A controller uses the circuit shown below. Inputs A and B are binary sensor signals. Gate G1 produces P. Gate G2 produces Q. The final output is X. A small ...
Question 1 Report
A controller uses the circuit shown below. Inputs A and B are binary sensor signals. Gate G1 produces P. Gate G2 produces Q. The final output is X. A small circle on the output of a gate represents inversion.
Use NOT, AND and OR, or equivalent Boolean notation, in your answers. The following table gives two combinations of the inputs.
| A | B | P | X |
|---|---|---|---|
| 0 | 0 | ||
| 1 | 0 |
(a) Identify G1. [1]
(b) State the value of P when A = 1 and B = 1. [1]
(c) Complete the P and X columns of the table. [4]
(d) Complete the expression P = NOT(A ___ B). [1]
(e) Give a Boolean expression for Q. [1]
(f) Give a Boolean expression for X before simplification. [2]
(g) Convert your expression for X into its simplest Boolean expression. Show your working. [2]
(h) Explain why NAND gates are described as universal gates. [3]
Answer Details
(a) G1 is a NAND gate [1]: it is an AND gate followed by inversion.
(b) With A = 1 and B = 1, \(P=\operatorname{NOT}(1\text{ AND }1)=\textbf{0}\) [1].
(c)
| A | B | P | X |
|---|---|---|---|
| 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 |
For the final AND, P is combined with \(Q=\operatorname{NOT}A\) [4].
(d) \(P=\operatorname{NOT}(A\text{ AND }B)\) [1].
(e) \(Q=\operatorname{NOT}A\) [1].
(f) \(X=\operatorname{NOT}(A\text{ AND }B)\text{ AND }\operatorname{NOT}A\) [2].
(g) By De Morgan's law, \[X=(\operatorname{NOT}A\text{ OR }\operatorname{NOT}B)\text{ AND }\operatorname{NOT}A=\operatorname{NOT}A\] The simplest expression is \(\textbf{NOT }A\) [2].
(h) NAND can make NOT by joining its inputs: \(\operatorname{NAND}(A,A)=\operatorname{NOT}A\). It can then form AND by inverting a NAND output, and OR using inverted inputs and De Morgan's law. Thus any Boolean circuit can be built from NAND gates [3].
Download The App On Google Playstore
Everything you need to excel in your exams
