Slide Number 1Multiplication and Division made easyBinary to BCDBinary to BCDConversion optionsBCD MultiplicationWhy Use a Comparator? Shift-Add-3 No Adders!No Shifters!No Recursion Full 8-bit conversionLab AssignmentDepartment of Electrical and Compute EngineeringMississippi State UniversitySherif Abdelwahed Shift-add-3 AlgorithmSeptember 5, 2007Computer Aided Digital Systems Design - EE 4743/6743Multiplication and Division made easy To multiply by 2, shift left To divide by 2, shift right Don’t need to use shifter though! B = A / 2A3A2A1A0B3B2B1B0‘0’Binary to BCD Seven-segment display on the Spartan 3 board Currently is set up to display hexadecimal 0-9, A-F Lab 2 is set up to display input switches in hexadecimal on seven-segment display Exercise is to build circuit to change to decimal Input is 8 switches -> 28= 256  Three segments neededBinary to BCD Easy to change from four-bit binary to BCD Anything above 9 is added to the next positionDecimal Digit Binary Representation0 00001 00012 00103 00114 01005 0101601107 01118 10009 1001Conversion options Given 4-bit number : 1100¾ Equivalent to decimal 12 Options: ¾ Subtract 10 from number to get 2 Then add one to the ten digit¾ Add 6, then check the remainderOriginal number: 1100 12Add 6 0110 6Final number 0001 0010 12BCD Multiplication If we want to multiply by 2, we can keep the number in BCD by thinking ahead If starting number is 5 or larger, then we will end up with a number that must be converted¾ 4 * 2 = 8 (1 BCD character)¾ 5 * 2 = 10 (2 BCD characters)¾ 6 * 2 = 12 (2 BCD characters) Add 3 if number is >4, then shift left for the multiply Like adding 6, but with no carryOriginal number 0111 7Add 3 1010 10Shift left 0001 0100 14Why Use a Comparator?  Use a recursive program to compute numbers without a comparator For every bit, we will check & shift once¾ 4-bit number require 4 shifts¾ 4 shifts mean our final answer is “multiplied” by 16¾ Use 4 bits to the left to “divide” by 16¾ Ones must be checked for >4 Add 3 if so Called “Shift-Add-3” formula Doesn’t necessarily need to be recursive…Shift-Add-3  Note for the largest 4-bit number, only one Add-3 is ever needed Can shorten procedure by doing a 3-bit shift to startOperation Tens Ones BinaryStart0000 00001111Shift (1)00000001 1110Shift (2)00000011 1100Shift (3)00000111 1000Add-300001010 1000Shift (4) 0001 01010000Decimal 15No Adders! Always adding by a constant (3) Instead of using an Adder, we can use a look-up table Only 4-bits input and output (since our value will always be in BCD) Verilog will want to make an adder out of S <= A + 3;A3 A2 A1 A0 S3 S2 S1 S00000 00000001 00010010 00100011 00110100 01000101 10000110 10010111 10101000 10111001 11001010 XXXX1011 XXXX1100 XXXX1101 XXXX1110 XXXX1111 XXXXNo Shifters! Use multiply-by-2 trick Instead of shifting, we can just connect our output one place to the left B = A * 2A3A2A1A0B4B3B2B1B0‘0’No Recursion A3A2A1A0S3S2S1S0Add-3Onesbcd(0)bcd(1)bcd(2)bcd(3)bcd(4)bcd(5)bcd(6)bcd(7)Tensbin(0)bin(1)bin(2)bin(3)‘0’‘0’‘0’‘0’Full 8-bit conversionA3A2A1A0S3S2S1S0Add-3A3A2A1A0S3S2S1S0Add-3Onesbcd(0)bcd(1)bcd(2)bcd(3)bcd(4)bcd(5)bcd(6)bcd(7)bcd(8)TensA3A2A1A0S3S2S1S0Add-3A3A2A1A0S3S2S1S0Add-3 bcd(9)bcd(10)bcd(11)HundredsA3A2A1A0S3S2S1S0Add-3A3A2A1A0S3S2S1S0Add-3A3A2A1A0S3S2S1S0Add-3bin(0)bin(1)bin(2)bin(3)bin(4)bin(5)bin(6)bin(7)‘0’‘0’‘0’‘0’Lab Assignment First task is to build the Add-3 circuit¾ All combinational¾ Just implement truth table¾ Four equations: S3, S2, S1, S0 Put the Add-3 module in the top-level schematic¾ Use previous circuit diagram For B grade: use all schematic For A grade: use Verilog for Add-3 circuit Use simulator to check Add-3