CHAPTER 12 SOLUTIONS in fog even if he has very little fuel as long as he has no passengers on board. He will also use it (for practise) even if it is clear, providing he has sufficient

Essential Software Development

© David Dawson and Adrian Janson Cambridge University Press 1

CHAPTER 12 SOLUTIONS

Hardware Design Project 1 – Aeroplane Landing Device

1 An aeroplane is fitted with an automatic landing device which is still in the

experimental stage and uses a lot of fuel. Whether the captain uses it or not

depends on three factors:

i fog conditions

ii whether he is carrying passengers

iii whether he has sufficient fuel.

If the runway is foggy and he has enough fuel, he will use it. He can afford to use

it in fog even if he has very little fuel as long as he has no passengers on board.

He will also use it (for practise) even if it is clear, providing he has sufficient fuel

and no passengers.

In all other conditions he will land manually.

Design a logic circuit which decides whether the captain should land manually or

on automatic. Your circuit should have three inputs:

i Fog (1) / No Fog (0)

ii Passengers (1) / No Passengers (0)

iii Enough Fuel (1) / Not Enough Fuel (0)

and one output:

Auto (1) / Manual (0)

Minimise the number of gates, where possible.

Hand in all your working out and the final circuit design. Please include written

explanations along with this, and tests on the circuit to ensure that it behaves

correctly.

Answer F: Fog (1), No Fog (0)

P: Passengers (1), No Passengers (0)

E: Enough Fuel (1), Not Enough Fuel (0)

X: Auto (1), Manual (0)

F P E X

0 0 0 0

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 1



Logic Statement: (F AND E) OR (F AND NOT P) OR (E AND NOT P)

2 The minimum number of binary digits used to represent one decimal digit is 4

(some are redundant). It is intended to display each decimal digit in an LCD

display, and select each element of the display depending of the position of four

switches (each switch representing one binary digit). The LCD display looks like

the diagram below:

E (1)

E (0)

F (1) F (1)

P (1) P (0)

1

1

1 1 0

0 0 0

F (0)

P

F

E

X



a b f g c e d

Select one digit, draw the truth table, work out the Karnaugh map and the

expression which describes when that digit is illuminated. Draw the logic circuit.

You may wish to work out the logic circuits for other LCD elements, but be

warned - it is a tedious process.

Answer: LED element ‘a’

1-4: Switch on (1), Switch off (0)

A: Display element (1), Do not display (0)

1 2 3 4 X

0 0 0 0 0

0 0 0 1 0

0 0 1 0 1

0 0 1 1 1

0 1 0 0 0

0 1 0 1 1

0 1 1 0 1

0 1 1 1 1

1 0 0 0 1

1 0 0 1 1



Logic Statement: 1 OR 3 OR (2 AND 4)

3 (1)

3 (0)

1 (1) 1 (1)

2 (1) 2 (0)

0

1

0

1

4 (1)

4 (0)

4 (0)

1 (0)

0

1 1 1

1

1

1 1 1

1 1

1

2

4

1 X

3



Hardware Design Project 2 – Vault Door

The Pressure Pad The pressure pad will detect whether a person is standing in front of the vault door or

not.

Keycard Slot When a keycard is inserted in the keycard slot, the vault door can be opened by turning

the handle - as long as someone is standing on the pressure pad. If an invalid card is

entered and the handle is turned, an alarm will sound.

Other situations The alarm will be sounded if the handle is tried without there being an access card in the

slot or a person on the pressure pad.

Design a logic circuit which will operate the vault door. You may set all of the input and

output conditions. Include all working, that is, Karnaugh map, groupings and logic

statement. After you have completed the circuit, include a brief written explanation of

how it works.

Answer:

H: Handle turned (1), Handle not turned (0)

P: Pressure pad (1), No one on pad (0)

K: Keycard present (1), No keycard (0)

O: Vault door opens (1), Vault door closed (0)

A: Alarm sounds (1), Alarm does not sound (0).

As the vault door will only open when all three

conditions are met, this will not be placed into a

Karnaugh map (but will be represented in the logic

circuit). The Karnaugh map below has been drawn

for the alarm condition.

H P K O A

0 0 0 0 0

0 0 1 0 0

0 1 0 0 0

0 1 1 0 0

1 0 0 0 1

1 0 1 0 1

1 1 0 0 1

1 1 1 1 0



Logic Statement: (H AND NOT P) OR (H AND NOT K)

P (1)

P (0)

H (1) H (1)

K (1) K (0)

0

1

1 0 0

1 0 0

H (0)

P

H

K

A

O



Test your knowledge

1

‘m’: 109, 01101101

‘q’: 113, 01110001

‘<space>’: 32, 00100000

2

255 / 2: 11111111 / 10 = 111111.1

468 / 4: 111010100 / 100 = 1110101

76 / 8: 1001100 / 1000 = 1001.1

123 / 17: 1111011 / 10001 = 111.001111 (approx).

87 / 5: 1001111 / 101 = 10001.011 (approx).

3 There are two types of flip-flop circuits: a NAND gate flip-flop and a NOR gate

flip-flop.

