ECE/CS 352 Summer 2000 Homework Set #1

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Department of Electrical and Computer Engineering University of Wisconsin - Madison

ECE/Comp Sci 352 Digital System Fundamentals

Homework #1 (Summer 2000) Solution

1. (Number representations, conversions) List the binary, octal, and hexadecimal numbers from 25610 to 27210.

Decimal Binary Octal Hexadecimal

256 100000000 400 100

257 100000001 401 101

258 100000010 402 102

259 100000011 403 103

260 100000100 404 104

261 100000101 405 105

262 100000110 406 106

263 100000111 407 107

264 100001000 410 108

265 100001001 411 109

266 100001010 412 10A

267 100001011 413 10B

268 100001100 414 10C

269 100001101 415 10D

270 100001110 416 10E

271 100001111 417 10F

272 100010000 420 110

2. (Binary code) What is the decimal representation for the highest number that can be represented by: a). 12 binary digits? 212 -1 = 4095 b). 6 octal digits? 86 -1 = 262143 c). 4 hexadecimal digits? 164 -1 = 65535

3. Book problem 1-7

Decimal Binary Octal Hexadecimal

369.3125 101110001.0101 561.24 171.5

189.625 10111101.101 275.5 BD.A

214.625 11010110.101 326.5 D6.A

62407.625 1111001111000111.101 171707.5 F3C7.A

ECE/CS 352 Summer 2000 Homework Set #1

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4. (Radix conversion) Find the two bases r and s such that:

12r = 111s (Hint: r > 2, s > 1)

There are an infinite number of them. To see this, set up the following equation:

1*r1 + 2 = 1*s2 + 1*s1 + 1 ; thus r = s2 + s - 1

Here are a few pairs that work:

r s 5 2 11 3 19 4 29 5 41 6 55 7 71 8 89 9

5. (Arithmetic) Perform the listed arithmetic on A and B without converting to decimal. (Let A= 00110111, B= 10110000) A+B, B-A, A-B, A*B, B*A

0 0 1 1 0 0 0 0 - 0 0 1 1 0 1 1 1 + 1 0 1 1 0 0 0 0 1 1 1 0 0 1 1 1

0 1 1 1 1 1 1 1 - 1 0 1 1 0 0 0 0 + 0 0 1 1 0 1 1 1 0 1 1 1 1 0 0 1

For A - B, "B" is larger in magnitude than A, so we perform A - B as above, and use a "negative" sign.

A - B =

− 0 1 1 1 1 0 0 1

cyA

+BC

bB

-AC

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Note: the "c" denotes carries from one column over. The "d" denotes carries from two columns over (when four to seven ones are added in a column). It is possible to have a carry from three columns over (when 8 or more ones are added in a column -- none happen here).

0 0 1 1 0 1 1 1 X 1 0 1 1 0 0 0 0 d - c c c c c c 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 1 0 0 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 1 0 0 1 0 0 1 0 1 1 1 0 1 0 0 0 0

1 0 1 1 0 0 0 0 X 0 0 1 1 0 1 1 1 d - c c c c c c 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 1 1 1 0 1 0 0 0 0

CHECK the work == convert them to decimal and compare the results -- I did!!!

AxBC

BxAC

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6. (Coding and Arithmetic) Convert the decimal digit strings to BCD coding and perform BCD arithmetic by using binary addition and correction (see page 19 in the Text) Show all carries. Let A = 123 and B = 489. Perform A + B.

Bit cy #1 - - - 0 0 0 b 0 0 0 a 0 1 1 - A 1 2 3 0 0 0 1 0 0 1 0 0 0 1 1 +B 4 8 9 0 1 0 0 1 0 0 0 1 0 0 1 Digit Cy 1 1 - - - 1 - - - 1 - - - 0 Bit Cy #2 - - - 0 0 1 - 1 1 0 - 1 0 0 - Bit Sum 5 11 12 0 1 0 1 1 0 1 1 1 1 0 0 Correctio 0 6 6 0 0 0 0 0 1 1 0 0 1 1 0 Sum 6 1 2 0 1 1 0 0 0 0 1 0 0 1 0

Notes: 1). The Bit Sum row is the initial binary sum, digit by digit. Bit carries in the top row are the binary bit carries for the binary sum.

