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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="m0062"> 

  <name>Signal Power</name>

  <metadata>
  <md:version>2.10</md:version>
  <md:created>2000/08/10</md:created>
  <md:revised>2007/02/28 16:38:00.224 US/Central</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="rainking">
      <md:firstname>Doug</md:firstname>
      
      <md:surname>Daniels</md:surname>
      <md:email>ddaniels888@gmail.com</md:email>
    </md:maintainer>
  </md:maintainerlist>
  
  <md:keywordlist>
    <md:keyword>average</md:keyword>
    <md:keyword>power</md:keyword>
    <md:keyword>rms</md:keyword>
    <md:keyword>root-mean-square</md:keyword>
    <md:keyword>Signal</md:keyword>
    <md:keyword>spectrum</md:keyword>
  </md:keywordlist>

  <md:abstract>This module examines signal power, looking at instantaneous and average power.  It uses orthogonality properties to derive a simple expression for average power.  It also defines and displays a power spectrum.</md:abstract>
</metadata>
  <content>


    <para id="para1">An interesting question you could ask about a signal is its
      average <term>power</term>. A signal's instantaneous power is
      defined to be its square, as if it were a voltage or current passing through a 1 Ω resistor. The average power is the average of
      the instantaneous power over some time interval. For a periodic
      signal, the natural time interval is clearly its period; for
      nonperiodic signals, a better choice would be entire time or
      time from onset.  For a periodic signal, the average power is
      the square of the root-mean-squared (rms) value. We define the
      rms value of a periodic signal to be

      <equation id="eqn1">
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:apply>
	      <m:ci type="fn">rms</m:ci>
	      <m:ci>s</m:ci>
	    </m:apply>
	    <m:apply>
	      <m:root/>
	      <m:degree><m:cn>2</m:cn></m:degree>
	      <m:apply><m:times/>
		<m:apply><m:divide/>
		  <m:cn>1</m:cn>
		  <m:ci>T</m:ci>
		</m:apply>
		<m:apply>
		  <m:int/>
		  <m:bvar><m:ci>t</m:ci></m:bvar>
		  <m:lowlimit><m:cn>0</m:cn></m:lowlimit>
		  <m:uplimit><m:ci>T</m:ci> </m:uplimit>
		  <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:apply>
	  </m:apply>
	</m:math>
      </equation>

      and thus its average power is
      <m:math>
	<m:apply>
	  <m:power/>
	  <m:apply>
	    <m:ci type="fn">rms</m:ci>
	    <m:ci>s</m:ci>
	  </m:apply>
	  <m:cn>2</m:cn>
	</m:apply>
      </m:math>.

      <equation id="eqn2">
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:apply>
	      <m:ci type="fn">power</m:ci>
	      <m:ci>s</m:ci>
	    </m:apply>
	    <m:apply>
	      <m:power/>
	      <m:apply>
		<m:ci type="fn">rms</m:ci>
		<m:ci>s</m:ci>
	      </m:apply>
	      <m:cn>2</m:cn>
	    </m:apply>
	    <m:apply><m:times/>
	      <m:apply><m:divide/>
		<m:cn>1</m:cn>
		<m:ci>T</m:ci>
	      </m:apply>
	      <m:apply>
		<m:int/>
		<m:bvar><m:ci>t</m:ci></m:bvar>
		<m:lowlimit><m:cn>0</m:cn></m:lowlimit>
		<m:uplimit><m:ci>T</m:ci> </m:uplimit>
		<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:apply>
	</m:math>
      </equation>
</para>

    <exercise id="exer1">
      <problem>
	<para id="probpara1">
	  What is the rms value of the half-wave rectified
	  sinusoid?
	</para>
      </problem>
      <solution>
	<para id="solpara1">A half-wave rectified sinusoid has half the average power of the original sine wave since it is zero half the time.
A sine wave's average power equals
	  <m:math>
	    <m:apply><m:divide/>
	      <m:apply><m:power/>
                 <m:ci>A</m:ci>
	         <m:cn>2</m:cn>
              </m:apply>
	      <m:cn>2</m:cn>
	    </m:apply>
	  </m:math>,
	  making the rms value of the half-wave rectified signal
	  <m:math>
	    <m:apply><m:divide/>
	      <m:ci>A</m:ci>
		<m:cn>2</m:cn>
	    </m:apply>
	  </m:math>.
	</para>
      </solution>
    </exercise>

    <para id="para2">
      To find the average power in the frequency domain, we need to
      substitute the spectral representation of the signal into this
      expression.

