2a.1 2a.2 EE 457 Unit 2a Unsigned 2’s Complement Sign and Zero Extension Fixed Point Systems and Arithmetic Hexadecimal Representation SIGNED AND UNSIGNED SYSTEMS 2a.3 2a.4 Signed Systems Unsigned and Signed Variables • Several systems have been used • Unsigned variables use unsigned binary (normal power-of-2 place values) to represent numbers – 2’s complement system – 1’s complement system 1 0 0 1 0 0 1 1 – Sign and magnitude 128 64 32 16 8 4 2 1 • Signed variables use the ____________ system (____________ weight) to represent numbers 1 0 0 1 0 0 1 1
2a.5 2a.6 2’s Complement System Zero and Sign Extension • Extension is the process of increasing the number of bits used • MSB has negative weight to represent a number without changing its value • MSB determines sign of the number Unsigned = Zero Extension (Always add leading 0’s): – 1 = negative – 0 = positive 111011 = 00111011 • To take the negative of a number Increase a 6-bit number to 8-bit number by zero extending (e.g. -7 => +7 or +2 => -2), requires __________ ____________________ 2’s complement = Sign Extension (Replicate sign bit): – This is accomplished by ______________________ 011010 = 00011010 pos. Sign bit is just repeated as many times as necessary x = -7 1001 110011 = 11110011 neg. 0110 Bit flip (1’s comp.) Add 1 + 1 0111 -x = -(-7) = +7 2a.7 2a.8 Zero and Sign Truncation Arithmetic & Sign • Truncation is the process of decreasing the number of bits used • You learned the addition (carry-method) and to represent a number without changing its value subtraction (borrow-method) algorithms in grade school Unsigned = Zero Truncation (Remove leading 0’s): • Consider A + B…do you definitely use the addition Decrease an 8-bit number to 6-bit number by truncating 0’s. Can’t 00111011 = 111011 algorithm? remove a ‘1’ because value is changed 2’s complement = Sign Truncation (Remove copies of sign bit): – What if A=(2), B=(-5)? pos. 00011010 = 011010 Any copies of the MSB can be removed without changing the numbers value. Be careful not to neg. 11110011 = 10011 change the sign by cutting off ALL the sign bits. • Human add/sub algorithm depends on ________!!
2a.9 2a.10 Unsigned and Signed Arithmetic Unsigned and Signed Subtraction • Addition/subtraction process is the same for • Subtraction process is the same for both both unsigned and signed numbers unsigned and signed numbers – Add columns right to left – Convert A – B to A + Comp. of B – Drop any final carry out – Drop any final carry out • This is the KEY reason we use 2’s complement • Examples: system to represent signed numbers If unsigned If signed • Examples: 1100 (12) (-4) - 0010 (2) (2) 1 1 If unsigned If signed 1001 + 0011 If unsigned If signed 1100 2a.11 2a.12 Overflow Unsigned Overflow • Overflow occurs when the result of an Overflow occurs when you cross arithmetic operation is too ____________ this discontinuity 0 +15 __________ with the given number of bits +1 0000 1111 0001 +14 +2 1110 0010 – Unsigned overflow (___) occurs when adding or +13 +3 1101 0011 subtracting unsigned numbers Plus 7 10 + 7 = 17 +12 1100 – Signed (2’s complement overflow) overflow ( ___ ) 0100 +4 4 - 6 = 14 occurs when adding or subtracting 2’s With 4-bit unsigned numbers we 10 1011 0101 +11 +5 can only represent 0 – 15. Thus, complement numbers 1010 0110 we say overflow has occurred. +10 1001 0111 +6 1000 +7 +9 +8
