www.elsolucionario.org Solutions to Problems Marked with a * in Logic and Computer Design Fundamentals, 4th Edition Chapter © 2008 Pearson Education, Inc 1-3* Decimal, Binary, Octal and Hexadecimal Numbers from (16)10 to (31)10 Dec Bin Oct Hex 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 20 21 22 23 24 25 26 27 30 31 32 33 34 35 36 37 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 1-7* ( 1001101 ) = + + 2 + = 77 ( 1010011.101 ) = + + + + – + – = 83.625 ( 10101110.1001 ) = + + + 2 + + –1 + – = 174.5625 1-9* Decimal Binary Octal 369.3125 101110001.0101 561.24 Hexadecimal 171.5 189.625 10111101.101 275.5 BD.A 214.625 11010110.101 326.5 D6.A 62407.625 1111001111000111.101 171707.5 F3C7.A 1-10* a) 8|7562 8|945 8|118 8|14 8|1 b) c) 6 0.45 × = 3.6 => 0.60 × = 4.8 => 0.80 × = 6.4 => 0.20 × =3.2 => 16612 3463 (7562.45)10 = (16612.3463)8 (1938.257)10 = (792.41CB)16 (175.175)10 = (10101111.001011)2 1-11* a) (673.6)8 b) (E7C.B)16 c) (310.2)4 = (110 111 011.110)2 = (1BB.C)16 = (1110 0111 1100.1011)2 = (7174.54)8 = (11 01 00.10)2 = (64.4)8 1-16* a) (BEE)r = (2699)10 11 × r + 14 × r + 14 × r = 2699 11 × r + 14 × r – 2685 = By the quadratic equation: r = 15 or ≈ –16.27 ANSWER: r = 15 www.elsolucionario.org Problem Solutions – Chapter b) (365)r = (194)10 × r + × r + × r = 194 × r + × r – 189 = By the quadratic equation: r = – or ANSWER: r = 1-18* a) (0100 1000 0110 0111)BCD b) (0011 0111 1000.0111 0101)BCD = (4867)10 = (1001100000011)2 = (378.75)10 = (101111010.11)2 1-19* (694)10 = (0110 1001 0100)BCD (835)10 = (1000 0011 0101)BCD 0001 0110 1001 0100 +1000 +0011 +0101 1111 1100 1001 +0110 +0110 +0000 0101 0010 1001 1-20* (a) 101 100 0111 1000 Move R 011 1100 100 column > 0111 Subtract -0011 011 1001 Subtract -0011 01 1001 Move R 1100 110 100 column > 0111 Subtract -0011 1001 110 Move R 0100 1110 Move R 010 01110 Move R 01 001110 Move R 1001110 Leftmost in BCD number shifted out: Finished 102 101 100 0011 1001 0111 Move R 001 1100 1011 Subtract -0011 -0011 001 1001 1000 Move R 00 1100 1100 Subtract -0011 -0011 00 1001 1001 Move R 0100 1100 Subtract -0011 0100 1001 Move R 0010 0100 Move R 001 0010 Move R 00 1001 Subtract -0011 00 0110 Move R 0011 Move R 0001 Move R 000 ished (b) 101 and 100 columns > 0111 01 101 and 100 columns > 0111 01 101 100 column > 0111 1101 01101 001101 100 column > 0111 001101 0001101 10001101 110001101Leftmost in BCD number shifted out: Fin- www.elsolucionario.org Problem Solutions – Chapter 1-25* a) (11111111)2 b) (0010 0101 0101)BCD c) 011 0010 011 0101 011 0101ASCII d) 0011 0010 1011 0101 1011 0101ASCII with Odd Parity www.elsolucionario.org Solutions to Problems Marked with a * in Logic and Computer Design Fundamentals, 4th Edition Chapter © 2008 Pearson Education, Inc 2-1.