Question

c) Simplify the expression f(a,b,c)=a.b+a.(b+c)+(a.b.(c+b.d)+a.b).c.d using rules of Boolean algebra
d) Compute the value of 3AB16-43510-6178 using 1’s complement arithmetic leaving your final answer in Octal.

Answer 1

(d) $3AB16-43510-6178$

Changing to binary form

$3AB16 \to 00111010101100010110$

$43510 \to 0100001101011010$

$6178 \to 0110000101111000$ c

1's component of $3AB16 \to 11000101010011101001$

1's component of $43510 \to 1011110010100101$

1's component of $6178 \to 1001111010000111$

Now $3AB16-43510-6178$ in octal form,

$652351-117207-60570$

(c) $f(a,b,c)=a.b+a.(b+c)+(a.b.(c+b.d)+a.b).c.d$

$=a.b.c+\bar c + a.(b +\bar b).(c + \bar c)((a+\bar a)+b+c)+a.b.(c+\bar c)(a+\bar a).(b+\bar b).c + (a+\bar a).b.(c+\bar c).d$

solving it, we get+

= $a.b.c.d+a.b.c.\bar d + a.b.\bar c.\bar d +a.b.\bar c.d + a.b.\bar c .d + a.b.\bar c .\bar d + a.\bar b.c.d + a.\bar b.c.\bar d + a.b.c.\bar d+a.b.\bar c. d + a.b.\bar c.d + a.b.c.\bar d + a.b.\bar c.d + a.b.\bar c.\bar d+a.b.c.d + \bar a .b.c.d + a.\bar b.c.d+\bar a.\bar b.c.d$

= $a.b.c.d+a.b.c.\bar d+\bar a.b.c.d+a.b.\bar c.\bar d+a.\bar b.c.\bar d+\bar a.\bar b.c.d$