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                  NPP               Number System, Boolean Algebra and Logic Circuits              201


                       Problem 3.54                                àíZ 3.54
                      Draw the truth table for the following      {ZåZ hoVw gË` Vm{bH$m ~ZmAmo…
                  Boolean functions:
                                              (a)  F =  C . B . A  +  C . B . A  +  C . B . A
                                              (b)  F =  ∑ m ( 1,0  ) 3 ,

                                              (c)  Q =  C . B . A  +  B . A  +  C . B . A

                  Solution:                                   hc:
                      (a) The given expression is:                (a)  {X`m J`m ì`§OH$ h¡…
                                                 F =    C . B . A  +  C . B . A  +  C . B . A
                      As we can see, there are three minterms     O¡gm {H$ ñnîQ> h¡, VrZ {_ÝQ>_© {ZåZmZwgma VrZ _mZm|
                  which correspond to the following input set of  Ho$ gmnoj h¡…
                  values:

                                                         C . B . A    010 ,
                                                            →
                                                         C . B . A    110 ,
                                                            →
                                                            →
                                                         C . B . A    111
                      Therefore, the truth table will contain three  Bg{bE gË` Vm{bH$m _|  A, B VWm  C Ho$ D$na
                  1s corresponding to the above mentioned val-  ~Vm`o J`o _mZm| go gå~pÝYV VrZ 1s hm|JoŸ& F Ho$ AÝ`
                  ues of A, B and C. All other values for F will be  g^r _mZ 0 hm|JoŸ& dm§{N>V gË` Vm{bH$m 8 n§p³V¶m| VWm
                  ‘0’. The desired truth table can be drawn with
                  the help of eight rows and four columns.    4 H$m°b_ H$s ghm`Vm go ~Zm`r Om gH$Vr h¡Ÿ&
                                                  Truth Table
                                      A              B           C           F
                                      0              0            0          0
                                      0              0            1          0
                                      0              1            0          1
                                      0              1            1          0
                                      1              0            0          0
                                      1              0            1          0
                                      1              1            0          1
                                      1              1            1          1
                      (b) The given expression is  F = Σm (0, 1,  (b) g_rH$aU  F = Σm (0, 1, 3)  Xmo Ma am{e`m|
                  3). This is a two variable function because the  dmbm ì`§OH$ h¡ Š`m|{H$ g~go ~‹S>r g§»`m 3 H$mo ~mBZar _|
                  highest number 3 can be represented as 11.  11 {bIm Om gH$Vm h¡ {Og_| {g\©$ Xmo {~Q>| h¢Ÿ& VrZ
                  There are three 1s in the Truth Table for the fol-
                  lowing combinations:                        {_ÝQ>_© {ZåZ ~mBZar g§»`mAm| hoVw h¢…
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