Priority Encoders MCQ Quiz - Objective Question with Answer for Priority Encoders - Download Free PDF

The correct answer is: The encoder will output the binary code for the highest priority active input.

Explanation:

A priority encoder is a combinational circuit that:

• Accepts multiple input lines, but only one output is generated at a time.

• If two or more inputs are active simultaneously, the input with the highest priority (usually the one with the highest input number) is encoded and reflected at the output.

This ensures unambiguous output even when multiple inputs are high.

🔍 Example:

For a 4-to-2 priority encoder with inputs D3 to D0 (D3 has highest priority):

Here, X represents the priority encoder output.

Y represents the valid output corresponding to the highest order active input.

• In a four-line to two-line priority encoder, the goal is to encode the highest order active input among four lines into a two-bit output.
• X = D₂+D₃
• This equation means that the output X is active (HIGH) when either D₂ or D₃ is active.
• It indicates the encoding of the higher-priority input.
• Y = D₁\(\overline D_2\)+D₃
• Y is the two-bit output. It is formed by combining D₁D₂ if both D1 and D₂ are active, or by using D₃ if D₃ is active alone. This represents the highest-order active input.

• X = D2+D3  
• Y = D1\(\overline D_2\)+D3

Here, option 1 is correct.

Concept-

An Encoder is a combinational circuit that performs the reverse operation of Decoder.

It has maximum of 2n input lines and ‘n’ output lines. It will produce a binary code equivalent to the input, which is active High.

Therefore, the encoder encodes 2n input lines with ‘n’ bits.

A 4 to 2 priority encoder has four inputs Y3, Y2, Y1 & Y0 and two outputs A1 & A0.

Here, the input, Y3 has the highest priority, whereas the input, Y0 has the lowest priority.

In this case, even if more than one input is ‘1’ at the same time, the output will be the binary code corresponding to the input, which is having higher priority.

Encoder (4 ∶ 2)

Truth table:-

A1 = y3 + y2

A₀ = y3 + y1

Priority encoder:-

If more than one input is high then encoder produce an output which may not be correct to overcome this we use priority encoder.

We considered one more output, V in order to know, whether the code available at outputs is valid or not.

If at least one input of the encoder is ‘1’, then the code available at outputs is a valid one. In this case, the output, V will be equal to 1.

If all the inputs of encoder are ‘0’, then the code available at outputs is not a valid one. In this case, the output, V will be equal to 0.

Analysis:-

Logic shown in given figure will work as

Priority encoder (4 × 2)

Inputs = D₃, D₂, D₁, D₀

Output:

y = D₃ + D₂’ D₁

X = D₃ + D₂

From the given logic circuit,

V = D₀ + D₁ + D₂ + D₃

y₀ = D₃ + D₁D̅₂

y₁ = D₂ + D₃

The truth table for the given logic circuit is given below.

From the above truth table, it is clear that When D₃ = 1, the output (Y₁ Y₀) = (11)₂ = (3)₁₀ irrespective of other inputs.

For D₃ = 0 and D₂ = 1, the output Y₁ Y₀ = (10)₂ = (2)₁₀ irrespective of other inputs.

For D₃ = D₂ = 0 and D₁ = 1, the output

Y₁Y₀ = (01)₂ = (1)₁₀ irrespective of other inputs.

The output will be Y₁Y₀ = (00)₂ = (0)₁₀ only

When D₃ = D₂ = D₁ = 0 and D₀ = 1.

Therefore, the given logic circuit is a priority encoder with a priority order D₃ > D₂ > D₁ > D₀

A 4 to 2 Priority Encoder takes 4 input bits and produces 2 Output bits. The truth Table of a 4 to 2 Priority Encoder is as shown:

We use Karnaugh Map to minimize the logic for B as shown:

B = I̅₂I₁ + I₃

Similarly the expression for A in obtained using K-map as shown:

A = I₂ + I₃

So, the given circuit is a priority Encoder only.

A 4 to 2 Priority Encoder takes 4 input bits and produces 2 Output bits. The truth Table of a 4 to 2 Priority Encoder is as shown:

We use Karnaugh Map to minimize the logic for B as shown:

B = I̅₂I₁ + I₃

Similarly the expression for A in obtained using K-map as shown:

A = I₂ + I₃

So, the given circuit is a priority Encoder only.

The correct answer is: The encoder will output the binary code for the highest priority active input.

Explanation:

A priority encoder is a combinational circuit that:

• Accepts multiple input lines, but only one output is generated at a time.

• If two or more inputs are active simultaneously, the input with the highest priority (usually the one with the highest input number) is encoded and reflected at the output.

This ensures unambiguous output even when multiple inputs are high.

🔍 Example:

For a 4-to-2 priority encoder with inputs D3 to D0 (D3 has highest priority):

Here, X represents the priority encoder output.

Y represents the valid output corresponding to the highest order active input.

• In a four-line to two-line priority encoder, the goal is to encode the highest order active input among four lines into a two-bit output.
• X = D₂+D₃
• This equation means that the output X is active (HIGH) when either D₂ or D₃ is active.
• It indicates the encoding of the higher-priority input.
• Y = D₁\(\overline D_2\)+D₃
• Y is the two-bit output. It is formed by combining D₁D₂ if both D1 and D₂ are active, or by using D₃ if D₃ is active alone. This represents the highest-order active input.

