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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¢…