Paper-Based Practice: Answer these questions on paper, then use the "Show Mark Scheme" button to check your work.
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Question 1
[4 marks]
This Karnaugh map represents a logic circuit.
Use this Karnaugh map to find the simplified expression for this circuit.
You should annotate the Karnaugh map to show the groups that you have used to find the simplified expression.
Use this Karnaugh map to find the simplified expression for this circuit.
You should annotate the Karnaugh map to show the groups that you have used to find the simplified expression.
Question 2
[4 marks]
Eve has designed a logic circuit. The expression she has created for the logic circuit is:
Complete the Karnaugh map for this expression and state the simplified expression.
You should draw groups around the cells you are using to simply this expression.
Show your working.
Q = (¬C ∧ D) ∨ (¬C ∧ ¬D) ∨ (A ∧ C ∧ ¬B)Complete the Karnaugh map for this expression and state the simplified expression.
You should draw groups around the cells you are using to simply this expression.
Show your working.
Question 3
[4 marks]
Elliott has designed a logic circuit. The expression he has created for the logic circuit is:
Complete the Karnaugh Map below to simplify this expression. Show your working.
Q = (A ∧ ¬B) ∨ (¬A ∧ C ∧ D) ∨ (A ∧ B)Complete the Karnaugh Map below to simplify this expression. Show your working.
Question 4
[5 marks]
Complete the Karnaugh map below for the Boolean expression (¬ A ∧ ¬ B) ∨ ( A ∧ ¬ B)
Use the Karnaugh map to find a simplified Boolean expression that is equivalent to (¬ A ∧ ¬ B) ∨ ( A ∧ ¬ B)
Use the Karnaugh map to find a simplified Boolean expression that is equivalent to (¬ A ∧ ¬ B) ∨ ( A ∧ ¬ B)
Question 5
[5 marks]
Find the Boolean expression represented in the Karnaugh Map below. Show your working.
Question 6
[3 marks]
Draw a Logic diagram for the following expression:
Q = ¬(A ∧ B) ∨ (C ∧¬ D)
Question 7
[3 marks]
Draw the logic gate diagram for this expression:
Q = (A v ¬B) ∧ C
Question 8
[4 marks]
Using the rules for manipulating Boolean expressions simplify the following:
A ∧B ∨ A∧(B∨C) ∨ B∧(B∨C)
A ∧B ∨ A∧(B∨C) ∨ B∧(B∨C)
Question 9
[2 marks]
State the purpose of a D-type flip-flop circuit.
Purpose of D-type flip-flop (1 mark each, max 2):
• A D-type flip-flop stores a single bit of data / acts as a 1-bit memory (1 mark)
• It captures the value of the D (data) input at the moment of a clock pulse / rising or falling edge and holds that value at the output Q until the next clock pulse (1 mark)
• A D-type flip-flop stores a single bit of data / acts as a 1-bit memory (1 mark)
• It captures the value of the D (data) input at the moment of a clock pulse / rising or falling edge and holds that value at the output Q until the next clock pulse (1 mark)
Question 10
[3 marks]
All users of a computer system have a unique username and password. The computer system has implemented two-factor authentication so that users must respond to either an email or text message containing a secret code to be able to access the system.
Let:
• A be a Boolean value for if a user enters a valid username
• B be a Boolean value for if a user enters a password that matches their username
• C be a Boolean value for if a user is able to respond to an email containing a secret code
• D be a Boolean value for if a user is able to respond to a text message containing a secret code
• Q be a Boolean value for if entry to the computer system is allowed
Complete the Boolean expression below:
Q ≡ .................................................................................
Let:
• A be a Boolean value for if a user enters a valid username
• B be a Boolean value for if a user enters a password that matches their username
• C be a Boolean value for if a user is able to respond to an email containing a secret code
• D be a Boolean value for if a user is able to respond to a text message containing a secret code
• Q be a Boolean value for if entry to the computer system is allowed
Complete the Boolean expression below:
Q ≡ .................................................................................
Answer: Q ≡ A ∧ B ∧ (C ∨ D)
Question 11
[3 marks]
A Boolean expression for a logic system is shown below:
Q ≡ ¬ (¬ A ∧ ¬ B)
Simplify this Boolean expression so that it does not include any negation.
