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Please Help? Confused on how to do this ?
Sep 24, 2014 at 5:10pm
a) A circuit has four inputs, P, Q, R, and S, representing the natural binary numbers 0000 = 0, to 1111 = 15. P is the most-significant bit.The circuit has one output, X, which is true if the number represented by the input is divisible by three (regard zero as being indivisible by three.) Design a truth table for this circuit and hence obtain an expression for X in terms of P, Q, R, and S. Give the circuit diagram of an arrangement of AND, OR, and NOT gates to implement this circuit. Design a second circuit to implement this function using NAND gates only.
b) A logic circuit has three inputs C, B, and A, where A is the least-significant bit.The circuit has three outputs R, Q, and P. For any binary code applied to the input terminals (A, B, and C) the output is given by the input plus 1; for example, if C, B,A = 0,1, 1, the output R, Q, P is 1,0,0. Note that 111 + 1 = 000 (i.e. there is no carry out). Design a circuit to implement this system.
b) A logic circuit has three inputs C, B, and A, where A is the least-significant bit.The circuit has three outputs R, Q, and P. For any binary code applied to the input terminals (A, B, and C) the output is given by the input plus 1; for example, if C, B,A = 0,1, 1, the output R, Q, P is 1,0,0. Note that 111 + 1 = 000 (i.e. there is no carry out). Design a circuit to implement this system.
Last edited on Sep 24, 2014 at 5:10pm
Sep 24, 2014 at 5:56pm
a) Truth table
If the table would be just that, then
However, there are more lines and thus more circuits.
PQRSX 00000 00010 00100 00111 01000 ... |
If the table would be just that, then
X = AND( AND( R, S ), AND( NOT(P), NOT(Q) ) )However, there are more lines and thus more circuits.
Sep 24, 2014 at 6:54pm| The circuit has one output, X, which is true if the number represented by the input is divisible by three (regard zero as being indivisible by three.) |
Whenever your number is divisible by 3, X is zero. Here's the truth table expanded past 6 to show that both 3 and 6 have a 1 for X
PQRS X 0 0000 0 1 0001 0 2 0010 0 3 0011 1 4 0100 0 5 0101 0 6 0110 1 7 0111 0 8 .... . etc. |
Sep 24, 2014 at 8:01pm
Same thing: create a truth table, create a boolean formuals for it, simplify it using De Morgan's laws and implement it using NAND logic.
Sep 24, 2014 at 8:12pm
So CBA will have 000 and then so on but when will the output be 1 or 0?
Sep 24, 2014 at 8:19pm
You have 3 outputs now.
CBA represent 3 digits of binary number, RQP represent digits of another binary number 1 lagger than input, so truth table will begin like:
CBA represent 3 digits of binary number, RQP represent digits of another binary number 1 lagger than input, so truth table will begin like:
╔═╤═╤═╦═╤═╤═╗ ║C│B│A║R│Q│P║ ╠═╪═╪═╬═╪═╪═╣ 0║0│0│0║0│0│1║1 ╟─┼─┼─╫─┼─┼─╢ 1║0│0│1║0│1│0║2 ╟─┼─┼─╫─┼─┼─╢ 2║0│1│0║0│1│1║3 ╟─┼─┼─╫─┼─┼─╢ 3║0│1│1║1│0│0║4 ╟─┼─┼─╫─┼─┼─╢ |
Last edited on Sep 24, 2014 at 8:19pm
Sep 24, 2014 at 8:32pmTopic archived. No new replies allowed.