<?xml version="1.0" encoding="utf-8"?>
<!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="m10280"> 
  
  <name>Binary Phase Shift Keying</name>

  <metadata>
  <md:version>2.13</md:version>
  <md:created>2001/08/09</md:created>
  <md:revised>2007/06/03 17:10:40.535 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="jac3">
      <md:firstname>John</md:firstname>
      <md:othername>Austin</md:othername>
      <md:surname>Cottrell</md:surname>
      <md:email>jac3@rice.edu</md:email>
    </md:maintainer>
  </md:maintainerlist>
  
  <md:keywordlist>
    <md:keyword>bandwidth</md:keyword>
    <md:keyword>binary phase shift keying</md:keyword>
    <md:keyword>bit interval</md:keyword>
    <md:keyword>BPSK</md:keyword>
    <md:keyword>datarate</md:keyword>
    <md:keyword>digital communication</md:keyword>
    <md:keyword>information communication</md:keyword>
    <md:keyword>signal set</md:keyword>
    <md:keyword>transmission bandwidth</md:keyword>
  </md:keywordlist>

  <md:abstract>Introduces a method for representing bits with an analog signal known as binary phase shift keying.</md:abstract>
</metadata>
  

  <content>
    
    <para id="para1">
      A commonly used example of a signal set consists of pulses that
      are negatives of each other (<cnxn target="fig1000"/>).

      <equation id="eq0000">
	<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: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:minus/>
	      <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:apply>
	  </m:apply>
	</m:math>

    <figure id="fig1000">
      <media type="image/png" src="sig26.png"/>
    </figure>
      Here, we have a baseband signal set suitable for wireline
      transmission. The entire bit stream 

      <m:math>
	<m:apply>
	  <m:ci type="fn">b</m:ci>
	  <m:ci>n</m:ci>
	</m:apply>
      </m:math>

      is represented by a sequence of these signals. Mathematically,
      the transmitted signal has the form

      <equation id="eq0001">
	<m:math display="block">
	  <m:apply>
	    <m:eq/>
	    <m:apply>
	      <m:ci type="fn">x</m:ci>
	      <m:ci>t</m:ci>
	    </m:apply>
	    <m:apply>
	      <m:sum/>
	      <m:bvar><m:ci>n</m:ci></m:bvar>
	      <m:domainofapplication>
		<m:ci>n</m:ci>
	      </m:domainofapplication>
	      <m:apply>
		<m:times/>
		  <m:apply>
                    <m:power/>
                    <m:apply><m:minus/><m:cn>1</m:cn></m:apply>
                    <m:apply><m:ci type="fn">b</m:ci><m:ci>n</m:ci></m:apply>
                  </m:apply>
		  <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:apply>
		    <m:minus/>
		    <m:ci>t</m:ci>
		    <m:apply>
		      <m:times/>
		      <m:ci>n</m:ci>
		      <m:ci>T</m:ci>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:apply>
	</m:math>
      </equation>

      and graphically <cnxn target="fig1001" strength="8"/> shows what
      a typical transmitted signal might be.
    </para>
    

    <figure id="fig1001" orient="vertical">
      <subfigure id="sub1100">
	<media type="image/png" src="sig27.png"/>
      </subfigure>
      <subfigure id="sub1101">
	<media type="image/png" src="sig29.png"/>
      </subfigure>
      <caption>The upper plot shows how a baseband signal set for
	transmitting the bit sequence <code>0110</code>. The lower one
	shows an amplitude-modulated variant suitable for wireless
	channels. 
      </caption>
    </figure>
    

    <para id="para1.2">
      This way of representing a bit stream---changing the bit changes
      the sign of the transmitted signal---is known as <term>binary
      phase shift keying</term> and abbreviated BPSK. The name comes from concisely expressing this popular way of communicating digital information. The word "binary" is clear
      enough (one binary-valued quantity is transmitted during a bit
      interval). Changing the sign of sinusoid amounts to
      changing---shifting---the phase by <m:math><m:pi/></m:math>
      (although we don't have a sinusoid yet). The word "keying"
      reflects back to the first electrical communication system, which happened to be digital as well: the telegraph.
    </para>

