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chap3 - 43 43 3 Block diagrams and operators Two new...

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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 discrete-time systems: block diagrams and operators; to introduce the whole-signal abstraction and to exhort you to use abstraction; to start manipulating operator expressions; to compare operator with difference-equation and block-diagram manipulations. The preceding chapters explained the verbal-description 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 whole-signal abstraction and
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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 micro-manipulations 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 three-equation 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. Block diagrams are one such representation.
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3 Block diagrams and operators: Two new representations 35 Exercise 12. Although this section pointed out a disadvantage of difference equations, it is also important to appreci- ate their virtues. Therefore, invent a verbal descrip- tion (a story) to represent the single equation d [ n ] = a [ n ] − 3a [ n 1 ] + 3a [ n 2 ] − a [ n 3 ] and then a verbal description to represent the equivalent set of three equations. Now have fun
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