Thirteen questions on the mathematics behind digital control — Boolean proofs and
simplification, Karnaugh-map minimisation into SOP and POS form, and conversions across
binary, BCD, octal, hexadecimal and Gray code. Each question opens to a full worked solution.
Work each problem fully — build the truth table, draw the Karnaugh map, show every conversion step — then open the worked solution to check your method.
Q1
Proving a four-variable Boolean identity
Using the laws of Boolean algebra (with A′ denoting the complement of A), prove the
identity: AD′ + A′B + C′D + B′C = (A′+B′+C′+D′)(A+B+C+D)
ShowHide worked solution
Truth-table proof
The clearest beginner-friendly proof evaluates both sides for all 16 input combinations and shows they always agree.
Both the LHS and the RHS are 0 only when A = B = C = D; they are 1 for every other row.
#
A
B
C
D
AD′
A′B
C′D
B′C
LHS
RHS
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
1
0
0
1
0
1
1
2
0
0
1
0
0
0
0
1
1
1
3
0
0
1
1
0
0
0
1
1
1
4
0
1
0
0
0
1
0
0
1
1
5
0
1
0
1
0
1
1
0
1
1
6
0
1
1
0
0
1
0
0
1
1
7
0
1
1
1
0
1
0
0
1
1
8
1
0
0
0
1
0
0
0
1
1
9
1
0
0
1
0
0
1
0
1
1
10
1
0
1
0
1
0
0
1
1
1
11
1
0
1
1
0
0
0
1
1
1
12
1
1
0
0
1
0
0
0
1
1
13
1
1
0
1
0
0
1
0
1
1
14
1
1
1
0
1
0
0
0
1
1
15
1
1
1
1
0
0
0
0
0
0
Why it works
By De Morgan’s law, (A′+B′+C′+D′) = (ABCD)′. So the RHS is 1 whenever at least one variable is 0 and at least one is 1 — that is, when the four variables are not all equal. Each LHS term detects a 0→1 transition around the cycle A→B→C→D→A; if the variables are not all equal there must be at least one such transition, so the LHS is also 1.
Final AnswerThe identity is true. Both sides equal 1 unless A = B = C = D (all-zeros or all-ones), where both equal 0.
Q2
Minimised SOP and POS with don’t-care terms
For F(A,C,E,G) = Σ(m0,m4,m5,m6,m11,m14,m15) with don’t-cares
D3, D7, D8, D13: (a) derive the minimised SOP expression, and (b) derive the
minimised POS expression.
ShowHide worked solution
With A the MSB and G the LSB, each minterm number is n = 8A + 4C + 2E + G. Plotting the 1s and don’t-cares (D) gives the map below.
K-map of F(A,C,E,G) — rows AC, columns EG.
AC \ EG
00
01
11
10
00
1
0
D
0
01
1
1
D
1
11
0
D
1
1
10
D
0
1
0
(a) Minimum SOP
Grouping the 1s, using the don’t-cares where they help:
A′C — the whole AC = 01 row (m4, m5, m6, m7).
CE — the 2×2 block covering m6, m7, m14, m15.
EG — the EG = 11 column, using don’t-cares D3 and D7.
A′E′G′ — the pair needed to cover m0.
FSOP = A′C + CE + EG + A′E′G′
(b) Minimum POS
Grouping the 0s, again using the don’t-cares:
FPOS = (A′ + E)(A + C + G′)(C + E′ + G)
Final AnswerMinimum SOP: F = A′C + CE + EG + A′E′G′ (4 terms). Minimum POS: F = (A′+E)(A+C+G′)(C+E′+G) (3 sum-terms — slightly cheaper).
Q3
Number-system conversions
Perform the following conversions: (a) (0011 0111 1000.0111 0101)BCD to decimal;
(b) octal (7174.54)8 to hexadecimal; (c) (1BB)16 to decimal;
(d) (1100)2 to octal; (e) (2000)8 to hexadecimal;
(f) (F3C7.A)16 to octal.
ShowHide worked solution
(a) BCD → Decimal
Each 4-bit group is one decimal digit: 0011=3, 0111=7, 1000=8, ·0111=7, 0101=5 → 378.75
Key idea: octal–hex conversions are easiest through binary, since 1 octal digit = 3 bits and 1 hex digit = 4 bits. Re-group the integer part from the right and the fractional part from the left.
Key idea: BCD is not pure binary — each decimal digit is encoded separately, so the same value takes more bits in BCD than in straight binary.
Final Answer(0001 0100)BCD = (1110)2 = (E)16.
Q7
A full Gray-code conversion chain
Complete the chain of conversions:
(1001)gray = (?)2 = (?)BCD = (?)10 = (?)16.
ShowHide worked solution
Gray→binary rule: the MSB is copied unchanged, then each next binary bit is the XOR of the previous binary bit with the current Gray bit — Bₖ = Bₖ₊₁ ⊕ Gₖ.
The carry beyond bit 7 is discarded because the word is only 8 bits wide. No overflow occurs, because the operands have opposite signs — overflow is only possible when adding two numbers of the same sign.
Final Answer(−1) + (+1) = 0000 00002 = 010.
