<?xml version="1.0" encoding="utf-8"?>
<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/" xmlns:q="http://cnx.rice.edu/qml/1.0" id="m0545" module-id="" cnxml-version="0.6"> 

  <title>Frequency Shift Keying</title>

  <metadata xmlns:md="http://cnx.rice.edu/mdml/0.4">
  <!-- WARNING! The 'metadata' section is read only. Do not edit below.
       Changes to the metadata section in the source will not be saved. -->
  <md:content-id>m0545</md:content-id>
  <md:title>Frequency Shift Keying</md:title>
  <md:version>2.12</md:version>
  <md:created>2000/08/14</md:created>
  <md:revised>2009/06/09 17:42:37.821 GMT-5</md:revised>
  <md:authorlist>
    <md:author id="dhj">
        <md:firstname>Don</md:firstname>
        <md:surname>Johnson</md:surname>
        <md:fullname>Don Johnson</md:fullname>
        <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:fullname>Don Johnson</md:fullname>
        <md:email>dhj@rice.edu</md:email>
    </md:maintainer>
    <md:maintainer id="jac3">
        <md:firstname>John</md:firstname>
        <md:othername>Austin</md:othername>
        <md:surname>Cottrell</md:surname>
        <md:fullname>John Cottrell</md:fullname>
        <md:email>jac3@rice.edu</md:email>
    </md:maintainer>
  </md:maintainerlist>
  <md:license href="http://creativecommons.org/licenses/by/1.0"/>
  <md:licensorlist>
    <md:licensor id="dhj">
        <md:firstname>Don</md:firstname>
        <md:surname>Johnson</md:surname>
        <md:fullname>Don Johnson</md:fullname>
        <md:email>dhj@rice.edu</md:email>
    </md:licensor>
  </md:licensorlist>
  <md:keywordlist>
    <md:keyword>bandwidth</md:keyword>
    <md:keyword>digital communication</md:keyword>
    <md:keyword>frequency shift keying</md:keyword>
    <md:keyword>FSK</md:keyword>
    <md:keyword>information communication</md:keyword>
  </md:keywordlist>
  <md:subjectlist>
    <md:subject>Science and Technology</md:subject>
  </md:subjectlist>
  <md:abstract>Frequency Shift Keying uses the bit to affect the frequency of a carrier sinusoid.</md:abstract>
  <md:language>en</md:language>
  <!-- WARNING! The 'metadata' section is read only. Do not edit above.
       Changes to the metadata section in the source will not be saved. -->
</metadata>

<content>
     <para id="para1">
      In <term>frequency-shift keying</term> (FSK), the bit
      affects the frequency of a carrier sinusoid.

      <equation id="eq0003">
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:apply>
	      <m:ci type="fn">
		<m:msub>
		  <m:mi>s</m:mi>
		  <m:mn>0</m:mn>
		</m:msub>
	      </m:ci>
	      <m:ci>t</m:ci>
	    </m:apply>
	    <m:apply>
	      <m:times/>
	      <m:ci>A</m:ci>
	      <m:apply>
		<m:ci type="fn">
		  <m:msub>
		    <m:mi>p</m:mi>
		    <m:mi>T</m:mi>
		  </m:msub>
		</m:ci>
		<m:ci>t</m:ci>
	      </m:apply>
	      <m:apply>
		<m:sin/>
		<m:apply>
		  <m:times/>
		  <m:cn>2</m:cn>
		  <m:pi/>
		  <m:ci>
		    <m:msub>
		      <m:mi>f</m:mi>
		      <m:mn>0</m:mn>
		    </m:msub>
		  </m:ci>
		  <m:ci>t</m:ci>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:apply>
	</m:math>
      </equation>

	<m:math display="block">
	  <m:apply>
	    <m:eq/>
	    <m:apply>
	      <m:ci type="fn">
		<m:msub>
		  <m:mi>s</m:mi>
		  <m:mn>1</m:mn>
		</m:msub>
	      </m:ci>
	      <m:ci>t</m:ci>
	    </m:apply>
	    <m:apply>
	      <m:times/>
	      <m:ci>A</m:ci>
	      <m:apply>
		<m:ci type="fn">
		  <m:msub>
		    <m:mi>p</m:mi>
		    <m:mi>T</m:mi>
		  </m:msub>
		</m:ci>
		<m:ci>t</m:ci>
	      </m:apply>
	      <m:apply>
		<m:sin/>
		<m:apply>
		  <m:times/>
		  <m:cn>2</m:cn>
		  <m:pi/>
		  <m:ci>
		    <m:msub>
		      <m:mi>f</m:mi>
		      <m:mn>1</m:mn>
		    </m:msub>
		  </m:ci>
		  <m:ci>t</m:ci>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:apply>
	</m:math>

