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w.gif"></A></P>
<HR>
<P>Draw the Bode Diagram for the transfer function:</P>
<P class=3Dequation><IMG alt=3D"Transfer Function"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/eqs/BodeHo=
Eq1.gif"></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 1 polynomial, the denominator =
is order=20
2.=20
<P class=3Dequation><IMG height=3D80 alt=3D"Rewritten Transfer =
Function"=20
=
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/Example2/i=
ndex.7.gif"=20
width=3D349></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 0.1 </P>
<LI>
<P>A pole at s=3D-10 </P>
<LI>
<P>A pole at s=3D-100 </P>
<LI>
<P>A zero at s=3D-1 </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 0.1 is equal to -20=20
dB). The phase is constant at 0 degrees. </P>
<LI>
<P>The pole at 10 rad/sec is the green line. It is 0 dB up to =
the=20
break frequency, then drops off with a slope of -20 dB/dec. =
The phase=20
is 0 degrees up to 1/10 the break frequency (1 rad/sec) then drops =
linearly=20
down to -90 degrees at 10 times the break frequency (100 rad/sec). =
</P>
<LI>
<P>The pole at 100 rad/sec is the blue line. It is 0 dB up to =
the=20
break frequency, then drops off with a slope of -20 dB/dec. =
The phase=20
is 0 degrees up to 1/10 the break frequency (10 rad/sec) then drops =
linearly=20
down to -90 degrees at 10 times the break frequency (1000 rad/sec). =
</P>
<LI>
<P>The zero at 1 rad/sec is the red line. It is 0 dB up to the =
break=20
frequency, then rises at 20 dB/dec. The phase is 0 degrees up =
to 1/10=20
the break frequency (0.1 rad/sec) then rises linearly to 90 =
degrees at=20
10 times the break frequency (10 rad/sec). =
</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 overall asymptotic plot is the translucent pink line, the exact =
response is the black line. </P></BLOCKQUOTE>
<P class=3Dfigure><IMG height=3D660 alt=3D"Bode Plot"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/Example2/i=
ndex.8.gif"=20
width=3D648></P>
<P> </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> &n=
bsp; =20
<A=20
href=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/Example3/=
index.html"><IMG=20
alt=3DNext=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/RightArr=
ow.gif">Example=20
3<IMG alt=3DNext=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/RightArr=
ow.gif"></A></P>
<HR>
<P> </P><!-- #EndEditable -->
<HR class=3DfullWidth>
<DIV id=3DsiteInfo>
<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>
<P class=3Dequation>
<SCRIPT type=3Dtext/javascript>
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</P><A href=3D"http://www.swarthmore.edu/NatSci/echeeve1">Erik=20
Cheever</A> <A=20
href=3D"http://www.engin.swarthmore.edu/">Department of=20
Engineering</A> <A=20
href=3D"http://www.swarthmore.edu/">Swarthmore College</A> =
</DIV><!--webbot bot=3D"Include" i-checksum=3D"64403" endspan -->
<HR class=3DfullWidth>
<A name=3Dex3></A><!--webbot bot=3D"Include" =
U-Include=3D"Example3/index.html" TAG=3D"BODY" startspan -->
<DIV id=3DoverDiv=20
style=3D"Z-INDEX: 1000; VISIBILITY: hidden; POSITION: =
absolute"></DIV><!-- #BeginEditable "MainText" -->
<H1>Example 3</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=3D39 alt=3D"Transfer Function"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/Example3/B=
odeEx3Eq.gif"=20
width=3D104></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 1 polynomial, the denominator =
is order=20
2.=20
<P class=3Dequation><IMG height=3D80 alt=3D"Rewritten Transfer =
Function"=20
=
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/Example3/i=
ndex.11.gif"=20
width=3D229> </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 33.3 </P>
<LI>
<P>A pole at s=3D-3 </P>
<LI>
<P>A pole at s=3D0 </P>
<LI>
<P>A zero at s=3D-10 </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 33.3 is equal to 30=20
dB). The phase is constant at 0 degrees. </P>
<LI>
<P>The pole at 3 rad/sec is the green line. It is 0 dB up to =
the break=20
frequency, then drops off with a slope of -20 dB/dec. The =
phase is 0=20
degrees up to 1/10 the break frequency (0.3 rad/sec) then drops =
linearly=20
down to -90 degrees at 10 times the break frequency (30 rad/sec). =
</P>
<LI>
<P>The pole at the origin. It is a straight line with a slope =
of -20=20
dB/dec. It goes through 0 dB at 1 rad/sec. The phase is =
-90=20
degrees. </P>
<LI>
<P>The zero at 10 rad/sec is the red line. It is 0 dB up to =
the break=20
frequency, then rises at 20 dB/dec. The phase is 0 degrees up =
to 1/10=20
the break frequency (1 rad/sec) then rises linearly to 90 =
degrees at=20
10 times the break frequency (100 rad/sec). =
</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 overall asymptotic plot is the translucent pink line, the exact =
response is the black line. </P></BLOCKQUOTE>
<P class=3Dfigure><IMG height=3D660 alt=3D"Bode Plot"=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/Example3/i=
ndex.12.gif"=20
width=3D648></P>
<P> </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> &n=
bsp; =20
<A=20
href=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Bode/Example4/=
index.html"><IMG=20
alt=3DNext=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/RightArr=
ow.gif">Example=20
4<IMG alt=3DNext=20
src=3D"http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/images/RightArr=
ow.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>
<P> </P><!-- #EndEditable -->
<HR class=3DfullWidth>
<DIV id=3DsiteInfo>
<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>
<P class=3Dequation>
<SCRIPT type=3Dtext/javascript>
document.write("<a =
href=3D\"mailto:erik_cheever@swarthmore.edu?subject=3D");
document.write(document.title);
document.write(" =
(Comments)\">Comments?</a> ");
document.write("<a =
href=3D\"mailto:erik_cheever@swarthmore.edu?subject=3D");
document.write(document.title);
document.write(" =
(Questions)\">Questions?</a> ");
document.write("<a =
href=3D\"mailto:erik_cheever@swarthmore.edu?subject=3D");
document.write(document.title);
document.write(" =
(Suggestions)\">Suggestions?</a> ");
document.write("<a =
href=3D\"mailto:erik_cheever@swarthmore.edu?subject=3D");
document.write(document.title);
document.write(" (Corrections)\">Corrections?</a>");
</SCRIPT>
</P><A href=3D"http://www.swarthmore.edu/NatSci/echeeve1">Erik=20
Cheever</A> <A=20
href=3D"http://www.engin.swarthmore.edu/">Department of=20
Engineering</A> <A=20
href=3D"http://www.swarthmore.edu/">Swarthmore College</A> =
</DIV><!--webbot bot=3D"Include" i-checksum=3D"22310" endspan -->
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