1 / 26

Binary Addition Binary Multiplication

Binary Addition Binary Multiplication. Section 4.5 and 4.7 . Topics. Calculations Examples Signed Binary Number Unsigned Binary Number Hardware Implementation Overflow Condition Multiplication. Unsigned Number. (2-bit example). Unsigned Addition. 1+2=. Unsigned Addition. 1+3=.

baxter
Download Presentation

Binary Addition Binary Multiplication

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Binary AdditionBinary Multiplication Section 4.5 and 4.7

  2. Topics • Calculations Examples • Signed Binary Number • Unsigned Binary Number • Hardware Implementation • Overflow Condition • Multiplication

  3. Unsigned Number (2-bit example)

  4. Unsigned Addition • 1+2=

  5. Unsigned Addition • 1+3= (Indicates Overflow) (Carry Out)

  6. Unsigned Subtraction (1) • 1-2= (1’s complement) (2’s complement)

  7. Unsigned Subtraction (2) • 2-1=

  8. Summary for Unsigned Addition/Subtraction • Overflow can be an issue in unsigned addition • Unsigned Subtraction (M-N) • If M≥N, and end carry will be produced. The end carry is discarded. • If M<N, • Take the 2’s complement of the sum • Place a negative sign in front

  9. Signed Binary Numbers • 4-bit binary number • 1 bit is used as a signed bit • -8 to +7 • 2’s complement

  10. Signed Addition (70+80) (Indicates a negative number) 70=21+22+26=2+4+64 80=24+26=16+64 10010110→01101001 →01101010 21+23+25+26=2+8+32+64=106 10010110↔-106 010010110 010010110↔ 21+22+24+27=2+4+16+128=150 Conclusion: There is a problem of overflow Fix: Use the end carry as the sign bit, and let b7 be the extra bit.

  11. Signed Subtraction (70-80) (Indicates a negative number) 70=21+22+26=2+4+64 80=24+26=16+64 11110110→00001001 →00001010 21+23=10 11110110↔-10 (No Problem)

  12. Signed Subtraction (-70-80) (Indicates a positive number! A negative number expected.) 70=21+22+26=2+4+64 80=24+26=16+64 101101010 →010010101 → 010010110 010010110 ↔21+22+24+27=2+4+16+128=150 101101010 ↔-150 Conclusion: There is a problem of overflow Fix: Use the end carry as the sign bit, and let b7 be the extra bit.

  13. Observations • Given the similarity between addition and subtraction, same hardware can be used. • Overflow is an issue that needs to be addressed in the hardware implementation • A signed number is not processed any different from an unsigned number. The programmer must interpret the results of addition and subtraction appropriately.

  14. Four-Bit Adder-Subtractor

  15. The Mode Input (1) If M=0, = If M=1, = B0

  16. The Mode Input (2) If M=0, If M=1,

  17. M=0 B3 B2 B0 B1 0

  18. M=1 1 2’s complement is generated of B is generated!

  19. Unsigned Addition When two unsigned numbers are added, an overflow is detected from the end carry.

  20. Detect Overflow in Signed Addition Observe The cary into the sign bit The carry out of the sign bit If they are not equal, they indicate an overflow.

  21. Two-Bit Binary Multiplier (multiplicand) (multiplier)

  22. Use an AND gate to multiply A0 and B0

  23. Hardware Correlation

  24. A Four-Bit Adder

  25. Four-bit by three-bit Binary Multiplier S10=A0B1+A1B0 S11=A0B2+A1B1+C1 S12=A0B3+A1B2+C2 S13=0+A1B3+C3 (S1X, where 1 is the first 4-bit adder)

  26. Four-bit by three-bit Binary Multiplier

More Related