Question

Here are the LED versions of the digits from 0 to 9. And the versions of A, b, Cd, E and F. 23455183 Beyond Karnaugh reductions: is 1 NAND gate,p+ q is 2 NAND gates Beyond Karnaugh reductions: pq + p is 4 NAND gates, p + pq is 5 NAND gates (9 points) Fill in the Karnaugh map below for the upper right LED for the decimal numbers. The DC values have already been filled in. Then use Karnaugh map techniques to reduce to a minimal circuit. (Hint: Three terms is minimal.) 84 83 84 84 21 DC DC 21 DC DC 21 DC 21 DC Minimal circuit: - Number of NAND gates: (14 points) The Karnaugh map below is for the upper right LED for the hexadecimal case. This still includes the values for 0 through 9 and there are no -don't care options. (Hint Five terms is minimal.) 84 8 84 84 Minimal circuit 21 21 21 21 Number of NAND gates:

Answer 1

The Required LED = 'b' A = 10 b = 11 c = 12 d = 13 E = 14 F = 15 For hexadecimal numbers, b = sigma_m (10, 13) + sigma_m (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) b = 8^bar 4^bar + 4^bar 1^bar + 4^bar 2^bar + 8 2^bar 1 + 8^bar 2^bar 1^bar + 8^bar 2 1 = 4^bar (8^bar + 1^bar) + 4^bar 2^bar + 2^bar(8 1^bar + 8^bar 1^bar) + 8^bar 2 1 = 46bar (8^bar + 1^bar) + 4^bar 2^bar + 2^bar (8^bar exclus^bar 1) + 8^bar 2 1 No. of nand gates = (2 + 1) + (1 + 1) + (1 + 1) + 1 + 1 + 1 + 1 = 3 + 2 + 2 + 4 = 11 (gate)