2a.13 2a.14 2’s Complement Overflow Testing for Overflow • Most fundamental test 0 – Check if answer is wrong (i.e. Positive + Positive yields a -1 +1 0000 1111 0001 negative) -2 +2 1110 0010 • Unsigned overflow ( C ) test -3 +3 1101 0011 – If _____________________________ 5 + 7 = +12 -4 1100 0100 +4 • Signed (2’s complement) overflow ( V ) test -6 + -4 = -10 – Only occurs if 1011 0101 -5 +5 With 4-bit 2’s complement 1010 • ____________________________ … or… 0110 numbers we can only represent -6 1001 0111 +6 • ____________________________ -8 to +7. Thus, we say overflow 1000 +7 -7 – Alternate test: See following slides has occurred. -8 Overflow occurs when you cross this discontinuity 2a.15 2a.16 Alternate Signed Overflow Test Overflow in Addition • Overflow occurs when the result of the A & B A3 B3 S3 C3 C4 V addition cannot be represented with the given 0 Both Positive 0 0 1 number of bits. 0 0 1 • Tests for overflow: 1 One Positive & One Negative – Unsigned: if Cout = 1 0 1 0 1 – Signed: if p + p = n or n + n = p 0 Both Negative 1 1 1 If unsigned If signed If unsigned If signed 1101 0110 + 0100 + 0101 • Check if _______________________________
2a.17 2a.18 Overflow in Subtraction Addition – Full Adders • Show the following addition operation • Overflow occurs when the result of the subtraction cannot be represented with the given number of bits. 0110 = X • Tests for overflow: + 0111 = Y – Unsigned: if Cout = 0 – Signed: if addition is p + p = n or n + n = p If unsigned If signed 0111 X Y X Y X Y X Y - 1000 Full Full Full Full C out C in C out C in C out C in C out C in Adder Adder Adder Adder S S S S If unsigned If signed 2a.19 2a.20 XOR Gate Review Performing Subtraction w/ Adders • To subtract X 0101 0101 = X Z – Flip bits of Y Y + 1100 - 0011 = Y 1 XOR 1101 – Add 1 0010 Z = X ⊕ Y X Y Z 0 0 0 0 1 1 X Y X Y X Y X Y 1 0 1 Full Full Full Full 1 1 0 C out C in C out C in C out C in C out C in Adder Adder Adder Adder S S S S True if an odd # of inputs are true 2 input case: True if inputs are different
2a.21 2a.22 XOR Conditional Inverter Adder/Subtractor • If one input (say, X) to an XOR gate is 0, • Using XOR gates X Y Z then Z=_____ before one set of 0 0 0 0 1 1 • If one input (say, X) to an XOR gate is 1, adder inputs we can 1 0 1 then Z=_____ 1 1 0 – Selectively pass or • Use one input as a control input which invert Y can conditionally pass or invert the – Add an extra ‘1’ via other input the Carry-in • If SUB/~ADD=0, – Z = X+Y • If SUB/~ADD=1, – Z = X-Y 2a.23 2a.24 Adder/Subtractor ALU Design • Exercise: Add Complete the ALU appropriate logic to design given the function table produce below – C (unsigned overflow) – V (signed overflow) OP[2:0] Z flags 000 X+Y 001 X-Y 011 SLT: Z=1, if X<Y Z=0, other 100 AND 110 OR Others Z = und.
2a.25 2a.26 Hexadecimal Representation • Since values in modern computers are many bits, we use hexadecimal as a shorthand notation (4 bits = 1 hex digit) – 11010010 = D2 hex – 0111011011001011 = 76CB hex • To interpret the value of a hex number, you must know what underlying binary system is NON-REQUIRED MATERIAL assumed (unsigned, 2’s comp. etc.) 2a.27 2a.28 Translating Hexadecimal Taking the 16’s Complement • Hex place values (16 2 , 16 1 , 16 0 ) can ONLY be used if • Taking the 2’s complement of a binary number yields the number is positive. its negative and is accomplished by finding the 1’s • If hex represents unsigned binary simply apply hex complement (bit flip) and adding 1 place values • Taking the 16’s complement of a hex number yields – B2 hex = 11*16 1 + 2*16 0 = 178 10 its negative and is accomplished by finding the 15’s • If hex represents signed value (2’s comp.) complement and adding 1 – First determine the sign to be pos. or neg. • Convert the MS-hex digit to binary to determine the MSB (e.g. for – 15’s complement is found by subtracting each digit of the B2 hex, B=1011 so since the MSB=1, B2 is neg.) hex number from F 16 • In general, hex values starting 0-7 = pos. / 8-F = neg. – If pos., apply hex place values (as if it were unsigned) Original value B2: FF - B2 Subtract each digit from F – If neg., take the 16’s complement and apply hex place 4D 15’s comp. of B2 values to find the neg. number’s magnitude + 1 Add 1 16’s comp. of B2: 4E 16’s comp. of B2
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