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Unformatted text preview: 3 Block diagrams and operators: Two new representations 3.1 Disadvantages of difference equations 34 3.2 Block diagrams to the rescue 35 3.3 The power of abstraction 40 3.4 Operations on whole signals 41 3.5 Feedback connections 45 3.6 Summary 49 The goals of this chapter are: • to introduce two representations for discretetime systems: block diagrams and operators; • to introduce the wholesignal abstraction and to exhort you to use abstraction; • to start manipulating operator expressions; • to compare operator with differenceequation and blockdiagram manipulations. The preceding chapters explained the verbaldescription and difference equation representations. This chapter continues the theme of multiple representations by introducing two new representations: block diagrams and operators. New representations are valuable because they suggest new thoughts and often provide new insight; an expert engineer values her representations the way an expert carpenter values her tools. This chapter first introduces block diagrams, discusses the wholesignal abstraction and 34 3.1 Disadvantages of difference equations the general value of abstraction, then introduces the operator representa tion. 3.1 Disadvantages of difference equations Chapter 2 illustrated the virtues of difference equations. When compared to the verbal description from which they originate, difference equations are compact, easy to analyze, and suited to computer implementation. Yet analyzing difference equations often involves chains of micromanipulations from which insight is hard to find. As an example, show that the difference equation d [ n ] = a [ n ] − 3a [ n − 1 ] + 3a [ n − 2 ] − a [ n − 3 ] is equivalent to this set of equations: d [ n ] = c [ n ] − c [ n − 1 ] c [ n ] = b [ n ] − b [ n − 1 ] b [ n ] = a [ n ] − a [ n − 1 ] . As the first step, use the last equation to eliminate b [ n ] and b [ n − 1 ] from the c [ n ] equation: c [ n ] = ( a [ n ] − a [ n − 1 ])  {z } b [ n ] −( a [ n − 1 ] − a [ n − 2 ])  {z } b [ n − 1 ] = a [ n ]− 2a [ n − 1 ]+ a [ n − 2 ] . Use that result to eliminate c [ n ] and c [ n − 1 ] from the d [ n ] equation: d [ n ] = ( a [ n ] − 2a [ n − 1 ] + a [ n − 2 ])  {z } c [ n ] −( a [ n − 1 ] − 2a [ n − 2 ] + a [ n − 3 ])  {z } c [ n − 1 ] = a [ n ] − 3a [ n − 1 ] + 3a [ n − 2 ] − a [ n − 3 ] . Voilà: The threeequation system is equivalent to the single difference equa tion. But what a mess. Each step is plausible yet the chain of steps seems random. If the last step had produced d [ n ] = a [ n ] − 2a [ n − 1 ] + 2a [ n − 2 ] − a [ n − 3 ] , it would not immediately look wrong. We would like a representation where it would look wrong, perhaps not immediately but at least quickly....
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This note was uploaded on 08/24/2011 for the course EECS 6.003 taught by Professor Dennism.freeman during the Spring '11 term at MIT.
 Spring '11
 DennisM.Freeman

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