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<!DOCTYPE document PUBLIC "-//CNX//DTD CNXML 0.5 plus MathML//EN" "http://cnx.rice.edu/cnxml/0.5/DTD/cnxml_mathml.dtd">
<document xmlns="http://cnx.rice.edu/cnxml" xmlns:m="http://www.w3.org/1998/Math/MathML" xmlns:md="http://cnx.rice.edu/mdml/0.4" xmlns:bib="http://bibtexml.sf.net/" id="m10110">

  <name>The Laplace Transforms</name>

  <metadata>
  <md:version>2.11</md:version>
  <md:created>2001/06/19</md:created>
  <md:revised>2003/07/09 14:05:19.358 GMT-5</md:revised>
  <md:authorlist>
      <md:author id="richb">
      <md:firstname>Richard</md:firstname>
      <md:othername>G.</md:othername>
      <md:surname>Baraniuk</md:surname>
      <md:email>richb@rice.edu</md:email>
    </md:author>
  </md:authorlist>

  <md:maintainerlist>
    <md:maintainer id="jrom">
      <md:firstname>Justin</md:firstname>
      
      <md:surname>Romberg</md:surname>
      <md:email>jrom@rice.edu</md:email>
    </md:maintainer>
    <md:maintainer id="richb">
      <md:firstname>Richard</md:firstname>
      <md:othername>G.</md:othername>
      <md:surname>Baraniuk</md:surname>
      <md:email>richb@rice.edu</md:email>
    </md:maintainer>
    <md:maintainer id="mjhaag">
      <md:firstname>Michael</md:firstname>
      
      <md:surname>Haag</md:surname>
      <md:email>mjhaag@rice.edu</md:email>
    </md:maintainer>
    <md:maintainer id="mariyah">
      <md:firstname>Mariyah</md:firstname>
      
      <md:surname>Poonawala</md:surname>
      <md:email>mariyah@rice.edu</md:email>
    </md:maintainer>
    <md:maintainer id="prash">
      <md:firstname>Prashant</md:firstname>
      
      <md:surname>Singh</md:surname>
      <md:email>prash@ece.rice.edu</md:email>
    </md:maintainer>
  </md:maintainerlist>
  
  <md:keywordlist>
    <md:keyword>continuous time</md:keyword>
    <md:keyword>laplace transform</md:keyword>
  </md:keywordlist>

  <md:abstract>Describes Laplace transforms.</md:abstract>
</metadata>

  <content>

    <para id="para1">
      The Laplace transform is a generalization of the <cnxn strength="5" document="m10098">Continuous-Time Fourier
      Transform</cnxn>.  However, instead of using <cnxn strength="5" document="m10784">complex sinusoids</cnxn> of the form
      <m:math>
	<m:apply>
	  <m:exp/>
	  <m:apply>
	    <m:times/>
	    <m:imaginaryi/>
	    <m:ci>ω</m:ci>
	    <m:ci>t</m:ci>
	  </m:apply>
	</m:apply>
      </m:math>,
      as the CTFT does, the Laplace transform uses the more general, 
      <m:math>
	<m:apply>
	  <m:exp/>
	  <m:apply>
	    <m:times/>
	    <m:ci>s</m:ci>
	    <m:ci>t</m:ci>
	  </m:apply>
	</m:apply>
      </m:math>,
      where
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>s</m:ci>
	  <m:apply>
	    <m:plus/>
	    <m:ci>σ</m:ci>
	    <m:apply>
	      <m:times/>
	      <m:imaginaryi/>
	      <m:ci>ω</m:ci>
	    </m:apply>
	  </m:apply>
	</m:apply>
      </m:math>.
    </para>

    <para id="para2">
      Although Laplace transforms are rarely solved using integration
      (<cnxn strength="5" document="m10111">tables</cnxn> and
      computers (<foreign>e.g.</foreign> Matlab) are much more
      common), we will provide the <term>bilateral Laplace transform
      pair</term> here.  These define the forward and inverse Laplace
      transformations.  Notice the similarities between the forward
      and inverse transforms.  This will give rise to many of the same
      symmetries found in <cnxn strength="5" document="m10096">Fourier
      analysis</cnxn>.