2). Bits "a" and "b" are the low digit and middle digit binary carries for the respective digit sums (ie, "a" is one if the bit sum for the lower digit is 16 or greater, as it is in this case once the "correction" is added.). These are the same as the digit carry.

3). The correction of "6" is added whenever the binary sum is TEN or greater.

7. (Character Codes) Convert the string "madam I am adam" to ASCII characters.

Character m a d a m I a m a d a m Decimal 109 97 100 97 109 32 73 32 97 109 32 97 100 97 109

Octal 155 141 144 141 155 040 111 040 141 155 040 141 144 141 155

Hex 6D 61 64 61 6D 20 49 20 61 6D 20 61 64 61 6D

8. (Parity) enumerate all patterns of five binary bits. Add a sixth bit to make the code words have EVEN parity. Given the even parity bit, how would you generate an ODD parity bit?

# Bin Pe # Bin Pe # Bin Pe # Bin Pe 0 00000 0 8 01000 1 16 10000 1 24 11000 0 1 00001 1 9 01001 0 17 10001 0 25 11001 1 2 00010 1 10 01010 0 18 10010 0 26 11010 1 3 00011 0 11 01011 1 19 10011 1 27 11011 0 4 00100 1 12 01100 0 20 10100 0 28 11100 1 5 00101 0 13 01101 1 21 10101 1 29 11101 0 6 00110 0 14 01110 1 22 10110 1 30 11110 0 7 00111 1 15 01111 0 23 10111 0 31 11111 1

Given EVEN parity, complement the parity bit to get ODD parity codewords.

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9. Book problem 2-1 ____ __ __ __ a) XYZ = X + Y + Z Verification of De Morgan’s theorem

b) X +YZ = (X+Y) . (X + Z) The second distribution Law

c) XY+YZ+XZ = XY+YZ+XZ

10. Book Problem 2-2 _ _ _ _ a) X Y + X Y + X Y = X + Y _ _ _ _ = (X Y + X Y ) + (X Y + XY) _ _ _ _ = X (Y + Y ) + Y(X + X) = X + Y

X Y Z XYZ XYZ X + Y + Z0 0 0 0 1 10 0 1 0 1 10 1 0 0 1 10 1 1 0 1 11 0 0 0 1 11 0 1 0 1 11 1 0 0 1 11 1 1 1 0 0

X Y Z XY YZ XZ XY+YZ+XZ XY YZ XZ XY+YZ+XZ0 0 0 0 0 0 0 0 0 0 00 0 1 0 1 0 1 0 0 1 10 1 0 1 0 0 1 0 1 0 10 1 1 1 0 0 1 0 0 1 11 0 0 0 0 1 1 1 0 0 11 0 1 0 1 0 1 1 0 0 11 1 0 0 0 1 1 0 1 0 11 1 1 0 0 0 0 0 0 0 0

X Y Z YZ X+YZ X+Y X+Z (X+Y)(X+Z)0 0 0 0 0 0 0 00 0 1 0 0 0 1 00 1 0 0 0 1 0 00 1 1 1 1 1 1 11 0 0 0 1 1 1 11 0 1 0 1 1 1 11 1 0 0 1 1 1 11 1 1 1 1 1 1 1