    </para>

    <figure id="fig1004">
      <media id="id8334936" alt="">
        <image src="sig31.png" mime-type="image/png"/>
        <image src="sig31.eps" mime-type="application/postscript"/>
      </media>
    </figure> 

    <para id="para1.1">
      The frequencies
      <m:math display="inline">
	<m:ci>
	  <m:msub>
	    <m:mi>f</m:mi>
	    <m:mn>0</m:mn>
	  </m:msub>
	</m:ci>
      </m:math>, 

      <m:math display="inline">
	<m:ci>
	  <m:msub>
	    <m:mi>f</m:mi>
	    <m:mn>1</m:mn>
	  </m:msub>
	</m:ci>
      </m:math>

      are usually harmonically related to the bit interval.  In the
      depicted example,

      <m:math display="inline">
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>f</m:mi>
	      <m:mn>0</m:mn>
	    </m:msub>
	  </m:ci>
	  <m:apply>
	    <m:divide/>
	    <m:cn>3</m:cn>
	    <m:ci>T</m:ci>
	  </m:apply>
	</m:apply>
      </m:math>
      and 

      <m:math display="inline">
	<m:apply>
	  <m:eq/>
	  <m:ci>
	    <m:msub>
	      <m:mi>f</m:mi>
	      <m:mn>1</m:mn>
	    </m:msub>
	  </m:ci>
	  <m:apply>
	    <m:divide/>
	    <m:cn>4</m:cn>
	    <m:ci>T</m:ci>
	  </m:apply>
	</m:apply>
      </m:math>.  As can be seen from the transmitted signal for our
      example bit stream (<link target-id="fig1005" strength="3"/>), the
      transitions at bit interval boundaries are smoother than those
      of <link document="m0543" target-id="fig1001" strength="2">BPSK</link>.
    </para>
    
    <figure id="fig1005">
      <media id="id8328186" alt="">
        <image src="sig32.png" mime-type="image/png"/>
        <image src="sig32.eps" mime-type="application/postscript"/>
      </media>
      <caption>
	This plot shows the FSK waveform for same bitstream used in
	the <link document="m10280" target-id="fig1001" strength="3">BPSK
	example</link>.
      </caption>
    </figure>

    <para id="para2">To determine the bandwidth required by this signal set, we again
      consider the alternating bit stream.  Think of it as two signals
      added together: The first comprised of the signal
      <m:math display="inline">
	<m:apply>
	  <m:ci type="fn">
	    <m:msub>
	      <m:mi>s</m:mi>
	      <m:mn>0</m:mn>
	    </m:msub>
	  </m:ci>
	  <m:ci>t</m:ci>
	</m:apply>
      </m:math>, the zero signal,

      <m:math display="inline">
	<m:apply>
	  <m:ci type="fn">
	    <m:msub>
	      <m:mi>s</m:mi>
	      <m:mn>0</m:mn>
	    </m:msub>
	  </m:ci>
	  <m:ci>t</m:ci>
	</m:apply>
      </m:math>, zero, <foreign>etc.</foreign>, and the second having
      the same structure but interleaved with the first and containing
<m:math display="inline">
	<m:apply>
	  <m:ci type="fn">
	    <m:msub>
	      <m:mi>s</m:mi>
	      <m:mn>1</m:mn>
	    </m:msub>
	  </m:ci>
	  <m:ci>t</m:ci>
	</m:apply>
      </m:math> (<link target-id="fig1006" strength="3"/>).
    </para>
    
    <figure id="fig1006">
      <media id="id8319922" alt="">
        <image src="sig33.png" mime-type="image/png"/>
        <image src="sig33.eps" mime-type="application/postscript"/>
      </media>
      <caption>
	The depicted decomposition of the FSK-modulated alternating
	bit stream into its frequency components simplifies the
	calculation of its bandwidth.
      </caption>
    </figure>

    <para id="para2.1">
      Each component can be thought of as a fixed-frequency sinusoid
      multiplied by a square wave of period
      <m:math display="inline">
	<m:apply><m:times/>
	    <m:cn>2</m:cn>
	    <m:ci>T</m:ci>
	</m:apply>
      </m:math>

      that alternates between one and zero.  This baseband square wave
      has the same Fourier spectrum as our BPSK example, but with the
      addition of the constant term
      <m:math display="inline">
	<m:ci>
	  <m:msub>
	    <m:mi>c</m:mi>
	    <m:mn>0</m:mn>
	  </m:msub>
	</m:ci>
      </m:math>.  This quantity's presence changes the number of
      Fourier series terms required for the 90% bandwidth: Now we need
      only include the zero and first harmonics to achieve it.  The
      bandwidth thus equals, with

