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Digital Logic Design: System Specification and Implementation, Study notes of Electrical and Electronics Engineering

Information on digital system design and synthesis in the context of enee244 digital logic design. It covers system specification from high-level to binary-level, including input and output encoding, boolean functions, and gate implementation using and, not, or, nand, and nor gates. Examples of decimal digit systems and code converters are given.

Typology: Study notes

Pre 2010

Uploaded on 07/30/2009

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Download Digital Logic Design: System Specification and Implementation and more Study notes Electrical and Electronics Engineering in PDF only on Docsity! 1 E N E E 24 4: D ig it al L og ic D es ig n D ig it al S ys te m D es ig n/ Sy nt he si s • S ys te m S pe ci fi ca ti on Fr om h ig h- le ve l t o bi na ry -l ev el – In pu t e nc od in g – O ut pu t e nc od in g – E xp re ss o ut pu t a s B oo le an f un ct io ns o n th e in pu t ( e. g. tr ut h ta bl e, a lg eb ra ic e xp re ss io n, d on ’t c ar e co nd iti on s) • S ys te m I m pl em en ta ti on – G at e im pl em en ta tio n of th e B oo le an o ut pu t f un ct io ns . – A ll B oo le an f un ct io ns c an b e im pl em en te d by th e se t o f {A N D , N O T , O R }. – A ll B oo le an f un ct io ns c an a ls o be im pl em en te d by 2 - in pu t N A N D g at e or 2 -i np ut N O R g at e. E N E E 24 4: D ig it al L og ic D es ig n Sy st em S pe ci fi ca ti on • T he s pe ci fi ca ti on of a s ys te m is a de sc ri pt io n of it s fu nc tio n an d of o th er ch ar ac te ri st ic s re qu ir ed fo r its u se . • H ig h- le ve l: a fu nc tio n (t ab le , a lg eb ra ic , co nd iti on , e tc .) o n a fi ni te s et o f i np ut . • B in ar y- le ve l: al l i np ut /o ut pu t v ar ia bl es a re in b in ar y. E N E E 24 4: D ig it al L og ic D es ig n E xa m pl e 1 A s ys te m ta ke s on e de ci m al d ig it an d ou tp ut s 0 on e ve n nu m be r an d 1 on o dd n um be r. • H ig h- le ve l: – In pu t: x ∈ {0 ,1 ,2 ,3 ,4 ,5 ,6 ,7 ,8 ,9 } – O ut pu t: z ∈ {0 ,1 } – F un ct io n: z = F (x ) = 1 0 1 0 1 0 1 0 1 0 z 9 8 7 6 5 4 3 2 1 0 x 0 x= 0, 2, 4, 6, 8 1 x= 1, 3, 5, 7, 9 E N E E 24 4: D ig it al L og ic D es ig n 1 1 0 0 1 9 0 1 0 0 0 8 1 0 1 1 1 7 0 0 1 1 0 6 1 0 1 0 1 5 0 0 1 0 0 4 1 0 0 1 1 3 0 0 0 1 0 2 1 0 0 0 1 1 0 0 0 0 0 0 z x 3 x 2 x 1 x 0 x E xa m pl e 1 (c on t’ d) • B in ar y- le ve l ( us in g B C D ) – In pu t v ar ia bl es : x 3 , x 2, x 1, x 0 – O ut pu t v ar ia bl es : z – F un ct io ns : z( x 3 ,x 2, x 1 ,x 0) = x 0 S x 1x 0 x 2 x 3 z= x 0 2 E N E E 24 4: D ig it al L og ic D es ig n E xa m pl e 2 A s ys te m ta ke s on e de ci m al d ig it an d ou tp ut s it s 9’ s co m pl em en t. • H ig h- le ve l: – In pu t: x ∈ {0 ,1 ,2 ,3 ,4 ,5 ,6 ,7 ,8 ,9 } – O ut pu t: z ∈ {0 ,1 ,2 ,3 ,4 ,5 ,6 ,7 ,8 ,9 } – F un ct io n: z = F (x ) = 9 – x 0 1 2 3 4 5 6 7 8 9 z 9 8 7 6 5 4 3 2 1 0 x E N E E 24 4: D ig it al L og ic D es ig n SE xa m pl e 2 (c on t’ d) • B in ar y- le ve l ( us in g B C D ) – In pu t v ar ia bl es : x 0 , x 1, x 2, x 3 – O ut pu t v ar ia bl es : z 0, z 1, z 2, z 3 – Fu nc tio ns : ( su m o f m in te rm s) z 3 (x 3, x 2 ,x 1, x 0 ) = m (0 ,1 ) z 2 (x 3, x 2 ,x 1, x 0 ) = m (2 ,3 ,4 ,5 ) z 1 (x 3, x 2 ,x 1, x 0 ) = m (2 ,3 ,6 ,7 ) z 0 (x 3, x 2 ,x 1, x 0 ) = m (0 ,2 ,4 ,6 ,8 ) z 1 = F( x) x 1x 0 x 2 x 3 z 0 = F( x ) z 2 = F( x) z 3 = F( x) 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1 0 0 1 z 3 z 2 z 1 z 0 0 1 0 0 1 9 1 1 0 0 0 8 2 0 1 1 1 7 3 0 1 1 0 6 4 0 1 0 1 5 5 0 1 0 0 4 6 0 0 1 1 3 7 0 0 1 0 2 8 0 0 0 1 1 9 0 0 0 0 0 z x 3 x 2 x 1 x 0 x E N E E 24 4: D ig it al L og ic D es ig n D es ig n E xa m pl e: C od e C on ve rt er • A n A -t o- B c od e co nv er te r tr an sl at es in fo rm at io n fr om b in ar y co de A to b in ar y co de B . – In pu ts : b it co m bi na tio n in c od e A . – O ut pu ts : b it co m bi na ti on in c od e B . – Fu nc tio n: in pu t a nd o ut pu t h av e th e sa m e in fo rm at io n. • D es ig n co nc er ns : – B in ar y co de s: B C D , 8 4- 2- 1, E xc es s- 3, a nd m or e. – D on ’t c ar e co nd iti on s. • E xa m pl es : – B C D to E xc es s- 3. – 84 -2 -1 to B C D . E N E E 24 4: D ig it al L og ic D es ig n Sy st em I m pl em en ta tio n • Fr om a lg eb ra ic e xp re ss io n to g at e im pl em en ta tio n – w (x 3, x 2 ,x 1, x 0 ) = x 1x 0 + x 2x 0+ x 3x 0 + x 2x 1 – S (x ,y ,z ) = x⊕ y⊕ z, C (x ,y ,z ) = x y+ z( x⊕ y) x 0 w x 3x 2x 1 x y S C z
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