% & 5.3 PRACTICAL APPLICATIONS. Given system, (49) , determine the Boolean Function, , in such a way that we always have expression: " Y1 = Y2

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1 5.3 PRACTICAL APPLICATIONS st EXAMPLE: Gven system, (49) & K K Y XvX 3 ( 2 & X ), determne the Boolean Functon, Y2 X2 & X 3 v X " X3 (X2,X)", n such a way that we always have expresson: " Y Y2 " (50). Therefore, n these gven condtons, we have: 39

2 (5) X 3 v ( X & X ) X & X v X Ths expresson (5) s a "Type 3" Boolean Equaton of expressons (3) or (6), as we have seen n Secton 5., Chapter 4; thus we have: ( [ )] ( ) [ ] ( X v X & X v X & X v X & X v(x & X ) X & X v X ) 0 f Then, we fnd the fnal expresson of the Solvng Equaton of the System of Boolean Equaton (SESBE): (52) f X 3 vx2 vx& X3 vx2& X3 vx 0 Accordng to (34) f "p ", then: 2 Ps Fs, where Po X3 s 0 That s: (53) P o Fo & P F, or we have: st. Condton: P o Fo (54) nd 2. Condton: P F (55) P X3 However, as we have seen, " F s ", where s {0, ( 2 ) }, represents all Boolean Functons ( 2 2) whch depend on the other remanng non-man varables (3-2), { X 2, X }. These Boolean Functons ", F 0 F " can be obtaned f t s assumed that the man ncognto " X 3 " corresponds to the "s {0, }" values. Ths "s {0, }" values must have a correspondng bnary expresson n " s". t Then we have: s t X 3 f X v X v X & X v X & X v X t 0 0 F 0 X 2 & X (56) t F X 2 v X (57) TABLE Suppose that n expresson (35) the mposed soluton s: (58) X 3 C F & C, F 0, 0, where C 0, C, are arbtrary constant whose values ("0/") must be determned by the above condtons (54) (55). 40

3 For the st Condton (54): P 0 F 0, where by expressons (32) (58), we have: P0 X3 C0, F0 & C, F C0, & F0 C, & F by expressons (56) (57), we have: F 0 X 2 & X ; F 0 X 2 v X ; F X 2 v X ; F X 2 & X Thus, we have: P0 F0 C0 X2 X C X X " 2 X2 X C0 C C0 X2 C0 X X2 X 0!, $#,, 4 & 9 4 & & 9 & 6 4, &, &, 9 & 6 C0, C, C0, (59) &K K Then, we have a frst condton: C0, C, Φ (that s, 0 or ) For the 2 nd Condton (55): P F 0, where through expressons (32) (58), we have: P0 X3 C0, F0 & C, F through expressons (56) (57), we have: F 0 X 2 & X ; F 0 X 2 v X ; F X 2 v X ; F X 2 & X Thus we have: P F C0 X2 X C X2 X " X2 X C X2 X 0! $#,,, & 6 & Then, we have a second condton: ( 60) C, ( 6) &K From the deduced condtons, (59) (60), we may conclude the followng: C 0, K C, Therefore, the fnal soluton of the proposed example s the followng: 4

4 (58) X 3 C F & C, F 0, 0, where ( 6) &K C 0, K C, (56) F 0 X 2 & X (57) F X 2 v X Thus, we have: X3 X2& X & 0 X2 X X2& X or ( 62) X3 X2& X (fnal soluton) 2 nd EXAMPLE: Suppose that we have a system that t s a part of a program gven graphcally n a State Transton Dagram (STD). There s a generc state "S k " whch corresponds to the Internal Stable State (ISS). From " S k " arses the Transfer Functon, "f, f 2 f 3 " to the correspondng ISS "S, S 2 S 3 ". The values of these functons n Hndrance are: f X 4 & X 2, from " S k " to "S "; f 2 X 4 & X 2, from " S k " to "S 2 "; f 3 X 3 & X, from " S k " to "S 3 ". Thus, determne the feedback functon "f 4 ", where "f 4 (X 4, X 3, X 2, X )". Ths feedback functon represents the permanence of the system n ISS "Sk". Besdes the feedback functon "f m ", around the ISS "S k " (k,2,...,2 p ), there are "m-" Transfer Functons of "n" varables. These "m-" Transfer Functons go to the other ISS programmed n the State Transton Dagram (STD). Havng hndrance as logcal unt, the followng condtons must be obeyed: 42

