select_ch5

# select_ch5 - Solutions Manual - Introduction to Digital...

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Unformatted text preview: Solutions Manual - Introduction to Digital Design - October 3, 2000 Exercise 5.4 (a) Kmap and prime implicants f (w; x; y; z) = P m(0; 1; 2; 3; 5; 7; 8; 10; 11; 15) z 1 0 0 1 0 0 59 w 1 1 0 0 1 1 1 1 y 1 0 0 1 x prime implicants: w x ; w z; x y; yz; x z (b) essential prime implicants: w z; yz; x z (c)The minimal sum of products for this function is unique: 0 0 0 0 0 0 0 f (w; x; y; z) = w z + yz + x z 0 0 0 Solutions Manual - Introduction to Digital Design - October 3, 2000 61 Exercise 5.6 (a) E (w; x; y; z ) = M (1; 3; 4; 7; 10; 13; 14; 15) = Q P m(0; 2; 5; 6; 8; 9; 11; 12) z 1 0 1 1 0 1 0 1 0 0 0 1 y 1 1 0 0 x w minimal sum of products: wy z + wx z + w x z + w yz + w xy z 0 0 0 0 0 0 0 0 0 0 x 1 0 1 1 0 1 0 1 0 0 0 1 y 1 1 0 0 w z minimal product of sums: (w + x + z )(w + y + z )(x + y + z )(w + x + z )(w + x + y + z ) 0 0 0 0 0 0 0 0 0 0 62 (b)E (w; x; y; z ) = P m(0; 4; 5; 9; 11; 14; 15); dc(w; x; y; z) = P m(2; 8) z 1 1 0 0 1 0 1 0 0 1 1 y 0 1 0 x Solutions Manual - Introduction to Digital Design - October 3, 2000 w minimal SP: w y z + w y x + wx z + wxy 0 0 0 0 0 0 z 1 1 0 0 1 0 1 0 0 1 1 y minimal PS: (w+x+z')(w+y')(x+y'+z)(w'+x'+y) P Q (c) E (x; y; z ) = m(0; 1; 4; 6) = M (2; 3; 5; 7) z x minimal sum of products: x y + xz 0 0 0 w 0 1 0 x 1100 1001 y z 1100 1001 y 0 0 x 0 minimal product of sums: (x + y )(x + z ) Solutions Manual - Introduction to Digital Design - October 3, 2000 79 Input: binary code represented as b = (b3b2 b1b0), where b 2 f0; 1g Output: Gray code represented as g = (g3g2 g1g0), where g 2 f0; 1g Function: g is the Gray code that corresponds to b. The correspondence between binary and Gray codes is shown in the following table: i i Exercise 5.15 b3 b2b1 b0 Binary 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 g3g2 g1g0 Gray 0000 0001 0011 0010 0110 0111 0101 0100 1100 1101 1111 1110 1010 1011 1001 1000 The correspoding Kmaps are as follows: g3 0 0 1 1 0 0 1 1 b0 0 0 1 1 0 0 b2 1 1 g2 0 1 0 1 0 1 0 1 b0 0 1 0 1 0 1 b2 0 1 b3 b3 b1 g1 0 1 1 0 0 1 1 0 b1 g0 b0 1 0 0 1 1 0 b2 0 1 b0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 b2 1 1 b3 b3 b1 b1 80 Solutions Manual - Introduction to Digital Design - October 3, 2000 To obtain a NOR-NOR network we produce the minimal product of sums: g3 g2 g1 g0 = = = = b3 (b2 + b3)(b2 + b3 ) (b1 + b2)(b1 + b2 ) (b0 + b1)(b0 + b1 ) 0 0 0 0 0 0 The gate network is presented in Figure 5.8. b0 b1 b0’ b1’ b1 b2 g1 b1’ b2’ b2 b3 b2’ b3’ b3 g3 g0 g2 Figure 5.8: Exercise 5.15 Solutions Manual - Introduction to Digital Design - October 3, 2000 81 Minterms 3-literal prods 2-literal prods 1-literal prods 1000 N 100- N 1-0- N 1-- 10-0 N 10- - N 1001 N 1-00 N 1- -0 N 1010 N 1-0- N 1100 N 10-1 N 1-01 N 1- -1 N g3 1011 N 101- N 11- - N 1101 N 1-10 N 1-1- N 1110 N 110- N 11-0 N 1111 N 1-11 N 11-1 N 111- N Table 5.5: Exercise 5.16 - synthesis of g3 output