10 Boolean logic

In this chapter, you will learn about:

+ the identification, definition, symbols and functions of the standard logic gates:
NOT, AND, OR, NAND, NOR and XOR
+ how to use logic gates to create logic circuits from:
– a given problem
– a logic expression
– a truth table
+ how to complete truth tables from:
– a given problem
– a logic expression
– a logic circuit
+ how to write a logic expression from:
– a given problem
– a logic circuit
– a truth table.

10.1 Standard logic gate symbols

Electronic circuits in computers, solid state drives and controlling devices
ess
are made up of thousands of logic gates. Logic gates take binary inputs
and produce a binary output. Several logic gates combined together form a
logic circuit and these circuits are designed to carry out a specific function.
The checking of the output from a logic gate or logic circuit is done using a
truth table.
This chapter will consider the function and role of logic gates, logic
circuits and truth tables. Also a number of possible applications of logic
circuits will be considered. A reference to Boolean algebra will be made
throughout the chapter; but this is really outside the scope of this text book.
However, Boolean algebra will be seen on many logic gate websites and
is included here for completeness, since many students may prefer this
notation to logic statements.

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 10.1 Standard logic gate symbols

10.1.1 Logic gate symbols

Six different logic gates will be considered in this chapter:

NOT gate AND gate

OR gate NAND gate

NOR gate XOR gate

V Figure 10.1 Logic gate symbols

Truth tables
Truth tables are used to trace the output from a logic gate or logic circuit. The
NOT gate is the only logic gate with one input; the other five gates have two
inputs (see Figure 10.1).
Although each logic gate can only have one or two inputs, the number of
inputs to a logic circuit can be more than 2; for example, three inputs give a
possible 23 (=8) binary combinations. And for four inputs, the number of possible
binary combinations is 24 (=16). It is clear that the number of possible binary
combinations is a multiple of the number 2 in every case. The possible inputs in
a truth table can be summarised as shown in Table 10.1.
W Table 10.1 All possible inputs for truth tables with two, three and four inputs

Inputs Inputs Inputs

A B A B C A B C D
0 0 0 0 0 0 0 0 0
0 1 0 0 1 0 0 0 1
1 0 0 1 0 0 0 1 0
1 1 0 1 1 0 0 1 1
1 0 0 0 1 0 0
1 0 1 0 1 0 1
1 1 0 0 1 1 0
1 1 1 0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1

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10 BOOLEAN LOGIC

As we can see, a truth table will also list the output for every possible
combination of inputs.

10.2 The function of the six logic gates

10.2.1 NOT gate
Description: Truth table: How to write this:
A X
W Table 10.2

V Figure 10.2 The output, X, is 1 if: Input Output X = NOT A (logic notation)
A X —
the input, A, is 0 X = A (Boolean algebra)
0 1
1 0

Note the use of Boolean algebra to

represent logic gates. This is optional at
IGCSE but many students may prefer to
10.2.2 AND gate use this notation (see NOTE later).

Description: Truth table: How to write this:

A
X
B W Table 10.3

V Figure 10.3 Inputs Outputs

The output, X, is 1 if: A B X X = A AND B (logic notation)

0 0 0
both inputs, A and B, are 1 X = A . B (Boolean algebra)
0 1 0
1 0 0
1 1 1

10.2.3 OR gate
Description: Truth table: How to write this:
A
X
B W Table 10.4

V Figure 10.4 Inputs Output

The output, X, is 1 if: A B X X = A OR B (logic notation)

either input, A or B, or 0 0 0
both, are 1 0 1 1 X = A + B (Boolean algebra)
1 0 1
1 1 1

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 10.2 The function of the six logic gates

10.2.4 NAND gate (NOT AND)

A Description: Truth table: How to write this:
X
B
W Table 10.5
V Figure 10.5
Inputs Output
The output, X, is 1 if: A B X
X = A NAND B (logic notation)
input A AND 0 0 1
X = A . B (Boolean algebra)
input B are NOT both 1 0 1 1
1 0 1
1 1 0

10.2.5 NOR gate (NOT OR)

A Description: Truth table: How to write this:
X
B
W Table 10.6
V Figure 10.6
Inputs Output
X = A NOR B (logic notation)
The output, X, is 1 if: A B X
neither input A nor 0 0 1
X = A + B (Boolean algebra)
input B is 1 0 1 0
1 0 0
1 1 0

