<?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="m0048">

  <name>Linear Time Invariant Systems</name>

  <metadata>
  <md:version>2.18</md:version>
  <md:created>2000/10/11</md:created>
  <md:revised>2007/05/22 09:50:47.852 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:maintainer id="bfite">
      <md:firstname>Benjamin</md:firstname>
      
      <md:surname>Fite</md:surname>
      <md:email>bfite@rice.edu</md:email>
    </md:maintainer>
  </md:maintainerlist>
  
  <md:keywordlist>
    <md:keyword>fourier series</md:keyword>
    <md:keyword>output spectrum</md:keyword>
    <md:keyword>time invariant</md:keyword>
  </md:keywordlist>

  <md:abstract>Application of periodic input to linear, time-invariant systems.
</md:abstract>
</metadata>

  <content>
    <para id="spectrum"> When we apply a periodic input to a linear,
      time-invariant system, the output is periodic and has Fourier
      series coefficients equal to the product of the system's
      frequency response and the input's Fourier coefficients (<cnxn document="m0044" target="output" strength="9">Filtering Periodic
      Signals</cnxn>).  The way we derived the spectrum of
      non-periodic signal from periodic ones makes it clear that the
      same kind of result works when the input is not periodic:
      <emphasis>If</emphasis>
      <m:math>
	<m:apply>
	  <m:ci type="fn">x</m:ci>
	  <m:ci>t</m:ci>
	</m:apply>
      </m:math>
      <emphasis>serves as the input to a linear, time-invariant system
	having frequency response</emphasis>
      <m:math>
	<m:apply>
	  <m:ci type="fn">H</m:ci>
	  <m:ci>f</m:ci>
	</m:apply>
      </m:math>, 
      <emphasis>the spectrum of the output is</emphasis>  
      <m:math>
	<m:apply>
	  <m:times/>
	  <m:apply>
	    <m:ci type="fn">X</m:ci>
	    <m:ci>f</m:ci>
	  </m:apply>
	  <m:apply>
	    <m:ci type="fn">H</m:ci>
	    <m:ci>f</m:ci>
	  </m:apply>
	</m:apply>
      </m:math>.
    </para>


    <example id="example">
      <para id="sampleprobs"> 
	Let's use this frequency-domain input-output relationship for
	linear, time-invariant systems to find a formula for the
	<m:math>
	  <m:apply>
	    <m:times/> 
	    <m:ci>R</m:ci>
	    <m:ci>C</m:ci>
	  </m:apply>
	</m:math>-circuit's response to a pulse input. We have
	expressions for the input's spectrum and the system's
	frequency response.

	<equation id="freqresponse">
	  <m:math>
	    <m:apply>
	      <m:eq/>
	      <m:apply>
		<m:ci type="fn">P</m:ci>
		<m:ci>f</m:ci>
	      </m:apply>
	      <m:apply>
		<m:times/>
		<m:apply>
		  <m:exp/>
		  <m:apply>
		    <m:minus/>
		    <m:apply>
		      <m:times/>
		      <m:imaginaryi/>
		      <m:pi/>
		      <m:ci>f</m:ci>
		      <m:ci>Δ</m:ci>
		    </m:apply>
		  </m:apply>
		</m:apply>
		<m:apply>
		  <m:divide/>
		  <m:apply>
		    <m:sin/>
		    <m:apply>
		      <m:times/>
		      <m:pi/>
		      <m:ci>f</m:ci>
		      <m:ci>Δ</m:ci>
		    </m:apply>
		  </m:apply>
		  <m:apply>
		    <m:times/>
		    <m:pi/>
		    <m:ci>f</m:ci>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:apply>    
	  </m:math>
	</equation>
	<equation id="freqresponse2">  
	  <m:math> 
	    <m:apply>
	      <m:eq/>
	      <m:apply>
		<m:ci type="fn">H</m:ci>
		<m:ci>f</m:ci>
	      </m:apply>
	      <m:apply>
		<m:divide/>
		<m:cn>1</m:cn>
		<m:apply>
		  <m:plus/>
		  <m:cn>1</m:cn>
		  <m:apply>
		    <m:times/>
		    <m:imaginaryi/>
		    <m:cn>2</m:cn>
		    <m:pi/>
		    <m:ci>f</m:ci>
		    <m:ci>R</m:ci>
		    <m:ci>C</m:ci>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:math>
	</equation> 

