digital circuit output

Question

$\text{what is the o/p of this circuit ?}$

Answer 1

The circuit is 

The Output of the $1^{st}$ NOR Gate will be 

Truth Table :

$a$	$a$	$Output = \overline{a}$
$0$	$0$	$1$
$1$	$1$	$0$

The output of the $2^{nd}$ NOR Gate will be: 

Truth Table :

$b$	$c$	$Output = \overline{b+c}$
$0$	$0$	$1$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$0$

The output of the $3^{rd}$ NOR Gate will be 

Truth Table :

$a$	$b$	$c$	$\overline{a}$	$b+c$	$\overline{b+c}$	$\overline{a}+\overline{b+c}$	$\overline{\overline{a}+\overline{b+c}} = $ $a.(b+c)$
0	0	0	1	0	1	1	0
0	0	1	1	1	0	1	0
0	1	0	1	1	0	1	0
0	1	1	1	1	0	1	0
1	0	0	0	0	1	1	0
1	0	1	0	1	0	0	1
1	1	0	0	1	0	0	1
1	1	1	0	1	0	0	1

∴ The output of the circuit will be : $f = \overline{\overline{a}+\overline{b+c}} = a.(b+c)$

Comments

Hands down, thanks!

Answer 2

You could do it directly using $NOR$ Gate, Small circuits, Wouldn't take much time.

But must faster way to solve it is that It is a Two-level NOR-NOR Circuits, So, It can be Replaced by Two Level OR-AND circuit. i.e. Both NOR Gates at the first level can be replaced by OR gates and the NOR gate at the second level can be replaced by AND gate. Thus, We have $f = a(b+c)$

Comments

thanks

Answer 3

F= a(b+c)

Comments

Thanks! :)