10.2.6 XOR gate

Description: Truth table: How to write this:
A
X
B W Table 10.7
X = A XOR B (logic notation)
The output, X, is 1 if:
V Figure 10.7 Inputs Output
(input A is 1 AND input
A B X X = (A . B) + (A . B) (Boolean
B is 0)
0 0 0 algebra)
or
0 1 1
(input A is 0 AND input
1 0 1 NOTE: this is sometimes written as:
B is 1)
(A + B) . (A . B)
1 1 0

Activity 10.1
Show why X = (A AND NOT B) OR (NOT A AND B)
and
Y = (A OR B) AND (NOT (A AND B)) both represent the same logic gate.

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10 BOOLEAN LOGIC

You will notice in the Boolean algebra, three new symbols; these have the
following meaning:
» . represents the AND operation
» + represents the OR operation
» a bar (above the letter or letters, e.g. a) represents the NOT operation.

10.3 Logic circuits, logic expressions, truth

tables and problem statements
When logic gates are combined together to carry out a particular function, such
as controlling a robot, they form a logic circuit. The following eight examples
show how to carry out the following tasks:
» Create a logic circuit from a:
– problem statement (examples 6 and 7)
– logic or Boolean expression (examples 3 and 8)
– truth table (examples 4 and 5)
» Complete a truth table from a:
– problem statement (examples 6 and 7)
– logic or Boolean expression (examples 3 and 8)
– logic circuit (example 1)
» Write a logic or Boolean expression from a:
– problem statement (examples 6 and 7)
– logic circuit (example 2)
– truth table (examples 4 and 5).

Example 1
Produce a truth table for the following logic circuit (note the use of black circles at the
junctions between wires):

A
B P

Q X

Part 1 Part 2 Part 3

V Figure 10.8
There are three inputs to this logic circuit, therefore, there will be eight possible
binary values that can be input.
To show stepwise how the truth table is produced, the logic circuit has been split up
into three parts, as shown by the dotted lines, and intermediate values are shown as
P, Q and R.

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 10.3 Logic circuits, logic expressions, truth tables and problem statements

A
Part 1
P This is the first part of the logic circuit; the first task is to find the intermediate values
B
P and Q.
The value of P is found from the AND gate where the inputs are A and B. The value
of Q is found from the NOR gate where the inputs are B and C. An intermediate truth
table is produced using the logic function descriptions in Section 10.2.
W Table 10.8
Q
C
input values Output values
V Figure 10.9 A B C P Q
0 0 0 0 1
0 0 1 0 0
0 1 0 0 0
0 1 1 0 0
1 0 0 0 1
1 0 1 0 0
1 1 0 1 0
1 1 1 1 0

Part 2
The second part of the logic circuit has P and Q as inputs and the intermediate
P output, R:
R
Q
This produces the following intermediate truth table. (Note: even though there are
V Figure 10.10 only two inputs to the logic gate, we have generated eight binary values in part 1 and
these must all be used in this second truth table).
W Table 10.9

Inputs Output
P Q R
0 1 1
0 0 0
0 0 0
0 0 0
0 1 1
0 0 0
1 0 1
1 0 1

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10 BOOLEAN LOGIC

Part 3
The final part of the logic circuit has R and C as inputs and the final
R output, X:
X
C This gives the third intermediate truth table:
V Figure 10.11 W Table 10.10

Inputs Output
R C X
1 0 1
0 1 1
0 0 0
0 1 1
1 0 1
0 1 1
1 0 1
1 1 0

Putting all three intermediate truth tables together produces the final truth table,
which represents the original logic circuit:
W Table 10.11

Input values Intermediate values Output

A B C P Q R X
0 0 0 0 1 1 1
0 0 1 0 0 0 1
0 1 0 0 0 0 0
0 1 1 0 0 0 1
1 0 0 0 1 1 1
1 0 1 0 0 0 1
1 1 0 1 0 1 1
1 1 1 1 0 1 0

The intermediate values can be left out of the final truth table, but it is good practice
to leave them in until you become confident about producing the truth tables. The final
truth table would then look like this:
W Table 10.12

Input values Output

A B C X
0 0 0 1
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 0

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 10.3 Logic circuits, logic expressions, truth tables and problem statements