	Thus, the output's Fourier transform equals

	<equation id="outputtransform">
	  <m:math> 
	    <m:apply>
	      <m:eq/>
	      <m:apply>
		<m:ci type="fn">Y</m:ci>
		<m:ci>f</m:ci>
	      </m:apply>
	      <m:apply>
		<m:times/>
		<m:apply>
		  <m:exp/>
		  <m:apply>
		    <m:minus/>
		    <m:apply>
		      <m:times/>
		      <m:imaginaryi/>
		      <m:pi/>
		      <m:ci>f</m:ci>
		      <m:ci>Δ</m:ci>
		    </m:apply>
		  </m:apply>
		</m:apply>
		<m:apply>
		  <m:divide/>
		  <m:apply>
		    <m:sin/>
		    <m:apply>
		      <m:times/>
		      <m:pi/>
		      <m:ci>f</m:ci>
		      <m:ci>Δ</m:ci>
		    </m:apply>
		  </m:apply>
		  <m:apply>
		    <m:times/>
		    <m:pi/>
		    <m:ci>f</m:ci>
		  </m:apply>
		</m:apply>
		<m:apply>
		  <m:divide/>
		  <m:cn>1</m:cn>
		  <m:apply>
		    <m:plus/>
		    <m:cn>1</m:cn>
		    <m:apply>
		      <m:times/>
		      <m:imaginaryi/>
		      <m:cn>2</m:cn>
		      <m:pi/>
		      <m:ci>f</m:ci>
		      <m:ci>R</m:ci>
		      <m:ci>C</m:ci>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:math>
	</equation>

	You won't find this Fourier transform in our table, and the
	required integral is difficult to evaluate as the expression
	stands. This situation requires cleverness and an
	understanding of the Fourier transform's properties. In
	particular, recall Euler's relation for the sinusoidal term
	and note the fact that multiplication by a complex exponential
	in the frequency domain amounts to a time delay. Let's
	momentarily make the expression for
	<m:math>
	  <m:apply>
	    <m:ci type="fn">Y</m:ci>
	    <m:ci>f</m:ci>
	  </m:apply>
	</m:math> 
	more complicated.   

