boolean-algebradigital-logic
I am simplifying the sum part of a full adder. The equation I get is:
$$x \cdot \overline y \cdot \overline {cin} + \overline x \cdot \overline y \cdot cin + x \cdot y \cdot cin + \overline x \cdot y \cdot \overline {cin}$$
How do I simplify this to get the two xor gates that make up the sum part of a full adder?
For a single-bit adder, the worst-case propagation delay is the longest time it will take to get a stable output, which for this case is 16ns.
simulate this circuit – Schematic created using CircuitLab
However, for an N-bit adder, you should consider the path from the input's least significant bit (LSB) to a stable output's most significant bit (MSB)-- $$A_0 \rightarrow C_{out,0} \rightarrow C_{out,1} \rightarrow C_{out,2} \rightarrow S_3 (1)$$ (Cin, i is implied, but not shown).
In the case of your circuit, we're given the following paths $$A_i \rightarrow S_i = 16ns (2)$$ $$A_i \rightarrow C_{out,i} = 12ns (3)$$ $$C_{in,i} \rightarrow S_i = 8ns (4)$$ $$C_{in,i} \rightarrow C_{out,i} = 4ns (5)$$
From this, we calculate the longest path previously shown in (1). This calculates to be $$12ns + 8ns + 8ns + 8ns = 36ns$$
For more information, please refer to this lecture on adder circuits
One can not implement every logical function using only XOR gates.
Since XOR is a logical operator which obeys associativity, the function implemented using only XOR gates can always be written as $$f = a_1\oplus a_2 \oplus a_3 \oplus \cdots \oplus a_n $$ where \$a_1, a_2 \ldots\$ are the inputs.
So the only possible functions that can be implemented using XOR gates are:
$$\tag1 f= \begin{array}{|cl} 1 & \mathrm{if }\ N = odd \\0 & else \end{array}$$
Where N is the number of 1's in the input. \$f\$ can be implemented as $$f = a_1\oplus a_2 \oplus a_3 \oplus \cdots \oplus a_n $$
$$\tag2 f= \begin{array}{|cl} 1 & \mathrm{if }\ N = even \\0 & else \end{array}$$
can be implemented as $$f = a_1\oplus a_2 \oplus a_3 \oplus \cdots \oplus a_n \oplus 1 $$
$$\tag3 f = \overline{a}$$ can be implemented as $$f = a\oplus 1 $$
In your case, SUM can be implemented using XOR gates only (as given in (1)) but not COUT.
Best Answer
If you are trying to use the various laws of Boolean algebra to go from the equation you have to a pair of XOR gates, it is not too tricky, you just have to spot similarities.
Lets look at your equation, what can we do?
$$x \cdot \overline y \cdot \overline c + \overline x \cdot \overline y \cdot c + x \cdot y \cdot c + \overline x \cdot y \cdot \overline c$$
Try taking out factors:
Notice there are some \$c\$ terms in common:
$$\begin{align} x \cdot \overline y \cdot \overline c + \overline x \cdot y \cdot \overline c &= (x \cdot \overline y + \overline x \cdot y) \cdot \overline c\\ \overline x \cdot \overline y \cdot c + x \cdot y \cdot c &= (\overline x \cdot \overline y + x \cdot y) \cdot c\\ \end{align}$$
That would leave us with:
$$(\overline x \cdot \overline y + x \cdot y) \cdot c + (x \cdot \overline y + \overline x \cdot y) \cdot \overline c$$
Using the standard table of Boolean algebra functions, we can easily spot the XNOR and XOR gates:
$$(\overline x \cdot \overline y + x \cdot y) \cdot c + (x \cdot \overline y + \overline x \cdot y) \cdot \overline c = \overline{(x \oplus y)}\cdot c + (x \oplus y)\cdot\overline c$$
$$\overline{(x \oplus y)}\cdot c + (x \oplus y)\cdot\overline c = \overline d \cdot c + d \cdot \overline c$$
$$\overline d \cdot c + d \cdot \overline c = d \oplus c$$
$$d \oplus c = (a \oplus b) \oplus c $$