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Pancyprian TS Dig. Elec. II study materials
Pancyprian TS Dig. Elec. II study resources

Pancyprian TS Dig. Elec. II practice questions

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Question 1Number systems, codes and Boolean algebra

Ποια είναι η δεκαδική τιμή του δυαδικού αριθμού 1011₂;

  1. 13
  2. 11 Correct answer.
  3. 15
  4. 9

Explanation

Κάθε ψηφίο πολλαπλασιάζεται με τη δύναμη του 2 που αντιστοιχεί στη θέση του: 1·8 + 0·4 + 1·2 + 1·1. Το άθροισμα δίνει 8 + 2 + 1 = 11.

Question 2Number systems, codes and Boolean algebra

Ποιο είναι το δυαδικό ισοδύναμο του δεκαδικού αριθμού 25;

  1. 11011₂
  2. 10101₂
  3. 10011₂
  4. 11001₂ Correct answer.

Explanation

Με διαδοχικές διαιρέσεις με το 2 τα υπόλοιπα από κάτω προς τα πάνω είναι 1,1,0,0,1. Ελέγχοντας: 16 + 8 + 1 = 25, άρα 11001₂.

Question 3Number systems, codes and Boolean algebra

Πόσες διακριτές τιμές μπορεί να αναπαραστήσει ένα δυαδικό word των 6 bits;

  1. 12
  2. 32
  3. 64 Correct answer.
  4. 128

Explanation

Με n bits αναπαρίστανται 2ⁿ διαφορετικοί συνδυασμοί. Για n = 6 προκύπτει 2⁶ = 64 διακριτές τιμές, από 000000 έως 111111.

Question 4Number systems, codes and Boolean algebra

Ποια είναι η δεκαεξαδική αναπαράσταση του δυαδικού αριθμού 1111 1010₂;

  1. EA
  2. AF
  3. F5
  4. FA Correct answer.

Explanation

Η ομάδα 1111 ισούται με F και η ομάδα 1010 ισούται με A. Συνενώνοντας τις δύο ομάδες προκύπτει FA.

Question 5Number systems, codes and Boolean algebra

Ποια λογική πύλη υλοποιεί τη συνάρτηση Y = A·B, δηλαδή δίνει έξοδο 1 μόνο όταν και οι δύο είσοδοι είναι 1;

  1. NAND
  2. AND Correct answer.
  3. XOR
  4. OR

Explanation

Η πύλη AND παράγει έξοδο 1 μόνο στην περίπτωση A = 1 και B = 1. Σε κάθε άλλο συνδυασμό εισόδων η έξοδος είναι 0.

Question 6Number systems, codes and Boolean algebra

Στο δυαδικό σύστημα, ποια είναι η τιμή του αριθμού 1000₂;

  1. 4
  2. 8 Correct answer.
  3. 16
  4. 10

Explanation

Το μόνο άναμμα είναι στη θέση βάρους 2³, δηλαδή 8. Τα τρία μηδενικά στις θέσεις 2², 2¹ και 2⁰ δεν προσθέτουν τίποτα.

Question 7Number systems, codes and Boolean algebra

Ποιο είναι το αποτέλεσμα του λογικού αθροίσματος (OR) 1010₂ + 0110₂, ψηφίο προς ψηφίο;

  1. 0010₂
  2. 1110₂ Correct answer.
  3. 1100₂
  4. 1011₂

Explanation

Η πράξη OR δίνει 1 όταν τουλάχιστον ένα από τα αντίστοιχα bits είναι 1. Έτσι 1∨0=1, 0∨1=1, 1∨1=1 και 0∨0=0, δηλαδή 1110₂.

Question 8Number systems, codes and Boolean algebra

Ποια είναι η τιμή του δυαδικού αριθμού 1101₂ σε οκταδικό σύστημα;

  1. 17₈
  2. 15₈ Correct answer.
  3. 13₈
  4. 31₈

Explanation

Ομαδοποιούμε τα bits σε τριάδες από τα δεξιά: 001 και 101, που δίνουν 1 και 5 αντίστοιχα. Άρα ο αριθμός γράφεται 15₈, δηλαδή 8 + 5 = 13 σε δεκαδικό.

