1 / 23

CSE 311 Foundations of Computing I

Explore the use of Finite State Machines (FSMs) in hardware, along with the concept of Carry-Look-Ahead adders. Learn how to encode states and input/output signals, build combinational logic circuits, and speed up computation.

hopperm
Download Presentation

CSE 311 Foundations of Computing I

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. CSE 311 Foundations of Computing I Lecture 25 Circuits for FSMs, Carry-Look-Ahead Adders Autumn 2011 CSE 311

  2. Announcements • Nice overview of adder circuits at http://www.aoki.ecei.tohoku.ac.jp/arith/mg/algorithm.html CSE 311

  3. Last lecture highlights • Languages not recognized by any DFA • { 0n1n : n≥0 } • Binary Palindromes • Strings of Balanced Parentheses • Using DFAs for efficient pattern matching CSE 311

  4. FSMs in Hardware • Encode the states in binary: e.g. states 0,1,2,3 represented as 000,100, 010,001, or as 00,01,10,11. • Encode the input symbols as binary signals • Encode the outputs possible as binary signals • Build combinational logic circuit to compute transition function: Boolean Circuit for State Transition Function current state next state output signals input signals CSE 311

  5. FSMs in Hardware • Combine with sequential logic for • Registers to store bits of state • Clock pulse • At start of clock pulse, current state bits from registers and input signals are released to the circuit • At end of clock pulse, output bits are produced and next state bits are stored back in the same registers Boolean Circuit for State Transition Function current state next state output signals input signals CSE 311

  6. FSM for binary addition • Assume that the two integers are an-1...a2a1a0 and bn-1...b2b1b0 and bits arrive together as [a0,b0] then [a1,b1] etc. s [0,0] [1,1] [0,1],[1,0] [1,1] [1,1] [0,1],[1,0] [1,1] Cout=0 0 Cout=0 1 Cout=1 0 Cout=1 1 [0,0] [0,0] [0,1],[1,0] [0,0] [0,1],[1,0] [0,1],[1,0] [1,1] [1,1] Generate a carry of 1 [0,1],[1,0] Propagate a carry of 1 if it was already there [0,0]

  7. FSM for binary addition using output on edges • Assume that the two integers are an-1...a2a1a0 and bn-1...b2b1b0 and bits arrive together as [a0,b0] then [a1,b1] etc. [0,0]:0 [1,1]:1 [1,1]:0 Cout=0 Cout=1 [0,0]:1 [0,1],[1,0]:1 [0,1],[1,0]:0 [1,1] Generate a carry of 1 [0,1],[1,0] Propagate a carry of 1 if it was already there

  8. Example: 1-bit Full Adder A B input signals 1-Bit Full Adder current state next state Cin Cout output Sum CSE 311

  9. FSMs without sequential logic • What if the entire input bit-strings are available at all once at the start? • E.g. 64-bit binary addition • Don’t want to wait for 64 clock cycles to compute the output! • Suppose all input strings have length n • Can chain together n copies of the state transition circuit as one big combinational logic circuit CSE 311

  10. A 2-bit ripple-carry adder a b a0 b0 a1 b1 1-Bit Full Adder Cin Cout Cin Cout 0 Cin Cout Sum0 Sum1 Sum CSE 311

  11. Problem with Chaining Transition Circuits • Resulting Boolean circuit is “deep” • There is a small delay at each gate in a Boolean circuit • The clock pulse has to be long enough so that all combinational logic circuits can be evaluated during a single pulse • Deep circuits mean slow clock. CSE 311

  12. Speeding things up? • To go faster, need to work on both 1st half and 2nd half of the input at once • How can you determine action of FSM on 2nd half without knowing state reached after reading 1st half? b1b2...bn/2 bn/2+1 ...bn-1bn what state? • Idea: Figure out what happens in 2nd half for all possible values of the middle state at once CSE 311

  13. Transition Function Composition Transition table gives a function for each input symbol f0 f1 0 0 0 0 s0 s1 s2 s3 State reached on input b1...bnis fbn(fbn-1...(fb2(fb1(start)))...)= fbn∘ fbn-1...∘ fb2∘ fb1(start) 1 1 1 1 1 1 s0 s1 s2 s3 s0 s1 s2 S3 0 0 0 0,1 1 CSE 311

  14. Transition Function Composition b1 b2 b3 b4 b5 b6 b7 b8 • Constant size 2-level • Boolean logic to • convert input symbol to • bits for transition function • compute composition of • two transition functions • Total depth 2 log2n • and size ≈n fb1 fb2 fb3 fb4 fb5 fb6 fb7 fb8 f g ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ fb2∘fb1 fb4∘fb3 fb6∘fb5 fb8∘fb7 g∘f b fb4∘fb3∘fb2∘fb1 fb8∘fb7∘fb6∘fb5 fb fb8∘fb7∘fb6∘fb5∘fb4∘fb3∘fb2∘fb1 CSE 311

