Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.
Upcoming SlideShare
×

# Arithmetic logic units

• Full Name
Comment goes here.

Are you sure you want to Yes No
Your message goes here

### Arithmetic logic units

1. 1. BINARY ADDITION How to add binary numbers. A B 0+0=0 0 0 0 0+1=1 1 0 1 1+0=1 2 1 0 1 + 1 = 10 3 1 1 1 + 1 + 1 = 11
2. 2. 28 FOR EXAMPLE: Here 1+1+1 (carry) produces 11, 1 1 1 0 0 +26 +1 1 0 1 0 54 11 0 1 1 0 Here, 0+0 gives 0 recorded as 1 wit a carry to the next higher column . 0+1 gives 1 1+0 produces 1 1+1 equal 0 carry is 1
3. 3. BINARY SUBTRACTION How to subtract BINARY NUMBERS 0-0=0 0-1=1 1-1=0 10 - 1 = 1
4. 4. FOR EXAMPLE: 7 0111 -5 -0101 2 0010 Another example: 15 - 7 8 1111 - 0111 1000
5. 5. HALF-ADDERS A logic circuit that can adds 2 bits are called Half-Adders The out put is SUM AND CARRY. Half adder do not add three Bits. The Boolean equations are: SUM = A B (A XOR B) CARRY = AB (A AND B) A B CARRY SUM
6. 6. INPUT OUT PUT A B SUM CARRY 0 0 0 1 1 0 1 1 0 1 1 0 0 0 0 0 SUM is 1 when A and B are different CARRY is 1 when A and B are 1s
7. 7. FULL ADDERS A logic circuit that can add 3 bits. Again there are two out put SUM and CARRY SUM equals A XOR B XOR C CARRY equals AB OR BC OR AC
8. 8. The Booleans equations are SUM = A B C CARRY =AB + BC + AC LOGIC GATE FULL ADDER A B C CARRY SUM
9. 9. INPUT OUTPUT A B C 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 SUM 1 0 1 1 0 1 0 0 1 CARRY 0 0 0 1 0 1 1 1 FUNDAMENTAL PRODUCT FOR CARRY ABC ABC ABC ABC SUM is 1 when the number of input 1’s is odd Carry is one when two are more input are 1s
10. 10. KARNAUGH MAP FOR CARRY C AB AB 0 0 AB 1 AB C 0 AB + BC + AC 0 1 1 1 FUNDAMENTAL PRODUCT ABC+ABC+ABC+ABC
11. 11. BINARY ADDERS A logic circuit that can add two binary numbers. HA represent Half adder FA represent Full adder The input are A0 and B0 The out puts are S0 (SUM) and C1 (CARRY)
12. 12. FOR EXAMPLES 1110 +1 0 0 1 ? A1 C1 A1 B1 FA S0 C0 B0 FA S1 A1 C0 A0 B0 FA S0 C0 B1 HA S1
13. 13. SIGNED BINARY NUMBERS The negative decimal numbers are( -1, -2, -3…….) one way to code these as binary numbers is to convert the magnitude(1, 2, 3……)to its binary equivalent And prefix the sign. with this approach -1, -2 and -3 becomes -001,-010.and -011. Use 0 for + sign and 1 for – sign. There for -001,-010 and -011 are coded as 1001,1010,1011. the forgoing number have the sign bit followed by the magnitude bits number in this form are called sign binary numbers or sign magnitude numbers.
14. 14. Express each of the following as 16 bits signed binary numbers +7 and -7 4 21 +7 = 0000 0000 0000 0111 -7 = 1000 0000 0000 0111
15. 15. 2’S COMPLEMENT A high invert signal to a controlled inverter produces 1’s complement. If 1’S COMPLEMENT: A=0111 The 1’s complement is A=1000
16. 16. 2’s complement: The new word obtained by adding 1 to 1’s complement A=A+1 1’s complement is Where A = 2’s complement A = 1’s complement A=0111 A=1000 2’s complement is A=1000 +1 A =1 0 0 1
17. 17. Prove that A = A IF A = 0 1 1 1 1’s complement is 1’s complement is A=1000 2’s complement is A=0110 2’s complement is A=1000 +1 A=1001 A = A HENCE PROOVE A=0110 +1 A=0111
18. 18. 2’S COMPLEMENT ADDER SUBTRACTOR 2’s complement adder- subtracted , a logic circuit that can add or subtract binary numbers. ADDITION: When SUB is low ,the B bits pass through the controlled inverter with out inversion therefore the FULL ADDERs produce the sum S=A+B
19. 19. 4 +3 7 0100 0011 0111 0 0 0 1 0 0 0 0 1 1 0 0 FA 0 0 0 1 FA 1 0 0 0 FA 1 0 1 0 Sub=0 1 0 FA 1 0
20. 20. SUBTRACTION: When SUB is high, the control inverter produces the 1’s complement The high SUB adds a one to the FULLADDER.this addition of 1 to The 1’s complement forms the two complement of B.the control inverter produces B and adding 1 result in B . the out put of the full adders is S=A+B Which is equivalent to S=A-B Because the 2’s complement is equal to the sign change.
21. 21. 9 1001 -2 0010 7 0111 1 0 0 1 0 0 1 0 1 FA 0 0 0 1 0 1 1 1 1 1 FA 1 0 1 0 FA 1 1 SUB=1 1 1 FA 1 1