Question 9Number systems, codes and Boolean algebra

Ποια είναι η συμπληρωματική μορφή ως προς 1 (ones' complement) του δυαδικού αριθμού 0110₂ σε 4 bits;

  1. 1001 Correct answer.
  2. 1010
  3. 0101
  4. 1111

Explanation

Το συμπλήρωμα ως προς 1 προκύπτει αντιστρέφοντας κάθε bit: 0→1 και 1→0. Έτσι το 0110 γίνεται 1001.

Question 10Number systems, codes and Boolean algebra

Ποια είναι η τιμή του αριθμού 1111₂ όταν ερμηνεύεται ως προσημασμένος ακέραιος σε μορφή συμπληρώματος ως προς 2 (two's complement) με 4 bits;

  1. -8
  2. -1 Correct answer.
  3. 15
  4. -7

Explanation

Στο two's complement το πιο σημαντικό bit έχει βάρος -2³ = -8 και τα υπόλοιπα θετικά βάρη 4, 2 και 1. Έτσι -8 + 4 + 2 + 1 = -1.

Question 11Number systems, codes and Boolean algebra

Ποια είναι η απλοποιημένη μορφή της παράστασης Y = A·B + A·B̄;

  1. A Correct answer.
  2. A·B
  3. B
  4. A + B

Explanation

Με παραγοντοποίηση ως προς A προκύπτει A·(B + B̄). Επειδή B + B̄ = 1, η παράσταση ανάγεται στο A·1 = A.

Question 12Number systems, codes and Boolean algebra

Ποια είναι η δυαδική αναπαράσταση του δεκαδικού -6 σε two's complement με 4 bits;

  1. 0110
  2. 1110
  3. 1010 Correct answer.
  4. 1001

Explanation

Γράφουμε πρώτα το +6 ως 0110, αντιστρέφουμε τα bits σε 1001 και προσθέτουμε 1, δίνοντας 1010. Επαλήθευση: -8 + 2 = -6.

Question 13Number systems, codes and Boolean algebra

Ποια είναι η τιμή του BCD κωδικού 1001 0110;

  1. 69
  2. 15
  3. 150
  4. 96 Correct answer.

Explanation

Στον κώδικα BCD κάθε ομάδα 4 bits παριστάνει ένα δεκαδικό ψηφίο: 1001 = 9 και 0110 = 6. Άρα ο κωδικός παριστάνει τον δεκαδικό αριθμό 96.

Question 14Number systems, codes and Boolean algebra

Ποια διάσημη ταυτότητα της Boolean άλγεβρας εκφράζει ο νόμος του De Morgan για δύο μεταβλητές;

  1. A ⊕ B = A·B̄ + Ā·B
  2. (A·B)̄ = Ā + B̄ Correct answer.
  3. A + A·B = A
  4. (A + B)̄ = Ā + B̄

Explanation

Ο νόμος του De Morgan λέει ότι το συμπλήρωμα γινομένου ισούται με το άθροισμα των συμπληρωμάτων. Αντίστοιχα ισχύει και (A + B)̄ = Ā·B̄.

Question 15Number systems, codes and Boolean algebra

Ποια είναι η έξοδος μιας πύλης XOR όταν οι είσοδοι είναι A = 1 και B = 1;

  1. Αβεβαίως, εξαρτάται από την τάση τροφοδοσίας
  2. Ακολουθεί την είσοδο A
  3. 0 Correct answer.
  4. 1

Explanation

Η XOR δίνει έξοδο 1 μόνο όταν οι είσοδοι διαφέρουν. Όταν A = B = 1, οι είσοδοι είναι ίσες, άρα η έξοδος είναι 0.

Question 16Number systems, codes and Boolean algebra

Ποια είναι η ελάχιστη μορφή SOP της συνάρτησης Y = Σm(0,1,2,3) με δύο μεταβλητές A και B;

  1. 1 Correct answer.
  2. A
  3. A + B
  4. Ā

Explanation

Οι ελάχιστοι όροι 0,1,2,3 καλύπτουν και τους τέσσερις συνδυασμούς των A και B. Η συνάρτηση είναι επομένως σταθερά 1 σε όλο τον πίνακα αληθείας.