      <equation id="eqn1">
	<name>Laplace Transform</name>
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:apply>
	      <m:ci type="fn">F</m:ci>
	      <m:ci>s</m:ci>
	    </m:apply>
	    <m:apply>
	      <m:int/>
	      <m:bvar>
		<m:ci>t</m:ci>
	      </m:bvar>
	      <m:lowlimit>
		<m:apply>
		  <m:minus/>
		  <m:infinity/>
		</m:apply>
	      </m:lowlimit>
	      <m:uplimit>
		<m:infinity/>
	      </m:uplimit>
	      <m:apply>
		<m:times/>
		<m:apply>
		  <m:ci type="fn">f</m:ci>
		  <m:ci>t</m:ci>
		</m:apply>
		<m:apply>
		  <m:exp/>
		  <m:apply>
		    <m:minus/>
		    <m:apply>
		      <m:times/>
		      <m:ci>s</m:ci>
		      <m:ci>t</m:ci>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:apply>
	</m:math>
      </equation>

      <equation id="eqn2">
	<name>Inverse Laplace Transform</name>
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:apply>
	      <m:ci type="fn">f</m:ci>
	      <m:ci>t</m:ci>
	    </m:apply>
	    <m:apply>
	      <m:times/>
	      <m:apply>
		<m:divide/>
		<m:cn>1</m:cn>
		<m:apply>
		  <m:times/>
		  <m:cn>2</m:cn>
		  <m:pi/>
		  <m:imaginaryi/>
		</m:apply>
	      </m:apply>
	      <m:apply>
		<m:int/>
		<m:bvar>
		  <m:ci>s</m:ci>
		</m:bvar>
		<m:lowlimit>
		  <m:apply>
		    <m:minus/>
		    <m:ci>c</m:ci>
		    <m:apply>
		      <m:times/>
		      <m:imaginaryi/>
		      <m:infinity/>
		    </m:apply>
		  </m:apply>
		</m:lowlimit>
		<m:uplimit>
		  <m:apply>
		    <m:plus/>
		    <m:ci>c</m:ci>
		    <m:apply>
		      <m:times/>
		      <m:imaginaryi/>
		      <m:infinity/>
		    </m:apply>
		  </m:apply>
		</m:uplimit>
		<m:apply>
		  <m:times/>
		  <m:apply>
		    <m:ci type="fn">F</m:ci>
		    <m:ci>s</m:ci>
		  </m:apply>
		  <m:apply>
		    <m:exp/>
		    <m:apply>
		      <m:times/>
		      <m:ci>s</m:ci>
		      <m:ci>t</m:ci>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:apply>
	</m:math>
      </equation>
    </para>

    <section id="sect1">
      <name>Finding the Laplace and Inverse Laplace Transforms</name>
      <section id="sect1a">
	<name>Solving the Integral</name>
	<para id="para3">
	  Probably the most difficult and least used method for
	  finding the Laplace transform of a signal is solving the
	  integral.  Although it is technically possible, it is
	  extremely time consuming.  Given how easy the next two
	  methods are for finding it, we will not provide any more
	  than this.  The integrals are primarily there in order to
	  understand where the following methods originate from.
	</para>
      </section>

      <section id="sect1b">
	<name>Using a Computer</name>
	<para id="para4">
	  Using a computer to find Laplace transforms is relatively
	  painless.  Matlab has two functions,
	  <code>laplace</code> and
	  <code>ilaplace</code>, that are both part of the
	  symbolic toolbox, and will find the Laplace and inverse
	  Laplace transforms respectively.  This method is generally
	  preferred for more complicated functions.  Simpler and more
	  contrived functions are usually found easily enough by <cnxn strength="5" target="sect1c">using tables</cnxn>.
	</para>
      </section>

      <section id="sect1c">
	<name>Using Tables</name>
	<para id="para5">
	  When first learning about the Laplace transform, tables are
	  the most common means for finding it.  With enough practice,
	  the tables themselves may become unnecessary, as the common
	  transforms can become second nature.  For the purpose of
	  this section, we will focus on the inverse Laplace
	  transform, since most design applications will begin in the
	  Laplace domain and give rise to a result in the time domain.
	  The method is as follows:

