The effects on the systems performance adding the open zeros or open poles
Adding a open zero in the left s-plane
For example )3)(2(
)()(
)3)(2()( 12
11
sss
asKsGH
sss
KsGH
Generally, adding a open zero in the left s-plane will lead the root loci to be bended to the left. And the more closer to the imaginary axis the open zero is,
the more prominent the effect on the systems performance is.
Re
Im
23
Re
6
Im
23
Re5.2
Im
23
Re4
Im
23
5.0a
5.0a 25.1a
5.2a
Adding a open pole in the left s-plane
For example ))(2(
)3()(
)2(
)3()( 12
11
asss
sKsGH
ss
sKsGH
Generally, adding a open pole in the left s-plane will lead the root loci to be bended to the right. And the more closer to the imaginary axis the open
pole is, the more prominent the effect on the systems performance is.
The effects on the systems performance adding the open zeros or open poles
Re
Re
23
Im
Re
Im
5
Re
ImIm
23 23
1 23
2a
0a 5.0a
Dominant poles and zeros of transfer functions
The location of the poles and zeros of a transfer function in the s-plane greatly affects the transient response of the system.
For analysis and design purposes, it is important to sort out the poles that have a dominant effect on the transient response and call these
the dominant poles.
Because most control systems in practice are of orders higher than two, it would be useful to establish guidelines on the approximation of high-order systems by lower-order ones insofar as the transient response is concerned.
In design, we can use the dominant poles to control the dynamic performance of the system, whereas the insignificant poles are used for the purpose of ensuring that the controller transfer function can be realized by physical components.
For all practical purposes, we can divide the s-plane into regions in which the dominant and insignificant poles can lie.
We intentionally do not assign specific values to the coordinates, since these are all relative to a given system.
Dominant poles and zeros of transfer functions
Im
Re
unstable
region
unstable
region
region of
dominant
poles
region of
insignificant
poles
D
The poles that are close to the
imaginary axis in the left-half
s-plane give rise to transient
responses that will decay relatively
slowly, whereas the poles that are
far away from the axis (relative to
the dominant poles) correspond to
fast-decaying time responses.
The distance D between the
dominant region and the least
significant region will be subject to
discussion. Regions of dominant and insignificant poles in the s-plane.
The question is: How large a pole is considered to be really large? It has been recognized in practice and in the literature that if the magnitude of the real part of a pole is at least 5 to 10 times that of a dominant pole or a pair of complex dominant poles, then the pole may be regarded as insignificant insofar as the transient response is concerned.
The zeros that are close to the imaginary axis in the left-half s-plane
affect the transient responses more significantly, whereas the zeros that are far away from the axis (relative to the dominant poles) have a smaller effect on the time response.
Dominant poles and zeros of transfer functions
We must point out that the regions shown in Fig. are selected merely for the definitions of dominant and insignificant poles.
For design purposes, such as in pole placement design, the dominant poles and the insignificant poles should most likely be located in the red regions.
Again, we do not show any
absolute coordinates. except
that the desired region of the
dominant poles is centered
around the line that corresponds
to = 0.707.
It should also be noted that,
while designing, we cannot place the
insignificant poles arbitrarily far to the
left in the s-plane or these may require
unrealistic system parameter values.
Im
Re
unstable
region
unstable
region
region of
dominant
poles
region of
insignificant
poles
D
450
450
Regions of dominant and insignificant poles
in the s-plane for design purpose.
Dominant poles and zeros of transfer functions
The proper way of neglecting the insignificant poles with
consideration of the steady-state response
(a) )22)(10(
20
)(
)(2
ssssR
sCThe pole at s=-10 is 10 times the
real part of the complex conjugate
poles, which are at -1 j 1 . (b)
)22)(110/(10
20
)(
)(2
ssssR
sC
Then we reason that when the absolute value of s is much smaller than 10,
because of the dominant nature of the complex poles. The term s/10 can be neglected
when compared with 1. Then, Eq.(b) is approximated by
(c) )22(10
20)(
2
sssM
the third-order system described by Eq. (a) and the second-order system
approximated by Eq. (c) all have a final value of unity when a unit-step input is applied.
On the other hand, if we simply throw away the term (s + 10) in Eq. (a), the
approximating second-order system will have a different steady-state value when a
unit-step input is applied.
110/ s
Example
Conventional root locus
system. theof ) ( the toalproportion
is and system a of locusroot plot the toparameter
variable theas )( select the weGenerally
*
*
gainloopopenK
K
gainroot locusK
ed.investigat be to
need isparameter theof locus-root then thecases,
many in variableis system theofparameter other Maybe
*no-K
We illustrate the parameter root locus and its sketching approaches by following example:
Parameter root locus --- the variable parameter of the control
systems is another parameter besides K * .
Parameter root locus
Extension of The Root Locus
. system theof locus-root sketch the ,0 from varing If
1
4
: is system a offunction transfer loopopen The
))(ss(sG(s)H(s)
:Solution
0141 041
:is system theofequation sticcharacteri
2 )s(s)(ss))(ss(s
0)(12(2
11
41
11
22
sG
)ss)(s
)s(s
)(ss
)s(s eq
Loci versus Other Parameters
Example 1
functiontransfer
loop-open equivalent The
)ss)(s
)s(ssGeq
2(2
1)(
2
The procedure of sketching root locus is shown as following:
2 ,2
7
2
1
:poles loopOpen
1 0
:zeros loopOpen (1)
32,1
21
pjp
z,z Re
Im
12
2
7
2
1j
2
7
2
1j
Loci versus Other Parameters
Example 1
Loci versus Other Parameters
4 1
1
s s
1
TK
The open-loop transfer function of the system is
which is not in the standard form as:
n
j
j
m
i
i
ps
zs
KsHsG
1
1
)(
)(
)()(
)1(
4
TKss
Example 2
The characteristic equation of the system is
042 sKss T
Now, equation is in suitable form for a root-locus study. We
need to identify open-loop transfer function, which we do
by writing the equivalent to the equation as
04
12
ss
sKT
Loci versus Other Parameters
Thus, for root-locus purposes,
the zeros are at s=0, And the poles are at -1/2+j1.94 and -1/2-j1.94.
Example 2
Zero-Degree Root Loci
The open loop transfer function
n
j
j
m
i
i
ps
zs
KsHsG
1
1
)(
)(
)()(
The characteristic equation is
0
)(
)(
1
1
1
n
j
j
m
i
i
ps
zs
K 1
)(
)(
1
1
n
j
j
m
i
i
ps
zs
K
)(sG
)(sH
1)(
)(
)()(
1
1
n
j
j
m
i
i
ps
zs
KsHsG
The magnitude and angle requirement for the
zero-degree root locus are
criterionMagnitude
ps
zs
Kn
j
j
m
i
i
1
||
||
1
1
criterionAnglekkpszsn
j
j
m
i
i ,2 ,1 ,0 2)()(11
The characteristic equation is )(sG
)(sH
Zero-Degree Root Loci
The Zero-Degree (00) Root Locus
In summary, all rules are the same, except:
All 1800s become 00s.
Odd becomes even in Rule 4.
If we substitute k2 for , )12( k
the sketching rules of the conventional root locus
are also suitable to thezero degree root locus (only related to the rule 4, 5 and 9 ).
Zero-Degree Root Loci