Question 17Number systems, codes and Boolean algebra

Ποια είναι η δεκαδική τιμή του δεκαεξαδικού αριθμού 2F;

  1. 215
  2. 47 Correct answer.
  3. 31
  4. 45

Explanation

Το δεκαεξαδικό 2F σημαίνει 2·16 + 15, αφού το F αντιστοιχεί στο 15. Το άθροισμα 32 + 15 δίνει 47.

Question 18Number systems, codes and Boolean algebra

Ποια είναι η τιμή του δυαδικού κλάσματος 0.101₂ σε δεκαδικό;

  1. 0.75
  2. 0.625 Correct answer.
  3. 0.5
  4. 0.125

Explanation

Τα κλασματικά βάρη είναι 2⁻¹, 2⁻² και 2⁻³, δηλαδή 0.5, 0.25 και 0.125. Το άθροισμα 0.5 + 0.125 δίνει 0.625.

Question 19Number systems, codes and Boolean algebra

Ποια είναι η τιμή του αριθμού 1110₂ όταν ερμηνεύεται ως προσημασμένος ακέραιος σε two's complement με 4 bits;

  1. -6
  2. -1
  3. -2 Correct answer.
  4. 14

Explanation

Το πιο σημαντικό bit δίνει -8 και τα υπόλοιπα προσθέτουν 4 + 2 = 6. Το άθροισμα -8 + 6 ισούται με -2.

Question 20Number systems, codes and Boolean algebra

Ποια είναι η έξοδος της παράστασης Y = A + Ā·B;

  1. A·B
  2. Ā + B
  3. A + B Correct answer.
  4. B

Explanation

Αναπτύσσοντας με επιμεριστικό νόμο: A + Ā·B = (A + Ā)(A + B). Επειδή A + Ā = 1, μένει A + B.

Question 21Number systems, codes and Boolean algebra

Ποιος είναι ο αριθμός των bits που απαιτούνται για την αναπαράσταση των 100 διακριτών καταστάσεων σε δυαδικό κώδικα;

  1. 6
  2. 100
  3. 7 Correct answer.
  4. 8

Explanation

Χρειάζεται ο μικρότερος n με 2ⁿ ≥ 100. Επειδή 2⁶ = 64 < 100 και 2⁷ = 128 ≥ 100, αρκούν 7 bits.

Question 22Number systems, codes and Boolean algebra

Ποια είναι η έξοδος της λογικής πύλης NAND όταν οι είσοδοι είναι A = 1 και B = 0;

  1. 1 Correct answer.
  2. Απροσδιόριστη
  3. Ακολουθεί το B
  4. 0

Explanation

Η AND με εισόδους 1 και 0 δίνει 0, και η NAND αντιστρέφει το αποτέλεσμα. Άρα η έξοδος είναι 1.

Question 23Number systems, codes and Boolean algebra

Ποια είναι η ελάχιστη μορφή SOP της συνάρτησης Y = Σm(0,2,4,6) με τρεις μεταβλητές A, B και C;

  1. A
  2. A + C̄
  3. B̄·C̄
  4. C̄ Correct answer.

Explanation

Οι ελάχιστοι όροι 0,2,4,6 έχουν όλοι C = 0 και διαφέρουν μόνο ως προς A και B. Η κοινή μεταβλητή είναι το C̄, άρα Y = C̄.

Question 24Number systems, codes and Boolean algebra

Ποια είναι η τιμή του 8-bit two's complement αριθμού 1101 0011₂;

  1. -45 Correct answer.
  2. 45
  3. -211
  4. -35

Explanation

Το πιο σημαντικό bit δίνει -128 και τα υπόλοιπα προσθέτουν 64 + 16 + 2 + 1 = 83. Το άθροισμα -128 + 83 ισούται με -45.

Question 25Combinational logic design and minimisation

A Boolean function is expressed as F = A·B + A·B̄. Which single literal result does this expression simplify to?

