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# Digital electronics problem

A logic circuit has three Boolean inputs X,Y and Z. Its output is F(X,Y,Z) such that:

F(X,Y,Z) = 1 if aX + bY +cZ > d

= 0 otherwise.

a,b,c,d are real constants.

For which of the following values of a,b,c,d does this circuit represent an implementation

of a three-input NAND gate with inputs X,Y and Z?

I. a = b = c = 1; d = 2.5

II. a = b = c = 1; d = 1.5

III. a = b = c = -1; d = 0

IV. None of the above

F(X,Y,Z) = 1 if aX + bY +cZ > d

= 0 otherwise.

a,b,c,d are real constants.

For which of the following values of a,b,c,d does this circuit represent an implementation

of a three-input NAND gate with inputs X,Y and Z?

I. a = b = c = 1; d = 2.5

II. a = b = c = 1; d = 1.5

III. a = b = c = -1; d = 0

IV. None of the above

IV. None of the above

nowire9 can u please explain this answer

how did u get it?

how did u get it?

1. Refer to the 3-input NAND truth table (the only case where F(X,Y,Z)=1 is when X=Y=Z=1, else F(X,Y,Z)=0):

X,Y,Z | F(X,Y,Z)

-----------------------

0,0,0 | 1

... | 1

1,1,1 | 0

2. Test to rule out false answers...

I. if X=1,Y=1,Z=1, then aX+bY+cZ>d is satisfied, but F(X,Y,Z)=0, not 1

II. if X=0,Y=0,Z=1, then aX+bY+cZ>d is NOT satisfied, but F(X,Y,Z)=1, not 0

III. if X=0,Y=1,Z=1, then aX+bY+cZ>d is NOT satisfied, but F(X,Y,Z)=1, not 0

3. This only leaves IV as a possible answer

X,Y,Z | F(X,Y,Z)

-----------------------

0,0,0 | 1

... | 1

1,1,1 | 0

2. Test to rule out false answers...

I. if X=1,Y=1,Z=1, then aX+bY+cZ>d is satisfied, but F(X,Y,Z)=0, not 1

II. if X=0,Y=0,Z=1, then aX+bY+cZ>d is NOT satisfied, but F(X,Y,Z)=1, not 0

III. if X=0,Y=1,Z=1, then aX+bY+cZ>d is NOT satisfied, but F(X,Y,Z)=1, not 0

3. This only leaves IV as a possible answer