1. Lecture 2 Binary Arithmetic Topics Terminology Fractions Base-r to decimal Unsigned Integers Signed magnitude Two’s complement August 26, 2003 CSCE 211 Digital Design

2. – 2 – CSCE 211H Fall 2003 Overview Last Time: Readings 1.1, 1.2 2.1-2.3 Analog vs Digital Conversion Base-r to decimal Conversion decimal to Base-r New: Readings 2.4-2.10, 3.1 Conversion of Fractions base-r  decimal Unsigned Arithmetic Signed Magnitude Two’s Complement Excess-1023VHDL Introductory Examples

3. – 3 – CSCE 211H Fall 2003 Review Last Time: Base-r to decimal Converting base-r to decimal by definition d n d n-1 d n-2 …d 2 d 1 d 0 = d n r n + d n-1 r n-1 … d 2 r 2 +d 1 r 1 + d 0 r 0 Example 4F0C = 4*16 3 + F*16 2 + 0*16 1 + C*16 0 4F0C = 4*16 3 + F*16 2 + 0*16 1 + C*16 0 = 4*4096 + 15*256 + 0 + 12*1 = 16384 + 3840 + 12 =20236

4. – 4 – CSCE 211H Fall 2003 Review Last Time: decimal to Base-r Repeated division algorithm Justification: d n d n-1 d n-2 …d 2 d 1 d 0 = d n r n + d n-1 r n-1 … d 2 r 2 +d 1 r 1 + d 0 r 0 Dividing each side by r yields (d n d n-1 d n-2 …d 2 d 1 d 0 ) / r = d n r n-1 + d n-1 r n-2 … d 2 r 1 +d 1 r 0 + d 0 r -1 So d 0 is the remainder of the first division ((q 1 ) / r = d n r n-2 + d n-1 r n-3 … d 3 r 1 +d 2 r 0 + d 1 r -1 So d 1 is the remainder of the next division and d 2 is the remainder of the next division …

5. – 5 – CSCE 211H Fall 2003 Review Last Time: decimal to Base-r Repeated division algorithm Example Convert 4343 to hex 4343/16 = 271 remainder = 7 271/16 = 16 remainder = 15 271/16 = 16 remainder = 15 16/16 = 1 remainder = 0 16/16 = 1 remainder = 0 1/16 = 0 remainder = 1 1/16 = 0 remainder = 1 So 4343 10 = 10F7 16 To check the answer convert back to decimal 10F7 = 1*16 3 + 15*16 + 7*1 = 4096 + 240 + 7 = 4343

6. – 6 – CSCE 211H Fall 2003 Review Last Time: Hex to Binary One hex digit = four binary digits Example 3FAC = 0011 1111 1010 1100 (spaces just for readability) 3FAC = 0011 1111 1010 1100 (spaces just for readability) Binary to hex four digits = one (from right!!!) Example 111101001111010 = 0011 1101 0011 1010 111101001111010 = 0011 1101 0011 1010 = 3 D A

7. – 7 – CSCE 211H Fall 2003 Hex to Decimal Fractions.d -1 d -2 d -3 …d –(n-2) d –(n-1) d -n = d -1 r -1 + d -2 r -2 …d -(n-1) r -(n-1) +d 1 r -n Example.1EF 16 = 1*16-1 + E*16-2 + F*16-3.1EF 16 = 1*16-1 + E*16-2 + F*16-3 = 1*.0625 + 14*.003906025 + 15*2.4414e-4 =.117201… (probably close but not right) =.117201… (probably close but not right)

8. – 8 – CSCE 211H Fall 2003 Decimal Fractions to hex.d -1 d -2 d -3 …d –(n-2) d –(n-1) d -n = d -1 r -1 + d -2 r -2 …d -(n-1) r -(n-1) +d 1 r -n Multiplication by r yields r *(.d -1 r -1 + d -2 r -2 …d -(n-1) r -(n-1) +d 1 r -n = d -1 r 0 + d -2 r -1 …d -(n-1) r -(n-2) +d 1 r -(n-1) Whole number part = d -1 r 0 Multiplying again by r yields d -2 r 0 as the whole number part … till fraction = 0

9. – 9 – CSCE 211H Fall 2003 Example: Hex Fractions to decimal Convert.3FA to decimal.3FA16 = 3*16-1 + F*16-2 + A*16-3.3FA16 = 3*16-1 + F*16-2 + A*16-3 = 3*.0625 + 15*.00390625 +10* (1/4096) =.191162109

10. – 10 – CSCE 211H Fall 2003 Unsigned integers What is the biggest integer representable using n-bits(n digits)? What is its value in decimal? Special cases 8 bits 16 bits 16 bits 32 bits 32 bits

11. – 11 – CSCE 211H Fall 2003 Arithmetic with Binary Numbers 10010110 100101101101 +00110111 - 00110111 x 101 +00110111 - 00110111 x 101 Problems with 8 bit operations 10010110 +10010110 +10010110

12. – 12 – CSCE 211H Fall 2003 Signed integers How do we represent? Signed-magnitude Excess representations w bits  0 <= unsigned_value < 2 w In excess-B we subtract the bias (B) to get the value. In excess-B we subtract the bias (B) to get the value. example example

13. – 13 – CSCE 211H Fall 2003 Complement Representations of Signed integers One’s complement Two’s complement

14. – 14 – CSCE 211H Fall 2003 Two’s Complement Operation One’s complement + 1 or Find rightmost 1, complement all bits to the left of it. Examples 01001110000000010000000000000010 01001110000000010000000000000010

15. – 15 – CSCE 211H Fall 2003 Two’s Complement Representation Consider a two’s complement binary number d n d n-1 d n-2 …d 2 d 1 d 0 If d n, the sign bit = 0 the number is positive and its magnitude is given by the other bits. If d n, the sign bit = 1 the number is negative and take its two’s complement to get the magnitude. Weighted Sum Interpretation 0 1 1 2 … n-1 2 n-2 n -2 n-1

16. – 16 – CSCE 211H Fall 2003 Two’s Complement Representation Consider a two’s complement binary number d n d n-1 d n-2 …d 2 d 1 d 0 If d n, the sign bit = 0 the number is positive and its magnitude is given by the other bits. If d n, the sign bit = 1 the number is negative and take its two’s complement to get the magnitude. Weighted Sum 0 1 Example 10010011 = 1 2 … n-1 2 n-2 n -2 n-1

17. – 17 – CSCE 211H Fall 2003 Two’s Complement Representation What is the 2’s complement representation in 16 bits of –5? +7?-1?0-2

18. – 18 – CSCE 211H Fall 2003 Arithmetic with Signed Integers Signed Magnitude Addition if the signs are the same add the magnitude if the signs are different subtract the smaller from the larger and use the sign of the larger if the signs are different subtract the smaller from the larger and use the sign of the largerSubtraction? Two’s complement Just add signs take care of themselves

19. – 19 – CSCE 211H Fall 2003 Overflow in Two’s Complement

20. – 20 – CSCE 211H Fall 2003 Binary Code Decimal

21. – 21 – CSCE 211H Fall 2003 Representations of Characters

22. – 22 – CSCE 211H Fall 2003 Basic Gates ANDORNOT

23. – 23 – CSCE 211H Fall 2003 Basic Gates NANDNORXOR

24. – 24 – CSCE 211H Fall 2003 Half Adder Circuit

25. – 25 – CSCE 211H Fall 2003 Full Adder