Solution Manual - Digital Design and Computer Architecture Harris ARM Edition, Exercises of Digital Systems Design

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S OLUTIONS

Digital Design and Computer Architecture: ARM Edition

S O L U T I O N S 1

CHAPTER 1

Exercise 1.

(a) Biologists study cells at many levels. The cells are built from organelles

such as the mitochondria, ribosomes, and chloroplasts. Organelles are built of

macromolecules such as proteins, lipids, nucleic acids, and carbohydrates.

These biochemical macromolecules are built simpler molecules such as carbon

chains and amino acids. When studying at one of these levels of abstraction, bi-

ologists are usually interested in the levels above and below: what the structures

at that level are used to build, and how the structures themselves are built.

(b) The fundamental building blocks of chemistry are electrons, protons,

and neutrons (physicists are interested in how the protons and neutrons are

built). These blocks combine to form atoms. Atoms combine to form molecules.

For example, when chemists study molecules, they can abstract away the lower

levels of detail so that they can describe the general properties of a molecule

such as benzene without having to calculate the motion of the individual elec-

trons in the molecule.

Exercise 1.

(a) Automobile designers use hierarchy to construct a car from major as-

semblies such as the engine, body, and suspension. The assemblies are con-

structed from subassemblies; for example, the engine contains cylinders, fuel

injectors, the ignition system, and the drive shaft. Modularity allows compo-

nents to be swapped without redesigning the rest of the car; for example, the

seats can be cloth, leather, or leather with a built in heater depending on the

model of the vehicle, so long as they all mount to the body in the same place.

Regularity involves the use of interchangeable parts and the sharing of parts be-

tween different vehicles; a 65R14 tire can be used on many different cars.

Sarah L. Harris and David Money Harris Digital Design and Computer Architecture: ARM Edition

S O L U T I O N S 3

Exercise 1.

Each digit conveys log 2 60 = 5.91 bits of information. 4000 10 = 1 6 40 60 (

in the 3600 column, 6 in the 60’s column, and 40 in the 1’s column).

Exercise 1.

216 = 65,536 numbers.

Exercise 1.

Exercise 1.

(a) 2 16 -1 = 65535; (b) 2 15 -1 = 32767; (c) 2 15 -1 = 32767

Exercise 1.

(a) 232 -1 = 4,294,967,295; (b) 2 31 -1 = 2,147,483,647; (c) 231 -1 =

Exercise 1.

(a) 0; (b) -2^15 = -32768; (c) -(2 15 -1) = -

Exercise 1.

(a) 0; (b) -

= -2,147,483,648; (c) -(

Exercise 1.

(a) 10; (b) 54; (c) 240; (d) 2215

Exercise 1.

(a) 14; (b) 36; (c) 215; (d) 15,

Exercise 1.

(a) A; (b) 36; (c) F0; (d) 8A

Digital Design and Computer Architecture: ARM Edition

(a) 10100101; (b) 00111011; (c) 1111111111111111;

6 S O L U T I O N S c h a p t e r 1

(a) 52; (b) 77; (c) 345; (d) 1515

Exercise 1.

(a) 0o16; (b) 0o64; (c) 0o339; (d) 0o

Exercise 1.

(a) 100010 2 , 22 16 , 34 10 ; (b) 110011 2 , 33 16 , 51 10 ; (c) 010101101 2 , AD 16 ,

173 10 ; (d) 011000100111 2 , 627 16 , 1575 10

Exercise 1.

(a) 0b10011; 0x13; 19; (b) 0b100101; 0x25; 37; (c) 0b11111001; 0xF9;

249; (d) 0b10101110000; 0x570; 1392

Exercise 1.

15 greater than 0, 16 less than 0; 15 greater and 15 less for sign/magnitude

Exercise 1.

(26-1) are greater than 0; 26 are less than 0. For sign/magnitude numbers,

(26-1) are still greater than 0, but (26-1) are less than 0.

Exercise 1.

