📄 bodeexamples.mht
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href=3D"mailto:erik_cheever@swarthmore.edu?subject=3DLinear Physical =
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</P><A href=3D"http://www.swarthmore.edu/NatSci/echeeve1">Erik=20
Cheever</A> <A=20
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Engineering</A> <A=20
href=3D"http://www.swarthmore.edu/">Swarthmore College</A> =
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<HR class=3DfullWidth>
<A name=3Dex6></A><!--webbot bot=3D"Include" =
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<DIV id=3DoverDiv=20
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<H1>Example 6</H1>
<HR>
<P class=3Dequation><A href=3D"javascript:history.go(-1)"><IMG =
alt=3DBack=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif">=20
Back <IMG alt=3DBack=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif"></A></P>
<HR>
<P>Draw the Bode Diagram for the transfer function:</P>
<P class=3Dequation><IMG height=3D42 alt=3D"Transfer Function"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/Example6/i=
ndex.10.gif"=20
width=3D119></P>
<H2>Step 1: Rewrite the transfer function in proper =
form. </H2>
<BLOCKQUOTE>
<P>Make both the lowest order term in the numerator and denominator=20
unity. The numerator is an order 2 polynomial, the denominator =
is order=20
3.=20
<P class=3Dequation><IMG height=3D175 alt=3D"Rewritten Transfer =
Function"=20
=
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/Example6/i=
ndex.11.gif"=20
width=3D271> </P></BLOCKQUOTE>
<H2>Step 2: Separate the transfer function into its constituent =
parts.</H2>
<BLOCKQUOTE>
<P>The transfer function has 4 components:</P>
<UL>
<LI>
<P>A constant of 1 </P>
<LI>
<P>A pole at s=3D-100 </P>
<LI>
<P>A repeated pole at the origin (s=3D0) </P>
<LI>
<P>Complex conjugate zeros at the roots of =
s<SUP>2</SUP>+s+25, <BR>with=20
<IMG height=3D25 alt=3D"Transfer Function"=20
=
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/Example6/i=
ndex.12.gif"=20
width=3D143> </P></LI></UL></BLOCKQUOTE>
<H2>Step 3: Draw the Bode diagram for each part.</H2>
<BLOCKQUOTE>
<P>This is done in the diagram below.</P>
<UL>
<LI>
<P>The constant is the cyan line (A quantity of 1 is equal to 0 =
dB). =20
The phase is constant at 0 degrees. </P>
<LI>
<P>The pole at 100 rad/sec is the green line. It is 0 dB up to =
the=20
break frequency, then falls with a slope of -20 dB/dec. The =
phase is 0=20
degrees up to 1/10 the break frequency then falls linearly to -90 =
degrees at=20
10 times the break frequency. </P>
<LI>
<P>The repeated poles at the origin are shown with the blue =
line. The=20
slope is -40 dB/decade (because pole is repeated), and goes through =
0 dB at=20
1 rad/sec. The slope is -180 degrees (again because of double =
pole).=20
</P>
<LI>
<P>The complex zero is shown by the red line. The zeros give a =
dip in=20
the magnitude plot=20
=
of <BR> &=
nbsp; =20
<IMG height=3D58 alt=3D"Peak height"=20
=
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/Example6/i=
ndex.13.gif"=20
width=3D316><BR>at a frequency of 5 rad/sec. This is shown by =
the red=20
circle. The phase goes from the low frequency asymptote (0 =
degrees)=20
=
at<BR> &=
nbsp; =20
<IMG height=3D86 alt=3D"Phase Transition, Low"=20
=
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/Example6/i=
ndex.14.gif"=20
width=3D218><BR>to the high frequency asymptote=20
=
at<BR> &=
nbsp; =20
<IMG height=3D89 alt=3D"Phase Transition High"=20
=
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/Example6/i=
ndex.15.gif"=20
width=3D226> </P></LI></UL></BLOCKQUOTE>
<H2>Step 4: Draw the overall Bode diagram by adding up the results =
from=20
step 3.</H2>
<BLOCKQUOTE>
<P>The exact response is the black line. </P></BLOCKQUOTE>
<P class=3Dfigure><IMG alt=3D"Bode Plot"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/Example6/i=
ndex.16.gif"></P>
<HR>
<P class=3Dequation><A href=3D"javascript:history.go(-1)"><IMG =
alt=3DBack=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif">=20
Back <IMG alt=3DBack=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/LeftArro=
w.gif"></A></P>
<HR>
<P>Please <A href=3D"mailto:erik_cheever@swarthmore.edu">contact me</A> =
with any=20
comments, positive or negative. <A=20
href=3D"http://www.swarthmore.edu/NatSci/echeeve1">Erik Cheever's home=20
page.</A></P>
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<P class=3Dequation>=A9 Copyright 2005-2007 <A=20
href=3D"mailto:erik_cheever@swarthmore.edu?subject=3DLinear Physical =
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Cheever</A> <A=20
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<P class=3Dequation>=A9 Copyright 2005-2007 <A=20
href=3D"mailto:erik_cheever@swarthmore.edu?subject=3DLinear Physical =
System Analysis (Copyright notice)">Erik=20
Cheever</A> This page may be freely used for =
educational=20
purposes.<BR></P>
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Engineering</A> <A=20
href=3D"http://www.swarthmore.edu/">Swarthmore College</A> </DIV><!-- =
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