Example 2
Write logic expressions from the following logic circuits:

a Logic gate 1
A

Logic gate 3
X
B

C Logic gate 2

V Figure 10.12

The first action is to look at the gates connected to the inputs A, B and C:
logic gate 1: (A AND B)
logic gate 2: (B OR C)
We then join these together using logic gate 3:
[(A AND B)] XOR [(B OR C)] which gives us the required logic expression.
(Note: the square brackets “[ ]” in the expression are not necessary and are used here
just for clarity.)
This would be written as: (A AND B) XOR (B OR C)
b
A

Logic gate 3
Logic gate 1

Logic gate 4
B
X

Logic gate 2

V Figure 10.13

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10 BOOLEAN LOGIC

Again, we will do this in the order of logic gates 1 and 2 first (connected to the three
inputs):
logic gate 1: (A NAND C)
logic gate 2: (B AND C)
However, logic gate 3 is also connected to one of the inputs so that should be done next:
logic gate 3: (logic gate 1) NOR A
If we replace (logic gate 1) by the logic expression above, we get:
((A NAND C) NOR A)
Finally, we can join all these together using:
logic gate 4: ((A NAND C) NOR A) OR (B AND C)

Activity 10.2
1 Produce truth tables from the following logic circuits:
a A

V Figure 10.14

b A

V Figure 10.15

c A
B

V Figure 10.16

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 10.3 Logic circuits, logic expressions, truth tables and problem statements

d A
B

V Figure 10.17

e A
B

V Figure 10.18
2 Write logic expressions for the following logic circuits:

a A

X
B

C
V Figure 10.19

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10 BOOLEAN LOGIC

B
C
V Figure 10.20

Example 3
A logic circuit can be represented by the following logic expression: (A XOR C) OR
(NOT C NAND B)
Produce a logic circuit and a truth table from the above statement.
In this example we have a connecting logic gate which is OR.
So, if we produce one half of the circuit from (A XOR C) we get:

C
V Figure 10.21
The other half of the circuit is found from (NOT C NAND B):
B

V Figure 10.22

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 10.3 Logic circuits, logic expressions, truth tables and problem statements

If we now combine these together to form the final logic circuit:

A
Stage ‘P’

Output (X)

B Stage ‘R’

Stage ‘Q’
C

V Figure 10.23
The truth table is shown:
W Table 10.13
Input values Values at stages: Output
A B C ‘P’ ‘Q’ ‘R’ X
0 0 0 0 1 1 1
0 0 1 1 0 1 1
0 1 0 0 1 0 0
0 1 1 1 0 1 1
1 0 0 1 1 1 1
1 0 1 0 0 1 1
1 1 0 1 1 0 1
1 1 1 0 0 1 1

Example 4
Look at the two truth tables below; in each case produce a logic expression and the
corresponding logic circuit:
a
W Table 10.14

Inputs Output To produce the logic statement, we only con-

cern ourselves with the truth table row where
A B X
the output value is 1. In this case, A = 1 and B =
0 0 0
0 which gives the logical expression:
0 1 0
1 0 1 A AND NOT B
1 1 0
So we have the logic expression: A AND NOT B
—
(Note that this could be written as A . B in Boolean.) It is now possible to draw the
corresponding logic circuit:
A
X

V Figure 10.24
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10 BOOLEAN LOGIC

b
This time we have two rows where the output is 1;
W Table 10.15
this gives the following logical expressions:
Inputs Output (NOT A AND B)
A B X
0 0 0
(A AND B)
0 1 1 We now join these together using an OR gate to give:
1 0 0
(NOT A AND B) OR (A AND B)
1 1 1

So we have the logic expression: (NOT A AND B) OR (A AND B)

—
(This could be written as: A . B + A . B)
It is now possible to draw the corresponding logic circuit:
A

V Figure 10.25

Example 5
a Which Boolean expression is represented by the following truth table?
W Table 10.16

Input values Output

A B C X
0 0 0 1
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 1
1 1 1 0
We only need to consider those rows where the output is a 1. This gives us the
following three logic expressions:
(NOT A AND NOT B AND NOT C)
(A AND NOT B AND NOT C)
(A AND B AND NOT C)
If we now join the three expressions with an OR gate, we end up with the final
logic expression:
(NOT A AND NOT B AND NOT C) OR (A AND NOT B AND NOT C) OR
(A AND B AND NOT C)
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 10.3 Logic circuits, logic expressions, truth tables and problem statements

b i Which logic expression is represented by the following truth table?

