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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="m0047">

  <name>Parseval's Theorem</name>

  <metadata>
  <md:version>2.4</md:version>
  <md:created>2000/07/25</md:created>
  <md:revised>2004/08/10 10:00:56.397 GMT-5</md:revised>
  <md:authorlist>
      <md:author id="dhj">
      <md:firstname>Don</md:firstname>
      
      <md:surname>Johnson</md:surname>
      <md:email>dhj@rice.edu</md:email>
    </md:author>
  </md:authorlist>

  <md:maintainerlist>
    <md:maintainer id="dhj">
      <md:firstname>Don</md:firstname>
      
      <md:surname>Johnson</md:surname>
      <md:email>dhj@rice.edu</md:email>
    </md:maintainer>
    <md:maintainer id="mrshawn">
      <md:firstname>Shawn</md:firstname>
      
      <md:surname>Stewart</md:surname>
      <md:email>mrshawn@alumni.rice.edu</md:email>
    </md:maintainer>
  </md:maintainerlist>
  
  <md:keywordlist>
    <md:keyword>parseval</md:keyword>
    <md:keyword>fourier transform</md:keyword>
  </md:keywordlist>

  <md:abstract>Information about Parseval's Theorem.  The title says it all, really.</md:abstract>
</metadata>

  <content>

    <para id="introduction"> Properties of the Fourier transform and
      some useful transform pairs are provided in this <cnxn document="m0045" target="shorttable" strength="9">table</cnxn>.
      Especially important among these properties is
      <term>Parseval's Theorem</term>, which states that power
      computed in either domain equals the power in the other.
      
      <equation id="parseval">
	<m:math> 
	  <m:apply>
	    <m:eq/>
	    <m:apply>
	      <m:int/>
	      <m:bvar>
		<m:ci>t</m:ci>
	      </m:bvar>
	      <m:uplimit>
		<m:infinity/>
	      </m:uplimit>
	      <m:lowlimit>
		<m:apply>
		  <m:minus/>
		  <m:infinity/>
		</m:apply>
	      </m:lowlimit>
	      <m:apply>
		<m:power/>
		<m:apply>
		  <m:ci type="fn">s</m:ci>
		  <m:ci>t</m:ci>
		</m:apply>
		<m:cn>2</m:cn>
	      </m:apply>
	    </m:apply>
	    <m:apply>
	      <m:int/>
	      <m:bvar>
		<m:ci>f</m:ci>
	      </m:bvar>
	      <m:uplimit>
		<m:infinity/>
	      </m:uplimit>
	      <m:lowlimit>
		<m:apply>
		  <m:minus/>
		  <m:infinity/>
		</m:apply>
	      </m:lowlimit>
	      <m:apply>
		<m:power/>
		<m:apply>
		  <m:abs/>
		  <m:apply>
		    <m:ci type="fn">S</m:ci>
		    <m:ci>f</m:ci>
		  </m:apply>
		</m:apply>
		<m:cn>2</m:cn>
	      </m:apply>
	    </m:apply>
	  </m:apply>
	</m:math>
      </equation>
      Of practical importance is the conjugate symmetry property:  When  
      <m:math>
	<m:apply>
	  <m:ci type="fn">s</m:ci> 
	  <m:ci>t</m:ci> 
	</m:apply> 
      </m:math>
	  is real-valued, the spectrum at negative frequencies equals
	  the complex conjugate of the spectrum at the corresponding
	  positive frequencies.  Consequently, we need only plot the
	  positive frequency portion of the spectrum (we can easily
	  determine the remainder of the spectrum).
    </para>

    <exercise id="exs">

      <problem>
	<para id="frstprob">
	  How many Fourier transform operations need to be applied to
	  get the original signal back: <m:math>
	    <m:apply>
	      <m:eq/>
	      <m:apply>
		<m:ci type="fn">ℱ</m:ci>
		<m:apply>
		  <m:ci type="fn">⋯</m:ci>
		  <m:apply>
		    <m:ci type="fn">ℱ</m:ci>
		    <m:ci>s</m:ci>
		  </m:apply>
		</m:apply>
	      </m:apply>
	      <m:apply>
		<m:ci type="fn">s</m:ci>
		<m:ci>t</m:ci>
	      </m:apply>
	    </m:apply> 
	  </m:math>
	  ?
	</para>
      </problem>