Exercise 1.

Exercise 1.

Exercise 1.

(5 × 109 bits/second)(60 seconds/minute)(1 byte/8 bits) = 3.75 × 1010

bytes

Exercise 1.

46.566 gigabytes

S O L U T I O N S 7

Exercise 1.

2 billion

Exercise 1.

128 kbits

Exercise 1.

Exercise 1.

Exercise 1.

(a) 1101; (b) 11000 (overflows)

Exercise 1.

100 101 110 111 000 001 010 011 Two's Complement

100

111 110 101 000 001 010 011

000 001 010 011 100 101 110 111

Sign/Magnitude

Unsigned

10 11 00 01 Two's Complement

10

11

00 01

00 01 10 11

Sign/Magnitude

Unsigned

Digital Design and Computer Architecture: ARM Edition

S O L U T I O N S 9

Exercise 1.

(a) 3; (b) 01111111; (c) 00000000 2 = -127 10 ; 11111111 2 = 128 10

Exercise 1.

Exercise 1.

(a) 001010001001; (b) 951; (c) 1000101; (d) each 4-bit group represents

one decimal digit, so conversion between binary and decimal is easy. BCD can

also be used to represent decimal fractions exactly.

Exercise 1.

(a) 0011 0111 0001

(b) 187

(c) 95 = 1011111

(d) Addition of BCD numbers doesn't work directly. Also, the representa-

tion doesn't maximize the amount of information that can be stored; for example

2 BCD digits requires 8 bits and can store up to 100 values (0-99) - unsigned 8-

bit binary can store 28 (256) values.

Exercise 1.

Three on each hand, so that they count in base six.

Exercise 1.

Both of them are full of it. 42 10 = 101010 2 , which has 3 1’s in its represen-

tation.

Exercise 1.

Both are right.

Exercise 1.

#include <stdio.h>

(^000 001 010 011 100 101 110 111) Biased

Digital Design and Computer Architecture: ARM Edition

10 S O L U T I O N S c h a p t e r 1

void main(void) { char bin[80]; int i = 0, dec = 0;

printf("Enter binary number: "); scanf("%s", bin);

while (bin[i] != 0) { if (bin[i] == '0') dec = dec * 2; else if (bin[i] == '1') dec = dec * 2 + 1; else printf("Bad character %c in the number.\n", bin[i]); i = i + 1; } printf("The decimal equivalent is %d\n", dec); }

Exercise 1.

/* This program works for numbers that don't overflow the range of an integer. */

#include <stdio.h>

void main(void) { int b1, b2, digits1 = 0, digits2 = 0; char num1[80], num2[80], tmp, c; int digit, num = 0, j;

printf ("Enter base #1: "); scanf("%d", &b1); printf ("Enter base #2: "); scanf("%d", &b2); printf ("Enter number in base %d ", b1); scanf("%s", num1);

while (num1[digits1] != 0) { c = num1[digits1++]; if (c >= 'a' && c <= 'z') c = c + 'A' - 'a'; if (c >= '0' && c <= '9') digit = c - '0'; else if (c >= 'A' && c <= 'F') digit = c - 'A' + 10; else printf("Illegal character %c\n", c); if (digit >= b1) printf("Illegal digit %c\n", c); num = num * b1 + digit;

} while (num > 0) { digit = num % b2; num = num / b2; num2[digits2++] = digit < 10? digit + '0' : digit + 'A' - 10; } num2[digits2] = 0;

for (j = 0; j < digits2/2; j++) { // reverse order of digits tmp = num2[j]; num2[j] = num2[digits2-j-1]; num2[digits2-j-1] = tmp; }

printf("The base %d equivalent is %s\n", b2, num2); }

Exercise 1.

12 S O L U T I O N S c h a p t e r 1

Exercise 1.

Exercise 1.

Exercise 1.

