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Real number : X Machine representation: X M Due to limited digits Let ulp= unit of least position

Machine Representation. Real number : X Machine representation: X M Due to limited digits Let ulp= unit of least position Then X M  X  X M +ulp Error exists: e= X M  X < ulp Overflow or Underflow. Number System. Positive binary number system:

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Real number : X Machine representation: X M Due to limited digits Let ulp= unit of least position

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  1. Machine Representation • Real number : X • Machine representation: XM • Due to limited digits • Let ulp= unit of least position • Then XM  X  XM +ulp • Error exists: e= XM X < ulp • Overflow or Underflow

  2. Number System • Positive binary number system: • X=(xn-1, xn-2, …, x0)=xi2i, xi0,1 • Signed-magnitude: • xn-1 sign bit, others magnitude • Complement • 2’s Complement • 1’s Compelement

  3. Complement Representation • Let R= givenconstant, Y: n-bit • Complement of Y = RY • Check if (Y) = RY • then ((Y))=R (R Y)=Y O.K. • Complement • 2’s (Radix) Complement: R= 2n • 1’s (Diminished-Radix) Complement: R=2n ulp

  4. Radix Complement • 2’s Complement • R= 2n, Y: n-bit, Y= 2n Y=Y’+1 • Y’= (1,1, …,1)  Y • Example: Y=(0110), Y=(1010) • X=(xn-1, xn-2, …, x0) =  xn-12n-1 + xi2i for i= 0, 1, …, n-2

  5. Diminished-Radix Complement • 1’s Complement • R= 2n 1, Y= 2n Y  1=Y’ • Y’= (1,1, …,1)  Y • Example: Y=(0110), Y=(0101) • X=(xn-1, xn-2, …, x0) =  xn-1 (2n-1 1)+ xi2i for i= 0, 1, …, n-2

  6. Example

  7. Ranges • SM: (2n1 1)  Y  (2n 1 1) • Two Zeros: (00…0) & (10…0) • 1’s C: (2n1 1)  Y  (2n 1 1) • Two Zeros:(00…0) & (11…1) • 2’s C: (2n1 1)  Y  2n 1

  8. Addition/Subtraction • Operations: • ADD (M=0); SUB (M=1) • SM: Sign-bit (SA), Overflow (=OV), End-around-carry (=C), magnitude >0 • 1’s C: Overflow, End-around-carry • 2’s C: Overflow • Complement  Using XOR gate

  9. 2’C ADD/SUB (1) • Ex 1: A +B=2+1= (010)+(001)=3 • (010)+(001)M+M=(010)+(001)+0= (011) • Ex2: A  B=21= (0010)(0001)= 1 • (010)+(001)M+M = (010) + (110) + (001) = 1(001) • Ex3 A +B=2+2= 4=OV • (010)+(010)M+M = (100)= 4

  10. 2’C ADD/SUB Overflow • Occurs when Sign bits: • P add P = N • N add N = P • OV=Cn  Cn-1

  11. 2’C ADD/SUB Proof(1) • Only consider A+B case: A  B=A+( B) • Let |A|=E & |B|=F 2n-1> E, F 0 • Case 1: A, B 0  2n> E+F 0 • IF E+F  2n-1 OV=1 (Chk OV) • Case 2: A,B <0 A= 2n E, B = 2n F • A+B=2n+2n (E +F)  Cn=1 (neglect) • neglect Cn  A+B= 2n (E +F)  Chk OV

  12. 2’C ADD/SUB Proof(2) • Case 3: A  0, B<0 B = 2n F • A+B= 2n +(EF)= 2n  (F E) • E  F  Cn=1 (neglect) & Sign bit=0 • F  E  Cn=0 & Sign bit=1 • Case 4: B  0, A <0 Similar to Case 3 • A+B= 2n (EF)= 2n + (F E) • E  F  Cn=0 & Sign bit=1 • F  E  Cn=1 (neglect) & Sign bit=0