      <solution>
	<para id="frstprobsoln">
	  <m:math> 
	    <m:apply>
	      <m:eq/>
	      <m:apply>
		<m:ci type="fn">ℱ</m:ci>
		<m:apply>
		  <m:ci type="fn">ℱ</m:ci>
		  <m:apply>
		    <m:ci type="fn">ℱ</m:ci>
		    <m:apply>
		      <m:ci type="fn">ℱ</m:ci>
		      <m:apply>
			<m:ci type="fn">s</m:ci>
			<m:ci>t</m:ci>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	      <m:apply>
		<m:ci type="fn">s</m:ci>
		<m:ci>t</m:ci>
	      </m:apply>
	    </m:apply>
	  </m:math>
	  .  We know that
	  <m:math>
	    <m:apply>
	      <m:eq/>
	      <m:apply>
		<m:ci type="fn">ℱ</m:ci>
		<m:apply>
		  <m:ci type="fn">S</m:ci>
		  <m:ci>f</m:ci>
		</m:apply>
	      </m:apply>
	      <m:apply>
		<m:int/>
		<m:bvar>
		  <m:ci>f</m:ci>
		</m:bvar>
		<m:uplimit>
		  <m:infinity/>
		</m:uplimit>
		<m:lowlimit>
		  <m:apply>
		    <m:minus/>
		    <m:infinity/>
		  </m:apply>
		</m:lowlimit>
		<m:apply>
		  <m:times/>
		  <m:apply>
		    <m:ci type="fn">S</m:ci>
		    <m:ci>f</m:ci>
		  </m:apply>
		  <m:apply>
		    <m:exp/>
		    <m:apply>
		      <m:minus/>
		      <m:apply>
			<m:times/>         
			<m:imaginaryi/>
			<m:cn>2</m:cn>
			<m:pi/>
			<m:ci>f</m:ci>
			<m:ci>t</m:ci>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	      <m:apply>
		<m:int/>
		<m:bvar>
		  <m:ci>f</m:ci>
		</m:bvar>
		<m:uplimit>
		  <m:infinity/>
		</m:uplimit>
		<m:lowlimit>
		  <m:apply>
		    <m:minus/>
		    <m:infinity/>
		  </m:apply>
		</m:lowlimit>
		<m:apply>
		  <m:times/>
		  <m:apply>
		    <m:ci type="fn">S</m:ci>
		    <m:ci>f</m:ci>
		  </m:apply>
		  <m:apply>
		    <m:conjugate/>
		    <m:apply>
		      <m:exp/>
		      <m:apply>
			<m:times/>         
			<m:imaginaryi/>
			<m:cn>2</m:cn>
			<m:pi/>
			<m:ci>f</m:ci>
			<m:apply>
			  <m:minus/>
			  <m:ci>t</m:ci>
			</m:apply>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>    
	      <m:apply>
		<m:ci type="fn">s</m:ci>
		<m:apply>
		  <m:minus/>
		  <m:ci>t</m:ci>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:math>
	  Therefore, two Fourier transforms applied to  
	  <m:math> 
	    <m:apply>
	      <m:ci type="fn">s</m:ci>
	      <m:ci>t</m:ci>
	    </m:apply>
	  </m:math>
	  yields
	  <m:math> 
	    <m:apply>
	      <m:ci type="fn">s</m:ci>
	      <m:apply>
		<m:minus/>
		<m:ci>t</m:ci>
	      </m:apply>
	    </m:apply>
	  </m:math>
	  .  We need two more to get us back where we started.
	</para>
      </solution>
    </exercise> 


  </content>
</document>