	<equation id="four_fifteen">
	  <m:math> 
	    <m:apply>
	      <m:eq/>
	      <m:apply>
		<m:times/>
		<m:apply>
		  <m:exp/>
		  <m:apply>
		    <m:minus/>
		    <m:apply>
		      <m:times/>    
		      <m:imaginaryi/>
		      <m:pi/>
		      <m:ci>f</m:ci>
		      <m:ci>Δ</m:ci>
		    </m:apply>
		  </m:apply>
		</m:apply>
		<m:apply>
		  <m:divide/>
		  <m:apply>
		    <m:sin/>
		    <m:apply>
		      <m:times/>
		      <m:pi/>
		      <m:ci>f</m:ci>
		      <m:ci>Δ</m:ci>
		    </m:apply>
		  </m:apply>
		  <m:apply>
		    <m:times/>
		    <m:pi/>
		    <m:ci>f</m:ci>
		  </m:apply>
		</m:apply>
	      </m:apply>
	      <m:apply>
		<m:times/>
		<m:apply>
		  <m:exp/>
		  <m:apply>
		    <m:minus/>
		    <m:apply>
		      <m:times/>
		      <m:imaginaryi/>
		      <m:pi/>
		      <m:ci>f</m:ci>
		      <m:ci>Δ</m:ci>
		    </m:apply>
		  </m:apply>
		</m:apply>
		<m:apply>
		  <m:divide/>
		  <m:apply>
		    <m:minus/>
		    <m:apply>
		      <m:exp/>
		      <m:apply>
			<m:times/>
			<m:imaginaryi/>
			<m:pi/>
			<m:ci>f</m:ci>
			<m:ci>Δ</m:ci>
		      </m:apply>
		    </m:apply>
		    <m:apply>
		      <m:exp/>
		      <m:apply>
			<m:minus/>
			<m:apply>
			  <m:times/>
			  <m:imaginaryi/>
			  <m:pi/>
			  <m:ci>f</m:ci>
			  <m:ci>Δ</m:ci>
			</m:apply>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		  <m:apply>
		    <m:times/>
		    <m:imaginaryi/>
		    <m:cn>2</m:cn>
		    <m:pi/>
		    <m:ci>f</m:ci>
		  </m:apply>
		</m:apply>
	      </m:apply>
	      <m:apply>
		<m:times/>
		<m:apply>
		  <m:divide/>
		  <m:cn>1</m:cn>
		  <m:apply>
		    <m:times/>
		    <m:imaginaryi/>
		    <m:cn>2</m:cn>
		    <m:pi/>
		    <m:ci>f</m:ci>
		  </m:apply>
		</m:apply>
		<m:apply>
		  <m:minus/>
		  <m:cn>1</m:cn>
		  <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>Δ</m:ci>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:math>
	</equation>

	Consequently,

	<equation id="consequent">
	  <m:math> 
	    <m:apply>
	      <m:eq/>
	      <m:apply>
		<m:ci type="fn">Y</m:ci>
		<m:ci>f</m:ci>
	      </m:apply>
	      <m:apply>
		<m:times/>
		<m:apply>
		  <m:divide/>
		  <m:cn>1</m:cn>
		  <m:apply>
		    <m:times/>
		    <m:imaginaryi/>
		    <m:cn>2</m:cn>
		    <m:pi/>
		    <m:ci>f</m:ci>
		  </m:apply>
		</m:apply>
		<m:apply>
		  <m:minus/>
		  <m:cn>1</m:cn>
		  <m:apply>
		    <m:exp/>
		    <m:apply>
		      <m:minus/>
		      <m:apply>
			<m:times/>
			<m:imaginaryi/>
			<m:pi/>
			<m:ci>f</m:ci>
			<m:ci>Δ</m:ci>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		</m:apply>
		<m:apply>
		  <m:divide/>
		  <m:cn>1</m:cn>
		  <m:apply>
		    <m:plus/>
		    <m:cn>1</m:cn>
		    <m:apply>
		      <m:times/>
		      <m:imaginaryi/>
		      <m:cn>2</m:cn>
		      <m:pi/>
		      <m:ci>f</m:ci>
		      <m:ci>R</m:ci>
		      <m:ci>C</m:ci>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:math>
	</equation>

	The table of <cnxn document="m0046" target="table2" strength="9">Fourier transform properties</cnxn> suggests
	thinking about this expression as a
	<emphasis>product</emphasis> of terms.
	
