1. CSE 370 Sample Final Exam Questions

2. 1) Logic Minimization CD AB 00011110 00 01 11 10 F = Σm(0,6,7,8,9,11,15) + d(1,13)

3. 1) Logic Minimization 11 dd1 111 1 CD AB 00011110 00 01 11 10 F = Σm(0,6,7,8,9,11,15) + d(1,13)

4. 1) Prime Implicants 11 dd1 111 1 CD AB 00011110 00 01 11 10 Note: Prime Implicant = largest box covering a 1

5. 1) Essential PIs 11 dd1 111 1 CD AB 00011110 00 01 11 10 Note: Essential PI = PIs that cover a 1 not covered by any other PIs

6. 1) Logic Minimization 11 dd1 111 1 CD AB 00011110 00 01 11 10 Prime Implicants: BC, AD, ABC, BCD Essential Prime Implicants: BC, AD, ABC Minimal Cover: F = BC + AD + ABC

7. 2a) Gates to Functions back into Gates FG H

8. FG H Z = ~(H + G) Z = ~H & ~G Z = ~(~(C+D)) & ~(~(F+F)) Z = (C+D) & F Z = (C+D) & ~(A+B) Z = (C+D) & ~A & ~B Z = ABC + ABD

9. 2b) Gates to Functions back into Gates Z = ABC + ABD Using deMorgans Law… Z = ~(~(ABC) & ~(ABD)) C A B D

10. 2c) Gates to Functions back into Gates NOR Circuit 3 gate delays @ 7ns 21ns total NAND Circuit 2 gate delays @ 9ns 18ns total

11. 3a) Multiplexer Logic Z = BC + ABD + AB 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 ABCDZ

12. 3a) Multiplexer Logic Z = BC + ABD + AB 0000 0001 00101 00111 0100 0101 0110 0111 1000 1001 10101 10111 1100 1101 1110 1111 ABCDZ

13. 3a) Multiplexer Logic Z = BC + ABD + AB 0000 0001 00101 00111 0100 01011 0110 01111 1000 1001 10101 10111 1100 1101 1110 1111 ABCDZ

14. 3a) Multiplexer Logic Z = BC + ABD + AB 0000 0001 00101 00111 0100 01011 0110 01111 10001 10011 10101 10111 1100 1101 1110 1111 ABCDZ

15. 3a) Multiplexer Logic Z = BC + ABD + AB 00000 00010 00101 00111 01000 01011 01100 01111 10001 10011 10101 10111 11000 11010 11100 11110 ABCDZ

16. 3b) Multiplexer Logic 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 S3S2S1S0 16:1 MUX Z = BC + ABD + AB 00110101111100000011010111110000 ABCD

17. 3c) Multiplexer Logic 0123456701234567 S2S1S0 8:1 MUXZ = BC + ABD + AB 1111 ABC 1DD11DD1 1DD111DD11 01DD110001DD1100 Note: Requires no extra logic

18. 4) Decoder Logic 0000 0011 0101 0111 1001 1010 1101 1111 W = (A xor C) + B ABCW

19. 4) Decoder Implementation 0123456701234567 S2S1S0 3:8 DEC ABC W = (A xor C) + B EN1

20. 5) Programmable Logic X Y Note: 2 Common Prime Implicants can be used by X and Y Z = ABCD + ABCD X = Z + ABCD + ABCD Y = Z + ABC + ABC

21. 5) Programmable Logic Z = ABCD + ABCD X = Z + ABCD + ABCD Y = Z + ABC + ABC Z = ABCD + ABCD X = Z + ABCD + ABCD Y = Z + ABC + ABC

22. 6) Verilog to Circuit

23. 7) Circuit to Verilog

24. module maxValue (DataOut, strobe, Data); // Port Definitions output reg [7:0] DataOut; input strobe; input [7:0] Data; // Behavioral Statements for DataOut // other required internal signals endmodule

25. 7) Circuit to Verilog // Behavioral Statements for DataOut always @ (posedge clk) begin if (regEnable) DataOut <= Data; else DataOut <= DataOut; end

26. 7) Circuit to Verilog regEnable strobeNegedge prevStrobe[2] prevStrobe[1] prevStrobe[0]

27. 7) Circuit to Verilog // other required internal signals wire regEnable, strobeNegedge; reg prevStrobe[2:0]; assign regEnable = (Data > DataOut) && strobeNegedge assign strobeNegedge = prevStrobe[2] & ~prevStrobe[1]; always @ (posedge clk) prevStrobe = {prevStrobe[1:0],strobe};

28. 8) Carry Look-ahead Adder Recall: Sum = A xor B xor C Cout = AB + BC + AC Rewrite Cout: Cout = AB + C (A + B)

29. 8) Carry Look-ahead Adder 0000 0111 1011 1101 ABP XOR P OR 2 functions are only different if A and B are 1

30. 8) Carry Look-ahead Adder Need carry out if A and B are 1 carry is generated in this case dont need propagate to cover this case these are then functionally equivalent Recall: Sum = A xor B xor C P = A + B: have to calculate A xor B P = A xor B: reuse this to compute Sum P = A xor B uses less logic

31. 9) Verilog State Machine 0000000 0001001 0010010 0011010 0100011 0101100 0110010 0111100 1000110 1001110 1010101 1011100 1100000 1101110 1110000 1111111 ABS[1]NS[0]NS[1]S[0]out 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

32. 9) Verilog State Machine 1 1 1 1 S[1:0] AB 00011110 00 01 11 10 out 1 111 111 1 00011110 00 01 11 10 NS[1] 11 11 11 11 00011110 00 01 11 10 NS[0] S[1:0] AB S[1:0] AB NS[1]=S[0]B + S[1]S[0] NS[0] = S[1]S[0]B + S[1]S[0]A + S[1]S[0]B + S[1]S[0]A out = S[1]S[0]AB + S[1]S[0]AB + S[1]S[0]AB + S[1]S[0]AB

33. 9) Verilog State Machine

34. 10) Design: Parallel to Serial Shift Register Load parallel data Shift out serial data Counter How many bits have shifted out Controller Shift Enable Done Shifting

35. 10a) Design: Parallel to Serial Shifter dataIn load dataOut Counter reset clk count char[7:0] start reset clk Controller clk sdata done Datapath start reset count load reset_cnt done

36. 10b) Design: Parallel to Serial Can we collapse these to 1 state?

37. 10b) Design: Parallel to Serial

38. 10c) Design: Parallel to Serial Shifter dataIn[8:0] load dataOut[8:0] Counter reset clk count clk sdata= dataOut[0] done = state[1] {char,1b0} start clk dataOut[8:0] Register nextState[1] clk state Note: Shifter shifts in 1 in MSB clk count[3:0] clk state[1:0] start + state[1] + state[1:0]==0 + ((state==1)&&(count==4hA) ~start && (state==1) && (count==4hA) nextState[0] start + ((state==1)&&(count<4hA)) char[7:0] start reset clk

39. 11) x370 CALL Instruction CALL Requirements Load address into PC from instruction Store PC+1 into RD

40. 11) Implementation Load address into PC inst[5:0] muxed to input of PC Load PC asserted Store PC+1 into RD PC muxed to A input of ALU ALU inst set to INC ALU output muxed to reg write data Reg write address set to R D

41. Final Questions?