By Wannarat Computer System Design Lecture 3 Wannarat Suntiamorntut

By Wannarat Arithmetic for Computer Implementing the Architecture

By Wannarat The numbers Binary number (base 2) numbers are finite (overflow) fraction and real number negative number e.g., No MIPS subi instruction, addi can add a negative number

By Wannarat Possible Representations Sign MagnitudeOne’s complementTwo’s complement 000 = = = = = = = = = = = = = = = = = = = = = = = = -1 Negative and Invert are different!

By Wannarat Addition & Subtraction { } = 0001 two’s complement = 0001 Overflow n-bit does not yield an n-bit number

By Wannarat Detecting overflow No overflow when add positive&negative No overflow when signs are the same for subtraction Overflow occur when : - add two positives yield a negative - add two negative gives a positive - subtract negative from positive and get negative - subtract positive from negative get a positive

By Wannarat Effect of Overflow Exception occur (Interrupt) - control jumps to predefined address for exception - Interrupted address is saved Don’t always to detect overflow - New MIPS instructions: addu, addiu...

By Wannarat ALU Build ALU support andi, ori instructions

By Wannarat Multiplexor Select one of the inputs to be output, base on control input

By Wannarat ALU for Addition instruction Cout = ab + bc in + ac in Sum = a xor b xor c in

By Wannarat ALU for Subtraction instruction Two’s complement approach : just negative b and ADD

By Wannarat Supporting slt

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MIPS Arithmetic Instruction format

By Wannarat Conclusion ALU in MIPS Use multiplexor to select output we want efficiently perform subtraction using two’s complement replicate 1-bit ALU to 32-bit ALU

By Wannarat Computation Problem :fast adder 32-bit ALU faster than 1-bit ALU? Carry-lookahead adder g = ab,p = a + b c1 = g0 + p0c0 c2 = g1 + p1c1c2 c3 = g2 + p2c2c3...

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Part II : Lecture III

By Wannarat Multiplication 0010(multiplicand) x1011 (multiplier) ????

By Wannarat Unsign Combinational Multiplier

By Wannarat Multiplication : First Version (Unsign)

By Wannarat Multiplication : First Version (contd.)

By Wannarat Analyze First Version 1 clock per cycle 50 % of bit in multiplicand always = 0 => 64-bit adder is wasted 0’s inserted in left of multiplicand as shifted => lead significant bits of product never changed once formed

By Wannarat Multiplication : Second Version

By Wannarat Multiplication : Second Version (Contd.)

By Wannarat Analyze Second Version Product register wasted space that exactly matches size of multiplier Combine Multiplier register and Product register

By Wannarat Multiplication : Third Version

By Wannarat Multiplication : Third Version (Contd.)

By Wannarat Analyze Third Version 2 steps per bit because multiplier & product combined MIPS registers Hi, Lo are left and right half of product

By Wannarat Booth’s Algorithm

By Wannarat Example : 2 x 7

By Wannarat Example : 2 x -3

By Wannarat Shifter : 2 kinds

By Wannarat Part III : Lecture III

By Wannarat Divide

By Wannarat Divide : First Version

By Wannarat Divide : First Version (Contd.)

By Wannarat Analyze First Version 50%bits in divisor always 0 =>1/2 of 64-bit adder is wasted => 1/2 divisor is wasted 1 step cannot produce a 1 in quotient bit => Switch order to shift first

By Wannarat Divide : Second Version

By Wannarat Divide : Second Version (Contd.)

By Wannarat Analyze Second Version Eliminate Quotient register by combining with Remainder as shifted left

By Wannarat Divide : Third Version

By Wannarat Divide : Third Version (Contd.)

By Wannarat Analyze Third Version Do Analyze by yourself

By Wannarat Floating Point : IEEE754

By Wannarat Floating-point Representation = -3/4 = -3/2 2 = -11/ 2 2 = -0.11= -1.1x 2 -1 = (-1) s x (1 + signifiand) x 2 (exponent-127) = (-1) x ( ) x 2 ( ) S E M 1-bit 8-bit 23-bit

By Wannarat Floating-point Addition x x10 -1 Step1 : Change exponent as : x = x 10 1 Step2 : Add significands (10) (10) (10) Sum = x 10 1

By Wannarat Floating-point Addition x x10 -1 Step3 : correct it (normalization) : x 10 1 = x 10 2 Step4 : Four digits for significand x 10 2

By Wannarat Example Floating- point Addition ( ) 0.5 = 1/2 = 1/2 1 = 0.1 x 2 0 = 1.00 x = -7/16 = -7/2 4 = = x 2 -2 step 1 : x 2 -1 step 2 : 1.0x ( x 2 -1 )=0.001 x 2 -1 step 3 : 1.0 x 2 -4 step 4 :

By Wannarat Multiplication Floating-point Study in Text Book by yourself.

By Wannarat Next on Lecture 4