rootlocusexamples .mht

以Matlab为应用背景

axis.   These real pole and zero locations are highlighted on =
diagram,=20
along with the portion of the locus that exists on the real =
axis.<BR><BR>Root=20
locus exists on real axis between:<BR>0 and -2<BR>-3 and negative=20
infinity<BR><BR>... because on the real axis, we have 3 poles at s =3D =
-2, -3, 0,=20
and we have no zeros.<BR></P>
<HR>

<P=20
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src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
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<HR>

<H2><A name=3DRuleInf></A>Asymptotes as |s| goes to infinity</H2>
<P=20
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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=
ple2/RLAsymptotes.png"></P>
<P>In the open loop transfer function, G(s)H(s), we have n=3D3 finite =
poles, and=20
m=3D0 finite zeros, therefore we have q=3Dn-m=3D3 zeros at =
infinity.<BR><BR>Angle of=20
asymptotes at odd multiples of =B1180=B0/q, (i.e., =B160=B0<BR>, =
=B1180=B0)<BR><BR>There=20
exists 3 poles at s =3D 0, -3, -2, ...so sum of poles=3D-5.<BR>There =
exists 0 zeros,=20
...so sum of zeros=3D0.<BR>(Any imaginary components of poles and zeros =
cancel=20
when summed because they appear as complex conjugate =
pairs.)<BR><BR>Intersect of=20
asymptotes is at ((sum of poles)-(sum of zeros))/q =3D =
-1.67.<BR>Intersect is at=20
((-5)-(0))/3 =3D -5/3 =3D -1.67 (highlighted by five pointed =
star).<BR></P>
<HR>

<P=20
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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; =
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src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
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Back <IMG=20
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<HR>

<H2><A name=3DRuleBrk></A>Break-Out and In Points on Real Axis</H2>
<P=20
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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=
ple2/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<BR>3 s<SUP>2</SUP> + 10 s =
+ 6 =3D 0.=20
(details below*) <BR><BR>This polynomial has 2 roots at s =3D -2.5,=20
-0.78.<BR><BR>From these 2 roots, there exists 2 real roots at s =3D =
-2.5,=20
-0.78.&nbsp;&nbsp; These are highlighted on the diagram above (with =
squares or=20
diamonds.)<BR><BR>Not all of these roots are on the locus. Of these 2 =
real=20
roots, there exists 1 root at s =3D -0.78 on the locus (i.e., =
K&gt;0).&nbsp;&nbsp;=20
Break-away (or break-in) points on the locus are shown by =
squares.<BR><BR>(Real=20
break-away (or break-in) with K less than 0 are shown with =
diamonds).<BR><BR>*=20
N(s) and D(s) are numerator and denominator polylnomials of G(s)H(s), =
and the=20
tick mark, &amp;apos;, denotes differentiation.<BR>N(s) =3D =
1<BR>N&amp;apos;(s) =3D=20
0<BR>D(s)=3D s<SUP>3</SUP> + 5 s<SUP>2</SUP> + 6 s<BR>D&amp;apos;(s)=3D =
3=20
s<SUP>2</SUP> + 10 s + 6<BR>N(s)D&amp;apos;(s)=3D 3 s<SUP>2</SUP> + 10 s =
+=20
6<BR>N&amp;apos;(s)D(s)=3D 0<BR>N(s)D&amp;apos;(s)-N&amp;apos;(s)D(s)=3D =
3=20
s<SUP>2</SUP> + 10 s + 6<BR><BR>Here we used=20
N(s)D&amp;apos;(s)-N&amp;apos;(s)D(s)=3D0, but we could multiply by -1 =
and use=20
N&amp;apos;(s)D(s)-N(s)D&amp;apos;(s)=3D0.<BR></P>
<HR>

<P=20
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href=3D"javascript:history.go(-1)"><IMG=20
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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
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href=3D"javascript:history.go(-1)"><IMG=20
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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
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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=
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Back <IMG=20
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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
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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=
ple2/RLCrossImag.png"></P>
<P>Locus crosses imaginary axis at 2 values of K.&nbsp;&nbsp; These =
values are=20
normally determined by using Routh&amp;apos;s method.&nbsp;&nbsp; This =
program=20
does it numerically, and so is only an estimate.<BR><BR>Locus crosses =
where K =3D=20
0, 30.2, corresponding to crossing imaginary axis at s=3D0, =B12.45j,=20
respectively.<BR><BR>These crossings are shown on plot.<BR></P>
<HR>

<P=20
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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; =
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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
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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=
ple2/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>3</SUP> + 5 s<SUP>2</SUP> + 6 s+ K( 1 ) =3D 0<BR><BR>So, by =
choosing K we=20
determine the characteristic equation whose roots are the closed loop=20
poles.<BR><BR>For example with K=3D4.00188, then the characteristic =
equation=20
is<BR>D(s)+KN(s) =3D s<SUP>3</SUP> + 5 s<SUP>2</SUP> + 6 s + 4.0019( 1 ) =
=3D 0,=20
or<BR>s<SUP>3</SUP> + 5 s<SUP>2</SUP> + 6 s + 4.0019=3D 0<BR><BR>This =
equation has=20
3 roots at s =3D -3.7, -0.67=B1 0.8j.&nbsp;&nbsp; These are shown by the =
large dots=20
on the root locus plot<BR></P>
<HR>

<P=20
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BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
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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
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BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
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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
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BORDER-RIGHT-STYLE: none; BORDER-LEFT-STYLE: none; TEXT-ALIGN: center; =
TEXT-DECORATION: none; BORDER-BOTTOM-STYLE: none"><IMG=20
alt=3DRLFG=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Root_Locus/Exam=
ple2/RLFindGain.png"></P>
<P>Characteristic Equation is 1+KG(s)H(s)=3D0, or 1+KN(s)/D(s)=3D0, =
or<BR>K =3D=20
-D(s)/N(s) =3D -( s<SUP>3</SUP> + 5 s<SUP>2</SUP> + 6 s ) / ( 1 )<BR>We =
can pick a=20
value of s on the locus, and find K=3D-D(s)/N(s).<BR><BR>For example if =
we choose=20
s=3D -0.7 + 0.84j (marked by asterisk),<BR>then D(s)=3D-4.15 + -0.222j, =
N(s)=3D 1 +=20
0j,<BR>and K=3D-D(s)/N(s)=3D 4.15 + 0.222j.<BR>This s value is not =
exactly on the=20
locus, so K is complex,<BR>(see note below), pick real part of K (=20
4.15)<BR><BR>For this K there exist 3 closed loop poles at s =3D -3.7,=20
-0.66=B10.83j.<BR><BR>Note: it is often difficult to choose a value of s =
that is=20
precisely on the locus, but we can pick a point that is =
close.&nbsp;&nbsp; If=20
the value is not exactly on the locus, then the calculated value of K =