  1. A Correct answer.
  2. B
  3. A·B
  4. A + B

Explanation

Factor out the common literal A to obtain A·(B + B̄). Since B + B̄ = 1 by the complement law, the expression reduces to A·1, which equals A. This is a direct application of the distributive and complement identities.

Question 26Combinational logic design and minimisation

How many rows does the truth table of a Boolean function with four input variables have?

  1. 8
  2. 32
  3. 16 Correct answer.
  4. 4

Explanation

A truth table has one row for every possible combination of input values. With n inputs the number of combinations is 2^n, so four inputs give 2^4 = 16 rows. Each row corresponds to one minterm of the function.

Question 27Combinational logic design and minimisation

Which pair of gates forms a functionally complete set on its own?

  1. NAND and AND
  2. AND and OR
  3. NAND alone Correct answer.
  4. XOR and AND

Explanation

A set is functionally complete if every Boolean function can be built from it. NAND by itself is complete because NOT, AND and OR can all be constructed from NAND gates only. AND with OR cannot produce inversion, and AND with XOR cannot produce a constant 1.

Question 28Combinational logic design and minimisation

In a Karnaugh map, two adjacent cells that are both 1 can be grouped to eliminate how many variables from the corresponding product term?

  1. Two variables
  2. Three variables
  3. No variables
  4. One variable Correct answer.

Explanation

A group of 2^k adjacent cells removes k variables from the product term. A pair of adjacent cells therefore eliminates exactly one variable, the one that changes between the two cells. Larger groups of 4, 8 and so on remove two, three and more variables.

Question 29Combinational logic design and minimisation

What is the canonical sum-of-products form of the function F(A,B) = A XOR B?

  1. Ā·B + A·B̄ Correct answer.
  2. A·B + A·B̄
  3. A + B
  4. Ā·B̄ + A·B

Explanation

XOR is 1 when exactly one input is 1, that is, for minterms A=0,B=1 and A=1,B=0. The canonical sum of these minterms is Ā·B + A·B̄. Each product term contains every variable exactly once.

Question 30Combinational logic design and minimisation

Which logic gate produces a 1 output only when all of its inputs are 1?

  1. OR gate
  2. NOR gate
  3. AND gate Correct answer.
  4. XOR gate

Explanation

The AND function is defined as the logical product of its inputs, so its output is 1 only when every input is 1. Any single 0 input forces the AND output to 0. OR, XOR and NOR all give a 1 for at least one input combination containing a 0.

Question 31Combinational logic design and minimisation

A 2-input multiplexer has data inputs I0 = 1, I1 = 0 and select line S = 0. What value appears at the output Y?

  1. S
  2. 1 Correct answer.
  3. 0
  4. I1

Explanation

A multiplexer routes the selected data input to the output. When the select line is 0 the input I0 is chosen, and here I0 = 1, so Y = 1. The value of I1 is irrelevant while S = 0.

Question 32Combinational logic design and minimisation

What output does a half adder produce for the input combination A = 1, B = 1?

  1. Sum = 0, Carry = 1 Correct answer.
  2. Sum = 1, Carry = 0
  3. Sum = 1, Carry = 1
  4. Sum = 0, Carry = 0

Explanation

A half adder computes Sum = A XOR B and Carry = A AND B. With both inputs 1, XOR gives 0 and AND gives 1, so the result is Sum = 0, Carry = 1. This matches the binary addition 1 + 1 = 10.

Question 33Combinational logic design and minimisation

A Karnaugh map for a four-variable function uses Gray code ordering on its axes. Why is this ordering required?

  1. It allows decimal labels to be used instead
  2. It makes the map symmetrical about its centre
  3. It ensures adjacent cells differ in exactly one variable Correct answer.
  4. It reduces the number of cells in the map

Explanation

Grouping works by combining cells whose input values differ in only one variable, because that variable can then be eliminated. Gray code ordering guarantees that horizontally or vertically adjacent cells differ in exactly one bit. Without this property, visual adjacency would not correspond to algebraic adjacency.

Question 34Combinational logic design and minimisation

A function of three variables has 1s in minterms 0, 1, 2, 3, 4, 5, 6 and 7. What is its minimised expression?

  1. Ā
  2. A·B·C
  3. A + B + C
  4. 1 Correct answer.