XNOR

Y = A + B + C

B C Y

A B C A 0 0 0 0

NAND

Y = ABCDE

C D E

B

A

OR

Y = A+B+C+D

Y

(b)

A

B

C

D

B D Y

C

A

(a) 1

Y Y

A B C D E Y 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

(c) 1

S O L U T I O N S 13

Exercise 1.

Exercise 1.

Exercise 1.

B C Y

A

B C Y

A

B C Y

A

Digital Design and Computer Architecture: ARM Edition

S O L U T I O N S 15

Exercise 1.

The circuit functions as a buffer with logic levels V IL = 1.5; V IH = 1.8; V OL

= 1.2; VOH = 3.0. It can receive inputs from LVCMOS and LVTTL gates be-

cause their output logic levels are compatible with this gate’s input levels. How-

ever, it cannot drive LVCMOS or LVTTL gates because the 1.2 VOL exceeds

the V IL of LVCMOS and LVTTL.

Exercise 1.

(a) AND gate; (b) VIL = 1.5; VIH = 2. 2 5; VOL = 0; VOH = 3

Exercise 1.

(a) XOR gate; (b) V IL = 1.25; V IH = 2; V OL = 0; VOH = 3

Exercise 1.

Exercise 1.

Exercise 1.

A

B

C

D

Y

(a)

A B

C

A

B

C

Y

(b) (c)

Y

A

B C

A B

C

A

B

C

Y

(b) (c)

Y

A

B

A A B

B

C

Y

(a)

Digital Design and Computer Architecture: ARM Edition

16 S O L U T I O N S c h a p t e r 1

Exercise 1.

XOR

Exercise 1.

Exercise 1.

Exercise 1.

A B

C

B

A

Y

C

A

A

B B

A B Y

B C Y

A

(a) (b) (c)

A B C

Y

A

B

C

Y

A

B

C

Y

weak

weak

weak

S O L U T I O N S 11

CHAPTER 2

Exercise 2.

(a)

(b)

(c)

(d)

(e)

Exercise 2.

(a)

(b)

(c)

(d)

(e)

Exercise 2.

(a)

Y = AB + AB + AB

Y = ABC + ABC

Y = ABC + ABC + ABC + ABC + ABC

Y = ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD

Y = ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD

Y = AB + AB + AB

Y = ABC + ABC + ABC + ABC + ABC

Y = ABC + ABC + ABC

Y = ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD

Y = ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD

Y = A + B 

Digital Design and Computer Architecture: ARM Edition

12 S O L U T I O N S c h a p t e r 2

(b)

(c)

(d)

(e)

Exercise 2.

(a)

(b)

(c)

(d)

(e)

Exercise 2.

(a)

(b)

(c)

(d)

(e)

This can also be expressed as:

Exercise 2.

Y = A + B + C   A + B + C   A + B + C   A + B + C   A + B + C   A + B + C 

Y = A + B + C   A + B + C   A + B + C 

Y  A + B + C + D   A + B + C + D   A + B + C + D   A + B + C + D   A + B + C + D 

 A + B + C + D   A + B + C + D   A + B + C + D   A + B + C + D 

Y  A + B + C + D   A + B + C + D   A + B + C + D   A + B + C + D   A + B + C + D 

 A + B + C + D   A + B + C + D   A + B + C + D 

Y = A + B

Y = A + B + C   A + B + C   A + B + C 

Y = A + B + C   A + B + C   A + B + C   A + B + C   A + B + C 

Y  A + B + C + D   A + B + C + D   A + B + C + D   A + B + C + D 

 A + B + C + D   A + B + C + D   A + B + C + D   A + B + C + D 

 A + B + C + D 

Y  A + B + C + D   A + B + C + D   A + B + C + D   A + B + C + D 

 A + B + C + D   A + B + C + D   A + B + C + D   A + B + C + D 

 A + B + C + D 

Y = A + B

Y = ABC + ABC

Y = AC + AB + AC

Y = AB + BD + ACD

Y = ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD

Y = A  B   C  D  + A  B   C  D 