Revision Problems 2

1. Lecture 26 Revision Problems 2 Prof. Sin-Min Lee Department of Computer Science

2. Example of combinational and sequential logic • Combinational: • input A, B • wait for clock edge • observe C • wait for another clock edge • observe C again: will stay the same • Sequential: • input A, B • wait for clock edge • observe C • wait for another clock edge • observe C again: may be different A C B Clock

3. Basically • Combinational: • No internal state (or memory or history or whatever you want to call it) • Output depends only on input • Sequential: • Output depends on internal state • Probably not going to be on this midterm since formal lecture on it started last Thursday.

7. Some commonly used components • Decoders: n inputs, 2noutputs. • the inputs are used to select which output is turned on. At any time exactly one output is on. • Multiplexors: 2n inputs, n selection bits, 1 output. • the selection bits determine which input will become the output. • Adder: 2n inputs, 2n outputs. • Computer Arithmetic.

8. Multiplexer • “Selects” binary information from one of many input lines and directs it to a single output line. • Also known as the “selector” circuit, • Selection is controlled by a particular set of inputs lines whose # depends on the # of the data input lines. • For a 2n-to-1 multiplexer, there are 2n data input lines and n selection lines whose bit combination determines which input is selected.

9. MUX Enable 2n Data Inputs Data Output n Input Select

10. Remember the 2 – 4 Decoder? Sel(3) S1 Sel(2) Sel(1) S0 Sel(0) Mutually Exclusive (Only one O/P asserted at any time

11. 4 to 1 MUX DataFlow D3:D0 Dout 4 Control 4 2 - 4 Decoder Sel(3:0) 2 S1:S0

12. 4-to-1 MUX (Gate level) Control Section Three of these signal inputs will always be 0. The other will depend on the data value selected

13. Multiplexer (cont.) • Until now, we have examined single-bit data selected by a MUX. What if we want to select m-bit data/words? Combine MUX blocks in parallel with common select and enable signals • Example: Construct a logic circuit that selects between 2 sets of 4-bit inputs (see next slide for solution).

14. Example: Quad 2-to-1 MUX • Uses four 4-to-1 MUXs with common select (S) and enable (E). • Select line chooses between Ai’s and Bi’s. The selected four-wire digital signal is sent to the Yi’s • Enable line turns MUX on and off (E=1 is on).

15. Implementing Boolean functions with Multiplexers • Any Boolean function of n variables can be implemented using a 2n-1-to-1 multiplexer. A MUX is basically a decoder with outputs ORed together, hence this isn’t surprising. • The SELECT signals generate the minterms of the function. • The data inputs identify which minterms are to be combined with an OR.

16. Example • F(X,Y,Z) = X’Y’Z + X’YZ’ + XYZ’ + XYZ = Σm(1,2,6,7) • There are n=3 inputs, thus we need a 22-to-1 MUX • The first n-1 (=2) inputs serve as the selection lines

17. Efficient Method for implementing Boolean functions • For an n-variable function (e.g., f(A,B,C,D)): • Need a 2n-1 line MUX with n-1 select lines. • Enumerate function as a truth table with consistent ordering of variables (e.g., A,B,C,D) • Attach the most significant n-1 variables to the n-1 select lines (e.g., A,B,C) • Examine pairs of adjacent rows (only the least significant variable differs, e.g., D=0 and D=1). • Determine whether the function output for the (A,B,C,0) and (A,B,C,1) combination is (0,0), (0,1), (1,0), or (1,1). • Attach 0, D, D’, or 1 to the data input corresponding to (A,B,C) respectively.

18. Another Example • Consider F(A,B,C) = m(1,3,5,6). We can implement this function using a 4-to-1 MUX as follows. • The index is ABC. Apply A and B to the S1 and S0 selection inputs of the MUX (A is most sig, S1 is most sig.) • Enumerate function in a truth table.

19. MUX Example (cont.) When A=B=0, F=C When A=0, B=1, F=C When A=1, B=0, F=C When A=B=1, F=C’

20. MUX implementation of F(A,B,C) = m(1,3,5,6) A B C C F C C’

21. 1 input Decoder Decoder O0 I O1 Treat Ias a 1 bit integer i. The ith output will be turned on (Oi=1), the other one off.

