This summary looks at the work of the course and how it can be tested using multi-choice type questions. They will not be the same as the test itself, but will give some guidance on how to approach these type of questions.

a)110010

b)010101

c)110111

d)101111

e)000111

METHOD: Write down binary place values until they are bigger than the decimal, i.e. 1, 2, 4, 8, 16, 32, 64. Now start the conversion. Take the nearest lower place value to the decimal, in this case 32, from the decimal, i.e. 47 - 32 = 15. Take the nearest lower binary value from the remainder, i.e. 15 - 8 = 7. What place values make 7, well 4 + 2 + 1. So the conversion is 32 column = 1, 16 column = 0, 8 column = 1, 4 column = 1, 2 column = 1, and 1 column = 1. CHECK: 32 + 8 + 4 + 2 + 1 = 47, so binary pattern is 101111 which is answer (d) above.

a) 19

b) 24

c) 23

d) 37

e) 22

METHOD: Convert Binary using place values 16, 8, 4, 2 and 1. We only need five as that is the number of digits in the binary number, Conversion is simply each digit multiplied by the place value and added together. So in this case (1x16) + (0 x 8) + (1 x 4) + (1 x 2) + (1 X 1) = 16 + 0 + 4 + 2 + 1 which is 23.

a) (C+D).A.B

b) (A.C.D)+B

c) A.B.C.D

d) (A+B).C.D

METHOD: Go across the circuit from left to right, noting what is there as you pass over it. The first thing in the example is B in Parallel with A, so that is A OR B, written A+B. Then C and D follow in series, which is an AND, giving (A OR B) AND C AND D. This is written (A+B).C.D which is answer (d).

A B C D (a) (b) (c) (d) (e) 0 0 0 0 0 0 0 1 1 0 0 0 1 0 1 1 1 0 0 0 1 0 1 0 1 0 1 0 0 1 1 1 0 1 0 1 0 1 0 0 0 1 0 0 0 0 1 0 1 0 1 1 0 0 0 1 1 0 1 0 1 1 1 0 1 1 1 1 1 1 1 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 1 0 1 0 1 1 1 1 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 0 1 1 0 1 1 1 1 0 0 1 1 0 0 1 1 1 1 0 1 1 1 1

A B X A B X A B X A B X 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 0 1 1 0 1 1 0 1 0 1 0 1 1 0 1 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 1 1 (a) (b) (c) (d)

METHOD: EXCLUSIVE OR has a 1 out when either A=1 and B=0 OR when A=0 and B=1 but not otherwise. An EXCLUSIVE NOR is the INVERSE, i.e. it will be a 0 when either A=1 and B=0 OR A=0 and B=1 but not otherwise. The only truth table with this pattern is (d) so that is the correct answer.

a) B+A.NOT B

b) A+NOT A.B

c) A+B

d) A+NOT B

METHOD: Use the Boolean rules to find the obvious equivalents, i.e. take out the common factors as in the following.

(NOT A+A).B + A.NOT B, NOT A+A is always 1, so B+A.NOT B, (a), is OK. (B+NOT B).A + NOT A.B, NOT B+B is always 1, so A+NOT A.B, (b), is OK. Take (c), A+B is only 0 when A=0 and B=0. Is this true in the case of (a) and (b)? Put values in (a), 0+0.1 which is 0. So (c) is OK. This means that the answer is (d) NOT EQUAL TO L.

a) B.B.B

b) B.1

c) B+0

d) B+NOT B

e) B+B

METHOD: Just use the simple boolean rules!

a)(NOT A.B)+C

b)NOT(A.B)+C

c)(NOT A+B).C

d)NOT(A+B).C

METHOD: Start at the left hand side top and move across, noting gates as they are found. First gate is an inverter on line A, so signal is NOT A. Then this signal is connected to a 2 INPUT AND Gate with C, giving NOT A.B. Finally the output of the AND gate is connected to a 2 INPUT OR Gate with C giving an output X = (NOT A.B)+C.

X = (NOT A.B) + (A.NOT B)

METHOD: Start on the left of the first diagram and follow variable A say, find an INVERTER in the signal line. Does this go to an AND gate, if so is the other signal on the AND gate variable B. IF no repeat with next diagram until this sequence is found. Then check that the rest has A AND an INVERTER on B, both AND gates output being connected to an OR gate as in (a).

METHOD: Expand the "bracket", so we get NOT C.A + NOT C.NOT B. We then include some more terms for the missing variables, as in NOT C.A = NOT C.A.B + NOT C.A.NOT B. B+NOT B =1, so this expansion has not altered the expression. The same is done for NOT C.NOT B giving NOT C.NOT B.A + NOT C.NOT B.NOT A. We now have four terms but two of them are the same. So we are left with

NOT C.A.B

NOT C.A.NOT B

NOT C.NOT A.NOT B

Plotting these on the map gives (d) as the answer to this question.