	<list id="listone" type="bulleted"><item> 
	    Multiplication by   
	    <m:math>
	      <m:apply>
		<m:divide/>
		<m:cn>1</m:cn>
		<m:apply>
		  <m:times/>
		  <m:imaginaryi/>
		  <m:cn>2</m:cn>
		  <m:pi/>
		  <m:ci>f</m:ci>
		</m:apply>
	      </m:apply> 
	    </m:math>
	    means integration.
	  </item>
	  <item> 
	    Multiplication by the complex exponential   
	    <m:math> 
	      <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>Δ</m:ci>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:math>
	    means delay by   
	    <m:math> 
	      <m:ci>Δ</m:ci>
	    </m:math>
	    seconds in the time domain.
	  </item>
	  <item> 
	    The term   
	    <m:math> 
	      <m:apply>
		<m:minus/>		
		  <m:cn>1</m:cn>
                  <m:apply>
		  <m:times/>
		  <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>Δ</m:ci>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:math>
	    means, in the time domain, subtract the time-delayed signal from its original.
	  </item>
	  <item> 
	    The inverse transform of the frequency response is   
	    <m:math> 
	      <m:apply>
		<m:times/>
		<m:apply>
		  <m:divide/>
		  <m:cn>1</m:cn>
		  <m:apply>
		    <m:times/>
		    <m:ci>R</m:ci>
		    <m:ci>C</m:ci>
		  </m:apply>
		</m:apply>
		<m:apply>
		  <m:exp/>
		  <m:apply>
		    <m:minus/>
		    <m:apply>
		      <m:divide/>
		      <m:ci>t</m:ci>
		      <m:apply>
			<m:times/>
			<m:ci>R</m:ci>
			<m:ci>C</m:ci>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		</m:apply>
		<m:apply>
		  <m:ci type="fn">u</m:ci>
		  <m:ci>t</m:ci>
		</m:apply>
	      </m:apply>
	    </m:math>.
	  </item>
	</list>
	
	We can translate each of these frequency-domain products into
	time-domain operations <emphasis>in any order we
	like</emphasis> because the order in which multiplications
	occur doesn't affect the result. Let's start with the product
	of
	<m:math>
	  <m:apply>
	    <m:divide/>
	    <m:cn>1</m:cn>
	    <m:apply>
	      <m:times/>
	      <m:imaginaryi/>
	      <m:cn>2</m:cn>
	      <m:pi/>
	      <m:ci>f</m:ci>
	    </m:apply>
	  </m:apply>
	</m:math>
	(integration in the time domain) and the transfer function:  

	<equation id="anothertransferfunction">
	  <m:math> 
	    <m:apply>
	      <m:times/>
	      <m:apply>
		<m:divide/>
		<m:cn>1</m:cn>
		<m:apply>
		  <m:times/>
		  <m:imaginaryi/>
		  <m:cn>2</m:cn>
		  <m:pi/>
		  <m:ci>f</m:ci>
		</m:apply>
	      </m:apply>
	      <m:apply>
		<m:divide/>
		<m:cn>1</m:cn>
		<m:apply>
		  <m:plus/>
		  <m:cn>1</m:cn>
		  <m:apply>
		    <m:times/>
		    <m:imaginaryi/>
		    <m:cn>2</m:cn>
		    <m:pi/>
		    <m:ci>f</m:ci>
		    <m:ci>R</m:ci>
		    <m:ci>C</m:ci>
		  </m:apply>
		</m:apply>
	      </m:apply>
	      <m:ci>↔</m:ci>
	      <m:apply>
		<m:minus/>
		<m:cn>1</m:cn>
		<m:apply>
		  <m:exp/>
		  <m:apply>
		    <m:minus/>
		    <m:apply>
		      <m:divide/>
		      <m:ci>t</m:ci>
		      <m:apply>
			<m:times/>
			<m:ci>R</m:ci>
			<m:ci>C</m:ci>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	      <m:apply>
		<m:ci type="fn">u</m:ci>
		<m:ci>t</m:ci>
	      </m:apply>
	    </m:apply>
	  </m:math>
	</equation>

	The middle term in the expression for   
	<m:math> 
	  <m:apply>
	    <m:ci type="fn">Y</m:ci>
	    <m:ci>f</m:ci>
	  </m:apply>
	</m:math>
	consists of the difference of two terms: the constant  
	<m:math> 
	  <m:cn>1</m:cn>
	</m:math> 
	and the complex exponential   
	<m:math> 
	  <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>Δ</m:ci>
	      </m:apply>
	    </m:apply>
	  </m:apply>
	</m:math>.  
	Because of the Fourier transform's linearity, we simply
	subtract the results.