You must explain which Boolean algebra rule(s) you are using at each step.
Q ≡ ¬ (¬ A ∧ ¬ B)
Simplify this Boolean expression so that it does not include any negation.
You must explain which Boolean algebra rule(s) you are using at each step.
Simplify ¬(¬A ∧ ¬B) using De Morgan's Law (1 mark each, max 3):
• Apply De Morgan's Law: ¬(¬A ∧ ¬B) ≡ ¬(¬A) ∨ ¬(¬B) (1 mark - correct application of De Morgan's)
• Apply double negation: ¬(¬A) = A and ¬(¬B) = B (1 mark)
• Simplified expression: A ∨ B (1 mark)
• Apply De Morgan's Law: ¬(¬A ∧ ¬B) ≡ ¬(¬A) ∨ ¬(¬B) (1 mark - correct application of De Morgan's)
• Apply double negation: ¬(¬A) = A and ¬(¬B) = B (1 mark)
• Simplified expression: A ∨ B (1 mark)
Question 12
[2 marks]
Use the Karnaugh map to find a simplified Boolean expression that is equivalent to (¬ A ∧ ¬ B) ∨ ( A ∧ ¬ B)
You should highlight the map as appropriate and write the expression.
You should highlight the map as appropriate and write the expression.
Simplified expression from Karnaugh map (1 mark each, max 2):
• Group the two 1s in the B=0 column of the Karnaugh map (1 mark)
• Simplified expression: ¬B (1 mark)
The two terms (¬A ∧ ¬B) ∨ (A ∧ ¬B) both require B=0, and A can be either value, so the simplified result is simply ¬B.
• Group the two 1s in the B=0 column of the Karnaugh map (1 mark)
• Simplified expression: ¬B (1 mark)
The two terms (¬A ∧ ¬B) ∨ (A ∧ ¬B) both require B=0, and A can be either value, so the simplified result is simply ¬B.
Question 13
[3 marks]
Describe the inputs and outputs used by a D-type flip-flop circuit, explaining how the inputs are used to control the outputs.
D-type flip-flop inputs and outputs (1 mark each, max 3):
• D (Data) input - the binary value (0 or 1) that will be stored (1 mark)
• Clock input - a signal that triggers the flip-flop to capture the value on the D input (1 mark)
• Q output shows the stored value; when the clock signal transitions (rising/falling edge), the value on D is captured and appears on Q until the next clock transition (1 mark)
• Q̄ (Q-bar) output shows the complement / opposite of Q
• D (Data) input - the binary value (0 or 1) that will be stored (1 mark)
• Clock input - a signal that triggers the flip-flop to capture the value on the D input (1 mark)
• Q output shows the stored value; when the clock signal transitions (rising/falling edge), the value on D is captured and appears on Q until the next clock transition (1 mark)
• Q̄ (Q-bar) output shows the complement / opposite of Q
Question 14
[4 marks]
The following Karnaugh map represents another logic circuit.
Use this Karnaugh map to find the simplified expression for this circuit.
You should highlight the map as appropriate and write the expression.
Use this Karnaugh map to find the simplified expression for this circuit.
You should highlight the map as appropriate and write the expression.
Question 15
[1 mark]
Simplify the Boolean expression ¬A ∨ ¬B using De Morgan's First Law.
De Morgan's First Law simplification (1 mark):
• ¬A ∨ ¬B ≡ ¬(A ∧ B) (1 mark)
De Morgan's First Law states that the complement of a conjunction is the disjunction of the complements: ¬(A ∧ B) ≡ ¬A ∨ ¬B
• ¬A ∨ ¬B ≡ ¬(A ∧ B) (1 mark)
De Morgan's First Law states that the complement of a conjunction is the disjunction of the complements: ¬(A ∧ B) ≡ ¬A ∨ ¬B
Question 16
[1 mark]
Simplify the Boolean expression ¬(¬B) using double negation.
Double negation simplification (1 mark):
• ¬(¬B) ≡ B (1 mark)
Double negation eliminates two NOTs, returning to the original value.
