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Two-minute miniquiz problem

Problem 9-1

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

[ H int: the polynomial s 2 + 101 s + 1100 ( s + 88 . 5 ) ( s + 12 . 5 ) ] size 12{ \[ H"int:" matrix { {} # {}} ital "the" matrix { {} # {}} ital "polynomial" matrix { {} # {}} s rSup { size 8{2} } +"101"s+"1100" approx \( s+"88" "." 5 \) \( s+"12" "." 5 \) \] } {}

Solution

Substituting K = 1000 into the system function yields

Y ( s ) = X ( s ) H ( s ) = 1 s 1000 ( s + 1 ) ( s + 100 ) + 1000 size 12{Y \( s \) =X \( s \) H \( s \) = left [ { {1} over {s} } right ] left [ { {"1000"} over { \( s+1 \) \( s+"100" \) +"1000"} } right ]} {}

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

Y ( s ) = 1000 s ( s + 88 . 5 ) ( s + 12 . 5 ) = 1000 / ( 88 . 5 ) ( 12 . 5 ) s + 100 / ( 88 . 5 ) ( 76 ) s + 88 . 5 + 1000 / ( 12 . 5 ) ( 76 ) s + 12 . 5 = 0 . 9 s + 0 . 15 s + 88 . 5 + 1 . 05 s + 12 . 5 alignl { stack { size 12{Y \( s \) = { {"1000"} over {s \( s+"88" "." 5 \) \( s+"12" "." 5 \) } } } {} #size 12{ matrix { {} # {}} = { {"1000"/ \( "88" "." 5 \) \( "12" "." 5 \) } over {s} } + { {"100"/ \( - "88" "." 5 \) \( - "76" \) } over {s+"88" "." 5} } + { {"1000"/ \( - "12" "." 5 \) \( "76" \) } over {s+"12" "." 5} } } {} # size 12{ matrix {{} # {} } = { {0 "." 9} over {s} } + { {0 "." "15"} over {s+"88" "." 5} } + { {1 "." "05"} over {s+"12" "." 5} } } {}} } {}

The step response is

y ( t ) = ( 0 . 9 + 0 . 15 e 88 . 5t 1 . 05 e 12 . 5t ) u ( t ) size 12{y \( t \) = \( 0 "." 9+0 "." "15"e rSup { size 8{ - "88" "." 5t} } - 1 "." "05"e rSup { size 8{ - "12" "." 5t} } \) 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 ) = K s ( s + 1 ) ( s + 100 ) 1 + K s ( s + 1 ) ( s + 100 ) = K s ( s + 1 ) ( s + 100 ) + K size 12{H \( s \) = { { { {K} over {s \( s+1 \) \( s+"100" \) } } } over {1+ { {K} over {s \( s+1 \) \( s+"100" \) } } } } = { {K} over {s \( s+1 \) \( s+"100" \) +K} } } {}

The steady-state response to a unit step is simply the response to the complex exponential x ( t ) = 1 . e 0 . t = 1 size 12{x \( t \) =" 1" "." e rSup { size 8{0 "." t} } =" 1"} {} which is y ( t ) = 1 . H ( 0 ) . e 0 . t = 1 size 12{y \( t \) =" 1" "." H \( 0 \) "." e rSup { size 8{0 "." t} } =" 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 ) = K ( s ) 1 + β ( s ) K ( s ) size 12{H \( s \) = { {K \( s \) } over {1+β \( s \) 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.

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Source:  OpenStax, Signals and systems. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10803/1.1
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