	<equation id="result">
	  <m:math> 
	    <m:apply>
	      <m:ci type="fn">Y</m:ci>
	      <m:ci>f</m:ci>
	    </m:apply>
	    <m:ci>↔</m:ci>
	    <m:apply>
	      <m:minus/>
	      <m:apply>
		<m:times/>
		<m:apply>
		  <m:minus/>
		  <m:cn>1</m:cn>
		  <m:apply>
		    <m:exp/>
		    <m:apply>
		      <m:minus/>
		      <m:apply>
			<m:divide/>
			<m:ci>t</m:ci>
			<m:apply>
			  <m:times/>
			  <m:ci>R</m:ci>
			  <m:ci>C</m:ci>
			</m:apply>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		</m:apply> 
		<m:apply>
		  <m:ci type="fn">u</m:ci>
		  <m:ci>t</m:ci>
		</m:apply>
	      </m:apply>
	      <m:apply>
		<m:times/>
		<m:apply>
		  <m:minus/>
		  <m:cn>1</m:cn>
		  <m:apply>
		    <m:exp/>
		    <m:apply>
		      <m:minus/>
		      <m:apply>
			<m:divide/>
			<m:apply>
			  <m:minus/>
			  <m:ci>t</m:ci>
			  <m:ci>Δ</m:ci>
			</m:apply>
			<m:apply>
			  <m:times/>
			  <m:ci>R</m:ci>
			  <m:ci>C</m:ci>
			</m:apply>
		      </m:apply>
		    </m:apply>
		  </m:apply>
		</m:apply>
		<m:apply>
		  <m:ci type="fn">u</m:ci>
		  <m:apply>
		    <m:minus/>
		    <m:ci>t</m:ci>
		    <m:ci>Δ</m:ci>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:math>
	</equation> 

	Note that in delaying the signal how we carefully included the
	unit step. The second term in this result does not begin until
	<m:math>
	  <m:apply>
	    <m:eq/>
	    <m:ci>t</m:ci>
	    <m:ci>Δ</m:ci>
	  </m:apply>
	</m:math>.  Thus, the waveforms shown in the <cnxn document="m0044" target="filter" strength="9">Filtering
	Periodic Signals</cnxn> example mentioned above are
	exponentials.  We say that the <term>time constant</term> of
	an exponentially decaying signal equals the time it takes to
	decrease by
	<m:math>
	  <m:apply>
	    <m:divide/>
	    <m:cn>1</m:cn>
	    <m:exponentiale/>
	  </m:apply>
	</m:math> 	
	of its original value.  Thus, the time-constant of the rising
	and falling portions of the output equal the product of the
	circuit's resistance and capacitance.
      </para>
    </example>