Explanation

When every minterm of a function is present, the function is 1 for all input combinations. All cells of the map can be grouped together, eliminating every variable and leaving the constant 1. The complement of such a function is identically 0.

Question 35Combinational logic design and minimisation

Which expression is the correct De Morgan transformation of the function F = (A + B)·C?

  1. (Ā·B̄) + C̄
  2. Ā·B̄·C̄
  3. Ā·B̄ + C̄ Correct answer.
  4. Ā + B̄ + C̄

Explanation

De Morgan's theorem states that the complement of a product equals the sum of the complements, and the complement of a sum equals the product of the complements. Applying it to (A + B)·C gives Ā·B̄ for the first factor and C̄ for the second, joined by OR.

Question 36Combinational logic design and minimisation

A four-variable Karnaugh map has 1s in the four corner cells. Which product term represents this group?

  1. B·D
  2. B̄·D̄ Correct answer.
  3. Ā·C̄
  4. A·C

Explanation

The four corners of a four-variable map are adjacent through wrap-around and form a valid group of four. In the corner cells the variables B and D are always 0 while A and C take both values, so B and D are eliminated and B̄·D̄ remains. Wrap-around grouping is a standard feature of the map.

Question 37Combinational logic design and minimisation

What is the purpose of a don't-care condition in combinational logic minimisation?

  1. It doubles the number of minterms in the function
  2. It indicates a faulty gate in the circuit
  3. It forces the output to 0 for that input combination
  4. It marks an input combination that never occurs, allowing either output value to be chosen Correct answer.

Explanation

A don't-care condition corresponds to an input combination that cannot occur or whose output is irrelevant. The designer may treat it as 0 or 1, whichever yields a simpler expression. This flexibility often allows larger groups to be formed on the Karnaugh map.

Question 38Combinational logic design and minimisation

A combinational circuit has three inputs and one output. Its output is 1 only when an odd number of inputs are 1. Which function does this describe?

  1. Three-input AND
  2. Three-input NAND
  3. Three-input XOR Correct answer.
  4. Three-input OR

Explanation

The XOR function detects odd parity, producing 1 when an odd number of its inputs are 1. For three inputs this covers the minterms with one or three 1s. OR gives 1 for any 1 input, and AND only when all inputs are 1.

Question 39Combinational logic design and minimisation

Which minimised expression is equivalent to F = A·B + A·B̄ + Ā·B?

  1. A·B
  2. A + B Correct answer.
  3. Ā + B̄
  4. A XOR B

Explanation

The first two terms combine to A, and the third term Ā·B combines with A·B to give B. The result is A + B, which covers all minterms where at least one input is 1. This can also be seen directly by grouping the three 1-cells on the map.

Question 40Combinational logic design and minimisation

How many 2-input NAND gates are needed to implement the function F = A·B using only NAND gates?

  1. One Correct answer.
  2. Three
  3. Four
  4. Two

Explanation

A single 2-input NAND gate already computes the complement of A·B, but the required function is A·B. One NAND gate followed by an inverter built from a NAND gate with tied inputs gives the product, so two gates are needed in total.

Question 41Combinational logic design and minimisation

A priority encoder has inputs I3, I2, I1, I0 with I3 highest priority. If I3 = 0, I2 = 1, I1 = 1 and I0 = 0, what output code is produced?

  1. 10 Correct answer.
  2. 00
  3. 11
  4. 01

Explanation

A priority encoder outputs the code of the highest-priority input that is asserted. Here I3 is 0, so the highest active input is I2, whose binary code is 10. Lower-priority inputs such as I1 do not affect the result.

Question 42Combinational logic design and minimisation

Which statement correctly describes a hazard in a combinational logic circuit?

  1. A hazard is a temporary false output caused by unequal propagation delays along different paths Correct answer.
  2. A hazard is the result of using too many don't-care terms
  3. A hazard occurs only in sequential circuits with feedback
  4. A hazard is a permanent stuck-at fault in a gate

Explanation

A static hazard produces a brief glitch when a single input changes and two paths with different delays reconverge at a gate. The output momentarily takes the wrong value before settling. Adding a redundant consensus term can eliminate the hazard.

