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Unit 19 Solutions

245

Unit 19 Problem Solutions

19.1 See FLD p. 777 for solution. 19.2 See FLD p. 777 for solution.

19.3 See FLD p. 778 for solution. 19.4 See FLD p. 778 for solution.

19.11 19.12 (a)

19.12 (b)

Unit 19 Solutions

246

19.13

S0

0

Reset

X

19.14 (a)

19.14 (b) Let S0, S1, S2 and S3 be the four FF outputs, then

D0 = x'(S0 + S1 + S2 + S3),

D1 = xS0,

19.14 (c) Using state assignment S0 = 00, S1 = 01, S2 = 10

and S3 = 11, and denoting state variables Q1 and

Q0, D1 is the OR of D2 and D3 from Part b) so

Unit 19 Solutions

247

S1

0

K

0

1

Reset

St

So

0

1

So0

Sh, SI Sh

19.15 (a)

D1 = K'Q1 + K'SoQ0 , D0 = K'Q0 + StQ0'

Next State Table for Q1+ Q0+: S1 = 00, S2 = 01 and

S3 = 11.

St K So

Q1Q0000 001 011 010

00 00 00 00 00

01 01 11 00 00

11 11 11 00 00

10 -- -- -- --

Output Table for Clr Sh Er SI

St K So

Q1Q0000 001 011 010

00 0000 0000 0000 0000

Er = KSoQ1'Q0; SI = K'So'Q1 + KSoQ1'Q0

19.15 (b)

Label the three FF outputs S1, S2 and S3.

D1 = St'S1 + KS2 + KS3; D2 = K'So'S2 + StS1

D3 = K'SoS2 + K'S3; Clr = StS1

Sh = K'S2 + KSo'S2 + K'S3 = K'S2 + So'S2 + K'S3

19.15 (c) Label the two FF outputs Q1, Q0 and the decoder

outputs S1 = 00, S2 = 01 and S3 = 11*, then

D0 = K'So'S2 + StS1 + K'SoS2 + K'S3

= K'S2 + StS1 + K'S3

19.15 (d)

Unit 19 Solutions

248

St

M

Done

Load

Sh

19.18

1

X

Clock

State S0S2S0S2S1S2S0

S1

19.19 (a)

State Q0

Q1

St M K Q0

+ Q1

+ Ad Sh Load Done

S0 0 0 0 - - 0 0 0 0 0 0

S0 0 0 1 - - 0 1 0 0 1 0

19.19 (b)

A+ = A'BX2 + A'B'X2 (X1

' + X3 ) + {AB}

= BX2 + A'X2 (X1

' + X3 )

B+ = A'B' (X2

' + X1X3

' ) + AB'X1

' + A'BX2

' + {AB}

= AX1

' + A'B'X1X3

' + A'X2

'

19.16

S /

S /

S /

S1 /

S /

S /

S / Done

0 1 3 5

7 9

Clock

State

St

S0S1S2S2S3S3S4

19.17

Unit 19 Solutions

249

PLA table obtained by tracing link paths:

State AB X1X2X3 A+B+ Z1Z2Z3

S0

00 - 0 - 01 010

00 01 - 10 101

00 110 01 100

19.19 (c) 25 × 5 ROM

AB X1X2X3 A+B+ Z1Z2Z3

00 000 01 010

00 001 01 010

19.19 (d)

S0/LC, LR

s

S1/SR

01

Reset

19.20 (a) C is loaded with 0.

19.20 (b)

S0/LC, LR

s

S1/SR

01

Reset

19.21 (b)

19.21 (a) C is loaded with 0.

Unit 19 Solutions

250

S0/LC, LR

s

S1/IncC

01

Reset

S2/IncC

19.22 (a) C must be at least 4 bits and is loaded with 810

(10002).

19.22 (b)

S0/LC, LR

s

S1/SR, IncC

01

Reset

S2/SR, IncC

1

z

19.23 (a) C must be at least 4 bits and is loaded with 710

(01112).

19.23 (b)

S0/LC, LR

s

S1/SR, IncC

01

Reset

S2/SR, IncC

S3/SR, IncC

S4/SR, IncC

19.24 (a) C must be at least 4 bits and is loaded with 810

(10002).

19.24 (b)

Unit 19 Solutions

S0

s

0

Reset

19.25 (a) Let S0, S1, S2 and S3 be the four FF outputs, then

D0 = s'(S0 + S3); D1 = sS0 + (TC)'S2

D2 = S1; D3 = (TC)S2

LDN = S1 + S2 or LDN = S1 + S2 + S3

CE = S1 or CE = S1 + S3

z1 = S1; z2 = S2

P3 = 0; P2 = 0; P1 = 1; P0 = 1

19.25 (b)

Using state assignment S0 = 00, S1 = 01, S2 = 11,

S3 = 10, and denoting the state variables as Q1 Q0,

D1 is the OR of D2 and D3 from Part (b) so

D1 = S1 + (TC)S2 = Q1'Q0 + (TC)Q1Q0

19.25 (c)

Initial PU,PL: 0000 0000

1st Add Lower half PU, PL: 0000 1011

1st Add Upper half PU, PL: 0000 1011

2nd Add Lower half PU, PL: 0000 0110

19.26 (a)

Unit 19 Solutions

252

S0/CP, LA, LB, CC

S

0

1

S1/

Reset

1

19.26 (b)

Label the 4 FF outputs S0, S1, S2 and S3.

D0 = S'S0 + S'S3

D1 = S(S0) + S2

19.26 (c)

Assume two FFs Q1Q0 and the following encoding:

S0 = 00, S1 = 01, S2 = 11 and S3 = 10. (The

decoder outputs are labeled S0, S1, S2 and S3.)

19.26 (d)

S

0

1

LA,LB,CC,CP

SB*

S1/

S2/

19.27

Unit 19 Solutions

19.28 (a) See answer to 18.30 (b).

19.28 (b)

19.29 (a)

See answer to 16.26 (c).

19.29 (b)

Unit 19 Solutions

19.30

Unit 19 Solutions

255

19.31

19.32

1

X

S /

0

10

0

Unit 19 Solutions

256

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