ii Show that your logic expression in part i is the same as: (B AND C) OR (A AND
C) OR (A AND B)
W Table 10.17
A B C X
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
i We only need to consider those rows where the output is a 1. This gives us the
following four Boolean expressions:

(NOT A AND B AND C)

(A AND NOT B AND C)
(A AND B AND NOT C)
(A AND B AND C)

If we now join the four expressions with an OR gate we end up with the
following logic expression:

(NOT A AND B AND C) OR (A AND NOT B AND C) OR (A AND B AND NOT

C) OR (A AND B AND C)

ii To show that (B AND C) OR (A AND C) OR (A AND B) produces the same output

as that shown in part i we need to produce a new truth table and show that
the output is the same as the one in the given truth table:
W Table 10.18

A B C B AND C A AND C A AND B (B AND C) OR (A X

AND C) OR (A AND
B)
0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0
0 1 0 0 0 0 0 0
0 1 1 1 0 0 1 1
1 0 0 0 0 0 0 0
1 0 1 0 1 0 1 1
1 1 0 0 0 1 1 1
1 1 1 1 1 1 1 1
As the second truth table shows, the outputs from logic expressions are both the
same; thus the logic expression (B AND C) OR (A AND C) OR (A AND B) gives the
same output as the logic expression in part i.

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10 BOOLEAN LOGIC

Activity 10.3
1 Produce:
i a truth table
ii a logic circuit
from the following logic expression:
(NOT A AND B) AND (NOT B OR C)

2 Produce:
i a truth table
ii a logic circuit
from the following logic expression:
(A XOR B) OR ((B NOR C) AND B) A B X
0 0 1
3 Produce:
0 1 0
i a logic expression
1 0 0
ii a logic circuit
1 1 1
from the following truth table:
4 Write down a logic expression for each of the following truth tables:
a b
A B C X A B C X
0 0 0 1 0 0 0 0
0 0 1 0 0 0 1 1
0 1 0 0 0 1 0 1
0 1 1 0 0 1 1 0
1 0 0 1 1 0 0 1
1 0 1 1 1 0 1 1
1 1 0 0 1 1 0 0
1 1 1 0 1 1 1 0

5 a Write down a logic expression corresponding to the following truth table:

A B C X
0 0 0 1
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 0

b Show that the following logic expression produces the same output as your
answer to part a above:
(NOT A AND NOT B) OR (A AND NOT B)

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 10.3 Logic circuits, logic expressions, truth tables and problem statements

Example 6
A safety system uses three inputs to a logic circuit. An alarm, X, sounds if input A
represents ON and input B represents OFF; or if input B represents ON and input C
represents OFF.
Produce a logic circuit and truth table to show the conditions that cause the output X to be 1.
The first thing to do is to write down the logic statement representing the scenario
in this example. To do this, it is necessary to recall that ON = 1 and OFF = 0 and also
that 0 is usually considered to be NOT 1.
So we get the following logic expression:
(A AND NOT B) OR (B AND NOT C)

this equates to: the two parts are this equates to

A is ON and B is OFF connected by the OR gate B is ON AND C is OFF

Note: this expression can also be written as follows (Boolean expression):

(A . — —)
B ) + (B . C

The logic circuit is made up of two parts as shown in the logic expression. We will produce
the logic gate for the first second part. Then join both parts together with the OR gate.
A B

B C

Part 1 Part 2
V Figure 10.26
Now combining both parts with the OR gate gives us:
A

Part 1

B X

Part 2

V Figure 10.27
In order to produce the truth table, there are two ways to do this:
» trace through the logic circuit using the method described in example 1 (Section 10.3)
» produce the truth table using the original logic expression; this second method
has the advantage that it allows you to check that your logic circuit is correct.
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10 BOOLEAN LOGIC

We will use the second method in this example:

W Table 10.19

Inputs Intermediate values Output

A B C (A AND NOT B) (B AND NOT C) X
0 0 0 0 0 0
0 0 1 0 0 0
0 1 0 0 1 1
0 1 1 0 0 0
1 0 0 1 0 1
1 0 1 1 0 1
1 1 0 0 1 1
1 1 1 0 0 0