Question 43Combinational logic design and minimisation

A four-variable function is given by F = Σm(0, 2, 8, 10). Which product term is the minimal result?

  1. Ā·C̄
  2. C̄·D̄
  3. B̄·D̄ Correct answer.
  4. Ā·B̄

Explanation

The minterms 0, 2, 8 and 10 all have B = 0 and D = 0, while A and C vary. They form a group of four corners that eliminates A and C, leaving B̄·D̄. This is the smallest single product term covering all four minterms.

Question 44Combinational logic design and minimisation

Which of these functions is not in a minimal sum-of-products form?

  1. A·B + A·C + B·C Correct answer.
  2. A·B + Ā·C
  3. A·B + Ā·B̄
  4. A + B·C

Explanation

In A·B + A·C + B·C the term B·C is the consensus of A·B and A·C and is redundant, so the expression is not minimal. Removing it leaves A·B + A·C, which covers the same minterms. The other options contain no redundant terms.

Question 45Combinational logic design and minimisation

A combinational circuit must output the majority value of three inputs A, B and C. Which expression implements this function correctly?

  1. A XOR B XOR C
  2. A·B·C
  3. A·B + A·C + B·C Correct answer.
  4. A + B + C

Explanation

The majority function is 1 when at least two inputs are 1. Each product term A·B, A·C and B·C represents a pair being 1, and their OR covers all cases with two or three 1s. This is the standard minimal majority expression.

Question 46Combinational logic design and minimisation

A Boolean function of four variables has exactly five minterms. What is the minimum number of product terms in a sum-of-products expression that could cover it, assuming no two minterms can be grouped?

  1. Two
  2. Five Correct answer.
  3. Three
  4. One

Explanation

If no two minterms are adjacent on the map, no grouping is possible and each minterm must be covered by its own product term. Five isolated minterms therefore require five product terms. Grouping can only reduce the count when adjacent cells exist.

Question 47Combinational logic design and minimisation

A function F(A,B,C,D) has don't-care conditions at minterms 3 and 7 and required 1s at minterms 2 and 6. If the don't-cares are treated as 1, which product term results after grouping?

  1. C·D
  2. A·C·D̄
  3. B̄·C Correct answer.
  4. Ā·B·C

Explanation

Minterms 2, 3, 6 and 7 all have B = 0 and C = 1, while A and D vary. Treating the don't-cares 3 and 7 as 1 allows a group of four that eliminates A and D, leaving B̄·C. Using don't-cares this way produces a simpler expression than covering only the required 1s.

Question 48Combinational logic design and minimisation

A static-1 hazard occurs in the circuit F = A·B + Ā·C when B = C = 1 and A changes. Which term should be added to eliminate the hazard?

  1. A·B̄
  2. Ā·B
  3. A·C
  4. B·C Correct answer.

Explanation

A static-1 hazard arises when two product terms are adjacent but not covered by a common term. The consensus of A·B and Ā·C is B·C, which holds the output at 1 while A changes. Adding this redundant term removes the glitch without changing the function.

Question 49Combinational logic design and minimisation

Which of the following functions cannot be reduced below three product terms in sum-of-products form, regardless of grouping?

  1. F = Σm(0, 1, 2, 3) for a four-variable function
  2. F = Σm(0, 1, 2, 3, 4, 5, 6, 7)
  3. F = Σm(0, 5, 10, 15) for a four-variable function Correct answer.
  4. F = A·B + A·B̄

Explanation

For Σm(0, 5, 10, 15) the minterms lie on a diagonal pattern with no two adjacent on the map, so each must be covered separately. That requires four product terms, which is more than three. The other functions contain large adjacent groups that reduce to one or two terms.

Question 50Combinational logic design and minimisation

A 2-input XOR gate has inputs A and B. Which expression correctly represents its output?

  1. A + B
  2. A ⊕ B = A·B̄ + Ā·B Correct answer.
  3. A·B
  4. A·B + Ā·B̄

Explanation

The XOR function is true when exactly one input is true, which is expressed as A·B̄ + Ā·B. The other options describe XNOR, OR, and AND respectively.