# 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.

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

Solution

Substituting K = 1000 into the system function yields

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

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

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

The step response is

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

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

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

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

The steady-state response to a unit step is simply the response to the complex exponential $x\left(t\right)=\text{1}\text{.}{e}^{0\text{.}t}=\text{1}$ which is $y\left(t\right)=\text{1}\text{.}H\left(0\right)\text{.}{e}^{0\text{.}t}=\text{1}$ . 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

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

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.

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