(Note: it is optional whether to leave in the intermediate values or simply remove them
giving a 4-column truth table with headings: A, B, C, X)

Example 7
A wind turbine has a safety system which uses three inputs to a logic circuit. A certain
combination of conditions results in an output, X, from the logic circuit being equal to 1.
When the value of X = 1 then the wind turbine is shut down.
The following table shows which parameters are being monitored and form the three
inputs to the logic circuit.
W Table 10.20

parameter parameter binary description of condition

description value
turbine speed S 0 turbine speed <= 1000 rpm
1 turbine speed > 1000 rpm
bearing temperature T 0 bearing temperature <= 80°C
1 bearing temperature > 80°C
wind velocity W 0 wind velocity <= 120 kph
1 wind velocity >120 kph

The output, X, will have a value of 1 if any of the following combination of conditions occur:
Either turbine speed <= 1000 rpm and bearing temperature > 80°C
Or turbine speed > 1000 rpm and wind velocity > 120 kph
Or bearing temperature <= 80°C and wind velocity > 120 kph

Design the logic circuit and complete the truth table to produce a value of X = 1 when
either of the three conditions above occur.
In this example, a real situation is given and it is necessary to convert the information
into a logic expression and then produce the logic circuit and truth table. It is
advisable in problems as complex as this to produce the logic circuit and truth table
separately (based on the conditions given) and then check them against each other to
see if there are any errors.

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 10.3 Logic circuits, logic expressions, truth tables and problem statements

Stage 1
The first thing to do is to convert each of the three statements into logic expressions.
Use the information given in the table and the three condition statements to find how
the three parameters S, T and W are linked. We usually look for the key words AND,
OR and NOT when converting actual statements into logic.
We end up with the following three logic expressions:
i turbine speed <= 1000 rpm and bearing temperature > 80°C
logic expression: (NOT S AND T)
ii turbine speed > 1000 rpm and wind velocity > 120 kph
logic expression: (S AND W)
iii bearing temperature <= 80°C and wind velocity > 120 kph
logic expression: (NOT T AND W)

Stage 2
This now produces three intermediate logic circuits:

i
S

V Figure 10.28

ii S
W

V Figure 10.29

iii
T

V Figure 10.30
Each of the three original statements were joined together by the word ‘OR’. Thus,
we need to join all of the three intermediate logic circuits by two OR gates to get the
final logic circuit.
We will start by joining i and ii together using an OR gate:

V Figure 10.31

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10 BOOLEAN LOGIC

Finally, we connect the logic circuit in Figure 10.31 to Figure 10.30 to obtain the
answer:

V Figure 10.32
The final part is to produce the truth table. We will do this using the original logic
statement. This method has the bonus of allowing an extra check to be made on the
logic circuit in Figure 10.32 to see whether or not it is correct. It is possible, however,
to produce the truth table straight from the logic circuit in Figure 10.32.
There were three parts to the problem, so the truth table will first evaluate each part.
Then, by applying OR gates, as shown below, the final value, X, is obtained:
i (NOT S AND T)
ii (S AND W)
iii (NOT T AND W)
We find the outputs from parts (i) and (ii) and then OR these two outputs together to
obtain a new intermediate, which we will now label part (iv).
We then OR parts (iii) and (iv) together to get the value of X.

Inputs Intermediate values Output

S T W (i) (ii) (iii) (iv) X
(NOT S (S AND (NOT T
AND T) W) AND W)
0 0 0 0 0 0 0 0
0 0 1 0 0 1 0 1
0 1 0 1 0 0 1 1
0 1 1 1 0 0 1 1
1 0 0 0 0 0 0 0
1 0 1 0 1 1 1 1
1 1 0 0 0 0 0 0
1 1 1 0 1 0 1 1

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 10.3 Logic circuits, logic expressions, truth tables and problem statements

Example 8
Consider the logic statement:

((A NOR B) AND C) NAND (A OR NOT B)

a Draw a logic circuit to represent the given logic statement.

b Complete the truth table for the given logic statement.
W Table 10.21

Input values Working space Output

A B C X

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

First, we need to break down the logic statement. Assign:

i P = (A NOR B)
ii Q = (A OR NOT B)
iii R = (P AND C)
And then note that X = R NAND Q
Now draw the logic gates for statements i to iii and connect them together.