5 p () I m 2 m () II (f ) 2 m ( III) ( δ, j & f & f j), where f j, then: δ,j, j or, f j, then: δ,j In ths second example, we have: m4; n4 varables. Thus, n condton (II), we have: (63) f v f 2 v f 3 v f 4 n condton (III) we have: f& f2 f& f3 f& f4 f2& f3 f2& f4 f3& f4, or, (64) f f f f 2 & & & & f 2 & f3 f2 f3 f4 f3 f4 f4 Then to fnd the fnal Solvng Equaton of the System of Boolean Equaton (SESBE), we must add expressons (63) (64). ( 65) f f2 f3 f 4 & f f2 & f f3 & f f4 & f 2 f3 & f 2 f4 & f 3 f 4 (SESBE), But, from the data, we have: f X 4 & X 2 ; f 2 X 4 & X 2 ; f 3 X 3 & X the feedback functon ncognto s: f4x4, X3, X2, X6. Then, each product element of expresson (6) wll be: & 9 & & 9 4 & 9 4 & 9 & 6 4 & 9 & 6 4X3& X9 f4 X3 X f4 f f2 f3 f4 X4& X2 X4& X2 X3& X f4 X4 X3& X4 X f4 f f2 X4 X2 X4 X2 f f3 X4 X2 X3 X X4 X3 X2 X f f4 X4 X2 f4 X4 X2 f4 f 2 f3 X4 X2 X3 X X4 X3 X2 X f 2 X4 X4 f4 X2 f4 X2 f4 f3 f4 43

6 Thus, we fnd the fnal expresson of the Solvng Equaton of the System of Boolean Equaton (SESBE): ( 66 ) ff 4, X 4, X 3, X6 X4 X 3 f 4 & X4 X f 4 & X 4 X3 X & & X4 f 4 & X3 X f 4 (SESBE) Accordng to (34), f "p ", then: 2 P s Fs, where Po f 4 s 0 P f4 (67) That s: P o Fo & P F, or we have: st. Condton: P o Fo (68) nd 2. Condton: P F (69) However, as we have seen, " F s ", where s {0, ( 2 ) }, represents all Boolean Functons ( 2 2) whch depend on the other remanng non-man varables (4-3), { X 4, X3, X }. These Boolean Functons " F 0 F " can be obtaned f t s assumed that, the man ncognto " f 4 " corresponds to the "s {0, }" values. Ths "s {0, }" values must have a correspondng bnary expresson n " s". t Then we have: st f 4 f X 4 X 3 f 4 &X 4 X f 4 &X 4 X 3 X &X 4 f 4 &X 3 X f 4 0 T 0 0 F 0 X 4 &X3 X (70) T F X4 X 3 &X4 X & X 4 X3 X (7) TABLE 2 Suppose that n expresson (54) the mposed soluton s: ( 72) f 4 C 0, F 0 & C, F, where C 0, C, are arbtrary constant whose values "0/" must be determned through condtons (68) (69). For the st Condton (68): P 0 F 0 0, where n expressons (67) (72), we have: 44

7 n expressons (70) (7), P f C F C F C0 F0,,, C, F & & & we have: F 0 X 4 &X 3 X ; F 0 X 4 (X3 & X); F X 4 X 3 &X 4 X & X 4 X 3 X ; F (X4 & X 3) (X4 & X ) (X4& X3 & X) P 0 F0 C0,& F0 C, & F F0 C0, C, F0 & C0, F F0 & C, F0 F0 & & F0 F F0 C0, C, F0 & C0, F F0 P 0 F0 C0, C, F0 & C0, F F 0 (73) ( 74) ( 75) & K K Therefore, from (73) we have: C0, C, C0, Then, the frst condton s: C, φ (that s: 0 or ) For the 2 nd Condton (69): P F 0, where n expressons (67) (72), we have: 4 9 ( 76) P F C0, F0 & C, F F C0, F0 F & C, F where, F 0 X 4 & K K C 0, &X 3 X F X 4 X 3 &X 4 X & X 4 X 3 X 45

8 P F ( C0, & C, ) X 4 &X 3 X &C X 4 X 3 &X 4 X & X 4 X 3 X, ( 77) ( 78) &K K &K Thus, we have: C 0, & C, C, Therefore, the second condton s: C 0, K C, From the st the 2 nd condtons, we have: st condton: (60) C 0, C, φ (0 or ); 2 nd condton: (63) C 0, 0 C, But the value of "C 0, " can not occur at the same tme, equal to "0" equal to "", that s, there s an INCOMPATIBILITY n the DATA of the problem. It means that we have a "bug" n that part of program. 3 rd EXAMPLE: Suppose that we have a system that t s a part of a program gven graphcally n a State Transton Dagram (STD). There s a generc state "S k " whch corresponds to the Internal Stable State (ISS). From " S k " arses the Transfer Functon, "f, f 2 f 3 " to the correspondng ISS "S, S 2 S 3 ". The values of these functons n Hndrance are: f X 4 & X 2, from " S k " to "S "; f 2 X 4 & X 2, from " S k " to "S 2 "; f 3 X 4 & X 3 & X, from " S k " to "S 3 ". Thus, determne the feedback functon "f 4 ", where "f 4 (X 4, X 3, X 2, X )". Ths feedback functon represents the permanence of the system n ISS "Sk". 46

