Digital Systems Section 12 Binary Adders

Lecture Digital Systems Binary Adders Addition of binary data is very fundamental in digital systems. The hardware implementation needs to be determined. The inputs are: single bit values, carry in The outputs are: sum, carry out. After creating a single-bit adder, we can chain multiple adders together. Overflow must also be considered. Overflow is the situation where the result of addition exceeds the magnitude which can be represented with the allocated number of bits.

Truth Table of Half Adder Lecture Digital Systems Half Adder A half adder adds two binary numbers. The inputs: A0, B0 (single bit inputs). The outputs: S0 (single bit sum) and C1 (carry out). Truth Table of Half Adder Circuit of Half Adder credential: bring a computer die photo wafer : This can be an hidden slide. I just want to use this to do my own planning. I have rearranged Culler’s lecture slides slightly and add more slides. This covers everything he covers in his first lecture (and more) but may We will save the fun part, “ Levels of Organization,” at the end (so student can stay awake): I will show the internal stricture of the SS10/20. Notes to Patterson: You may want to edit the slides in your section or add extra slides to taylor your needs.

Lecture Digital Systems Multiple Bit Addition Consider the addition of 2 binary numbers, A and B. Addition of each bit position Ai and Bi creates a sum Si and a carry Ci+1.

Truth Table of Half Adder Lecture Digital Systems Full Adder A full adder adds two binary numbers but also include a carry in. The inputs: Ai, Bi, Ci (single bit inputs). The outputs: Si (single bit sum) and Ci+1 (carry out). Truth Table of Half Adder K-Map for Si K-Map for Ci+1

Lecture Digital Systems Full Adder Let us now simplify Boolean function for S by using Boolean algebra. Si = Ai’Bi’Ci + Ai’BiCi’ + AiBi’Ci’ + AiBiCi Si = Ci·(Ai’Bi’ + AiBi) + Ci’·(Ai’Bi + AiBi’) Si = Ci·(Ai  Bi )’ + Ci’·(Ai  Bi ) Si = Ci  (Ai  Bi ) Hint: A  B = A·B’ + A’·B (A·B)’ = A’ + B’ (A+B)’ = A’ · B’

Lecture Digital Systems Full Adder Let us now simplify Boolean function for Ci+1 by using Boolean algebra. Ci+1 = AiBi + AiCi +BiCi (previous result) Ci+1 = AiBi + CiAi’Bi +CiAiBi’ (as shown by K-map below) Ci+1 = AiBi + Ci ·(Ai’Bi + AiBi’) Ci+1 = AiBi + Ci ·(Ai  Bi) Hint: A  B = A·B’ + A’·B (A·B)’ = A’ + B’ (A+B)’ = A’ · B’

Full Adder The logic circuit of the full adder can be shown as: Lecture Digital Systems Full Adder The logic circuit of the full adder can be shown as: Half adder Half adder Si = Ci  (Ai  Bi ) Ci+1 = AiBi + Ci ·(Ai  Bi) A full adder can be made from 2 half adders and an OR Gate. Such structure repetition simplifies circuit design.

Lecture Digital Systems Full Adder = This single bit full adder will be the building block of large adders.

n × Full Adder = n-bit Ripple Carry Adder Lecture Digital Systems n × Full Adder = n-bit Ripple Carry Adder MSB position LSB position 4-bit ripple-carry adder

Digital Systems Section 13 Signed Numbers

How to Represent Signed Numbers Lecture Digital Systems How to Represent Signed Numbers For decimal numbers, it is common to use the sign + and –, as for +25, –16, +433, –2775. For computers, where operations are done using binary digits, it is desirable to represent signed numbers also in bits. There are 3 representations of signed binary numbers: Signed magnitude 1’s complement 2’s complement In each case, the left-most bit indicates the sign: 0 means positive, 1 means negative.

