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EENG 2710 Chapter 2

EENG 2710 Chapter 2. Homework. Problem 2.1c. F(x,y,z) = yz ’(z’ + z’x) +( x’ + z’)(x’y + x’zz) [T4b) = yz’ (z’(1 + x) +( x’ + z’)(x’y + x’zz) = y z’z’ +( x’+ z’)(x’y + x’ zz ) [T1b]

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EENG 2710 Chapter 2

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  1. EENG 2710 Chapter 2 Homework

  2. Problem 2.1c • F(x,y,z) = yz’(z’ + z’x) +( x’ + z’)(x’y + x’zz) [T4b) = yz’ (z’(1 + x) +( x’ + z’)(x’y + x’zz) = yz’z’ +( x’+ z’)(x’y + x’zz) [T1b] = yz’ +( x’ + z’)(x’y + x’z) [P5b] = yz’ +( x’ + z’) x’ (y + z) [T4b] = yz’ + x’ (y + z) [P5b] = yz’ +x’y + x’z [T9a] = zx’ + z’y + x’y = zx’ + z’y Note: zx’ = ab, z’y = a’c, x’y = bc

  3. Problem 2.2 c • [AB’C + AB + (ABC)’ + AC’ +ABC’]’ [AB’C + AB + (ABC)’ + AC’(1 + B)]’ [T4A] [(AB’C + AB) + (ABC)’ + AC’]’ [T7A] [AC + AB + (ABC)’ + AC’]’ [AC + AC’ + AB + (ABC)’]’ [T6A] [A(C + C’) + AB + (ABC)’]’ [A + AB + (ABC)’]’ [T4A] [A (1 + B) + (ABC)’]’ [A + (ABC)’]’ [T8a] [ A’ABC] [P6B] 0

  4. Problem 2.3 • T4(b): a (a + b) = a • Proof: a (a + b) = aa + ab) = a + ab = a(1 + b) = a(1) = a

  5. Problem 2.4 • T5(b): a (a’+b) = ab • Proof: a (a’+b) = aa’ + ab = 0 + ab = ab

  6. Problem 2.5 • T9(b): (a + b)(a’ + c)(b + c) = (a + b)(a’ + c) Use T6(b): (a + b)(a + b’) = a [(a + b)(a + b’)+b][((a + b)(a + b’))’ + c] [(a + b)(a + b’) +b][(a + b)’ + (a + b’)’ + c] [(a + b)(a + b’) +b][a’b’ + a’b + c] [aa + ab + ab’ + bb’ + b][a’b’ + a’b + c] [a +a(b + b’) +0 + b][a’(b + b’) + c] (a +a +b)(a’ + c) (a + b)(a’ +c)

  7. Problem 2.6b • Simplify the following switching function:

  8. Problem 2.7a

  9. Problem 2.16b

  10. Problem 2.18b • m(1, 3, 6, 7) = M(0, 2, 4, 5)

  11. Problem 2.25

  12. Problem 2.18b (Using Boolean Algebra)

  13. Problem 2.29a

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