    <exercise id="exer1">
      <problem>
	<para id="exer1probtext">
	  Derive the filter's output by considering the terms in 
	  <cnxn target="four_fifteen" strength="9"/> 
	  in the order given. Integrate last rather than first. You
	  should get the same answer.
	</para>
      </problem>
      <solution>
	<para id="exer1probsoln">
	  The inverse transform of the frequency response is  
	  <m:math> 
	    <m:apply>
	      <m:times/>
	      <m:apply>
		<m:divide/>
		<m:cn>1</m:cn>
		<m:apply>
		  <m:times/>
		  <m:ci>R</m:ci>
		  <m:ci>C</m:ci>
		</m:apply>
	      </m:apply>
	      <m:apply>
		<m:exp/>
		<m:apply>
		  <m:minus/>
		  <m:apply>
		    <m:divide/>
		    <m:ci>t</m:ci>
		    <m:apply>
		      <m:times/>
		      <m:ci>R</m:ci>
		      <m:ci>C</m:ci>
		    </m:apply>
		  </m:apply>
		</m:apply>
	      </m:apply>
	      <m:apply>
		<m:ci type="fn">u</m:ci>
		<m:ci>t</m:ci>
	      </m:apply>
	    </m:apply>  
	  </m:math>.  
	  Multiplying the frequency response by   
	  <m:math> 
	    <m:apply>
	      <m:minus/>
	      <m:cn>1</m:cn>
	      <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>Δ</m:ci>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:math>
	  means subtract from the original signal its time-delayed
	  version. Delaying the frequency response's time-domain
	  version by
	  <m:math>
	    <m:ci>Δ</m:ci> 
	  </m:math> 
	  results in  
	  <m:math> 
	    <m:apply>
	      <m:times/>
	      <m:apply>
		<m:divide/>
		<m:cn>1</m:cn>
		<m:apply>
		  <m:times/>
		  <m:ci>R</m:ci>
		  <m:ci>C</m:ci>
		</m:apply>
	      </m:apply>
	      <m:apply>
		<m:exp/>
		<m:apply>
		  <m:divide/>
		  <m:apply>
		    <m:minus/>
		    <m:apply>
		      <m:minus/>
		      <m:ci>t</m:ci>
		      <m:ci>Δ</m:ci>
		    </m:apply>
		  </m:apply>
		  <m:apply>
		    <m:times/>
		    <m:ci>R</m:ci>
		    <m:ci>C</m:ci>
		  </m:apply>
		</m:apply>
	      </m:apply>
	      <m:apply>
		<m:ci type="fn">u</m:ci>
		<m:apply>
		  <m:minus/>
		  <m:ci>t</m:ci>
		  <m:ci>Δ</m:ci>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:math>.  
	  Subtracting from the undelayed signal yields   
	  <m:math> 
	    <m:apply>
	      <m:minus/>
	      <m:apply>
		<m:times/>
		<m:apply>
		  <m:divide/>
		  <m:cn>1</m:cn>
		  <m:apply>
		    <m:times/>
		    <m:ci>R</m:ci>
		    <m:ci>C</m:ci>
		  </m:apply>
		</m:apply>
		<m:apply>
		  <m:exp/>
		  <m:apply>
		    <m:divide/>
		    <m:apply>
		      <m:minus/>
		      <m:ci>t</m:ci>
		    </m:apply>
		    <m:apply>
		      <m:times/>
		      <m:ci>R</m:ci>
		      <m:ci>C</m:ci>
		    </m:apply>
		  </m:apply>
		</m:apply>
		<m:apply>
		  <m:ci type="fn">u</m:ci>
		  <m:ci>t</m:ci>
		</m:apply>
	      </m:apply>
	      <m:apply>
		<m:times/>
		<m:apply>
		  <m:divide/>
		  <m:cn>1</m:cn>
		  <m:apply>
		    <m:times/>
		    <m:ci>R</m:ci>
		    <m:ci>C</m:ci>
		  </m:apply>
		</m:apply>
		<m:apply>
		  <m:exp/>
		  <m:apply>
		    <m:divide/>
		    <m:apply>
		      <m:minus/>
		      <m:apply>
			<m:minus/>
			<m:ci>t</m:ci>
			<m:ci>Δ</m:ci>
		      </m:apply>
		    </m:apply>
		    <m:apply>
		      <m:times/>
		      <m:ci>R</m:ci>
		      <m:ci>C</m:ci>
		    </m:apply>
		  </m:apply>
		</m:apply>
		<m:apply>
		  <m:ci type="fn">u</m:ci>
		  <m:apply>
		    <m:minus/>
		    <m:ci>t</m:ci>
		    <m:ci>Δ</m:ci>
		  </m:apply>
		</m:apply>
	      </m:apply>
	    </m:apply>
	  </m:math>.  	  
	  Now we integrate this sum. Because the integral of a sum
	  equals the sum of the component integrals (integration is
	  linear), we can consider each separately.  Because
	  integration and signal-delay are linear, the integral of a
	  delayed signal equals the delayed version of the
	  integral. The integral is provided in the
	  <cnxn target="result" strength="9">example</cnxn>.
	</para>
      </solution>
    </exercise>