      <equation id="eqn3">
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:apply>
	      <m:ci type="fn">power</m:ci>
	      <m:ci>s</m:ci>
	    </m:apply>
	    <m:apply><m:times/>
	      <m:apply><m:divide/>
		<m:cn>1</m:cn>
		<m:ci>T</m:ci>
	      </m:apply>
	      <m:apply>
		<m:int/>
		<m:bvar><m:ci>t</m:ci></m:bvar>
		<m:lowlimit><m:cn>0</m:cn></m:lowlimit>
		<m:uplimit><m:ci>T</m:ci> </m:uplimit>
		<m:apply><m:power/>
		  <m:apply><m:plus/>
		    <m:ci><m:msub>
			<m:mi>a</m:mi>
			<m:mn>0</m:mn>
		      </m:msub></m:ci>
		    <m:apply>
		      <m:sum/>
		      <m:bvar><m:ci>k</m:ci></m:bvar>
		      <m:lowlimit><m:cn>1</m:cn></m:lowlimit>
		      <m:uplimit><m:infinity/></m:uplimit>
		      <m:apply><m:times/>
			<m:ci><m:msub>
			    <m:mi>a</m:mi>
			    <m:mi>k</m:mi>
			  </m:msub></m:ci>
			<m:apply><m:cos/>
			  <m:apply><m:divide/>
			    <m:apply><m:times/>
			      <m:cn>2</m:cn>
			      <m:pi/>
			      <m:ci>k</m:ci>
			      <m:ci>t</m:ci>
			    </m:apply>
			    <m:ci>T</m:ci>
			  </m:apply>
			</m:apply>
		      </m:apply>
		    </m:apply>
		    <m:apply>
		      <m:sum/>
		      <m:bvar><m:ci>k</m:ci></m:bvar>
		      <m:lowlimit><m:cn>1</m:cn></m:lowlimit>
		      <m:uplimit><m:infinity/></m:uplimit>
			<m:apply>
			<m:times/>
			<m:ci><m:msub>
			    <m:mi>b</m:mi>
			    <m:mi>k</m:mi>
			  </m:msub></m:ci>
			<m:apply><m:sin/>
			  <m:apply><m:divide/>
			    <m:apply><m:times/>
			      <m:cn>2</m:cn>
			      <m:pi/>
			      <m:ci>k</m:ci>
			      <m:ci>t</m:ci>
			    </m:apply>
			    <m:ci>T</m:ci>
			  </m:apply>
			</m:apply>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		  <m:cn>2</m:cn>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:apply>
	</m:math>
      </equation>
      The square inside the integral will contain all possible
      pairwise products. However, the <cnxn document="m0060" target="orthogonality" strength="5">orthogonality
      properties</cnxn> say that most of these crossterms integrate to
      zero. The survivors leave a rather simple expression for the
      power we seek.
      
      <equation id="power">
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:apply>
	      <m:ci type="fn">power</m:ci>
	      <m:ci>s</m:ci>
	    </m:apply>
	    <m:apply><m:plus/>
	      <m:apply>
		<m:power/>
		<m:ci><m:msub>
		    <m:mi>a</m:mi>
		    <m:mn>0</m:mn>
		  </m:msub></m:ci>
		<m:cn>2</m:cn>
	      </m:apply>
	      <m:apply><m:times/>
		<m:apply><m:divide/>
		  <m:cn>1</m:cn>
		  <m:cn>2</m:cn>
		</m:apply>
		<m:apply>
		  <m:sum/>
		  <m:bvar><m:ci>k</m:ci></m:bvar>
		  <m:lowlimit><m:cn>1</m:cn></m:lowlimit>
		  <m:uplimit><m:infinity/></m:uplimit>
		  <m:apply><m:plus/>
		    <m:apply>
		      <m:power/>
		      <m:ci><m:msub>
			  <m:mi>a</m:mi>
			  <m:mi>k</m:mi>
			</m:msub></m:ci>
		      <m:cn>2</m:cn>
		    </m:apply>
		    <m:apply>
		      <m:power/>
		      <m:ci><m:msub>
			  <m:mi>b</m:mi>
			  <m:mi>k</m:mi>
			</m:msub></m:ci>
		      <m:cn>2</m:cn>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:apply>
	</m:math>
      </equation>
    </para>