22. 1 input Decoder O0 I O1

23. 2 input Decoder Decoder O0 I0 O1 O2 I1 O3 Treat I0I1 as a 2 bit integer i. The ith output will be turned on (Oi=1), all the others off.

24. 2 input Decoder I0 I1 O0 = !I0 && !I1 O1 = !I0 && I1 O2 = I0 && !I1 O3 = I0 && I1

25. 3 Input Decoder Decoder O0 I0 O1 O2 I1 O3 O4 O5 I2 O6 O7

26. 3-Decoder Partial Implementation I0 I1 I2 O0 O1 . . .

27. 2 Input Multiplexor Inputs: I0 and I1 Selector: S Output: O If S is a 0: O=I0 If S is a 1: O=I1 Mux I0 O I1 S

28. 2-Mux Logic Design S I0 I1 I0 && !S O I1 && S

29. 4 Input Multiplexor Inputs: I0I1 I2 I3 Selectors: S0 S1 Output: O Mux I0 I1 O I2 I3 S0 S1

30. One Possible 4-Mux 2-Decoder S0 I0 S1 I1 O I2 I3

31. Adder • We want to build a box that can add two 32 bit numbers. • Assume 2s complement representation • We can start by building a 1 bit adder.

32. Addition • We need to build a 1 bit adder • compute binary addition of 2 bits. • We already know that the result is 2 bits. This is addition! A + B O0 O1

33. One Implementation A && B A O0 B !A (!A && B) || (A && !B) B O1 A !B

34. Binary addition and our adder 1 1 Carry What we really want is something that can be used to implement the binary addition algorithm. • O0 is the carry • O1 is the sum 01001 + 01101 10110

35. What about the second column? 1 1 Carry 01001 + 01101 • We are adding 3 bits • new bit is the carry from the first column. • The output is still 2 bits, a sum and a carry 10110

36. Truth Table for Addition

37. 1 bit adder (3 inputs!) • We can come up with a logic design: Carry Out = (A&&B) || (A&&CarryIn) || (B&&CarryIn) Sum = (!A && !B && CarryIn) || (!A && B && !CarryIn) || ( A && !B && !CarryIn) || ( A && B && CarryIn)

38. New Component: 1 Bit Adder Carry In adder A Sum B Carry Out

39. Building a 32 bit Adder • 64 inputs • 32 bit output A0 A1 … A31 B0 B1 … B31 … … … Result R0 R1 … R31

40. C3 C2 C1 C0 4 Bit Ripple Carry Adder A3A2A1A0 + B3B2B1B0 S3S2S1S0 B3 A3 B2 A2 B1 A1 B0 A0 C3 C2 C1 adder adder adder adder C0 S3 S2 S1 S0

41. 0 1 0 0 1 0 1 0 0 0 0 1 1 0 1 1 0 0 0 adder adder adder adder 0 1 1 0 4 Bit Ripple Carry Adder 0011 + 0010 adder adder adder adder 0

42. Subtraction • Compute A-B as A + (-B-1) + 1 • -B-1 is just all the bits of B inverted. • Add the +1 by setting C0 to 1

43. B inverted 1 0 0 0 1 0 1 1 0 0 0 1 1 1 0 1 adder adder adder adder 0 0 1 0 Subtraction 0101 - 0011 adder adder adder adder 1

47. Two’s Complement Numbers • Nothing is different! • This is the advantage of using 2’s complement representation. • Overflow: • For addition: sign of the result is different than the sign of the operands (and they have the same sign).

48. 1 1 1 0 1 0 0 0 1 1 1 1 0 1 1 1 1 1 1 adder adder adder adder 0 0 1 0 -3 + 7 1101 + 0111 adder adder adder adder 0

49. 0 0 1 1 1 0 0 1 1 0 1 0 0 1 1 1 0 1 0 adder adder adder adder 1 0 1 0 Overflow! -3 + -7 1101 + 1001 adder adder adder adder 0

50. Ripple Carry Timing • All the adders are actually operating all the time (they are just combinational circuits). • We wait long enough (until the last carry has been computed) and then pay attention to the complete answer. • It is likely that there are intermediate values that are wrong!