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201

Unit 16 Solutions

i

x

a

i

'

i

x

ai

i+1

b

i

b'

i+1

aZ

16.17 (b)

State

Next State

X = 0 X = 1

Z

S0 S0 S10

S0

0 1

0

1S2

S1

S3

ai

bi

State

aibi

ai+1bi+1

X = 0 X = 1

Z

S000 00 10 0

xi

ai bi0 1

00

0

1

bi+1 = bi

' + xiai

16.17 (c)

xi

ai bi0 1

00

01

11

10

0

1

0

1

0

1

1

1

Unit 16 Problem Solutions

16.15 See FLD p. 767 for solution.

16.17 (a) The state meanings are given in the following table:

Name Meaning

S0No 1’s have occurred

16.16 See FLD p. 768 for solution.

1

S3

1

S2

0

0

S0

0

S1

0

1 1

16.1 -

16.14

See Lab Solutions in this manual.

Since no 1’s have occurred, a1 and b1 are the same

as S0 or, a1 = 0; b1 = 0;

202

Unit 16 Solutions

DQ

x

i

a

i

a

i+1

16.17 (d)

ai

ai+1 yi

xi = 0 xi = 1 xi = 0 xi = 1

16.18 (a)

16.18 (a)

yi

aibi

ai+1bi+1

xi yi =

00

xi yi =

01

xi yi =

11

xi yi =

10

00 00 10 00 01

16.19 (a)

aibi

ai+1bi+1

xi yi =

00

xi yi =

01

xi yi =

11

xi yi =

10

00 00 10 00 01

16.20 (a)

16.20 (b) ai+1 = aixi' + aiyi + aibi + bi'xi'yi

bi+1 = bixi + biyi' + ai'bi + ai'xiyi'

16.20 (c) Z1 = an+1bn+1'

Si

Si+1

xi yi =

00

xi yi =

01

xi yi =

11

xi yi =

10

Initial S-1 S0S1S3S2

16.20 (d)

203

Unit 16 Solutions

16.21 (a)

S0S1

11

10

Present

State

Next State

X = 0 1

Output

X = 0 X = 1

S0 S2 S1 0 0

S1 S3 S0 0 1

S2 S0 S3 1 0

1S1

S2

The state meanings are given in the following table:

Name Meaning

S0even #0's and even #1's received

The output becomes 1 whenever an even #0's or an

even #1's (greater than 0) occurs.

A B

A+ B+

X = 0 1

Z

X = 0 X = 1

DA

X

A B 0 1

00

01

1

1

0

0

DB

X

A B 0 1

00

01

1

0

1

0

Z

X

A B 0 1

00

01

0

0

0

1

16.21 (b)

A B

JA KA

X = 0 1

0 0 1 X 0 X

0 1 1 X 0 X

A B

JB KB

X = 0 1

0 0 1 X 1 X

0 1 X 1 X 1

KAX

A B 0 1

00

X

X

JA

X

A B 0 1

00

1

0

204

Unit 16 Solutions

JB

X

A B 0 1

00

01

1

X

1

X

16.21 (b)

(cont.) KB

X

A B 0 1

00

01

X

1

X

1

The state meanings are given in the following table:

Name Meaning

S0reset state

S1even #0's and even #1's received

S2odd #0's and even #1's received

16.21 (c)

16.21 (c)

(cont.) 0

S0

0

S1

1

S5

0

S2

0

1

0

0

1

1

Present

State

Next State

X = 0 1 Z

S0 S2 S5 0

S1 S2 S5 1

S2 S1 S6 0

S3 S1 S6 1

Guidelines: I: (0, 1)2x, (2, 3)2x, (4, 5)2x

II: (2, 5)2x, (1, 6)4x, (3, 4)

An assignment is

A

B C 0 1

00

01

S0

S1

S5

S4

S0000 110 100 0

S1001 110 100 1

S6011 101 111 0

-- 010 --- --- -

205

Unit 16 Solutions

Z

X A

B C 00 01 11 10

00

01

0

1

0

1

0

1

0

1

TCX A

B C 00 01 11 10

00

01

0

1

1

0

0

1

1

0

TBX A

B C 00 01 11 10

00

01

1

1

1

1

0

0

0

0

TAX A

B C 00 01 11 10

00

01

1

1

1

1

1

1

1

1

16.22 (a)

S1

S0S2

1

1

1

0

0

0

0

01

1

carry

carry

carry

16.22 (b) JAX

A B 0 1

00

01

0

0

0

1

KAX

A B 0 1

00

01

X

X

X

X

JBX

A B 0 1

00

01

0

X

1

X

KBX

A B 0 1

00

01

X

1

X

0

Z

X

A B 0 1

00

0

1

A B X DA DB Z

- 1 1 1 0 0

16.22 (c)

16.21 (c)

(cont.)