    <para id="conclusion">In this example, we used the table extensively to find the
      inverse Fourier transform, relying mostly on what multiplication
      by certain factors, like
      <m:math>
	<m:apply>
	  <m:divide/>
	  <m:cn>1</m:cn>
	  <m:apply>
	    <m:times/>
	    <m:imaginaryi/>
	    <m:cn>2</m:cn>
	    <m:pi/>
	    <m:ci>f</m:ci>
	  </m:apply>
	</m:apply>
      </m:math> 
      and  
      <m:math> 
	<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>Δ</m:ci>
	    </m:apply>
	  </m:apply>
	</m:apply>
      </m:math>,       
      meant.  We essentially treated multiplication by these factors
      as if they were transfer functions of some fictitious circuit.
      The transfer function
      <m:math>
	<m:apply>
	  <m:divide/>
	  <m:cn>1</m:cn>
	  <m:apply>
	    <m:times/>
	    <m:imaginaryi/>
	    <m:cn>2</m:cn>
	    <m:pi/>
	    <m:ci>f</m:ci>
	  </m:apply>
	</m:apply>
      </m:math> 
      corresponded to a circuit that integrated, and   
      <m:math> 
	<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>Δ</m:ci>
	    </m:apply>
	  </m:apply>
	</m:apply>
      </m:math>
      to one that delayed.  We even implicitly interpreted the
      circuit's transfer function as the input's spectrum! This
      approach to finding inverse transforms -- breaking down a
      complicated expression into products and sums of simple
      components -- is the engineer's way of breaking down the problem
      into several subproblems that are much easier to solve and then gluing the results together.  Along the way we may make the
      system serve as the input, but in the rule
      <m:math>
	<m:apply>
	  <m:eq/>
	  <m:apply>
	    <m:ci type="fn">Y</m:ci>
	    <m:ci>f</m:ci>
	  </m:apply>
	  <m:apply>
	    <m:times/>
	    <m:apply>
	      <m:ci type="fn">X</m:ci>
	      <m:ci>f</m:ci>
	    </m:apply>
	    <m:apply>
	      <m:ci type="fn">H</m:ci>
	      <m:ci>f</m:ci>
	    </m:apply>
	  </m:apply>
	</m:apply>
      </m:math>, 
      which term is the input and which is the transfer function is
      merely a notational matter (we labeled one factor with an
      <m:math><m:ci>X</m:ci></m:math> and the other with an
      <m:math><m:ci>H</m:ci></m:math>).
    </para>


    <section id="TransferFunctions">
      <name>Transfer Functions</name> 
      <para id="TransferIntro">The notion of a transfer function applies well beyond linear
	circuits. Although we don't have all we need to demonstrate
	the result as yet, <emphasis>all</emphasis> linear,
	time-invariant systems have a frequency-domain input-output
	relation given by the product of the input's Fourier transform
	and the system's transfer function. Thus, linear circuits are
	a special case of linear, time-invariant systems. As we tackle
	more sophisticated problems in transmitting, manipulating, and
	receiving information, we will assume linear systems having
	certain properties (transfer functions)
	<emphasis>without</emphasis> worrying about what circuit has
	the desired property. At this point, you may be concerned that
	this approach is glib, and rightly so. Later we'll show that
	by involving software that we really don't need to be
	concerned about constructing a transfer function from circuit elements and op-amps.
      </para>
    </section>