How to Represent Signed Numbers Lecture Digital Systems How to Represent Signed Numbers bn–1 Magnitude MSB Unsigned number b1 b0 Magnitude Sign 0 denotes 1 denotes + – MSB Signed number bn–1 b1 b0 bn–2

Lecture Digital Systems Signed Numbers 4-bit signed binary number comparison 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 1 + – 2 3 4 5 6 7 8 Number circle for 4-bit 2’s complement numbers

4-bit signed binary number comparison Lecture Digital Systems Signed Numbers 4-bit signed binary number comparison

Signed Magnitude Representation Lecture Digital Systems Signed Magnitude Representation As mentioned before, in signed magnitude, the left-most bit is used to indicate the sign. 0 means positive, 1 means negative. 000011002 = 1210 Sign bit Magnitude 100011002 = –1210 Sign bit Magnitude By using signed magnitude, n bits can be used to represent integers N in the range of: 2n–1 – 1 ≤ N ≤ 2n–1 – 1 For example, the range of an unsigned 4-bit binary number is from 0 to 15. The range of a signed 4-bit binary number is –7 to + 7 (or 11112 to 01112) For signed magnitude, there are two representations for zero. For example, with n = 4, 0000 and 1000.

1’s Complement Representation Lecture Digital Systems 1’s Complement Representation The 1’s complement of a binary number involves inverting all bits. 1s become 0s, and 0s become 1s. As example, 1’s complement of 00110011 is 11001100 1’s complement of 10101010 is 01010101. Thus, for an n-bit number N, the 1’s complement is 2n – 1 – N. To find the negative of a number, take the 1’s complement of that number. For 1’s complement, there are two representations for zero. For example, with n = 4, 0000 and 1111. 000011002 = 1210 Sign bit Magnitude 111100112 = –1210 Sign bit Magnitude

1’s Complement Addition / Subtraction Lecture Digital Systems 1’s Complement Addition / Subtraction As Example 1, suppose we wish to add 1210 + 110. 1210 = 11002 = 011002 in 1’s complement 110 = 00012 = 000012 in 1’s complement Step 1: Add the binary numbers Step 2: Add the carry to lowest-order bit 011002 000012 011012 011012 = 1310

1’s Complement Addition / Subtraction Lecture Digital Systems 1’s Complement Addition / Subtraction As Example 2, suppose we wish to substract 1210 – 110. 1210 = 11002 = 011002 in 1’s complement –110 = –00012 = 111102 in 1’s complement Step 1: Add the binary numbers Step 2: Add the carry to lowest-order bit 011002 111102 1 010102 1 010112 = 1110

1’s Complement Addition / Subtraction Lecture Digital Systems 1’s Complement Addition / Subtraction As Example 3, suppose we wish to substract –510 – 410. –510 = –01012 = 110102 in 1’s complement –410 = –01002 = 110112 in 1’s complement Step 1: Add the binary numbers Step 2: Add the carry to lowest-order bit 110102 110112 1 101012 1 101102 = –10012 = –910

1’s Complement Addition / Subtraction Lecture Digital Systems 1’s Complement Addition / Subtraction As Example 4, suppose we wish to substract –510 + 410. –510 = –01012 = 110102 in 1’s complement –410 = –01002 = 001002 in 1’s complement Step 1: Add the binary numbers Step 2: Add the carry to lowest-order bit 110102 001002 111102 111102 = –00012 = –110

2’s Complement Representation Lecture Digital Systems 2’s Complement Representation The 2’s complement of a binary number involves inverting all bits and adding 1. As example, 2’s complement of 00110011 is 11001101 2’s complement of 10101010 is 01010110. Thus, for an n-bit number N, the 2’s complement is 2n – 1 – N + 1 = 2n – N. To find the negative of a number, take the 2’s complement of that number. For 2’s complement more negative numbers than positive. 000011002 = 1210 Sign bit Magnitude 111101002 = –1210 Sign bit Magnitude