• ¬(¬B) ≡ B (1 mark)
Double negation eliminates two NOTs, returning to the original value.
Question 17
[2 marks]
Simplify the Boolean expression (A ∨ B) ∧ (A ∨ C) using distribution.
Distribution simplification (1 mark each, max 2):
• (A ∨ B) ∧ (A ∨ C) (1 mark for identifying the common term A)
• Simplified: A ∨ (B ∧ C) (1 mark for correct application of distribution law)
Distribution law: (A ∨ B) ∧ (A ∨ C) ≡ A ∨ (B ∧ C)
• (A ∨ B) ∧ (A ∨ C) (1 mark for identifying the common term A)
• Simplified: A ∨ (B ∧ C) (1 mark for correct application of distribution law)
Distribution law: (A ∨ B) ∧ (A ∨ C) ≡ A ∨ (B ∧ C)
Question 18
[2 marks]
A computer scientist has created a logic circuit (Fig. 6) with inputs A, B, and C.
The circuit applies a NOT gate to input A, then ANDs the result with B. The output of this AND gate is then ORed with input C to produce the final output P.
Give the Boolean expression that represents the logic circuit.
Do not attempt to simplify the expression.
The circuit applies a NOT gate to input A, then ANDs the result with B. The output of this AND gate is then ORed with input C to produce the final output P.
Give the Boolean expression that represents the logic circuit.
Do not attempt to simplify the expression.
P = (NOT A AND B) OR C (1 mark for the NOT A AND B part, 1 mark for ORing with C)
Acceptable notation:
• P = (¬A ∧ B) ∨ C
• P = (A̅ . B) + C
• P = (NOT A AND B) OR C
Both parts must be correct for full marks. Accept any standard Boolean notation.
Acceptable notation:
• P = (¬A ∧ B) ∨ C
• P = (A̅ . B) + C
• P = (NOT A AND B) OR C
Both parts must be correct for full marks. Accept any standard Boolean notation.
Question 19
[3 marks]
A Boolean expression for a logic system is shown below:
Simplify this Boolean expression so that it does not include any negation.
You must explain which Boolean algebra rule(s) you are using at each step.
Q = ¬(¬A ∧ ¬B)Simplify this Boolean expression so that it does not include any negation.
You must explain which Boolean algebra rule(s) you are using at each step.
Step 1: Apply De Morgan's Law: ¬(¬A ∧ ¬B) = ¬(¬A) ∨ ¬(¬B)
De Morgan's Law states that NOT (X AND Y) = (NOT X) OR (NOT Y) (1 mark)
Step 2: Apply double negation: ¬(¬A) = A and ¬(¬B) = B
Double negation states that NOT NOT X = X (1 mark)
Step 3: Final simplified expression: Q = A ∨ B (1 mark)
Must show the rule used at each step for full marks.
De Morgan's Law states that NOT (X AND Y) = (NOT X) OR (NOT Y) (1 mark)
Step 2: Apply double negation: ¬(¬A) = A and ¬(¬B) = B
Double negation states that NOT NOT X = X (1 mark)
Step 3: Final simplified expression: Q = A ∨ B (1 mark)
Must show the rule used at each step for full marks.
Question 20
[8 marks]
A logic circuit has two inputs (A, B) and two outputs (S, C). The circuit is a half adder.
a) The logic circuit above has two inputs (A, B) and two outputs (S, C).
Give the Boolean expressions for the outputs S and C.
[2]
Give the Boolean expressions for the outputs S and C.
S = A XOR B / S = A ⊕ B (1 mark)
C = A AND B / C = A ∧ B (1 mark)Accept any standard Boolean notation for XOR and AND.
b) Complete the truth table for this logic circuit.
[2]
| A | B | S | C |
|---|---|---|---|
| 0 | 0 | ? | ? |
| 0 | 1 | ? | ? |
| 1 | 0 | ? | ? |
| 1 | 1 | ? | ? |
A=0, B=0: S=0, C=0
A=0, B=1: S=1, C=0
A=1, B=0: S=1, C=0
A=1, B=1: S=0, C=11 mark for all S values correct
1 mark for all C values correct
c) Describe how this logic circuit can be adapted to add together two 4-bit binary numbers.
