ME 4447/6405

1. ME 4447/6405 Microprocessor Control of Manufacturing Systems and Introduction to Mechatronics Instructor: Professor Charles Ume Lecture #6

2. Review Digital Arithmetic

3. Hexadecimal – Base 16

4. Octal – Base 8

5. - 64 34 32 2 32 10 - Remainder A Remainder 2 Remainder 2 Read Conversion from base 10  base 16  base 2 and back  base 10 Example 1: 67410 - Read Remainders backwards Therefore, 67410 = 2A216 = 0010 1010 0010

6. 2A216 to base 10: = 2 x 160 + A x 161 + 2 x 162 = 67410 0010 1010 00102 to base 10: =0 x 20 + 1 x 21 + 0 x 22 + 0 x 23 +0 x 24 + 1 x 25 + 0 x 26 + 1 x 27 +0 x 28 + 1 x 29 + 0 x 210 + 0 x 211 =67410 Example 1 Cont’d

7. Remainder 1 Remainder 0 Remainder 0 Remainder 0 Read Remainder 1 Remainder 1 Conversion from base 10 to 2 This method known as “repeated division by 2.” Example 2: Convert 3710 to base 2. Stop dividing when the answer is 0. The remainders are read in reverse order to form the answer. Therefore, 3710 = 00100101 = 2516(Note: 2 front bits added for 8-bit processor)

8. Note: 3710 = 7 x 100 + 3 x 101 = 7 + 30 = 37 Therefore, 1001012 = 1 x 20 + 0 x 21 + 1 x 22 + 0 x 23 + 0 x 24 + 1 x 25 + 0 x 26 + 0 x 27 = 1 + 0 + 4 + 0 + 0 +32 + 0 + 0 = 3710 Example 2 Cont’d

9. Convert 110.1112 to the decimal system Solution: 110.1112 = (1 x 2-3) + (1 x 2-2) + (1 x 2-1) + (0 x 20) + (1 x 21) + (1 x 22) = 0.125 + 0.25 + 0.5 + 2 + 4 = 6.87510 Example 3

10. Convert 2A2.516 to the decimal system Solution: 2A2.5 = (5 x 16-1) + (2 x 160) + (A x 161) + (2 x 162) = 0.3125 + 2 +160 +512 = 674.3125 Example 4

11. Example 5 Convert 0.7510 to Binary Solution: 0.75 x 2 = 1.50  1 0.50 x 2 = 1.00  1 Therefore, 0.7510 = 0.112 Read

12. Example 6 Convert 0.301710 to Binary Solution: 0.3017 x 2 = 0.6034  0 0.6034 x 2 = 1.2068  1 0.2068 x 2 = 0.4136  0 0.4136 x 2 = 0.8272  0 Therefore, 0.301710 = 0.0100…… Read

13. The circuitry required for computers to add binary numbers is relatively simple. However, since circuitry to subtract binary numbers is very difficult to design and build, another subtraction method would be desirable. The method is subtraction by addition of the 1’s or 2’s complements of the number. Subtraction by the addition of 1’s complement requires much more hardware to implement. Most computers on the market perform subtraction by the addition of 2’s complement. Therefore we will concentrate only on subtraction by 2’s complement. Binary number = 1011 1001 1’s complement = 0100 0110 2’s complement = 0100 0111 When subtraction is performed the answer can be positive or negative, depending upon the relative values of the minuend and subtrahend. We must, therefore, in some way account for the sign of the answer.

14. The most significant bit (MSB) is used as the sign bit. If the MSB is a ‘1’, the number represented by the remaining 7 bits is assumed to be negative and is represented by its 2’s complement. If the MSB is a ‘0’, the number represented in the remaining seven bits is positive and is equal to the binary equivalent of these seven bits. Example 1: 1710 – 610=(1116 – 616) 0000 0110 = 616 1111 1001 = 1’s complement of 616 1111 1010 = 2’s complement of 616 0001 0001 = 1116 1111 1010 = 616 in 2’s complement ------------- 1 0000 1011 1 0000 1011 = + 0B16 = + 1110

15. Example 2: 610 – 1710 = (616 – 1116) 0001 0001 = 1116 1110 1110 = 1’s complement of 1116 1110 1111 = 2’s complement of 1116 0000 0110 = 616 1110 1111 = 1116 in 2’s complement ------------ 1111 0101 1111 0101 = - (0000 1010 + 1) = - 0000 1011 = - 0B16 or - 1110 Note: C and N bits will be set. C is set to 1 because the absolute value of subtraction (17) Is larger than the absolute value of the minuend (6)

16. Since bit 7 serves as the sign bit. A ‘0’ in bit 7 indicates a positive number. Thus the maximum positive number an 8-bit processor can represent in 2’s complement is 0111 11112 or 7F16 or 12710. Likewise a ‘1’ in bit 7 indicates a negative number. The minimum negative number that an 8-bit processor can represent in 2’s complement is 8016 (1000 0000). In binary it is 2’s complement of 1000 0000 = - 1000 0000 = - 8016 = -12810 Its 2’s complement is – (0111 1111 + 1) = - (1000 0000) = - 8016 = - 12810 It is possible to (add) or subtract two numbers and get invalid answer, especially when the answer is out of the range of –128  +127 This situation is referred to as 2’s complement overflow. If a subtraction (or addition) is performed and 2’s complement overflow occurs the V bit will be set to a ‘1’ otherwise it will be cleared.

17. Example 3: - 8510 - 9010= -17510 8510 = 5516 = 0101 01012 1010 1010 = 1’s comp. of 5516 1010 1011 = 2’s comp. of 5516 9010 = 5A16 = 0101 10102 1010 0101 = 1’s comp. of 5A16 1010 0110 = 2’s comp. of 5A16 1010 1011 = (2’s comp. of 5516) 1010 0110 = (2’s comp. of 5A16) ------------- 1 0101 0001 = +5116 = +8110 V bit will be set. C bit will be set.

18. Example 4: 4D16 + 6616 • 01001101 positive • 01100110 positive • -------------- • 10110011 negative • V bit is set.

19. Logic Gates

20. Why do we study Logic Gates? Used to Determine Status or Occurrence of Events Logic Gates

21. Symbol Input A Input B Output C 0 0 1 1 0 1 0 1 0 0 0 1 Equation “AND” Function C = A ● B

22. Symbol Input A Input B Output C 0 0 1 1 0 1 0 1 0 1 1 1 Equation “OR” Function C = A + B

23. Symbol Input A Input B Output C 0 0 1 1 0 1 0 1 0 1 1 0 Equation C = A + B “Exclusive OR” Function (XOR)

24. Symbol Input A Output C 0 1 1 0 Equation C = A “Not” Function (Inverter)

25. Input A Input A Input B Input B Output C Output C 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 0 0 1 1 1 0 “NOR” Function “NAND” Function

26. QUESTIONS???