NAND gate flip-flop truth table

S R Q

0 0 forbidden

0 1 1

1 0 0

1 1 No change

NOR gate flip-flop truth table

S R Q

0 0 No change

1 0 1

0 1 0

1 1 Forbidden

4 The half-adder circuit allows no carry-in from previous adder circuits and,

therefore, would be suitable for only the least significant bit in a series of binary

digits. Apart from the least significant bit, every other column in a binary addition

problem will require there to be a carry-in and carry-out feature. The full adder

found at the most significant bit end of the problem will produce a carry-out for

the most significant bit and identify whether the leftmost digit is a one or zero.

5 A full-adder circuit is effective for adding together only one bit information. To

add together two 8-bit numbers, eight full adders will be required.

6 17



7 a ASCII control characters are not used in modern printers.

b A control stream is a structure of control characters that imparts information

such as the start of the transmission, end of the transmission and often a check bit.

c An escape sequence is a sequence of ASCII control characters beginning with

character ‘27’. It was used to output a number of commands directly to the printer

within the stream of characters that it was printing. The control characters were

not printed, but caused particular commands to be carried out.

8 Binary addition is a simple matter for two’s complement numbers. Once they

have been changed to two’s complement, the numbers can simply be added

together as two binary numbers normally would be.

9 Binary subtraction is a simple matter in two’s complement numbers. Once the

number to be subtracted is translated into two’s complement, the numbers can be

added together as for any two binary numbers.

10 Binary multiplication takes place in a very similar fashion to that of decimal

multiplication. As each digit is multiplied, if the multiplier is a 0 then the result is

a line of bits of 0 value. If the multiplier is a 1 then the multiplicand is simply

written down below the multiplier. The partial products are then added together to

give the end result.

11 Binary division in two’s complement numbers involves a shift and subtract

method, much like long-division.

12 Single precision floating point.

13 Double-precision floating point.

14 Binary digits are calculated by using powers of 2 based on their position as shown

below.

Binary 1 1 1 1 1 1 1 1

Power

of two

2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0

Decimal 128 64 32 16 8 4 2 1

15 There are a number of shortcomings with sign-magnitude representation. Firstly,

if numbers are to be added or subtracted, the value of the leftmost bit will have to

be considered each time. It is even more inefficient since there are two

representations of zero.



Extend your knowledge

1

75 + 39 = 01001011 +

00100111 =

01110010 (114)

139 + 59 = 010001011 +

000111011 =

011000100 (196)

123 + -56 = 01111011 +

11001000 =

01000011 (67)

15 + 7 = 00001111 +

00000111 =

00010110 (22)

-28 + 22 = 11100100 +

00010110 =

11111010 00000101 + 1 = 0000110 (-6)

2

Logic Statement: C OR (A AND B)

B (1)

B (0)

A (1) A (1)

C (1) C (0)

0

0

1 1 1

1 1 0

A (0)



3

4

5

Draw an AND gate, and draw lines from A and B into it.

Take a line from C and put it through a NOT gate.

B T I X

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1

O S L X

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1

A

B

A

B

C

X



Take another line from B and put it through another NOT gate.

Take both the outputs from the NOT gates and put them through another AND

gate.

Take the outputs from both of the AND gates and put them into an OR gate.

A

B

C

A

B

C

A

B

C



The output from the OR gate, is X (refer to the truth table).

244 = 11110100

45 = 101101

17 = 10001

117 = 1110101

76 = 1001100

7

10010000 = 144

1000010 = 66

100101 = 37

1101101 = 109

111011 = 59

8

ABCDh = 1010101111001101

87FEh = 1000011111111110

9Dh = 10011101

BC9Ch = 1011110010011100

BAC7h = 1011101011000111

9

0101 0100 1110 0111 = 54E7

1111 0001 = F1

0110 1110 = 6E

0111 0001 1110 1101 = 71ED

0001 1111 1101 0110 = 1FD6

10

44h = 68

BEh = 190

65h = 101

A

B

C X



E7h = 231

EBh = 235

11

241 = F1h

49 = 31h

171 = ABh

1117 = 45Dh

20376 = 4F98h

12

4 * 8 = 100 x 1000 = 100000 (32)

16 * 45 = 10000 x 101101 = 1011010000 (720)

23 * 16 = 10111 x 10000 = 101110000 (368)

255 * 8 = 11111111 x 1000 = 11111111000 (2040)

123 * 2 = 1111011 x 10 = 11110110 (246)

13

121 = 1111001

-34 = 1011101

59 = 111011

-251 = 100000100

199 = 11000111

14

123 + 45 = 1111011 +

101101 =

10101000 (168)

-49 + 33 = 1001110 +

100001 =

1101111 + 1 = 10010000 (-16)

136+ -29 = 10001000 +

11100010 =

??

233 + 67 = 11101001 +

1000011 =

100101100 (300)

-199 + -73 = 100111000 +

110110110 =

1011101110?? = 1100010000 (-272)

Documents

CHAPTER 12 SOLUTIONS in fog even if he has very little fuel as long as he has no passengers on board. He will also use it (for practise) even if it is clear, providing he has sufficient