    <para id="para2">
      The <term>datarate</term> <m:math><m:ci>R</m:ci></m:math> of a
      digital communication system is how frequently an information
      bit is transmitted. In this example it equals the reciprocal of
      the bit interval:

      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:ci>R</m:ci>
	  <m:apply>
	    <m:divide/>
	    <m:cn>1</m:cn>
	    <m:ci>T</m:ci>
	  </m:apply>
	</m:apply>
      </m:math>.  Thus, for a 1 Mbps (megabit per second)
      transmission, we must have
      <m:math>
        <m:apply><m:eq/>
           <m:ci>T</m:ci>
           <m:apply><m:times/>
             <m:cn>1</m:cn><m:ci>μs</m:ci>
           </m:apply>
        </m:apply>
      </m:math>. 
    </para>

    <para id="para2.1"> The choice of signals to represent bit values
    is arbitrary to some degree. Clearly, we do not want to choose
    signal set members to be the same; we couldn't distinguish bits if
    we did so. We could also have made the negative-amplitude pulse
    represent a 0 and the positive one a 1. This choice is indeed
    arbitrary and will have no effect on performance
    <emphasis>assuming </emphasis>the receiver knows which signal
    represents which bit. As in all communication systems, we design
    transmitter and receiver together.  </para>

  <para id="para2.2">
    A simple signal set for both wireless and wireline channels
      amounts to amplitude modulating a baseband signal set (more
      appropriate for a wireline channel) by a carrier having a
      frequency harmonic with the bit interval.
    </para>

    <equation id="eq0002"> 
      <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: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:math>
    </equation>
    <para id="missing_para">
      <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:minus/>
	    <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: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:math>
     </para>

    <figure id="fig1002">
      <media type="image/png" src="sig28.png"/>
    </figure>

    <exercise id="exer1">
      <problem>
	<para id="prob1">
	What is the value of <m:math><m:ci>k</m:ci></m:math> in this example?
        </para>
      </problem>
 
     <solution>
	<para id="sol1">  
	  <m:math>
	    <m:apply>
	      <m:eq/>
	      <m:ci>k</m:ci>
	      <m:cn>4</m:cn>
	    </m:apply>
	  </m:math>.
	</para>
      </solution>
    </exercise>

    <para id="para3">
      This signal set is also known as a BPSK signal set. We'll
      show later that indeed both signal sets provide identical performance
      levels when the signal-to-noise ratios are equal.
    </para>
    
    <exercise id="exer2">
      <problem>
	<para id="prob2">
	Write a formula, in the style of the <cnxn document="m0542" target="fig0001" strength="7">baseband signal set</cnxn>, for
	the transmitted signal as shown in the <cnxn document="m0542" target="fig1001" strength="7">plot of the baseband signal
	set</cnxn> that emerges when we use this modulated signal.
	</para>
      </problem>

      <solution>
	<para id="sol2">
	  <m:math display="block">
	    <m:apply>
	      <m:eq/>
	      <m:apply>
		<m:ci type="fn">x</m:ci>
		<m:ci>t</m:ci>
	      </m:apply>
	      <m:apply>
		<m:sum/>
		<m:bvar><m:ci>n</m:ci></m:bvar>
		<m:domainofapplication>
		  <m:ci>n</m:ci>
	      </m:domainofapplication>
		<m:apply>
		  <m:times/>
		  <m:apply>
                    <m:power/>
                    <m:apply><m:minus/><m:cn>1</m:cn></m:apply>
                    <m:apply><m:ci type="fn">b</m:ci><m:ci>n</m:ci></m:apply>
                  </m:apply>
		  <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:apply>
		      <m:minus/>
		      <m:ci>t</m:ci>
		      <m:apply>
			<m:times/>
			<m:ci>n</m:ci>
			<m:ci>T</m:ci>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		  <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:math>
	</para>
      </solution>
    </exercise>


