\ \begin{array}{cc}& \end{array}=\frac{0\text{.}9}{s}+\frac{0\text{.}\text{15}}{s+\text{88}\text{.}5}+\frac{1\text{.}\text{05}}{s+\text{12}\text{.}5}\end{array}$

The step response is

$y(t)=(0\text{.}9+0\text{.}\text{15}{e}^{-\text{88}\text{.}\mathrm{5t}}-1\text{.}\text{05}{e}^{-\text{12}\text{.}\mathrm{5t}})u(t)$

Note that y(∞) = 0.9 which fits with the result obtained from the steady-state analysis which gives 1000/1100 ≈ 0.9.

A plot of the step response for K = 1000 along with those for several values of K are shown next.

How does the step response change as K is increased?

• As the gain is increased, the steady-state error in position decreases.

• As the gain is increased, the step response becomes a damped oscillation. This could be disastrous in a position control system. Suppose we designed a system for doing microsurgery on the brain or the eye!

Thus, we cannot achieve an arbitrarily small position error without causing damped oscillations with this controller design.

2/ Simple position control system with zero position error

Consider a new design in which the error is integrated.

We can use Black’s formula to find H(s) as follows

$H(s)=\frac{\frac{K}{s(s+1)(s+\text{100})}}{1+\frac{K}{s(s+1)(s+\text{100})}}=\frac{K}{s(s+1)(s+\text{100})+K}$

The steady-state response to a unit step is simply the response to the complex exponential $x(t)=\text{1}\text{.}{e}^{0\text{.}t}=\text{1}$ which is $y(t)=\text{1}\text{.}H(0)\text{.}{e}^{0\text{.}t}=\text{1}$ . The position error ε = 1− 1 = 0. Hence, it appears that this position control system (with integral controller) has no position error for any value of K. So how do we pick K?

3/ Simple position control system with zero position error — step response

The step response is shown for several values of K.

• The steady-state error is zero for K<10100 (we will see how this value is determined later).

• The step response shows oscillations that are damped for K<10100 but shows oscillations whose amplitude grows exponentially for K>10100. Such a system is called unstable. When the system becomes unstable, the steady-state position error is not zero! Furthermore, a position control system that is unstable is even more disastrous than one that exhibits damped oscillations in response to a unit step.

4/ BIBO stability

There are many ways one can define stability of a system. We shall use the following. A system for which every bounded input yields a bounded output is called BIBO stable. A feedback system with closed loop system function

$H(s)=\frac{K(s)}{1+\beta (s)K(s)}$

is BIBO stable if its poles (the natural frequencies of the closed loop system) are in the left half of the s-plane. Thus, determining the conditions for which a system is stable reduces to finding whether the zeros of 1+β(s)K(s) = 0 are in the left-half s-plane. When β(s)K(s) is a rational function, this condition is tested by determining whether the roots of the numerator polynomial of 1+β(s)K(s) are located in the left-half s-plane.