	  <list id="list1" type="enumerated">
	    <item>
	      Write the function you wish to transform,
	      <m:math>
		<m:apply>
		  <m:ci type="fn">H</m:ci>
		  <m:ci>s</m:ci>
		</m:apply>
	      </m:math>,
	      as a sum of other functions,
	      <m:math>
		<m:apply>
		  <m:eq/>
		  <m:apply>
		    <m:ci type="fn">H</m:ci>
		    <m:ci>s</m:ci>
		  </m:apply>
		  <m:apply>
		    <m:sum/>
		    <m:bvar>
		      <m:ci>i</m:ci>
		    </m:bvar>
		    <m:lowlimit>
		      <m:cn>1</m:cn>
		    </m:lowlimit>
		    <m:uplimit>
		      <m:ci>m</m:ci>
		    </m:uplimit>
		    <m:apply>
		      <m:ci type="fn"><m:msub>
			  <m:mi>H</m:mi>
			  <m:mi>i</m:mi>
			</m:msub></m:ci>
		      <m:ci>s</m:ci>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:math>
	      where each of the 
	      <m:math>
		<m:ci><m:msub>
		    <m:mi>H</m:mi>
		    <m:mi>i</m:mi>
		  </m:msub></m:ci>
	      </m:math>
	      is known from a <cnxn strength="5" document="m10111">table</cnxn>.
	    </item>

	    <item>
	      Invert each
	      <m:math>
		<m:apply>
		  <m:ci type="fn"><m:msub>
		      <m:mi>H</m:mi>
		      <m:mi>i</m:mi>
		    </m:msub></m:ci>
		  <m:ci>s</m:ci>
		</m:apply>
	      </m:math>
	      to get its
	      <m:math>
		<m:apply>
		  <m:ci type="fn"><m:msub>
		      <m:mi>h</m:mi>
		      <m:mi>i</m:mi>
		    </m:msub></m:ci>
		  <m:ci>t</m:ci>
		</m:apply>
	      </m:math>.
	    </item>

	    <item>
	      Sum up the
	      <m:math>
		<m:apply>
		  <m:ci type="fn"><m:msub>
		      <m:mi>h</m:mi>
		      <m:mi>i</m:mi>
		    </m:msub></m:ci>
		  <m:ci>t</m:ci>
		</m:apply>
	      </m:math>
	      to get
	      <m:math>
		<m:apply>
		  <m:eq/>
		  <m:apply>
		    <m:ci type="fn">h</m:ci>
		    <m:ci>t</m:ci>
		  </m:apply>
		  <m:apply>
		    <m:sum/>
		    <m:bvar>
		      <m:ci>i</m:ci>
		    </m:bvar>
		    <m:lowlimit>
		      <m:cn>1</m:cn>
		    </m:lowlimit>
		    <m:uplimit>
		      <m:ci>m</m:ci>
		    </m:uplimit>
		    <m:apply>
		      <m:ci type="fn"><m:msub>
			  <m:mi>h</m:mi>
			  <m:mi>i</m:mi>
			</m:msub></m:ci>
		      <m:ci>t</m:ci>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:math>
	    </item>
	  </list>
	</para>

	<example id="exa1">
	  <para id="para6">
	    Compute
	    <m:math>
	      <m:apply>
		<m:ci type="fn">h</m:ci>
		<m:ci>t</m:ci>
	      </m:apply>
	    </m:math>
	    for
	    <m:math>
	      <m:apply>
		<m:forall/>
		<m:bvar><m:ci>s</m:ci></m:bvar>
		<m:condition>
		  <m:apply>
		    <m:gt/>
		    <m:apply>
		      <m:real/>
		      <m:ci>s</m:ci>
		    </m:apply>
		    <m:cn>-5</m:cn>
		  </m:apply>
		</m:condition>
		<m:apply>
		  <m:eq/>
		  <m:apply>
		    <m:ci type="fn">H</m:ci>
		    <m:ci>s</m:ci>
		  </m:apply>
		  <m:apply>
		    <m:divide/>
		    <m:cn>1</m:cn>
		    <m:apply>
		      <m:plus/>
		      <m:ci>s</m:ci>
		      <m:cn>5</m:cn>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:math>
	  </para>

	  <para id="para7">
	    This can be solved directly from the <cnxn strength="5" document="m10111">table</cnxn> to be
	    <m:math>
	      <m:apply>
		<m:eq/>
		<m:apply>
		  <m:ci type="fn">h</m:ci>
		  <m:ci>t</m:ci>
		</m:apply>
		<m:apply>
		  <m:exp/>
		  <m:apply>
		    <m:minus/>
		    <m:apply>
		      <m:times/>
		      <m:cn>5</m:cn>
		      <m:ci>t</m:ci>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:math>
	  </para>
	</example>