2’s Complement Addition / Subtraction Lecture Digital Systems 2’s Complement Addition / Subtraction As Example 5, suppose we wish to add –510 + 910. –510 = –01012 = 110112 in 2’s complement –910 = –10012 = 010012 in 2’s complement Step 1: Add the binary numbers Step 2: Discard the carry 110112 010012 1 001002 = 410

2’s Complement Addition / Subtraction Lecture Digital Systems 2’s Complement Addition / Subtraction As Example 6, suppose we wish to add –510 – 910. –510 = –01012 = 110112 in 2’s complement –910 = –10012 = 101112 in 2’s complement Step 1: Add the binary numbers Step 2: Discard the carry 110112 101112 1 100102 = –11102 = –1410

2’s Complement Addition / Subtraction Lecture Digital Systems 2’s Complement Addition / Subtraction As Example 7, suppose we wish to substract 1310 – 510. 1310 = –11012 = 011012 in 2’s complement –510 = –01012 = 110112 in 2’s complement Step 1: Add the binary numbers Step 2: Discard the carry 011012 110112 1 010002 = 810

2’s Complement Addition / Subtraction Lecture Digital Systems 2’s Complement Addition / Subtraction As Example 8, suppose we wish to substract 510 – 1210. 510 = –01012 = 001012 in 2’s complement –1210 = –11002 = 101002 in 2’s complement Step 1: Add the binary numbers Step 2: Discard the carry 001012 101002 110012 = –01112 = –710

Adder-Subtractor Circuit Lecture Digital Systems Adder-Subtractor Circuit The following circuit is called an adder-subtractor. This circuit is capable of adding and subtracting binary numbers. When D = 0, the circuit performs addition, S = A + B. When D = 1, the circuit performs subtraction, S = A – B. The XOR Gates invert the value of B to its 2’s complement (C0 = 1). D

Comparing the Signed Numbers Lecture Digital Systems Comparing the Signed Numbers Signed magnitude: Negating is very easy  Just change the sign bit Adding or subtracting is difficult  If the signs are the same, add the magnitudes and keep the sign. If the signs are different, subtract the smaller operand from the larger operand. The sign of the result is the same as the sign of the larger operand. Rather complex circuit is required. 1’s complement: Negating is easy  Invert the number but keep the sign bit. Adding and subtracting is much easier  Include the sign bits, add directly. If there is carry, add it to the sum Simple but must differentiate cases where carry is 0 or 1. 2’s complement: Negating is not easy  Invert the number, keep the sign, add 1. Adding and subtracting is easy  Include the sign bits, add directly. Ignore the carry, directly get the result. Simple circuit

Overflow in Binary Addition and Subtraction Lecture Digital Systems Overflow in Binary Addition and Subtraction When two numbers of the same sign are added, the answer may not fit the number of bits provided. In this case, the answer exceeds the magnitude which can be represented with the allotted number of bits. This is called overflow. In 2’s complement, overflow occurs when a transition is made from 2n–1 –1 to –2n–1 when adding or from –2n–1 to 2n–1 –1. 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 1 + – 2 3 4 5 6 7 8

Overflow in 4-bit 2 Complement Number Lecture Digital Systems Overflow in 4-bit 2 Complement Number 00 0010 0011 0101 2 3 5 01 0011 0110 1001 Overflow 3 6 –7 11 1110 1101 1011 –2 –3 –5 10 1101 1010 0111 Overflow –3 –6 7 00 0010 1100 1110 2 –4 –2 11 1110 0100 0010 –2 4 2

Homework 10 Create a full adder by using NOR Gates only. Lecture Digital Systems Homework 10 Create a full adder by using NOR Gates only. Convert 1 0111 01102 to decimal from signed magnitude 1’s complement 2’s complement Calculate the following equations using signed magnitude: –810 – 1010 = 2310 – 1710 = Calculate each of the following equations using 1’s and 2’s complement: –310 – 1010 = 2210 – 3010 = Please write your Class number after your Student ID. Deadline: 1 day before class. Monday, 27 November 2017 (Class 2). Tuesday, 28 November 2017 (Class 1).