    <para id="para3.1">
    What is the transmission bandwidth of these signal sets?  We need
      only consider the baseband version as the second is an
      amplitude-modulated version of the first. The bandwidth is
      determined by the bit sequence.  If the bit sequence is
      constant—always 0 or always 1—the transmitted signal
      is a constant, which has zero bandwidth. The
      worst-case—bandwidth consuming—bit sequence is the
      alternating one shown in <cnxn target="fig1003" strength="9"/>. In this case, the transmitted signal is a square
      wave having a period of

      <m:math>
	<m:apply>
	  <m:times/>
	  <m:cn>2</m:cn>
	  <m:ci>T</m:ci>
	</m:apply>
      </m:math>.
    </para>


    <figure id="fig1003">
      <media type="image/png" src="sig30.png"/>
      <caption>
      Here we show the transmitted waveform corresponding to an
      alternating bit sequence. </caption>
    </figure>


    <para id="para3.2">
      From our work in Fourier series, we know that this signal's
      spectrum contains odd-harmonics of the fundamental, which here
      equals

      <m:math>
	<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:math>.  Thus, strictly speaking, the signal's bandwidth is
      infinite. In practical terms, we use the 90%-power bandwidth to
      assess the effective range of frequencies consumed by the
      signal. The first and third harmonics contain that fraction of
      the total power, meaning that the effective bandwidth of our
      baseband signal is

      <m:math>
	<m:apply>
	  <m:divide/>
	  <m:cn>3</m:cn>
	  <m:apply>
	    <m:times/>
	    <m:cn>2</m:cn>
	    <m:ci>T</m:ci>
	  </m:apply>
	</m:apply>
      </m:math>

      or, expressing this quantity in terms of the datarate,

      <m:math>
	<m:apply>
	  <m:divide/>
	  <m:apply>
	    <m:times/>
	    <m:cn>3</m:cn>
	    <m:ci>R</m:ci>
	  </m:apply>
	  <m:cn>2</m:cn>
	</m:apply>
      </m:math>. Thus, a digital communications signal requires more
      bandwidth than the datarate: a 1 Mbps baseband system
      requires a bandwidth of at least 1.5 MHz. Listen carefully
      when someone describes the transmission bandwidth of digital
      communication systems: Did they say "megabits" or "megahertz"?
    </para>


    <exercise id="exer3">
      <problem>
	<para id="prob3">
	  Show that indeed the first and third harmonics contain 90%
	  of the transmitted power. If the receiver uses a front-end
	  filter of bandwidth

	  <m:math>
	    <m:apply>
	      <m:divide/>
	      <m:cn>3</m:cn>
	      <m:apply>
		<m:times/>
		<m:cn>2</m:cn>
		<m:ci>T</m:ci>
	      </m:apply>
	    </m:apply>
	  </m:math>, what is the total harmonic distortion of the
	  received signal?
	</para>
      </problem>

      <solution>
	<para id="sol3">
	  The harmonic distortion is 10%.
	</para>
      </solution>
    </exercise>


    <exercise id="exer4">
      <problem>
	<para id="prob4">
	  What is the 90% transmission bandwidth of the modulated signal set?
	</para>
      </problem>

      <solution>
	<para id="sol4">
	  Twice the baseband bandwidth because both positive and
	  negative frequencies are shifted to the carrier by the
	  modulation:
	  
	  <m:math>
	    <m:apply>
	      <m:times/>
	      <m:cn>3</m:cn>
	      <m:ci>R</m:ci>
	    </m:apply>
	  </m:math>.
	</para>
      </solution>
    </exercise>

  </content>
</document>