	<example id="exa2">
	  <para id="para8">
	    Find the time domain representation,
	    <m:math>
	      <m:apply>
		<m:ci type="fn">h</m:ci>
		<m:ci>t</m:ci>
	      </m:apply>
	    </m:math>,
	    of
	    <m:math>
	      <m:apply>
		<m:forall/>
		<m:bvar><m:ci>s</m:ci></m:bvar>
		<m:condition>
		  <m:apply>
		    <m:gt/>
		    <m:apply>
		      <m:real/>
		      <m:ci>s</m:ci>
		    </m:apply>
		    <m:cn>-10</m:cn>
		  </m:apply>
		</m:condition>
		<m:apply>
		  <m:eq/>
		  <m:apply>
		    <m:ci type="fn">H</m:ci>
		    <m:ci>s</m:ci>
		  </m:apply>
		  <m:apply>
		    <m:divide/>
		    <m:cn>25</m:cn>
		    <m:apply>
		      <m:plus/>
		      <m:ci>s</m:ci>
		      <m:cn>10</m:cn>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:math> 
	  </para>

	  <para id="para9">
	    To solve this, we first notice that
	    <m:math>
	      <m:apply>
		<m:ci type="fn">H</m:ci>
		<m:ci>s</m:ci>
	      </m:apply>
	    </m:math>
	    can also be written as
	    <m:math>
	      <m:apply>
		<m:times/>
		<m:cn>25</m:cn>
		<m:apply>
		  <m:divide/>
		  <m:cn>1</m:cn>
		  <m:apply>
		    <m:plus/>
		    <m:ci>s</m:ci>
		    <m:cn>10</m:cn>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:math>.  We can then go to the <cnxn strength="5" document="m10111">table</cnxn> to find
	    <m:math>
	      <m:apply>
		<m:eq/>
		<m:apply>
		  <m:ci type="fn">h</m:ci>
		  <m:ci>t</m:ci>
		</m:apply>
		<m:apply>
		  <m:times/>
		  <m:cn>25</m:cn>
		  <m:apply>
		    <m:exp/>
		    <m:apply>
		      <m:minus/>
		      <m:apply>
			<m:times/>
			<m:cn>10</m:cn>
			<m:ci>t</m:ci>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:math>
	  </para>
	</example>

	<example id="exa3">
	  <para id="para10">
	    We can now extend the two previous examples by finding
	    <m:math>
	      <m:apply>
		<m:ci type="fn">h</m:ci>
		<m:ci>t</m:ci>
	      </m:apply>
	    </m:math>
	    for
	    <m:math>
	      <m:apply>
		<m:forall/>
		<m:bvar><m:ci>s</m:ci></m:bvar>
		<m:condition>
		  <m:apply>
		    <m:gt/>
		    <m:apply>
		      <m:real/>
		      <m:ci>s</m:ci>
		    </m:apply>
		    <m:apply>
		      <m:minus/>
		      <m:cn>5</m:cn>
		    </m:apply>
		  </m:apply>
		</m:condition>
		<m:apply>
		  <m:eq/>
		  <m:apply>
		    <m:ci type="fn">H</m:ci>
		    <m:ci>s</m:ci>
		  </m:apply>
		  <m:apply>
		    <m:plus/>
		    <m:apply>
		      <m:divide/>
		      <m:cn>1</m:cn>
		      <m:apply>
			<m:plus/>
			<m:ci>s</m:ci>
			<m:cn>5</m:cn>
		      </m:apply>
		    </m:apply>
		    <m:apply>
		      <m:divide/>
		      <m:cn>25</m:cn>
		      <m:apply>
			<m:plus/>
			<m:ci>s</m:ci>
			<m:cn>10</m:cn>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:math>
	  </para>

	  <para id="para11">
	    To do this, we take advantage of the additive property of
	    linearity and the three-step method described above to
	    yield the result
	    <m:math>
	      <m:apply>
		<m:eq/>
		<m:apply>
		  <m:ci type="fn">h</m:ci>
		  <m:ci>t</m:ci>
		</m:apply>
		<m:apply>
		  <m:plus/>
		  <m:apply>
		    <m:exp/>
		    <m:apply>
		      <m:minus/>
		      <m:apply>
			<m:times/>
			<m:cn>5</m:cn>
			<m:ci>t</m:ci>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		  <m:apply>
		    <m:times/>
		    <m:cn>25</m:cn>
		    <m:apply>
		      <m:exp/>
		      <m:apply>
			<m:minus/>
			<m:apply>
			  <m:times/>
			  <m:cn>10</m:cn>
			  <m:ci>t</m:ci>
			</m:apply>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:math>
	  </para>
	</example>