  13. 2’C ADD/SUB Circuit

  14. 1’C ADD/SUB (1) • Ex 1: A +B=2+1= (010)+(001)=3 • (010)+(001)M+C=(010)+(001)+0= (011) • Ex2: A  B=21= (010)(001)= 1 • (010)+(001)M+C = (010) + (110) =1(000) = (000)+(001)=(001) • Ex3 A +B=2+2= 4=OV • (010)+(010)M+C = (100)= 3

  15. 1’C ADD/SUB Overflow • Same as 2’C • OV occurs when Sign bits: • P add P = N • N add N = P • OV=Cn  Cn-1

  16. 1’C ADD/SUB Proof(1) • Only consider A+B case: A  B=A+( B) • Let |A|=E & |B|=F 2n-1> E, F 0 • Case 1: A, B 0  Same as 2’C • IF E+F  2n-1 OV=1 (Chk OV) • Case 2: A,B <0 A= 2n E 1, B = 2n F 1 • A+B=2n+2n (E +F)  2  Cn=C=1 (neglect) • Add end-around carry A+B= 2n (E +F) 1 •  Chk OV

  17. 1’C ADD/SUB Proof(2) • Case 3: A  0, B<0 B = 2n F  1 • A+B= 2n +(EF)  1= 2n  (F E) 1 • E >F  C=1 (neglect 2n ,1) & A+B=E  F>0 • F  E  C=0 & Sign bit=1 (may have 0) • Case 4: B  0, A <0 Similar to Case 3 • A+B= 2n (EF)  1= 2n + (F E) 1 • E  F  C=0 & Sign bit=1 • F > E  C=1 (neglect 2n ,1) & A+B= F E

  18. 1’C ADD/SUB Circuit

  19. End-Around-Carry Oscillation?

  20. SM ADD/SUB (1) • Ex 1: A  B=21= (0010)(0001)= 1 • Init. Sign P= SA (SBM)=01=1 • Mag. (010)+(001)P=(010)+(110)=1(000) • Add C (end-arnd-crry): (000)+(001)=(001) • Final Mag. Indicator Q=P(C)’=0 • Final Sign S= (SA C)v((SBM)C’)= 0v0 =0 • Answer= (S, (Mag)  Q)=(0001)

  21. SM ADD/SUB (2) • Ex 2: A  B=12= (0001)(0010)= 1 • Init. Sign P= SA (SBM)= 01=1 • Mag. (001)+(010)P=(001)+(101)=0(110) • Add C (end-arnd-crry): (110)+(000)=(110) • Final Mag. Indicator Q=P(C)’=1 • Final Sign S= (SAC)v((SBM)C’)= 0 v 1 =1 • Answer= (S, (Mag)  Q)=(1001)

  22. SM ADD/SUB (3) • Ex 3: (0010)+(0111)= 2+7=9= OV • Init. Sign P=00=0 • Mag. (010)+(111)P=(010)+(111)=1(001) • Add C (end-arnd-crry): (001)+(001)=(010) • OV=P’C=1

  23. SM ADD/SUB Proof(1) • A  B=A+( B)  need (n-1)-bit Adder • Init Sign P=0 for (SA=(SBM)); P=1 otherwise • OV occurs only at P=0 • Let |A|=E and |B|=F (mag. E, F 0) C • Case 1: P=0, • If (E+F P)=E+F < 2n1  C=0, No OV • Otherwise C=1 and OV  OV=P’C=1

  24. SM ADD/SUB Proof(2) • Case 2: P=1, then • E+(FP)= E+F’= E+(2n F1)=2n+(EF)1 • if E > F  C=1  Sign=SA • (EF)1+C= (E F); Neglect 2n =C=1 • otherwise (F E, Sign =(SBM)) C=0 Q=PC’=1 • 2n+(EF)1+C= 2n(F E) 1 • (Mag)  Q= 2n (2n(F E) 1 ) 1= (F E)

  25. SM ADD/SUB Proof(3) • Sign= SAC v (SBM)C’

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