      <m:math display="inline">
	<m:apply>
	  <m:lt/>
	  <m:ci>
	    <m:msub>
	      <m:mi>f</m:mi>
	      <m:mn>0</m:mn>
	    </m:msub>
	  </m:ci>
	  <m:ci>
	    <m:msub>
	      <m:mi>f</m:mi>
	      <m:mn>1</m:mn>
	    </m:msub>
	  </m:ci>
	</m:apply>
      </m:math>,

      <m:math display="inline">
	<m:apply>
	  <m:eq/>
	  <m:apply>
	    <m:minus/>
	    <m:apply>
	      <m:plus/>
	      <m:ci>
		<m:msub>
		  <m:mi>f</m:mi>
		  <m:mn>1</m:mn>
		</m:msub>
	      </m:ci>
	      <m:apply>
		<m:divide/>
		<m:cn>1</m:cn>
		<m:apply>
		  <m:times/>
		  <m:cn>2</m:cn>
		  <m:ci>T</m:ci>
		</m:apply>
	      </m:apply>
	    </m:apply>
	    <m:apply>
	      <m:minus/>
	      <m:ci>
		<m:msub>
		  <m:mi>f</m:mi>
		  <m:mn>0</m:mn>
		</m:msub>
	      </m:ci>
	      <m:apply>
		<m:divide/>
		<m:cn>1</m:cn>
		<m:apply>
		  <m:times/>
		  <m:cn>2</m:cn>
		  <m:ci>T</m:ci>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:apply>
	  <m:apply>
	    <m:plus/>
	    <m:apply>
	      <m:minus/>
	      <m:ci>
		<m:msub>
		  <m:mi>f</m:mi>
		  <m:mn>1</m:mn>
		</m:msub>
	      </m:ci>
	      <m:ci>
		<m:msub>
		  <m:mi>f</m:mi>
		  <m:mn>0</m:mn>
		</m:msub>
	      </m:ci>
	    </m:apply>
	    <m:apply>
	      <m:divide/>
	      <m:cn>1</m:cn>
	      <m:ci>T</m:ci>
	    </m:apply>
	  </m:apply>
	</m:apply>
      </m:math>.  If the two frequencies are harmonics of the
      bit-interval duration,
    <m:math>
      <m:apply><m:eq/>
	<m:ci>
	  <m:msub>
	    <m:mi>f</m:mi>
	    <m:mn>0</m:mn>
	  </m:msub>
	</m:ci>
	<m:apply><m:divide/>
	  <m:ci>
	    <m:msub>
	      <m:mi>k</m:mi>
	      <m:mn>0</m:mn>
	    </m:msub>
	  </m:ci>
	  <m:ci>T</m:ci>
	</m:apply>
      </m:apply>
    </m:math> and
    <m:math>
      <m:apply><m:eq/>
	<m:ci>
	  <m:msub>
	    <m:mi>f</m:mi>
	    <m:mn>1</m:mn>
	  </m:msub>
	</m:ci>
	<m:apply><m:divide/>
	  <m:ci>
	    <m:msub>
	      <m:mi>k</m:mi>
	      <m:mn>1</m:mn>
	    </m:msub>
	  </m:ci>
	  <m:ci>T</m:ci>
	</m:apply>
      </m:apply>
    </m:math> with
    <m:math>
      <m:apply><m:gt/>
	<m:ci>
	  <m:msub>
	    <m:mi>k</m:mi>
	    <m:mn>1</m:mn>
	  </m:msub>
	</m:ci>
	<m:ci>
	  <m:msub>
	<m:mi>k</m:mi>
	    <m:mn>0</m:mn>
	  </m:msub>
	</m:ci>
      </m:apply>
    </m:math>, the bandwidth equals
    <m:math>
      <m:apply><m:divide/>
        <m:apply><m:plus/>
	  <m:ci>
	    <m:msub>
	      <m:mi>k</m:mi>
	      <m:mn>1</m:mn>
	    </m:msub>
	  </m:ci>
	  <m:apply><m:minus/>
            <m:ci>
              <m:msub>
	        <m:mi>k</m:mi>
	        <m:mn>0</m:mn>
	      </m:msub>
	    </m:ci>
          </m:apply>
          <m:cn>1</m:cn>
         </m:apply>
         <m:ci>T</m:ci>
      </m:apply>
    </m:math>.  If the difference between harmonic numbers is
     <m:math><m:cn>1</m:cn></m:math>, then the FSK bandwidth is
     <emphasis>smaller</emphasis> than the BPSK bandwidth.  If the
     difference is <m:math><m:cn>2</m:cn></m:math>, the bandwidths are
     equal and larger differences produce a transmission bandwidth
     larger than that resulting from using a BPSK signal set.
    </para>

  </content>
</document>