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_ _ _ _ b) A B + B C + AB + B C = 1 _ _ _ _ = (A B + AB) + (B C + B C) _ _ _ = B (A + A) + B (C + C) _ = B + B = 1 _ _ c) Y + X Z + X Y = X + Y + Z _ _ = Y + X Y + X Z _ _ = (Y + X) (Y + Y ) + X Z _ = Y + X + X Z _ = Y + ( X + X ) ( X + Z ) =X + Y + Z _ _ _ _ _ _ _ d) X Y + Y Z + XZ + XY + Y Z = X Y + XZ + Y Z _ _ _ _ _ =X Y + Y Z ( X + X ) + XZ + XY + YZ _ _ _ _ _ _ = X Y + X Y Z + X Y Z + XZ +XY+ Y Z _ _ _ _ = X Y(1 + Z) + X Y Z + X Z + X Y + Y Z _ _ _ _ = X Y + XZ (1 + Y) + XY + Y Z _ _ _ _ = X Y + XZ +XY (Z + Z) + Y Z _ _ _ = X Y + XZ + XYZ + Y Z (1 + X) _ _ _ = X Y + XZ (1 + Y) + Y Z _ _ _ = X Y + XZ + Y Z 11. Book Problem 2-6 _ _ _ _ _ a) A B C + A B C + A B = AB (C + C) + AB = B ( A + A) = B

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______ _ _ _ _ _ _ _ _ b) ( A + B) (A + B) = A B (A + B) = A B _ _ c) A B C + A C = (A B + A) C = (B + A) C = AC + BC _ _ _ d) BC + B(AD + A D ) = BC + ABD + A B D = BC + AB (D + D) = AB + BC _ _ _ _ _ _ _ _ e) (A + B + AB)(AB + AC + BC) = (A + B)(AB + AC) = AB(A + B) + AC(A + B) _ _ _ _ _ _ = AB + A B B + A A C + A B C = AB + A B C 12. Book Problem 2-11 a) E = ∑m (0,1,2,5), ΠM(3,4,6,7), F = ∑m(2,3,6,7), ΠM(0,1,4,5) _ _ b) E = ∑<3,4,6,7>, F = ∑(0,1,4,5) c) E + F = ∑m (0,1,2,3,5,6,7), E . F = ∑m (2) _ _ _ _ _ _ _ _ _ _ _ _ d) E = X Y Z + X Y Z + X Y Z + X Y Z , F = X Y Z + X Y Z + X Y Z + X Y Z _ _ _ e) E = X Z + Y Z , F = Y 13. Book Problem 2-12 _ _

a) (AB + C)(B + CD) = AB + BC + ABCD = AB + BC s.o.p. = B(A + C) p.o.s.

_ _ _ _ _ _ _ b) X + X ((X + Y)(Y + Z)) =(X + X)(X + (X + Y)(Y + Z))

_ _ _ _ _ _ = (X + X +Y)(X + Y + Z) = X + Y + Z s.o.p. and p.o.s.

_ _ _ _ _ c) (A + BC + CD)(B + EF) = (A + B + C)(A + B + D)(A + C + D)(B + E)(B + F)

p.o.s. _ _ _ _ _ _

(A + BC + CD)(B + EF) = A(B + EF) + BC(B + EF) + CD(B + EF) _ _ _

= AB + AEF + BCEF + BCD + CDEF s.o.p.

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14. Book Problem 2-8 _ _ _ _ a) F = ABC + AC + AB ____ _ _ _ _ _ _ _ ______ ______

F = ABC + AC + AB = ((A + B + C ) + (A + C) + ( A + B )) _ _ _ _ b) F = ABC +AC + AB

____ _ _ _ _ = ABC . AC . AB

15. (Canonical forms) Given F(A,B,C) shown as the Canonical Sum of minterms below: F A B C( , , ) ( , , , , )= ∑ 1 2 3 6 7 a). express F(A,B,C) and F'(A,B,C) (the complement of F) as a truth table

A B C F F ' 0 0 0 0 1 0 0 1 1 0 0 1 0 1 0 0 1 1 1 0 1 0 0 0 1 1 0 1 0 1 1 1 0 1 0 1 1 1 1 0

b). express F(A,B,C) and F'(A,B,C) in Canonical Product of Maxterms Form.

F(A,B,C) =ΠM(0,4,5) = (A+B+C)*(A'+B+C)*(A'+B+C')

F '(A,B,C) =ΠM(1,2,3,6,7) = (A+B+C')*(A+B'+C)*(A+B'+C')*(A'+B'+C)*(A'+B'+C')

c). express F '(A,B,C) (the complement of F) in Canonical Sum of minterms Form.

