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📄 rootlocusexamples .mht

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style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
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href=3D"javascript:history.go(-1)"><IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif">=20
Back <IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif"></A></P>
<HR>

<H2><A name=3DRuleReal></A>Locus on Real Axis</H2>
<P=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"><IMG=20
alt=3DRLAx=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Root_Locus/Exam=
ple1/RLRealAxis.png"></P>
<P>The root locus exists on real axis to left of an odd number of poles =
and=20
zeros of open loop transfer function, G(s)H(s), that are on the real =
axis.=20
&nbsp; These real pole and zero locations are highlighted on diagram, =
along with=20
the portion of the locus that exists on the real axis.</P>
<P>Root locus exists on real axis between:<BR>0 and -3<BR><BR>... =
because on the=20
real axis, we have 2 poles at s =3D -3, 0, and we have no zeros.<BR></P>
<HR>

<P=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"><A=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
href=3D"javascript:history.go(-1)"><IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif">=20
Back <IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif"></A></P>
<HR>

<H2><A name=3DRuleInf></A>Asymptotes as |s| goes to infinity</H2>
<P=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"><IMG=20
alt=3DRLAsym=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Root_Locus/Exam=
ple1/RLAsymptotes.png"></P>
<P>In the open loop transfer function, G(s)H(s), we have n=3D2 finite =
poles, and=20
m=3D0 finite zeros, therefore we have q=3Dn-m=3D2 zeros at =
infinity.<BR><BR>Angle of=20
asymptotes at odd multiples of =B1180=B0/q, (i.e., =
=B190=B0)<BR><BR>There exists 2 poles=20
at s =3D 0, -3, ...so sum of poles=3D-3.<BR>There exists 0 zeros, ...so =
sum of=20
zeros=3D0.<BR>(Any imaginary components of poles and zeros cancel when =
summed=20
because they appear as complex conjugate pairs.)<BR><BR>Intersect of =
asymptotes=20
is at ((sum of poles)-(sum of zeros))/q =3D -1.5.<BR>Intersect is at =
((-3)-(0))/2=20
=3D -3/2 =3D -1.5 (highlighted by five pointed star).<BR></P>
<HR>

<P=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"><A=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
href=3D"javascript:history.go(-1)"><IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif">=20
Back <IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif"></A></P>
<HR>

<H2><A name=3DRuleBrk></A>Break-Out and In Points on Real Axis</H2>
<P=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"><IMG=20
alt=3DRLBOI=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Root_Locus/Exam=
ple1/RLBreakOutIn.png"></P>
<P>Break Out (or Break In) points occur where=20
N(s)D&amp;apos;(s)-N&amp;apos;(s)D(s)=3D0, or 2 s + 3 =3D 0. (details =
below*)=20
<BR><BR>This polynomial has 1 root at s =3D -1.5.<BR><BR>From these 1 =
root, there=20
exists 1 real root at s =3D -1.5.&nbsp; These are highlighted on the =
diagram above=20
(with squares or diamonds.)<BR><BR>These roots are all on the locus =
(i.e.,=20
K&gt;0), and are highlighted with squares.<BR><BR>* N(s) and D(s) are =
numerator=20
and denominator polynomials of G(s)H(s), and the tick mark, &amp;apos;, =
denotes=20
differentiation.<BR>N(s) =3D 1<BR>N&amp;apos;(s) =3D 0<BR>D(s)=3D =
s<SUP>2</SUP> + 3=20
s<BR>D&amp;apos;(s)=3D 2 s + 3<BR>N(s)D&amp;apos;(s)=3D 2 s +=20
3<BR>N&amp;apos;(s)D(s)=3D 0<BR>N(s)D&amp;apos;(s)-N&amp;apos;(s)D(s)=3D =
2 s +=20
3<BR><BR>Here we used N(s)D&amp;apos;(s)-N&amp;apos;(s)D(s)=3D0, but we =
could=20
multiply by -1 and use =
N&amp;apos;(s)D(s)-N(s)D&amp;apos;(s)=3D0.<BR></P>
<HR>

<P=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"><A=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
href=3D"javascript:history.go(-1)"><IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif">=20
Back <IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif"></A></P>
<HR>

<H2><A name=3DRuleDep></A>Angle of Departure</H2>
<P>No complex poles in loop gain, so no angles of departure.<BR></P>
<HR>

<P=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"><A=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
href=3D"javascript:history.go(-1)"><IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif">=20
Back <IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif"></A></P>
<HR>

<H2><A name=3DRuleArv></A>Angle of Arrival</H2>
<P>No complex zeros in loop gain, so no angles of arrival.<BR></P>
<HR>

<P=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"><A=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
href=3D"javascript:history.go(-1)"><IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif">=20
Back <IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif"></A></P>
<HR>

<H2><A name=3DRuleImag></A>Cross Imag. Axis</H2>
<P=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"><IMG=20
alt=3DRLImag=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Root_Locus/Exam=
ple1/RLCrossImag.png"></P>
<P>Locus crosses imaginary axis at 1 value of K.&nbsp; These values are =
normally=20
determined by using Routh&amp;apos;s method.&nbsp; This program does it=20
numerically, and so is only an estimate.<BR><BR>Locus crosses where K =
=3D 0,=20
corresponding to crossing imaginary axis at s=3D0.<BR><BR>These =
crossings are=20
shown on plot.<BR></P>
<HR>

<P=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"><A=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
href=3D"javascript:history.go(-1)"><IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif">=20
Back <IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif"></A></P>
<HR>

<H2><A name=3DRuleFindPole></A>Changing K Changes Closed Loop Poles</H2>
<P=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"><IMG=20
alt=3DRLFR=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Root_Locus/Exam=
ple1/RLLocPos.png"></P>
<P>Characteristic Equation is 1+KG(s)H(s)=3D0, or =
1+KN(s)/D(s)=3D0,<BR>or D(s)+KN(s)=20
=3D s<SUP>2</SUP> + 3 s+ K( 1 ) =3D 0<BR><BR>So, by choosing K we =
determine the=20
characteristic equation whose roots are the closed loop =
poles.<BR><BR>For=20
example with K=3D2.25225, then the characteristic equation =
is<BR>D(s)+KN(s) =3D=20
s<SUP>2</SUP> + 3 s + 2.2522( 1 ) =3D 0, or<BR>s<SUP>2</SUP> + 3 s + =
2.2522=3D=20
0<BR><BR>This equation has 2 roots at s =3D -1.5=B10.047j.&nbsp; These =
are shown by=20
the large dots on the root locus plot<BR></P>
<HR>

<P=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"><A=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
href=3D"javascript:history.go(-1)"><IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif">=20
Back <IMG=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif"></A></P>
<HR>

<H2><A name=3DRuleFindK></A>Choose Pole Location and Find K</H2>
<P=20
style=3D"MARGIN: 0px; COLOR: red; BORDER-TOP-STYLE: none; =
BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"><IMG=20
12345

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