206

Unit 16 Solutions

16.23 (a)

S0

00

S1

00

S2

00

11-,

1-1

00-,

0-0

00-,

011,

100

11-, 1-1

00-, 0-0

011,

100

011,

100

Inputs: X2X1X0Outputs: ID

D0 = X2'X1X0 + X2X1'X0'

D1 = (X2'X1' + X2'X0')(Q0 + Q2 + Q4 + Q6)

D3 = (X2'X1' + X2'X0')Q1

D5 = (X2'X1' + X2'X0')(Q3 + Q5)

D2 = (X2X1 + X2X0)(Q0 + Q1 + Q3 + Q5)

D4 = (X2X1 + X2X0)Q2

D6 = (X2X1 + X2X0)(Q4 + Q6)

I = Q5

D = Q6

16.23 (b)

Using the assignment S0 = 000, S1 = 001, S2 = 010, S3 = 011, S4 = 100, S5 = 101,

S6 = 110, S7 = 111:

D0 = (X2'X1X0 + X2X1'X0')(S0 + S1 + S2 + S3 + S4 + S5 + S6)

= (X2'X1X0 + X2X1'X0')(Q0' + Q1' + Q2')*

= X2'X1X0 + X2X1'X0'*

16.23 (c)

16.24 (a)

11-,1-1/0000-,0-0/00

Inputs: X2X1X0Outputs: ID

S0

011,100/00

011,

100

011,

100

D0 = X2'X1X0 + X2X1'X0'

D1 = (X2'X1' + X2'X0')(Q0 + Q2 + Q4)

D3 = (X2'X1' + X2'X0')(Q1 + Q3)

D2 = (X2X1 + X2X0)(Q0 + Q1 + Q3)

D4 = (X2X1 + X2X0)(Q2 + Q4)

I = (X2'X1' + X2'X0')Q3

D = (X2X0 + X2 X1)Q4

16.24 (b)

207

Unit 16 Solutions

Using the assignment S0 = 000, S1 = 001, S2 = 010, S3 = 011, S4 = 100:

D2 = X2X0Q1Q0' + X2X1Q1Q0' + X2X0Q2 + X2X1Q2 or

16.24 (c)

16.25 (a) S0

000

S1

001

S6

110

01,

10

00,

11

01, 10

00, 11

Inputs: XY Outputs: Z2Z1Z0

10

D0 = 0

D1 = (X'Y' + XY)(Q0 + Q4 + Q5 + Q6)

D2 = (X'Y' + XY)Q1

D3 = (X'Y' + XY)(Q2 + Q3)

D6 = (X'Y + XY')(Q0 + Q1 + Q2 + Q3)

D5 = (X'Y + XY')Q6

16.25 (b)

Using the assignment S0 = 000, S1 = 001, S2 = 010, S3 = 011, S4 = 100, S5 = 101, S6 = 110:

D0 = (X'Y' + XY)(S0 + S4 + S5 + S6 + S2 + S3) + (X'Y + XY')S6

= (X'Y' + XY)(Q0' + Q2'Q1 + Q2Q1') + (X'Y + XY')Q2Q1Q0'

16.25 (c)

16.26 (a)

01,10/110

00,11/001

Inputs: XY Outputs: Z2Z1Z0

S0

S6

S1

01,10/110

D0 = 0

D1 = (X'Y' + XY)(Q0 + Q5 + Q6)

D2 = (X'Y' + XY)(Q1 + Q2)

D6 = (X'Y + XY')(Q0 + Q1 + Q2)

D5 = (X'Y + XY')(Q6 + Q5)

16.26 (b)

208

Unit 16 Solutions

Using the assignment S0 = 000, S1 = 001, S2 = 101,

S6 = 111, S5 = 011

D2 = Q1'Q0 + XY'Q1' + X'YQ1' or

D0 = 1

Z2 = X'Y + X Y'

Z1 = Q1'Q0 + XY'Q1' + X'YQ1' or

= Q1'Q0 + XY'Q0' + X'YQ1' or

16.26 (c) 16.27 (a) S0

000

S2

S6

01,

00,

Inputs: XY Outputs: Z2Z1Z0

01,

10

01,

10

00,

11

00,

11

10

D0 = 0

D1 = (X'Y' + XY)(Q0 + Q4 + Q5 + Q6 + Q7)

D2 = (X'Y' + XY)Q1

D3 = (X'Y' + XY)(Q2 + Q3)

16.27 (b) Using the assignment S0 = 000, S1 = 001, S2 = 010,

S3 = 011, S4 = 100, S5 = 101, S6 = 110, S7 = 111:

D0 = (X'Y' + XY)(S0 + S4 + S5 + S6 + S7 + S2 +

S3) + (X'Y + XY')(S0 + S1 + S2 + S3 + S6)

16.27 (c)

16.28 (a)

01,10/111

00,11/001

Inputs: XY Outputs: Z2Z1Z0

S0

209

Unit 16 Solutions

Using the assignment S0 = 000, S1 = 001, S2 = 100,

S5 = 011, S6 = 010, S7 = 111:

D2 = X'YQ1' + XY'Q1' + Q1'Q0

16.28 (c)