Output Q

X
B
Output P

Output R
C

V Figure 10.33

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10 BOOLEAN LOGIC

We can then fill out the truth table in stages, starting with P, Q and then R, followed
finally by X:
Input Values Working Space
Output
Q = (A OR R = (P AND C)
A B C P = (A NOR B) X
NOT B)
0 0 0 1 1 0 1

0 0 1 1 1 1 0

0 1 0 0 0 0 1

0 1 1 0 0 0 1

1 0 0 0 1 0 1

1 0 1 0 1 0 1

1 1 0 0 1 0 1

1 1 1 0 1 0 1

Find out more Activity 10.4

1 Draw the logic circuits and complete the truth tables for the following logic or
Boolean algebra
Boolean expressions:
appears throughout
this chapter and is a X = (A OR B) OR (NOT A AND B)
the official method b Y = (NOT A AND NOT B) AND (NOT B OR C)
for depicting logic
statements. Try using c T = 1 if (switch K is ON or switch L is ON) OR (switch K is ON and switch M
truth tables to prove is OFF) OR (switch M is ON)
the following pairs d X = (A AND NOT B) OR (NOT B AND C)
of Boolean (logic) e R = 1 if (switch A is ON and switch B is ON) AND (switch B is ON or switch C
statements are the
is OFF)
same:
i A + (A . B) = A 2 Produce the logic circuit and complete a truth table to represent the following scenario.
−
ii A + A .B = A + B A chemical process is protected by a logic circuit. There are three inputs to the
iii (A + B) . (A + C) = logic circuit representing key parameters in the chemical process. An alarm,
A+B.C X, will give an output value of 1 depending on certain conditions in the chemical
− −
iv (A.B) = A+ B process. The following table describes the process conditions being monitored:
− −
v A + B = A+ B
Parameter description Parameter Binary value Description of condition
chemical reaction rate R 0 reaction rate < 40 mol/l/sec
1 reaction rate >= 40 mol/l/sec
process temperature T 0 temperature > 115°C
1 temperature <= 115°C
concentration of C 0 concentration <= 4 mol
chemicals 1 concentration > 4 mol

An alarm, X, will generate the value 1 if:

either reaction rate < 40 mol/l/sec

or concentration > 4 mol AND temperature > 115°C
or reaction rate >= 40 mol/l/sec AND temperature > 115°C

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 10.3 Logic circuits, logic expressions, truth tables and problem statements

3 Produce the logic circuit and complete a truth table to represent the following
scenario.
A power station has a safety system controlled by a logic circuit. Three inputs
to the logic circuit determine whether the output, S, is 1. When S = 1 the power
station shuts down.
The following table describes the conditions being monitored.

Parameter Parameter Binary value Description of condition

description
gas temperature G 0 gas temperature <= 160°C
1 gas temperature > 160°C
reactor pressure R 0 reactor pressure <= 10 bar
1 reactor pressure > 10 bar
water W 0 water temperature <= 120°C
temperature 1 water temperature > 120°C

Output, S, will generate a value of 1, if:

either gas temperature > 160°C AND water temperature <= 120°C
or gas temperature <= 160°C AND reactor pressure > 10 bar
or water temperature > 120°C AND reactor pressure > 10 bar
4 A car’s engine management system uses three sensors A, B and C. The data from
these sensors forms the input to a logic circuit. When the output (X) from the
logic circuit is 1 a signal is sent to a warning light on the dash board of the car.
The following table describes the conditions being monitored.

Parameter Parameter Binary value Description of condition

description
exhaust E 0 exhaust temperature > 400°C
temperature 1 exhaust temperature <= 400°C
oil pressure P 0 oil pressure <= 10 bar
1 oil pressure > 10 bar
water temperature W 0 water temperature <= 110°C
1 water temperature > 110°C

Output, X, will generate a value of 1, if:

either exhaust temperature > 400°C AND oil pressure > 10 bar
or oil pressure <= 10 bar AND water temperature > 110 °C
or oil pressure > 10 bar AND water temperature > 110 °C
Produce:
i a truth table,
ii a logic expression and
iii a logic circuit
to represent the above scenario.
Also confirm that the output from your logic circuit matches the output from
your truth table in part i.

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