  15. Carry-Look-Ahead Adder Compute generate Gi= ai ∧ bi [1,1] propagate Pi=ai⊕ bi [0,1],[1,0] These determine transition and output functions • Carry Ci=Gi∨ (Pi∧ Ci-1 ) also written Ci=Gi+PiCi-1 • Sumi = Pi⊕ Ci-1 Unwinding, we get C0=G0 C1=G1+G0P1 C2=G2+G1P2+G0P1P2 C3=G3+G2P3+G1P2P3+G0P1P2P3 C4=G4+G3P4+G2P3P4+G1P2P3P4+G0P1P2P3P4 etc. CSE 311

  16. Carry-Look-Ahead Adder Compute all generate Gi= ai ∧ bi [1,1] propagate Pi=ai⊕ bi [0,1],[1,0] Then compute all: C0=G0 C1=G1+G0P1 C2=G2+G1P2+G0P1P2 C3=G3+G2P3+G1P2P3+G0P1P2P3 C4=G4+G3P4+G2P3P4+G1P2P3P4+G0P1P2P3P4 etc. Finally, use these to compute Sum0 = P0 Sum1 = P1⊕ C0Sum2 = P2⊕ C1 Sum3 = P3⊕ C2 Sum4 = P4⊕ C3 Sum5 = P5⊕ C4 etc If all Ci are computed using 2-level logic, total depth is 4. CSE 311

  17. Adders using Composition Carry-look-ahead circuit for carry Cn-1 has 2 + 3 +...+ n = (n+2)(n-1)/2 gates • a lot more than ripple-carry adder circuit. Composition: (G,P) followed by (G’,P’) gives the same effect as (G’’,P’’) where • G’’= G’ ∨ (G ∧ P’) and P’’= P’ ∧ P also written as G’’= G’+G P’ and P’’= P’P Composition gives an alternative approach CSE 311

  18. Adders using Composition Composition: (G,P) followed by (G’,P’) gives the same effect as (G’’,P’’) where • G’’= G’ ∨ (G ∧ P’) and P’’= P’ ∧ P also written as G’’= G’+G P’ and P’’= P’P Use this for the circuit component in transition function composition tree • Computes Cn-1 in depth 2 log2 n and size ≈n • But we need all of C0, C1, ..., Cn-1 not just Cn-1 ∘ CSE 311

  19. Transition Function Composition b0 b1 b2 b3 b4 b5 b6 b7 • Constant size 2-level • Boolean logic to • convert input symbol to • bits for transition function • compute composition of • two transition functions • Total depth 2 log2n • and size ≈n fb0 fb1 fb2 fb3 fb4 fb5 fb6 fb7 f g ∘ ∘ ∘ ∘ ∘ ∘ ∘ ∘ fb1∘fb0 fb3∘fb2 fb5∘fb4 fb7∘fb6 g∘f b fb3∘fb2∘fb1∘fb0 fb7∘fb6∘fb5∘fb4 fb fb7∘fb6∘fb5∘fb4∘fb3∘fb2∘fb1∘fb0 CSE 311

  20. Computing all the values • We need to compute all of fb7∘fb6∘fb5∘fb4∘fb3∘fb2∘fb1∘fb0 Already computed fb6∘fb5∘fb4∘fb3∘fb2∘fb1∘fb0=fb6∘(fb5∘fb4)∘(fb3∘fb2∘fb1∘fb0) fb5∘fb4∘fb3∘fb2∘fb1∘fb0= (fb5∘fb4) ∘ (fb3∘fb2∘fb1∘fb0) fb4∘fb3∘fb2∘fb1∘fb0=fb4∘(fb3∘fb2∘fb1∘fb0 ) fb3∘fb2∘fb1∘fb0Already computed fb2∘fb1∘fb0=fb2∘ (fb1∘fb0) fb1∘fb0Already computed fb0 Already computed CSE 311

  21. Parallel Prefix Circuit • The general way of doing this efficiently is called a parallel prefix circuit • Designed and analyzed by Michael Fischer and Richard Ladner (University of Washington) • Uses the adder composition operation that sets G’’= G’+G P’ and P’’= P’P • we just show it for the part for computing P’’ which is a Parallel Prefix AND Circuit CSE 311

  22. The Parallel Prefix AND Circuit n/2 AND gates per level log2n levels Parallel Prefix n inputs Pn Parallel Prefix n/2 inputs P1P2...Pn Pn/2+1 P1P2...Pn/2Pn/2+1 P1P2...Pn/2 Pn/2 Parallel Prefix n/2 inputs P1P2 P1 P1

  23. Parallel Prefix Adder • Circuit depth 2 log2 n • Circuit size 4 n log2 n • Actual adder circuits in hardware use combinations of these ideas and more but this gives the basics • Nice overview of adder circuits at http://www.aoki.ecei.tohoku.ac.jp/arith/mg/algorithm.html CSE 311

More Related