9 Besdes the feedback functon "f m ", around the ISS "S k " (k,2,...,2 p ), there are "m-" Transfer Functons of "n" varables. These "m-" Transfer Functons go to the other ISS programmed n the State Transton Dagram (STD). Havng hndrance as logcal unt, the followng condtons must be obeyed: p () I m 2 m () II (f ) 2 m ( III) ( δ, j & f & f j), where f j, then: δ,j, j or, f j, then: δ,j In ths thrd example, we also have: m4; n4 varables. Thus, n condton (II), we have: (79) f v f 2 v f 3 v f 4 n condton (III) we have: f& f2 f& f3 f& f4 f2& f3 f2& f4 f3& f4, or, (80) f f f f 2 & & & & f 2 & f3 f2 f3 f4 f3 f4 f4 Thus, to fnd the fnal Solvng Equaton of the System of Boolean Equaton (SESBE), we must add expressons (79) (80): (8) f f2 f3 f 4 & f f2 & f f3 & f f4 & f 2 f3 & f 2 f4 & f 3 f4 (SESBE), However, from the data, we have: f X 4 & X 2 ; f 2 X 4 & X 2 ; f 3 X 4 & X 3 & X the feedback functon ncognto s: f4x4, X3, X2, X6. Then, each product element of expresson (8) wll be: 47

10 & 9 & & & 9 & & 6 4X & X & X 9 f4 4 3 X f4 f f2 f3 f 4 X4& X2 X4& X2 X4 & X3& X f4 X4 X3 f4 & X4 X f4 f f2 X4 X2 X4 X2 f f3 X4 X2 X 4 & X 3 & X f f4 X4 X2 f4 X 4 X2 f4 f 2 X 4 & X 3 f3 X4 X2 & X f 2 f X4 X f X X2 f4 f3 f4 4 3 Thus, we fnd the fnal expresson of the Solvng Equaton of the System of Boolean Equaton (SESBE): (82) ff 4, X 4, X 3, X6 X4 X 3 f 4 & X4 X f 4 & & X4 f 4 & X3 X f 4 (SESBE) Accordng to (34), f "p ", then: 2 P s Fs, where Po f 4 s 0 P f4 (83) That s: (84) P o Fo & P F, or we have: st. Condton: P o Fo (85) nd 2. Condton: P F (86) However, as we have seen, " F s ", where s {0, ( 2 ) }, represents all Boolean Functons ( 2 2) whch depend on the other remanng non-man varables (4-3), { X 4, X3, X }. These Boolean Functons " F 0 F " can be obtaned f t s assumed that, the man ncognto " f 4 " corresponds to the "s {0, }" values. These "s {0, }" values must have a correspondng bnary expresson n " s". t Then we have: 48

11 s t f 4 f X 4 X 3 f 4 &X 4 X f 4 &X 4 f X 3 X f 4 4 & 0 t 0 0 F 0 X 4 &X3 X (87) t F X4 X 3 &X4 X (88) TABLE 3 Suppose that n expresson (34) the mposed soluton s: (89) f 4 C F & C, F 0, 0 where C 0, C, are arbtrary constant whose values, "0/", must be determned through condtons (85) (86). For the st Condton (85): P 0 F 0, where n expressons (83) (87), we have: P f C F C F C0 F0,,, C, F & & & n expressons (87) (88), we have: F 0 X 4 &X 3 X ; F 0 X 4 (X3 & X); F X 4 X 3 &X 4 X ; F (X4& X 3) (X4 & X ) P 0 F0 C0,& F0 C, & F F0 C0, C, F0 & C0, F F0 & C, F0 F0 & & F0 F F0 C0, C, F0 & C0, F F0 P 0 F0 C0, C, F0 & C0, F F 0 (90) ( 9) & K K Therefore, from (90) we have: C0, C, C0, Then, the frst condton s: 49

12 ( 92) & K K C 0, C, φ (that s: 0 or ) For the 2 nd Condton (86): P F 0, where, n expressons (83) (89), we have: 4 9 ( 93) P F C0, F0 & C, F F C0, F0 F & C, F where, n (87) (88) we have: F 0 X 4 &X 3 X F X 4 X 3 &X 4 X Then, we have: P F C0, X 4 &X 3 X X 4 X 3 &X 4 X & C X 4 X, 3 &X 4 X or, (94) P F C X 4 X, 3 &X 4 X 0 Therefore, the 2 nd Condton s gven n expresson (94), whch s ndependent of C o,, that s: ( 95) &K K C 0, φ (. e., "0" or "") C, From the st the 2 nd condtons, we have: st condton: (92) C 0, C, φ (0 or ); 2 nd condton: (95) C 0, φ (0 or ) C, have: Now we have the COMPATIBILITY for the fnal values of the constants, that s, we (96) C 0, C, Thus n expresson (89), the feedback functon ncognto, we found the followng result: 0, 50

13 (89) f 4 C F & C, F 0, 0 where: C 0, C, ; F 0 X 4 Then, we have: (97) f 4 X 4 &X 3 X &X 3 X ; F X 4 X 3 &X 4 X Ths 3 rd Example s very smlar to the 2 nd, but now we do not fnd any INCOMPATIBILITY n the DATA of the problem. 5

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