[4]
• A half adder can only add two single bits without a carry input (1 mark)
• To add multi-bit numbers, a full adder is needed which has an additional carry input (Cin) (1 mark)
• A full adder can be made from two half adders and an OR gate (1 mark)
• Four full adders would be chained together / connected in series, with the carry output (Cout) of each full adder connected to the carry input (Cin) of the next full adder (1 mark)
• The first adder (least significant bit) can be a half adder as there is no carry in, with the remaining three being full adders / OR all four can be full adders with the first carry input set to 0 (1 mark)
Award up to 4 marks
• To add multi-bit numbers, a full adder is needed which has an additional carry input (Cin) (1 mark)
• A full adder can be made from two half adders and an OR gate (1 mark)
• Four full adders would be chained together / connected in series, with the carry output (Cout) of each full adder connected to the carry input (Cin) of the next full adder (1 mark)
• The first adder (least significant bit) can be a half adder as there is no carry in, with the remaining three being full adders / OR all four can be full adders with the first carry input set to 0 (1 mark)
Award up to 4 marks
Question 21
[3 marks]
Describe the inputs and outputs used by a D-type flip-flop circuit, explaining how the inputs are used to control the outputs.
Inputs (up to 2 marks):
• D (Data) input: the value (0 or 1) that is to be stored / captured (1 mark)
• Clock (CLK) input: a signal that controls when the D input is read / sampled. The value of D is only captured on the rising edge (or falling edge) of the clock signal (1 mark)
Outputs (up to 1 mark):
• Q output: holds the stored value / the value of D that was captured at the last clock edge (1 mark)
• Q̅ (NOT Q) output: always the complement / inverse of Q (1 mark)
How inputs control outputs:
• When the clock signal transitions (rising/falling edge), the current value of D is transferred to Q (1 mark)
• Between clock edges, the output Q remains stable / does not change regardless of changes to D (1 mark)
Award up to 3 marks total.
• D (Data) input: the value (0 or 1) that is to be stored / captured (1 mark)
• Clock (CLK) input: a signal that controls when the D input is read / sampled. The value of D is only captured on the rising edge (or falling edge) of the clock signal (1 mark)
Outputs (up to 1 mark):
• Q output: holds the stored value / the value of D that was captured at the last clock edge (1 mark)
• Q̅ (NOT Q) output: always the complement / inverse of Q (1 mark)
How inputs control outputs:
• When the clock signal transitions (rising/falling edge), the current value of D is transferred to Q (1 mark)
• Between clock edges, the output Q remains stable / does not change regardless of changes to D (1 mark)
Award up to 3 marks total.
Question 22
[4 marks]
(a) Simplify the Boolean expression ¬A ∨ ¬B using De Morgan's First Law.
(b) Simplify the Boolean expression ¬(¬B) using double negation.
(c) Simplify the Boolean expression (A ∨ B) ∧ (A ∨ C) using distribution.
(b) Simplify the Boolean expression ¬(¬B) using double negation.
(c) Simplify the Boolean expression (A ∨ B) ∧ (A ∨ C) using distribution.
(a) De Morgan's First Law: ¬A ∨ ¬B = ¬(A ∧ B) (1 mark)
De Morgan's First Law states: NOT X OR NOT Y = NOT (X AND Y)
(b) Double negation: ¬(¬B) = B (1 mark)
Double negation states: NOT NOT X = X
(c) Distribution: (A ∨ B) ∧ (A ∨ C) = A ∨ (B ∧ C) (2 marks)
Distribution law states: (X OR Y) AND (X OR Z) = X OR (Y AND Z)
1 mark for identifying the distribution law
1 mark for the correct simplified expression
De Morgan's First Law states: NOT X OR NOT Y = NOT (X AND Y)
(b) Double negation: ¬(¬B) = B (1 mark)
Double negation states: NOT NOT X = X
(c) Distribution: (A ∨ B) ∧ (A ∨ C) = A ∨ (B ∧ C) (2 marks)
Distribution law states: (X OR Y) AND (X OR Z) = X OR (Y AND Z)
1 mark for identifying the distribution law
1 mark for the correct simplified expression