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EEE342 Digital Electronics

EEE342 Digital Electronics. Lecture 2: Codes & Arithmetic. Ian McCrum Room 5B18, 02890366364 IJ.McCrum@ulster.ac.uk http://www.eej.ulst.ac.uk/~ian/modules/EEE342. Binary Arithmetic - Addition. So 0 0 1 1 +0 +1 +0 +1 -- -- -- -- 0 1 1 10 which is ‘0’ and carry ‘1’

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EEE342 Digital Electronics

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  1. EEE342 Digital Electronics Lecture 2:Codes & Arithmetic Ian McCrum Room 5B18, 02890366364 IJ.McCrum@ulster.ac.uk http://www.eej.ulst.ac.uk/~ian/modules/EEE342

  2. Binary Arithmetic - Addition So 0 0 1 1 +0 +1 +0 +1 -- -- -- -- 0 1 1 10 which is ‘0’ and carry ‘1’ If you end up adding three ones then you get 11, which is a ‘1’ and a carry ‘1’ Remember to carry when you get a 2, not a ten. 0 0 0 0 1 0 1 0 1 1 1 1 0 0 1 = 1401 1 0 1 1 1 0 1 0 1 1 1 0 0 0 1 = +23921 +1 ------------------------------ ------ 0

  3. Binary arithmetic - Subtraction So 0 1 1 0 10 2 -0 -1 -0 -1 -1 -1 1 -- -- -- -- -- 0 0 1 can’t do so borrow ‘1’ 1 E.G 1010 = 1010 -0101 = 510 ----- I.e, read this as 1 from 0 won’t go so 1 from 2 is 1 and carry 1, then we have 1 from 1 which is 0 then 1 from 0, can’t go so borrow 1. 1 from 2 is 0 and carry 1, last subtraction is 1 from 1, which is 0 1 1 ----- 0101 = 510

  4. Binary Multiplication 0 0 1 1 *0*1 *0*1 -- -- -- -- 0 001

  5. Long Binary multiplication As for decimal, you add a zero (equivalent to shifting left one digit position, but since you only multiply by ‘1’ or by zero it becomes an addition problem 10100 = 16 + 4 = 2010 * 101 = 4 + 1 = 510 --------- 10100 000000 1010000 ------------ = 64+32+4 = 10010 1100100

  6. Binary Division 0 0 1 1 ÷0 ÷1 ÷0 ÷1 -- -- -- -- 0 001

  7. Binary Long Division • Beware the Internet! There are actually two ways of doing “division” described on the net • The conventional way • A simple way that gives the wrong answer BUT it is consistent. It is used in cryptography to protect binary data - a process known as adding a CRC. This is a strange thing, we will not deal with it here. • In calculating CRC and other cryptographic things we throw the answer away – it is only the remainder that is used! (not on course…) COM181 Computer Hardware:IMcC

  8. Binary Long Division For decimal long division we subtract the divisor or a multiple of it, for binary this becomes simpler, we subtract the divisor or all zeroes, . 1 0 1 0. 101 / 1 1 10 1 0 1 32+16+4+1= 53 1 01 1 divide by 5 0 0 1 1 0 is 10 rem 3 1 01 1 0 0 1 1

  9. Two’s complement • There is NO subtraction in the two’s complement system • You “take the two’s complement” of a number to change its sign • You then add it (to subtract it!) • i.e two numbers 5 and 3. • 5-3 is actually 5 + (-3) so to subtract 3 from 5, take the two’s complement of 3 and add it to 5.

  10. in a 4 bit two’s complement system The “weights” are (-8) (+4) (+2) and (+1) Hence 5 is 0101 and 3 is 0011 (-3) is (-8)+(4)+(1) = 1101 Adding 0101 = +5 1101 = (-3) ------- 1 0010 = +2 in two’s complement we ignore bits beyond the most significant bit. COM181 Computer Hardware:IMcC

  11. Alternative way to take the two’s complement • E.g to convert 3 to -3. • Write 3 in binary, at the correct length (4 bits in our example here, could be bigger) 0011 NOW INVERT THE BITS 1100 THIS GIVES THIS +1 NOW ADD ONE ---- 1101 AS BEFORE. (inverting the bits is called taking the one’s complement)

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