    <figure id="fig1">
      <name>Power Spectrum of a Half-Wave Rectified Sinusoid</name>
      <media type="image/png" src="fourier3.png"/>
      <caption>Power spectrum of a half-wave rectified sinusoid.</caption>
    </figure>
    
    <para id="para3">
      It could well be that computing this sum is easier than
      integrating the signal's square. Furthermore, the contribution
      of each term in the Fourier series toward representing the
      signal can be measured by its contribution to the signal's
      average power. Thus, the power contained in a signal at its
      <m:math>
	<m:ci>k</m:ci>
      </m:math>
      th harmonic is
      <m:math>
	<m:apply><m:divide/>
	  <m:apply><m:plus/>
	    <m:apply>
	      <m:power/>
	      <m:ci><m:msub>
		  <m:mi>a</m:mi>
		  <m:mi>k</m:mi>
		</m:msub></m:ci>
	      <m:cn>2</m:cn>
	    </m:apply>
	    <m:apply>
	      <m:power/>
	      <m:ci><m:msub>
		  <m:mi>b</m:mi>
		  <m:mi>k</m:mi>
		</m:msub></m:ci>
	      <m:cn>2</m:cn>
	    </m:apply>
	  </m:apply>
	  <m:cn>2</m:cn>
	</m:apply>
      </m:math>

      .  The <term>power spectrum</term>

      <m:math>
	<m:apply>
	  <m:ci type="fn"><m:msub>
	      <m:mi>P</m:mi>
	      <m:mi>s</m:mi>
	    </m:msub></m:ci>
	  <m:ci>k</m:ci>
	</m:apply>
      </m:math>, such as shown in <cnxn target="fig1" strength="7"/>,
      plots each harmonic's contribution to the total power.
    </para>

    <exercise id="exer2">
      <problem>
	<para id="probpara2">
	  In stereophonic systems, deviation of a sine wave from the
	  ideal is measured by the total harmonic distortion, which
	  equals the total power in the harmonics higher than the
	  first compared to power in the fundamental. Find an
	  expression for the total harmonic distortion for any
	  periodic signal.  Is this calculation most easily performed
	  in the time or frequency domain?
	</para>
      </problem>
      <solution>
	<para id="solpara2">
	  Total harmonic distortion equals 
	  <m:math>
	    <m:apply>
	      <m:divide/>
	      <m:apply>
		<m:sum/>
		<m:bvar><m:ci>k</m:ci></m:bvar>
		<m:lowlimit><m:cn>2</m:cn></m:lowlimit>
		<m:uplimit><m:infinity/></m:uplimit>
		<m:apply><m:plus/>
		  <m:apply>
		    <m:power/>
		    <m:ci><m:msub>
			<m:mi>a</m:mi>
			<m:mi>k</m:mi>
		      </m:msub></m:ci>
		    <m:cn>2</m:cn>
		  </m:apply>
		  <m:apply>
		    <m:power/>
		    <m:ci><m:msub>
			<m:mi>b</m:mi>
			<m:mi>k</m:mi>
		      </m:msub></m:ci>
		    <m:cn>2</m:cn>
		  </m:apply>
		</m:apply>
	      </m:apply>
	      <m:apply><m:plus/>
		<m:apply>
		  <m:power/>
		  <m:ci><m:msub>
		      <m:mi>a</m:mi>
		      <m:mn>1</m:mn>
		    </m:msub></m:ci>
		  <m:cn>2</m:cn>
		</m:apply>
		<m:apply>
		  <m:power/>
		  <m:ci><m:msub>
		      <m:mi>b</m:mi>
		      <m:mn>1</m:mn>
		    </m:msub></m:ci>
		  <m:cn>2</m:cn>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:math>
	  . Clearly, this quantity is most easily computed in the
	  frequency domain. However, the numerator equals the the
	  square of the signal's rms value minus the power in the
	  average and the power in the first harmonic.
	</para>
      </solution>
    </exercise>

  </content>
</document>
