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Topic 1.2.2 – System simplification using Boolean Algebra Learning Objectives: At the end of this topic you will be able to;  Generate the Boolean expression for a system [with up to 3 inputs] from a truth table;  Generate the Boolean expression for a system [with up to 4 inputs] from a logic diagram;  Recall and use the following identities o A.1  A o A.0  0 o A.A  0 o A  1 1 o A0  A o AA 1
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Topic 1.2.2 – System simplification using Boolean Algebra Learning Objectives: At the end of this topic you will be able to;  Generate the Boolean expression for a system [with up to 3 inputs] from a truth table;  Generate the Boolean expression for a system [with up to 4 inputs] from a logic diagram;  Recall and use the following identities o  A1A  = . o  00A  = . o  0AA  = . o  11A  =+ o  A0A  =+ o  1AA  =+  elect and use the following identities o  BABAA  +=+  . o  A1BAABA  =+=+  )(.  Apply !e organ#s theorem to simplify a logic system [with up to 3 inputs] 1  o!ule T1 #ntro!uction to Analogue an! \$igital Systems.Boolean Algebra. \$n our pre%ious unit we disco%ered that logic systems can become &uite complex if we don#t use any simplification techni&ues' \$n this topic you will be introduced to the first method of simplification( Boolean Algebra')e ha%e already introduced some of the ideas behind Boolean Algebra in simple terms in *opic +'+'+( and +','+' -ere is a brief summary of the idea#s we ha%e co%ered so far'  A bar( on top of an input %ariable( is used to represent the ./* function' 0'g' ./* C  1 C  A dot 2'#( is used to represent the A.! function' 0'g' A  and B  1 BA .  A plus 2#( is used to represent the /R function' 0'g' E  /R F  1 FE +  A plus with a circle 2  #( is used to represent the 0xclusi%e /R 05/R6 function' 0'g' C  05/R D  1 DC ⊕ %ercise 1:  )rite down the Boolean 0xpressions for the following examples7+'./* X 8888888888888888,' R  A.! T 88888888888888883' S  /R R 88888888888888884' C  05/R G 88888888888888889'./* B  A.! C 8888888888888888:'./* X  /R ./*  Y 8888888888888888' P  05/R ./* B 8888888888888888<'./*  A  A.! B 68888888888888888=' X  ./R  Y 8888888888888888 2   Topic 1.2.2 – System simplification using Boolean Algebra *he following table should remind you of the wor> we did in *opic +'+'+'GateymbolBoolean 0&uation./*   AQ = A.!  BAQ  . = /R  BAQ  += .A.!  BAQ  . = ./R  BAQ  += 05/R BABAQ or BAQ ..  +=⊕= 05./R BABAQ or BAQ ..  +=⊕= .ow let us put this into practice' *here are two ways in which Boolean expressions for a logic system can be formed( either from a truth table or from a logic circuit diagram' )e will now consider each of these in turn starting with the easiest( which is to complete a Boolean expression from a truth table' 3  A Q AB Q ABQ ABQ AQ ABQ ABQ  o!ule T1 #ntro!uction to Analogue an! \$igital Systems.\$eriving a Boolean %pression from a Trut& Table ?onsider the following truth table of a particular logic system' .ote 7 at thisstage it does not matter what the system is meant to do6' CBAQ @@@@@@++@+@@@++++@@++@++++@@+++@\$n this table there are four combinations of inputs that will produce an output at Q ' \$n order to write down the Boolean 0xpression for the whole system we first ha%e to write down the Boolean 0&uation for each line in the truth table where the output is a +' All input %ariables three in this case6 must be included in each Boolean 0&uation on each line( as shown below CBAQ Boolean   'uation @@@@@@++  CBA  .. @+@@@+++  CBA  .. +@@+  CBA  .. +@++  CBA  .. ++@@+++@*o obtain the Boolean expression for the whole system we simply ta>e each of these terms and 2/R# them together' 4 CBACBACBACBAQ  ........  +++=
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