16.29 (b) Name Meaning

S0Staying on first oor

S1Moving from first to second oor

S2Staying on second oor

S3Moving from second to first oor

S1

FS'2UP

FS2R2DO

N2R2DO,

N1R1DO,

N1'N2DC

UP

16.29 (a) Ni = Qi

+ = (Qi + FBi + CALLi)Ri

'

= QiRi

' + FBiRi

' + CALLiRi

'

16.30 (a)

S0

000000

L'R'H'

H'

H'

L'H' LR'H'

L'RH' R'H'

210

Unit 16 Solutions

16.30 (b) First, assign LC = Q1, LB = Q2, LA = Q3, RA = Q4, RB = Q5, RC = Q6. So S0 = 000000, S1 = 001000, S2 = 011000, etc.

This state machine has too many state variables to use Karnaugh maps. Instead, we will write down equations for

each ip-op by inspection.

First consider Q1. Q1 = 1 in states S3 or S7 only.

I. (S0, S1, S2, S3, S4, S5, S6) for S7 in LRH = 001, 011, 101

(S1, S2, S3, S6, S7) for S0 in LRH = 010

(S3, S4, S5, S6, S7) for S0 in LRH = 100

II. Every state matches S0 and S7. But S0 and S7 match the best, so (S0, S7)×(many times)

III. (S1, S2, S3, S7) (S4, S5, S6, S7) etc.

From LogicAid:

16.30 (c) State LRH = 000 001 010 011 100 101 110 111 LC LB LA RA RB RC

S0 S0 S7 S4 S7 S1 S7 - - 0 0 0 0 0 0

S1 S0 S7 S0 S7 S2 S7 - - 0 0 1 0 0 0

S2 S0 S7 S0 S7 S3 S7 - - 0 1 1 0 0 0

Q1

Q2 Q30 1

00

01

11

S0

S2

S6

S7

S3

S5

211

Unit 16 Solutions

16.31

IDLE

0

RE' FF' PL'

ST, FF ST, RE

RE

FF

PL

ST, FF, RE

Note: This state graph assumes that only one of the buttons ST, PL, RE, and FF can be pressed at any given time.

The graph is incompletely specified and must be augmented before using LogicAid. For example, the arc from REW

to PLAY should be labeled PL ST' FF'.

DQ1

Q1+

ST

RE

FF

PL

M

P

R

F

D1 = ST' FF PS Q1

' Q2

' Q3 + ST' RE PL Q1

' Q2

' Q3

212

Unit 16 Solutions

16.32 (a) 0000 0000 0000

1 0000 0000 0001

1 0000 0000 0011

0 0000 0000 0110

16.32 (c)

x3x2x1x0y3y2y1y0

0000 0000

0001 0001

The tables of combinations for the BCD multiply

by 2 circuit are

x3x2x1x0y3y2y1y0

1000 1011

1001 1100

16.32 (c)

cont.

16.33 (a)

0

S0

S1

S2

0

S6

S4

0

1

1

0

0, 1

1

0

00 01 11 10

00

01

0

0

0

1

1

1

X

X

x0

x1

x3x2

y3

00 01 11 10

00

01

0

0

1

0

0

1

X

X

x0

x1

x3x2

y2

00 01 11 10

00

01

0

0

0

0

1

0

X

X

x3x2

y1

x0

x1

00 01 11 10

00

01

0

1

0

0

1

0

X

X

x0

x1

x3x2

y0

213

Unit 16 Solutions

Present

State

Next State

X = 0 1 Z

S0 S0 S1 0

S1 S2 S3 0

S2 S0 S4 0

ABC

A+B+C+

X = 0 1 Z

S0000 000 001 0

S1001 010 011 0

S2010 000 101 0

16.33 (b)

ai+1 xi ai

bi ci00 01 11 10

00

01

0

0

0

0

0

0

1

1

bi+1 xi ai

bi ci00 01 11 10

00

01

0

1

0

1

0

1

1

1

ci+1 xi ai

bi ci00 01 11 10

00

01

0

0

0

0

1

1

0

0

16.33 (c) Z

an

bn cn0 1

00

01

0

0

0

0

16.34 (a) S2

1

S0

0

S1

1

0

1

1

0

1

0

214

Unit 16 Solutions

DQ

CK

x

i

a

i

a

i+1

Typical Cell

Z

16.34 (c)

16.34 (d)

a1 = b1 = 0

a2 = (x1 + 0)(x1' + 1) = x1

b2 = (x1 + 1)(x1 + 0) = x1

16.34 (b)

ai bi0 1

00

01

0

0

1

0

ai bi0 1

00

01

0

1

1

1

0

0 1

0

10

1

1

bn+1

S0

0 1

0

1S3

S2

S1

ai

bi

State

Next State

xi = 0 xi = 1

Z

S0 S0 S10

S1 S2 S31

aibi

ai+1bi+1

xi = 0 xi = 1

Z

00 00 11 0

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