* a) XYZ = X + Y + Z Verification of DeMorgan’s Theorem X Y Z XYZ XYZ X+Y+Z 0 0 1 0 1 1 0 1 1 1 0 1 1 1 1 0 1 1 1 0 X + YZ = ( X + Y ) ⋅ ( X + Z ) b) The Second Distributive Law X Y Z YZ X+YZ X+Y X+Z (X+Y)(X+Z) 0 0 0 0 0 0 0 0 0 1 1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 XY + YZ + XZ = XY + YZ + XZ c) 2-2.* X Y Z XY YZ 0 0 0 0 0 0 1 0 1 1 0 1 1 1 0 0 1 0 0 1 0 1 1 1 0 1 0 1 1 1 0 0 0 0 a) XZ XY+YZ+XZ XY XY + XY + XY YZ XZ XY+YZ+XZ = X+Y = = ( XY + XY ) + ( XY + XY ) = X(Y + Y) + Y( X + X ) = X+Y b) AB + BC + AB + BC = ( AB + AB ) + ( BC + BC ) = B(A + A) + B(C + C) www.elsolucionario.org Problem Solutions – Chapter B+B = c) Y + XZ + XY = X+Y+Z = XY + XZ + YZ = Y + XY + XZ = ( Y + X ) ( Y + Y ) + XZ = Y + X + XZ = Y + ( X + X )( X + Z ) = X+Y+Z d) XY + YZ + XZ + XY + YZ = X Y + YZ ( X + X ) + XZ + XY + YZ = XY + XYZ + XYZ + XZ + XY + YZ = XY ( + Z ) + XYZ + XZ + XY + YZ = XY + XZ ( + Y ) + XY + YZ = XY + XZ + XY ( Z + Z ) + YZ = XY + XZ + XYZ + YZ ( + X ) = XY + XZ ( + Y ) + YZ = XY + XZ + YZ 2-7.* a) XY + XYZ + XY = X + XYZ = ( X + XY ) ( X + Z ) = ( X + X ) ( X + Y ) ( X + Z ) = ( X + Y ) ( X + Z ) = X + YZ b) X + Y ( Z + X + Z ) = X + Y ( Z + XZ ) = X + Y ( Z + X ) ( Z + Z ) = X + YZ + XY c) WX ( Z + YZ ) + X ( W + WYZ ) = WXZ + WXYZ + WX + WXYZ = ( X + X ) ( X + Y ) + YZ = X + Y + YZ = X + Y = WXZ + WXZ + WX = WX + WX = X d) ( AB + AB ) ( CD + CD ) + AC = ABCD + ABCD + ABCD + ABCD + A + C = ABCD + A + C = A + C + A ( BCD ) = A + C + C ( BD ) = A + C + BD 2-9.* a) F = (A + B )( A + B ) b) F = ( ( V + W )X + Y )Z c) F = [ W + X + ( Y + Z ) ( Y + Z ) ] [ W + X + YZ + YZ ] d) F = ABC + ( A + B )C + A ( B + C ) 2-10.* Truth Tables a, b, c X Y Z a A B C b W X Y Z c 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 1 1 0 1 0 0 0 0 1 1 0 1 1 1 0 1 1 1 1 1 1 1 0 0 www.elsolucionario.org Problem Solutions – Chapter Truth Tables a, b, c a) Sum of Minterms: 0 1 1 1 1 0 1 1 1 1 1 1 1 XYZ + XYZ + XYZ + XYZ Product of Maxterms: ( X + Y + Z ) ( X + Y + Z ) ( X + Y + Z ) ( X + Y + Z ) b) ABC + ABC + ABC + ABC Sum of Minterms: Product of Maxterms: ( A + B + C ) ( A + B + C ) ( A + B + C ) ( A + B + C ) c) WXYZ + WXYZ + WXYZ + WXYZ + WXYZ + WXYZ Sum of Minterms: + W XYZ Product of Maxterms: ( W + X + Y + Z ) ( W + X + Y + Z ) ( W + X + Y + Z ) ( W + X + Y + Z )( W + X + Y + Z ) ( W + X + Y + Z ) ( W + X + Y + Z )( W + X + Y + Z ) ( W + X + Y + Z ) 2-12.