F(A,B,C) =∑m(0,4,5) = A'B'C' + AB'C' + AB'C

d) write an algebraic expression for F(A,B,C) in Product of Sums and Sum of Products form. Use Boolean algebra to attempt to minimize the number of literals.

ECE/CS 352 Summer 2000 Homework Set #1

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Product of Sums Form F(A,B,C) = (A+B+C)*(A'+B+C)*(A'+B+C')

= (A+B+C)*[(A'+B) + CC'] = (A+B+C)*(A'+B) = (A+B)*(A'+B) +C(A'+B) = (B+C)*(A'+B)

Sum of Products Form F(A,B,C) = A'B'C + A'BC' + A'BC + ABC' + ABC

= A'B'C + A'BC + A'BC' + A'BC + ABC' + ABC = A'C(B' + B) + A'B((C' + C) + AB(C' + C) = A'C + A'B + AB = A'C + B

16. (Canonical forms) Given F(A,B,C) shown as the Canonical Product of Maxterms below: F = ∏ ( , , , )0 2 4 5

a). express F(A,B,C) and F '(A,B,C) (the complement of F) as a truth table.

A B C F F ' 0 0 0 0 1 0 0 1 1 0 0 1 0 0 1 0 1 1 1 0 1 0 0 0 1 1 0 1 0 1 1 1 0 1 0 1 1 1 1 0

b). express F(A,B,C) and F '(A,B,C) in Canonical Sum of minterms Form.

F(A,B,C) =∑m(1,3,6,7)

F '(A,B,C) =∑m(0,2,4,5) c). express F'(A,B,C) (the complement of F) in Canonical Product of Maxterms Form.

ECE/CS 352 Summer 2000 Homework Set #1

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F(A,B,C) =ΠM(1,3,6,7)

d) write an algebraic expression for F(A,B,C) in Product of Sums and Sum of Products form. Use Boolean algebra to attempt to minimize the number of literals.

Product of Sums Form F(A,B,C) = (A+B+C)*(A+B'+C)*(A'+B+C)*(A'+B+C')

= [(A+C) + BB']*[(A'+B) + CC'] = (A+C)*(A'+B)

Sum of Products Form F(A,B,C) = A'B'C + A'BC + ABC' + ABC

= A'C(B' + B) + AB(C' + C) = A'C + AB

17. Book Problem 2-12(b) _ _ _ _ _ _ _ X + X((X + Y)(Y + Z)) = (X + X)(X + (X + Y)(Y + Z)) _ _ _ _ _ _ = (X + X + Y)(X + Y + Z) = X + Y + Z s.o.p and p.o.s form 18. (K-Map) Map F(A,B,C) from problems 15 and 16 above. Use 3-variable K-Maps. Problem 15:

A

B

C

0 1 3 2

4 5 7 6

A=0

BC=00 =01 =10=11

A=1

F(A,B,C)

1 1 1

1 1

0

0 0

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Problem 16:

19 (Logic Diagram) Draw the logic diagram for the minimized expressions in problem 15(d). Draw both the Product of Sums and Sum of Product forms.

20. (Bonus problem -- will be used on the Project, part 1) Convert the last five digits of your student number to a binary number. Truncate the result to the least significant 16 bits (Modulo 216). Use those bits to form a truth table of four variables (A,B,C,D) (eg. If b0 is a 1, minterm 0 is present, if b1 is 1, minterm 1 is present, etc). Express the function in Sum of Minterm and Product of Maxterm forms. Draw the 4-variable K-Map of the function.

B

C

B

A

F(A,B,C)

A

F(A,B,C)C

B

Product of Sums

Sum of Products

A

B

C

0 1 3 2

4 5 7 6

A=0

BC=00 =01 =10=11

A=1

F(A,B,C)

1 1

1 1

0

0 0

0

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Let Student_ID = 12345 Converting to HEX we get: 303916 which is 0011 0000 0011 10012.

