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assign2_soln - =xz y’z =z(x y’ 4d F=xy’z x’y’z...

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1a. Yz x 00 01 11 10 0 1 1 1 1 1 F=xy+x’ z 1b BC A 00 01 11 10 0 1 1 1 1 1 1 F=C’+A’B 2a yz wx 00 01 11 10 00 1 1 1 01 1 1 11 1 1 10 1 1 1 F=x’y+z 2b yz wx 00 01 11 10 00 1 1 01 1 1 11 1 1 1 10 1 1 F=xz+wy+x’y 3a yz x 00 01 11 10 0 1 1 X 1 1 1 1 X X F=1= Σ (0,1,2,3,4,5,6,7) 3b CD AB 00 01 11 10 00 1 1 01 X 1 1 X 11 X X 1 10 1 F=A’D+BD+C’D= Σ (1,3,5,7,9,13,15)
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4a F=xy’z+x’y’z+xyz x y Z F 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 1 4b 4c F=xy’z+x’y’z+xyz =xz(y+y’)+y’z(x+x')
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Unformatted text preview: =xz+y’z =z(x+y’) 4d F=xy’z+x’y’z+xyz x y Z F 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 1 4e 4f 5 CD AB 00 01 11 10 00 1 1 0 X 01 0 0 0 0 11 0 0 X X 10 X 1 1 X F’=B+A’C F=(B+A’C)’=B’(A+C’) ((A+C’)’+B)’=(A+C’)B’ 6 Refer t o your notes. We covered the even parity bit generator. So odd parity bit generator is just the complement (this you should show with a table and then boolean algebra): P=(x ○ + y ○ + z)’ C=(x ○ + y ○ + z ○ + P)’...
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