• X = D2+D3  
• Y = D1\(\overline D_2\)+D3

Here, option 1 is correct.

A 4 to 2 Priority Encoder takes 4 input bits and produces 2 Output bits. The truth Table of a 4 to 2 Priority Encoder is as shown:

We use Karnaugh Map to minimize the logic for B as shown:

B = I̅₂I₁ + I₃

Similarly the expression for A in obtained using K-map as shown:

A = I₂ + I₃

So, the given circuit is a priority Encoder only.

From the given logic circuit,

V = D₀ + D₁ + D₂ + D₃

y₀ = D₃ + D₁D̅₂

y₁ = D₂ + D₃

The truth table for the given logic circuit is given below.

From the above truth table, it is clear that When D₃ = 1, the output (Y₁ Y₀) = (11)₂ = (3)₁₀ irrespective of other inputs.

For D₃ = 0 and D₂ = 1, the output Y₁ Y₀ = (10)₂ = (2)₁₀ irrespective of other inputs.

For D₃ = D₂ = 0 and D₁ = 1, the output

Y₁Y₀ = (01)₂ = (1)₁₀ irrespective of other inputs.

The output will be Y₁Y₀ = (00)₂ = (0)₁₀ only

When D₃ = D₂ = D₁ = 0 and D₀ = 1.

Therefore, the given logic circuit is a priority encoder with a priority order D₃ > D₂ > D₁ > D₀

For a simple decimal → BCD encoder, we find A is true for 1, 3, 5, 7, 9

where A → LSB of BCD code is:

A = 1 + 3 + 5 + 7 + 9B

= 2 + 3 + 6 + 7C

= 4 + 5 + 6 + 7 MSB D = 8 + 9

Now, with 7 given priority over 0, 2 & 5, the logic will be A is HIGH if 1 is HIGH & 2, 4, 6, 8 are LOW

A is HIGH if 3 is HIGH & 4, 6, 8 are LOW

A is HIGH if 5 is HIGH & 6, 8 are LOW

A is HIGH if 7 is HIGH & 8 is LOW

A is HIGH if 9 is HIGH

So, logic equation  \(\rm A =1\cdot\bar 2 \cdot\bar 4 \cdot\bar6 \cdot\bar8 + 3\cdot\bar4 \cdot\bar6\cdot\bar8 + 5\cdot\bar 6\cdot\bar8+ 7\cdot\bar8+ 9\)

Concept-

An Encoder is a combinational circuit that performs the reverse operation of Decoder.

It has maximum of 2n input lines and ‘n’ output lines. It will produce a binary code equivalent to the input, which is active High.

Therefore, the encoder encodes 2n input lines with ‘n’ bits.

A 4 to 2 priority encoder has four inputs Y3, Y2, Y1 & Y0 and two outputs A1 & A0.

Here, the input, Y3 has the highest priority, whereas the input, Y0 has the lowest priority.

In this case, even if more than one input is ‘1’ at the same time, the output will be the binary code corresponding to the input, which is having higher priority.

Encoder (4 ∶ 2)

Truth table:-

A1 = y3 + y2

A₀ = y3 + y1

Priority encoder:-

If more than one input is high then encoder produce an output which may not be correct to overcome this we use priority encoder.

We considered one more output, V in order to know, whether the code available at outputs is valid or not.

If at least one input of the encoder is ‘1’, then the code available at outputs is a valid one. In this case, the output, V will be equal to 1.

If all the inputs of encoder are ‘0’, then the code available at outputs is not a valid one. In this case, the output, V will be equal to 0.

Analysis:-

Logic shown in given figure will work as

Priority encoder (4 × 2)

Inputs = D₃, D₂, D₁, D₀

Output:

y = D₃ + D₂’ D₁

X = D₃ + D₂

The correct answer is: The encoder will output the binary code for the highest priority active input.

Explanation:

A priority encoder is a combinational circuit that:

• Accepts multiple input lines, but only one output is generated at a time.

• If two or more inputs are active simultaneously, the input with the highest priority (usually the one with the highest input number) is encoded and reflected at the output.

This ensures unambiguous output even when multiple inputs are high.

🔍 Example:

For a 4-to-2 priority encoder with inputs D3 to D0 (D3 has highest priority):

Here, X represents the priority encoder output.

Y represents the valid output corresponding to the highest order active input.

• In a four-line to two-line priority encoder, the goal is to encode the highest order active input among four lines into a two-bit output.
• X = D₂+D₃
• This equation means that the output X is active (HIGH) when either D₂ or D₃ is active.
• It indicates the encoding of the higher-priority input.
• Y = D₁\(\overline D_2\)+D₃
• Y is the two-bit output. It is formed by combining D₁D₂ if both D1 and D₂ are active, or by using D₃ if D₃ is active alone. This represents the highest-order active input.

• X = D2+D3  
• Y = D1\(\overline D_2\)+D3

Here, option 1 is correct.