Page 93 - Practical Control Engineering a Guide for Engineers, Managers, and Practitioners

68 Chapter Three

The numerator in Eq. (3-40), namely,gks+ gi, has one zero. That
is, the value s =-I I k causes this term to be zero, so the zero of this
factor is -I I k.
The denominator in Eq. (3-40), namely,

2
(~s +(1 + gk)s+ gi) s
has the same form as the quadratic in Eq. (3-25) with one extra factor.
Therefore, the denominator in Eq. (3-40) has three zeros (values at
which a quantity equals zero). Conventionally, we say that Eq. (3-40)
has three poles (values at which the quantity becomes infinite) and
one zero (the value at which the quantity becomes zero).
Partial Fractions and Poles
Applying the quadratic equation solver, the poles of Eq. (3-40) are
found to be

1+ gk ~(1+ gk) -4~gi
2
----2-~-±~---2-~--~- and o.o (3-44)

Two of the roots in Eq. (3-44) are the same as those obtained in
Eq. (3-30). Assume for the time being, that the argument of the radical
in Eq. (3-44) is positive so that the poles will all be zero or negative
real numbers.
To make the following partial fraction algebra a little easier I will
factor out ~ so that the coefficient of s is unity and Eq. (3-40)
2
becomes
(gks+ gl)Sc
Y= (gks+ gi)Sc = ~
(3-45)
(1 + gk) gi) (s- s 1 )(s- s 2 )s
2
~ s +---s+- s
~ ~

The resulting quadratic equation for poles is a little different

Question 3-10 Is this expression for the poles really different from Eq. (3-30)?

Answer No, a little algebra can show that they are identical.