Q9
Simplifying two Boolean expressions
Simplify the following, stating clearly which rules you apply: (a) C + B·C;
(b) A·B + A(B+C) + B(B+C).
ShowHide worked solution
(a) C + B·C
C + B·C
= C·(1 + B) [ distributive ]
= C·1 [ annulment: 1 + X = 1 ]
= C [ identity ]
This is the absorption law: C + B·C = C.
(b) A·B + A(B+C) + B(B+C)
Expand the brackets:
= AB + AB + AC + BB + BC
Apply idempotence (X+X=X, X·X=X):
= AB + AC + B + BC
Apply absorption (B + AB = B, B + BC = B):
= B + AC
Final Answer(a) C + B·C = C. (b) A·B + A(B+C) + B(B+C) = B + A·C.
Q10
Karnaugh-map minimisation of a four-variable function
Apply a Karnaugh map to minimise F(A,B,C,D) = Σm(3,4,6,7,9,12,13,14,15).
ShowHide worked solution
K-map of F(A,B,C,D) — rows AB, columns CD.
AB \ CD
00
01
11
10
00
0
0
1
0
01
1
0
1
1
11
1
1
1
1
10
0
1
0
0
Groups
A·B — the full AB = 11 row (m12, m13, m14, m15).
B·D′ — the 2×2 wrap-around block over m4, m6, m12, m14.
A′·C·D — the pair m3, m7 (essential for m3).
A·C′·D — the pair m9, m13 (essential for m9).
F = A·B + B·D’ + A’·C·D + A·C’·D
Final AnswerMinimum SOP: F = A·B + B·D′ + A′·C·D + A·C′·D (4 terms, 10 literals).
Q11
SOP and POS from a maxterm list with don’t-cares
For F(A,B,C,D) = ΠM(1,3,4,6,9,11) with don’t-care set
D(A,B,C,D) = Σm(0,2,5,10,12,14), obtain the minimised SOP and the
minimised POS expressions.
ShowHide worked solution
F is 0 at the six maxterms (1, 3, 4, 6, 9, 11) and don’t-care at (0, 2, 5, 10, 12, 14). The remaining cells — m7, m8, m13, m15 — are the 1s.
K-map of F — rows AB, columns CD. D marks a don’t-care.
AB \ CD
00
01
11
10
00
D
0
0
D
01
0
D
1
0
11
D
1
1
D
10
1
0
0
D
Minimum SOP — group the 1s and useful Ds
FSOP = B·D + A·D’ (2 terms, 4 literals)
Minimum POS — group the 0s and useful Ds
FPOS = (B + D’)(B’ + D) (2 terms, 4 literals)
Final AnswerMinimum SOP: F = B·D + A·D′. Minimum POS: F = (B+D′)(B′+D). The don’t-cares let both forms collapse to just two terms.
Q12
Canonical SOP and POS forms of a function
Take a Boolean equation of your choice and express it first in maxterm (POS) form and then
in minterm (SOP) form. Then show how a partial expression is converted into standard SOP
and standard POS form.
ShowHide worked solution
Take the three-variable function F(A,B,C) = AB + C and build its truth table.
F = 1 for minterms 1, 3, 5, 6, 7; F = 0 for maxterms 0, 2, 4.
A partial expression is brought to standard form by re-introducing every missing variable: for SOP, multiply each term by (X + X′) for the missing variable X; for POS, add (X·X′) to each sum-term. Expanding the brackets and removing duplicates with idempotence then gives the full canonical expression.
Final AnswerFor F = AB + C: SOP = Σm(1,3,5,6,7) and POS = ΠM(0,2,4). Standard form is reached by expanding each missing variable as (X + X′) for SOP or (X·X′) for POS, then applying idempotence.
Q13
From truth table to efficient ladder logic
Given the truth table of a Boolean function, derive the most efficient ladder-logic
implementation. Use a structured method — Boolean algebra or a Karnaugh map —
to obtain the simplified expression first.
ShowHide worked solution
As a representative example, take the three-input XOR (odd-parity) function and walk through the full procedure: table → map → equation → ladder.
Three-input XOR — output is 1 when an odd number of inputs are 1.
A
B
C
F = A⊕B⊕C
0
0
0
0
0
0
1
1
0
1
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1
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1
1
0
1
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1
1
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1
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1
1
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0
1
1
1
1
K-map — the 1s are isolated, so no grouping is possible: XOR cannot be reduced.
A \ BC
00
01
11
10
0
0
1
0
1
1
1
0
1
0
Because no cells are adjacent, the minimum form keeps all four product terms:
F = A’B’C + A’BC’ + AB’C’ + ABC
Each product term becomes one rung: series contacts for the AND, the four rungs joined in parallel for the OR, all feeding a single coil F.
Four parallel branches — one per minterm — feed a single coil F. A slashed contact is normally-closed, representing the complemented input.
Final AnswerThe procedure is: mark the F = 1 rows, plot a K-map and group, write the minimised SOP, then turn each product term into a rung — series contacts for AND, parallel branches for OR — with one output coil. For three-input XOR this gives four rungs into coil F.