* ( AB + C ) ( B + CD ) = AB + ABCD + BC = AB + BC s.o.p a) = B ( A + C ) p.o.s X + X ( X + Y ) ( Y + Z ) = ( X + X ) ( X + ( X + Y) ( Y + Z ) ) b) = ( X + X + Y ) ( X + Y + Z ) p.o.s = ( + Y ) ( X + Y + Z ) = X + Y + Z s.o.p ( A + BC + CD ) ( B + EF ) = ( A + B + C ) ( A + B + D ) ( A + C + D ) ( B + EF ) c) = ( A + B + C ) ( A + B + D ) ( A + C + D ) ( B + E ) ( B + F ) p.o.s ( A + BC + CD ) ( B + EF ) = A ( B + EF ) + BC ( B + EF ) + CD ( B + EF ) = A B + AEF + BCEF + BCD + CDEF s.o.p 2-15.* a) b) Y 1 X 1 c) B 1 1 A B 1 A 1 C A + CB Z XZ + XY C B+C 2-18.* a) b) 1 1 Z Σm ( 3, , , ) W 1 1 C 1 Y X c) Y X Z Σm ( 3, 4, 5, 7, 9, 13, 14, 15 ) 1 A 1 B D Σm ( 0, , 6, , , 10, 13, 15 ) www.elsolucionario.org Problem Solutions – Chapter 2-19.* a) Prime = XZ, WX, XZ, WZ b) Prime = CD, AC, BD, ABD, BC Essential = XZ, XZ c) Prime = AB, AC, AD, BC, BD, CD Essential = AC, BD , ABD Essential = AC, BC, BD 2-22.* a) s.o.p CD + AC + BD c) s.o.p BD + ABD + ( A BC or ACD ) b) s.o.p AC + BD + AD p.o.s ( C + D ) ( A + D ) ( A + B + C ) p.o.s ( C + D ) ( A + D ) ( A + B + C ) p.o.s ( A + B ) ( B + D ) ( B + C + D ) 2-25.* b) a) B X A X 1 W C Primes = AB, AC, BC, ABC Essential = AB, AC, BC F = AB + AC + BC c) Y 1 X X 1 C X X X X Z Primes = XZ, XZ, WXY, WXY, WYZ, WYZ Essential = XZ F = XZ + WXY + WXY A X 1 1 X X X B D Primes = AB, C, AD, BD Essential = C, AD F = C + AD + ( BD or AB ) 2-32.* X ⊕ Y = XY + XY Dual (X ⊕ Y ) = Dual ( XY + XY ) = (X + Y)(X + Y) = XY + XY = XY + XY = X⊕Y www.elsolucionario.org Solutions to Problems Marked with a * in Logic and Computer Design Fundamentals, 4th Edition Chapter © 2008 Pearson Education, Inc 3-2.* C A 1 1 1 B F = AB + AC D 3-24.* a) b) VDD F7 A G7 F6 A G6 F5 G5 F4 G4 F3 A G3 F2 A G2 F1 G1 F0 G0 www.elsolucionario.org Problem Solutions – Chapter 3-30.* D0 D1 D2 D3 D4 D5 D6 D7 D8 DECODER A0 A1 A2 A0 A1 A2 D9 D10 D11 D12 D13 A3 A4 DECODER D14 A0 A1 D15 D16 D17 D18 D19 D20 D21 D22 D23 D24 D25 D26 D27 D28 D29 D30 D31 3-35.* D3 X X X D2 X X D1 X 0 D0 0 A1 X 0 1 A0 X 1 V = D0 + D1 + D2 + D3 A0 = D0 ( D + D ) V 1 1 D1 A1 X D3 D2 D0 D1 A1 D2 A0 D0 V A0 D1 X 1 A1 = D0 D1 D3 D3 1 D2 D0 www.elsolucionario.org Problem Solutions – Chapter 7-51.* module register_4_bit (D, CLK, CLR, Q) ; input [3:0] D ; input CLK, CLR ; output [3:0] Q ; reg [3:0] Q ; always @(posedge CLK or negedge CLR) begin if (~CLR) //asynchronous RESET active low Q = 4'b0000; else //use CLK rising edge Q = D; end endmodule CLK CLR D 0000 1010 0101 Q 0000 1010 20 1111 0101 40 0000 60 7-53.* library IEEE; use IEEE.std_logic_1164.all; entity prob_7_53 is port (clk, RESET, W, X, Y : in STD_LOGIC; Z : out STD_LOGIC); end prob_7_53; architecture process_3 of prob_7_53 is type state_type is (STA, STB, STC); signal state, next_state: state_type; begin Process - state register state_register: process (clk, RESET) begin if (RESET = '1') then state