Minterms 3-literal prods 2-literal prods 0100 N 010- N 01- 1000 N 01-0 N 10- 100- N 0101 N 10-0 N g2 0110 N 1001 N 01-1 N 1010 N 011- N 10-1 N 0111 N 101- N 1011 N Table 5.6: Exercise 5.16 - synthesis of g2 output Using Quine-McCluskey minimization method to solve Exercise 5.15 we get Tables 5.5, 5.6, 5.7, and 5.8. Minterms are obtained from the switching function presented in Exercise 5.15. From the QM tables we obtain the expressions: Exercise 5.16 g3 g2 g1 g0 = = = = b3 b2b3 + b2b3 b1b2 + b1b2 b0b1 + b0b1 0 0 0 0 0 0 82 Solutions Manual - Introduction to Digital Design - October 3, 2000 Minterms 3-literal prods 2-literal prods 0010 N 001- N -010100 N -010 N -10010- N 0011 N -100 N g1 0101 N 1010 N -011 N 1100 N 101- N -101 N 1011 N 110- N 1101 N Table 5.7: Exercise 5.16 - synthesis of g1 output Minterms 3-literal prods 2-literal prods 0001 N 0-01 N - -01 0010 N -001 N - -10 0-10 N 0101 N -010 N g0 0110 N 1001 N -101 N 1010 N -110 N 1-01 N 1101 N 1-10 N 1110 N Table 5.8: Exercise 5.16 - synthesis of g0 output Solutions Manual - Introduction to Digital Design - October 3, 2000 85 x3 x2 x1 x0 OR Array 1 2 3 4 5 6 7 8 9 AND Array -- programmable connection -- connection made y6 y5 y4 y3 y2 y1 y0 Figure 5.9: PLA implementation for Exercise 5.18 Exercise 5.19 The high-level speci cation for this system is: Input: b is a decimal digit, represented in BCD Output: e is a decimal digit, represented in Excess-3 code Function: e = b The conversion of a BCD code, represented as b = (b3; b2; b1; b0) to an Excess-3 code, represented by e = (e3; e2 ; e1; e0), is de ned by the following K-maps: 86 0 0 1 Solutions Manual - Introduction to Digital Design - October 3, 2000 b0 0 1 1 0 1 0 1 b -2 0 1 0 1 0 1 b0 1 0 1 0 b -2 - b3 e3 : b3 e2 : b1 b0 1 1 1 0 0 0 1 1 0 0 b -2 - b1 b0 1 1 1 0 0 0 0 0 1 1 b -2 - b3 e1 : b3 e0 : b1 e3 e2 e1 e0 = = = = b1 The minimal sum of products are: b1b2 + b0b2 + b3 b1b2 + b0b2 + b2b1b0 b1b0 + b1b0 b0 0 0 0 0 0 0 0 The implementation of these expressions by a PLA is shown in gure 5.10. Solutions Manual - Introduction to Digital Design - October 3, 2000 87 b3 b2 b1 b0 OR Array 1 2 3 4 5 6 7 8 9 AND Array -- programmable connection -- connection made e3 e2 e1 e0 Figure 5.10: PLA implementation of a BCD to Excess-3 converter A high-level speci cation for this system is: Input: x is a decimal digit, represented in Excess-3 code Output: y is a decimal digit, represented in 2-out-of-5 code Function: y = x The table that shows the correspondence between the excess-3 code (x) and the 2-out-of-5 code (y ) is show in page 28 of the textbook (Table 2.3). From the table we get the following K-maps: Exercise 5.20 x0 0 1 0 0 0 1 0 1 0x 2 1 0 0 1 0 1 x0 1 1 0 0x 2 0 0 1 0 1 1 x0 0 0 0 1x 2 0 x4 y0 : x4 y1 : x4 y2 : x1 x1 x1 ...
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## This note was uploaded on 03/26/2010 for the course CS 187154200 taught by Professor Ercegovac,m.d. during the Winter '09 term at UCLA.

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