	<para id="para12">
	  For more complicated examples, it may be more difficult to
	  break up the transfer function into parts that exist in a
	  table.  In this case, it is often necessary to use <cnxn strength="5" document="m2111">partial fraction
	  expansion</cnxn> to get the transfer function into a more
	  usable form.
	</para>
      </section>
    </section>

    <section id="sect2">
      <name>Visualizing the Laplace Transform</name>
      <para id="para13">
	With the Fourier transform, we had a <term>complex-valued
	function</term> of a <emphasis>purely imaginary
	variable</emphasis>,
	<m:math>
	  <m:apply>
	    <m:ci type="fn">F</m:ci>
	    <m:apply>
	      <m:times/>
	      <m:imaginaryi/>
	      <m:ci>ω</m:ci>
	    </m:apply>
	  </m:apply>
	</m:math>.  This was something we could envision with two
	2-dimensional plots (real and imaginary parts or magnitude and
	phase).  However, with Laplace, we have a <term>complex-valued
	function</term> of a <emphasis>complex variable</emphasis>.
	In order to examine the magnitude and phase or real and
	imaginary parts of this function, we must examine
	3-dimensional surface plots of each component.
      </para>

      <figure id="fig1" orient="horizontal">
	<name>real and imaginary sample plots</name>
	<subfigure id="fig1a">
	  <media type="image/png" src="laplace1.png"/>
	  <caption>
	    The Real part of
	    <m:math>
	      <m:apply>
		<m:ci type="fn">H</m:ci>
		<m:ci>s</m:ci>
	      </m:apply>
	    </m:math>
	  </caption>
	</subfigure>
	<subfigure id="fig1b">
	  <media type="image/png" src="laplace2.png"/>
	  <caption>
	    The Imaginary part of
	    <m:math>
	      <m:apply>
		<m:ci type="fn">H</m:ci>
		<m:ci>s</m:ci>
	      </m:apply>
	    </m:math>
	  </caption>
	</subfigure>
	<caption>
	  Real and imaginary parts of 
	  <m:math>
	    <m:apply>
	      <m:ci type="fn">H</m:ci>
	      <m:ci>s</m:ci>
	    </m:apply>
	  </m:math>
	  are now each 3-dimensional surfaces.
	</caption>
      </figure>

      <figure id="fig2" orient="horizontal">
	<name>magnitude and phase sample plots</name>
	<subfigure id="fig2a">
	  <media type="image/png" src="laplace3.png"/>
	  <caption>
	    The Magnitude of
	    <m:math>
	      <m:apply>
		<m:ci type="fn">H</m:ci>
		<m:ci>s</m:ci>
	      </m:apply>
	    </m:math>
	  </caption>
	</subfigure>
	<subfigure id="fig2b">
	  <media type="image/png" src="laplace4.png"/>
	  <caption>
	    The Phase of
	    <m:math>
	      <m:apply>
		<m:ci type="fn">H</m:ci>
		<m:ci>s</m:ci>
	      </m:apply>
	    </m:math>
	  </caption>
	</subfigure>
	<caption>
	  Magnitude and phase of 
	  <m:math>
	    <m:apply>
	      <m:ci type="fn">H</m:ci>
	      <m:ci>s</m:ci>
	    </m:apply>
	  </m:math>
	  are also each 3-dimensional surfaces.  This representation
	  is more common than real and imaginary parts.
	</caption>
      </figure>

      <para id="para14">
	While these are legitimate ways of looking at a signal in the
	Laplace domain, it is quite difficult to draw and/or analyze.
	For this reason, a simpler method has been developed.
	Although it will not be discussed in detail here, the method
	of <cnxn strength="5" document="m10112">Poles and Zeros</cnxn>
	is much easier to understand and is the way both the Laplace
	transform and its discrete-time counterpart the <cnxn strength="5" document="m10549">Z-transform</cnxn> are
	represented graphically.
      </para>
    </section>

  </content>
</document>