    <section id="CommutativeTransferFunctions">
      <name>Commutative Transfer Functions</name>
      <para id="Interesting">
	Another interesting notion arises from the commutative
	property of multiplication (exploited in an
	<cnxn target="example" strength="5">example above</cnxn>): 	
	We can rather arbitrarily chose an order in which to apply
	each product. Consider a cascade of two linear, time-invariant
	systems. Because the Fourier transform of the first system's
	output is
	<m:math>
	  <m:apply>
	    <m:times/>
	    <m:apply>
	      <m:ci type="fn">X</m:ci>
	      <m:ci>f</m:ci>
	    </m:apply>
	    <m:apply>
	      <m:ci type="fn">
		<m:msub>	    
		  <m:mi>H</m:mi>
		  <m:mn>1</m:mn>
		</m:msub>
	      </m:ci>
	      <m:ci>f</m:ci>
	    </m:apply>
	  </m:apply>      
	</m:math>
	and it serves as the second system's input, the cascade's
	output spectrum is 
	<m:math>
	  <m:apply>
	    <m:times/>
	    <m:apply>
	      <m:ci type="fn">X</m:ci>
	      <m:ci>f</m:ci>
	    </m:apply>
	    <m:apply>
	      <m:ci type="fn">
		<m:msub>	    
		  <m:mi>H</m:mi>
		  <m:mn>1</m:mn>
		</m:msub>
	      </m:ci>
	      <m:ci>f</m:ci>
	    </m:apply>
	    <m:apply>
	      <m:ci type="fn">
		<m:msub>	    
		  <m:mi>H</m:mi>
		  <m:mn>2</m:mn>
		</m:msub>
	      </m:ci>
	      <m:ci>f</m:ci>
	    </m:apply>
	  </m:apply>      
	</m:math>. 
	Because this product also equals 
	<m:math>
	  <m:apply>
	    <m:times/>
	    <m:apply>
	      <m:ci type="fn">X</m:ci>
	      <m:ci>f</m:ci>
	    </m:apply>
	    <m:apply>
	      <m:ci type="fn">
		<m:msub>	    
		  <m:mi>H</m:mi>
		  <m:mn>2</m:mn>
		</m:msub>
	      </m:ci>
	      <m:ci>f</m:ci>
	    </m:apply>
	    <m:apply>
	      <m:ci type="fn">
		<m:msub>	    
		  <m:mi>H</m:mi>
		  <m:mn>1</m:mn>
		</m:msub>
	      </m:ci>
	      <m:ci>f</m:ci>
	    </m:apply>
	  </m:apply> </m:math>, the <emphasis>cascade having the
	linear systems in the opposite order yields the same
	result</emphasis>.  Furthermore, the cascade acts like a
	<emphasis>single</emphasis> linear system, having transfer
	function
	<m:math>
	  <m:apply>
	    <m:times/>
	    <m:apply>
	      <m:ci type="fn">
		<m:msub>	    
		  <m:mi>H</m:mi>
		  <m:mn>1</m:mn>
		</m:msub>
	      </m:ci>
	      <m:ci>f</m:ci>
	    </m:apply>
	    <m:apply>
	      <m:ci type="fn">
		<m:msub>	    
		  <m:mi>H</m:mi>
		  <m:mn>2</m:mn>
		</m:msub>
	      </m:ci>
	      <m:ci>f</m:ci>
	    </m:apply>
	  </m:apply> </m:math>.  This result applies to other
	configurations of linear, time-invariant systems as well; see
	<cnxn target="i9" document="m10350" strength="6">this
	Frequency Domain Problem</cnxn>.  Engineers exploit this
	property by determining what transfer function they want, then
	breaking it down into components arranged according to
	standard configurations.  Using the fact that op-amp circuits
	can be connected in cascade with the transfer function
	equaling the product of its component's transfer function
	(<cnxn target="i25" document="m10349" strength="6">see this
	analog signal processing problem</cnxn>), we find a ready way
	of realizing designs.  We now understand why op-amp
	implementations of transfer functions are so important.
      </para>
    </section>     

  </content>
</document>
