In: Computer Science
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Boolean Functions, Truth Tables, Logic Minimization, Two-Level Forms Consider a boolean function f (a, b, c, d). Suppose that the function is 1 if
• There is a single 1 among the inputs, or
• There is a single 0 among the inputs, or
• There are exactly two 1’s among the inputs
and it is 0 otherwise.
(a) Write down a truth table for the function
(b) Using a Karnaugh map, provide a minimal sum-of-products (AND-OR) expres- sion.
(c) Using a Karnaugh map, provide a minimal product-of-sums (OR-AND) expres- sion.
(d) Provide a minimal NAND-NAND expression
(e) Provide a minimal OR-NAND expression (f) Provide a minimal NOR-OR expression
(g) Provide a minimal NOR-NOR expression (h) Provide a minimal AND-NOR expression
(i) Provide a minimal NAND-AND expression
(j) Provide a AND-XOR expression (with no negations)
(a) Truth table
Input | Output | |||
a | b | c | d | f |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 1 |
0 | 0 | 1 | 0 | 1 |
0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 0 |
(b) minimal sum-of-products
f=ab' + cd' + bc' + da'
(c) minimal product-of-sums
f ' = abcd + a'b'c'd'
( f ')' = ( abcd + a'b'c'd' )'
f =( a' + b' + c' + d') ( a + b + c + d)
(d) minimal NAND-NAND expression
f = ab' + cd' + bc' + da'
f = ( (ab')' (cd')' (bc')' (da')' )'
(e) minimal OR-NAND expression
f = ab' + cd' + bc' + da'
f = ( (ab')' (cd')' (bc')' (da')' )'
f = ( (a' + b) (c' + d) (b' + c) (d' + a) )'
(f) NOR-OR expression
f = ab' + cd' + bc' + da'
f = ( (ab')' (cd')' (bc')' (da')' )'
f = ( (a' + b) (c' + d) (b' + c) (d' + a) )'
f = (a' + b)' + (c' + d)' + (b' + c)' + (d' + a)'
(g) minimal NOR-NOR expression
f = ( a' + b' + c' + d') ( a + b + c + d)
f =[ {( a' + b' + c' + d' )}' + {( a + b + c + d )}' ]'
(h) AND-NOR expression
f ' = abcd + a'b'c'd'
( f ')' = ( abcd + a'b'c'd' )'
f = ( abcd + a'b'c'd' )'
(i) NAND-AND expression
f ' = abcd + a'b'c'd'
( f ')' = ( abcd + a'b'c'd' )'
f = ( abcd + a'b'c'd' )'
f = ( abcd )' ( a'b'c'd')'