F(A,B,C,D) =∑m(0,3,4,5,12,13) The rest is left to the student.

21 Book Problem 2-14 (c)

_ _ _ C + A B

22 Book Problem 2-16 ( c )

_ _ _ _ B D + B D + A D 23 Book Problem 2-18 (b) & ( c ) (b)

∑m(3,4,5,7,9,13,14,15)

B

1 1 1

A 1 1

C

C

1 1

1 1 1 1 B

A 1 1

1 1

D

Y

1

1 1 1 X

W 1 1 1

1

Z

ECE/CS 352 Summer 2000 Homework Set #1

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(c)

∑m(0,2,6,7,8,10,13,15) 24 Book Problem 2-19 ( c ) _ _ _ Prime = AB, AC, AD, BC, BD, CD _ _ Essential = AC, BC, BD 25 Book Problem 2-24 (a) & ( c ) (a)

F = 1 (c)

_ _ F = AD + CD + BD

C

1 1

1 1 B

A 1 1

1 1

D

Y

1 1 X 1

X 1 1 X X

Z

C

1 1

X 1 1 X B

A X X 1 1

1

D

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26 (Table methods) Given function P(v,w,x,y,z) shown below:

P(v,w,x,y,z) = ∑ (0,1,4,6,7,13,15,23,31) + ∑ d(5,17,21).

(a). Use the tabular method to find all the Prime Implicants of the function P(v,w,x,y,z), where the second summation is the don't care minterms. (b). Use an implicant table to derive a minimum cover for the function P(v,w,x,y,z). (c). Write a simplified expression for the function P(v,w,x,y,z).

Answer: Use all minterms and dont' cares to generate the PIs:

mi Index Order √ 1-Cubes √ 2-Cubes √

0 00000 √ 0000- √ 00-0- PI 1 00001 √ 00-00 √ -0-01 PI

4 00100 √ 00-01 √ 001-- PI 5 00101 √ -0001 √ 0-1-1 PI

6 00110 √ 001-0 √ -01-1 PI

17 10001 √ 0010- √ -111 PI

7 00111 √ 001-1 √ 13 01101 √ 0-101 √

21 10101 √ -0101 √

15 01111 √ 0011- √

23 10111 √ 10-01 √

31 11111 √ 0-111 √ -0111 √

011-1 √

101-1 √

-1111 √ 1-111 √

PIs are v'w'y' w'y'z v'w'x v'xz w'xz xyz

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Implicant Table:

PI \ Minterm 0 1 4 6 7 13 15 19 31 Type of Implicant

00-0- X X X EPI: v'w'y'

-0-01 X Redundant

001-- X X X EPI: v'w'x

0-1-1 X X X EPI: v'xz

-01-1 X X Redundant

- -111 X X X X EPI: xyz

(Check) √ √ √ √ √ √ √ √ √ (For minterm usage)

P(v,w,x,y,z) = v'w'y' + v'w'x + v'xz + xyz

27 Book Problem 2-29 ( b )

28 Repeat book problem 2-29 (b) using NOR gate

A'

B

CD'E

BD'

A'

B'

A

B'

C'DE'

B'D

A

B

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29 Book Problem 2-32 using first NAND and the NOR gate (a)

(b)

30 Book Problem 2-35 Truth table: 3-bit Parity Generator and 4-bits Parity Checker

P = X ⊕ Y ⊕ Z , C = X ⊕ Y ⊕ Z ⊕ P

X Y Z P W X Y P C0 0 0 0 0 0 0 0 00 0 1 1 0 0 0 1 10 1 0 1 0 0 1 0 10 1 1 0 0 0 1 1 01 0 0 1 0 1 0 0 11 0 1 0 0 1 0 1 01 1 0 0 0 1 1 0 01 1 1 1 0 1 1 1 1

1 0 0 0 11 0 0 1 01 0 1 0 01 0 1 1 11 1 0 0 01 1 0 1 11 1 1 0 11 1 1 1 0

A

B

C

F

G

A'

B'

C'

F

G

X

Y

ZP

X

Y

Z

P

C