Page 79 - Practical Control Engineering a Guide for Engineers, Managers, and Practitioners

54 Chapter Three

Answer Use the quadratic equation root solver

If Eq. (3-27) holds then the argument of the square root will be positive and
the roots will be real. Also, the square root term will be less in magnitude than
(1 + gk) so the roots cannot be positive. As to the second question, the quadratic
equation root solver shows that if I < 0 then one of the roots would be positive
and in turn would lead to an unbounded response.

Underdamplng
Finally, consider the case when
2
(1 + gk) < 4-rgl

so, (3-29)
I> (1+ gk)2
4-rg

(Note how the integral gain is greater than that for critical damping.) The
argument inside the square root is now negative. But we know that

and "-4 = 2j

and because of the inequality in Eq. (3-29), the roots are

where a< 0 and fj > 0 are real numbers. 1his means that the solution will
have exponential terms with imaginary arguments (see App. B) as in

e(a+jfJ)t or

1
The e« term (with a < 0 ) means that the transient response will
die away, but what about the other factor? Euler's equation (see App. B)
can be useful here.
ei/JI = cos(fjt)+ jsin(fjt)

The eiflt factor implies sinusoidal or oscillatory behavior while
the eat factor decreases to zero at a rate depending on a. Both factors
promise an underdamped behavior where there are oscillations that