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A half adder is implemented with XOR and AND gates. A full adder is implemented with two half adders and one OR gate.
The propagation delay of an XOR gate is twice that of an AND/OR gate. The propagation delay of an AND/OR gate is 1.2 microseconds. A 4-bit-ripple-carry binary adder is implemented by using four full adders. The total propagation time of this 4-bit binary adder in microseconds is ______.
The propagation delay of an XOR gate is twice that of an AND/OR gate. The propagation delay of an AND/OR gate is 1.2 microseconds. A 4-bit-ripple-carry binary adder is implemented by using four full adders. The total propagation time of this 4-bit binary adder in microseconds is ______.
Correct answer is '12'. Can you explain this answer?
Verified Answer
A half adder is implemented with XOR and AND gates. A full adder is im...
S1 should wait for C1 to be ready. Delay for generating C is 1 EXOR + 1 AND + 1 OR = 2.4 + 1.2 + 1.2 = 4.8 ms
Delay for sum is XOR + XOR = 2.4 + 2.4 = 4.8 ms
But for the second adder, there the first EXOR can be done even before waiting for the previous output. So, we can get sum in another 2.4 ms and carry in another 2.4 ms. In this way, 4 bit sum can be obtained after
4.8 ms + 3 * 2.4 ms = 12 ms.
Ans is 12 ns
It took me a while but here's how it is : The first carry and sum will be available after 4.8 ns. This should be straight forward.
However, for the subsequent stages, you need to keep in mind that the output of half adders is already there at 2.4 ns. So in a sense, it is already computed. The remaining half adder for each full adder is just waiting for the previous carry, which when available from the previous stage can be processed in 2.4 ns. So each next stage will take only 2.4 ns each.
The catch here is that half of the output in each next stage is already computed, only half needs to be processed.
It took me a while but here's how it is : The first carry and sum will be available after 4.8 ns. This should be straight forward.
However, for the subsequent stages, you need to keep in mind that the output of half adders is already there at 2.4 ns. So in a sense, it is already computed. The remaining half adder for each full adder is just waiting for the previous carry, which when available from the previous stage can be processed in 2.4 ns. So each next stage will take only 2.4 ns each.
The catch here is that half of the output in each next stage is already computed, only half needs to be processed.
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Most Upvoted Answer
A half adder is implemented with XOR and AND gates. A full adder is im...
Total Propagation time of the 4-bit binary adder = Propagation time of one Full Adder * Number of Full Adders
Propagation time of one Full Adder = Propagation time of two Half Adders + Propagation time of one OR gate
Propagation time of two Half Adders = 2 * Propagation time of one XOR gate + 2 * Propagation time of one AND gate
Propagation time of one XOR gate = 2 * Propagation time of one AND/OR gate
Propagation time of one AND gate = Propagation time of one OR gate = 1.2 microseconds
Let's calculate each component step by step.
1. Propagation time of one XOR gate = 2 * Propagation time of one AND/OR gate = 2 * 1.2 microseconds = 2.4 microseconds
2. Propagation time of one Half Adder = 2 * Propagation time of one XOR gate + 2 * Propagation time of one AND gate = 2 * 2.4 microseconds + 2 * 1.2 microseconds = 4.8 microseconds + 2.4 microseconds = 7.2 microseconds
3. Propagation time of one Full Adder = Propagation time of two Half Adders + Propagation time of one OR gate = 2 * 7.2 microseconds + 1.2 microseconds = 14.4 microseconds + 1.2 microseconds = 15.6 microseconds
4. Total Propagation time of the 4-bit binary adder = Propagation time of one Full Adder * Number of Full Adders = 15.6 microseconds * 4 = 62.4 microseconds
However, we are given that the correct answer is 12 microseconds. This means there is an error in the calculation or the given information. Let's review the steps.
The correct calculation should be as follows:
1. Propagation time of one XOR gate = 2 * Propagation time of one AND/OR gate = 2 * 1.2 microseconds = 2.4 microseconds
2. Propagation time of one Half Adder = 2 * Propagation time of one XOR gate + 2 * Propagation time of one AND gate = 2 * 2.4 microseconds + 2 * 1.2 microseconds = 4.8 microseconds + 2.4 microseconds = 7.2 microseconds
3. Propagation time of one Full Adder = Propagation time of two Half Adders + Propagation time of one OR gate = 2 * 7.2 microseconds + 1.2 microseconds = 14.4 microseconds + 1.2 microseconds = 15.6 microseconds
4. Total Propagation time of the 4-bit binary adder = Propagation time of one Full Adder * Number of Full Adders = 15.6 microseconds * 4 = 62.4 microseconds
Since the given answer is 12 microseconds, it seems there might be a mistake in the given information or the answer itself. Please double-check the problem statement or provide further clarification to resolve the discrepancy.
Propagation time of one Full Adder = Propagation time of two Half Adders + Propagation time of one OR gate
Propagation time of two Half Adders = 2 * Propagation time of one XOR gate + 2 * Propagation time of one AND gate
Propagation time of one XOR gate = 2 * Propagation time of one AND/OR gate
Propagation time of one AND gate = Propagation time of one OR gate = 1.2 microseconds
Let's calculate each component step by step.
1. Propagation time of one XOR gate = 2 * Propagation time of one AND/OR gate = 2 * 1.2 microseconds = 2.4 microseconds
2. Propagation time of one Half Adder = 2 * Propagation time of one XOR gate + 2 * Propagation time of one AND gate = 2 * 2.4 microseconds + 2 * 1.2 microseconds = 4.8 microseconds + 2.4 microseconds = 7.2 microseconds
3. Propagation time of one Full Adder = Propagation time of two Half Adders + Propagation time of one OR gate = 2 * 7.2 microseconds + 1.2 microseconds = 14.4 microseconds + 1.2 microseconds = 15.6 microseconds
4. Total Propagation time of the 4-bit binary adder = Propagation time of one Full Adder * Number of Full Adders = 15.6 microseconds * 4 = 62.4 microseconds
However, we are given that the correct answer is 12 microseconds. This means there is an error in the calculation or the given information. Let's review the steps.
The correct calculation should be as follows:
1. Propagation time of one XOR gate = 2 * Propagation time of one AND/OR gate = 2 * 1.2 microseconds = 2.4 microseconds
2. Propagation time of one Half Adder = 2 * Propagation time of one XOR gate + 2 * Propagation time of one AND gate = 2 * 2.4 microseconds + 2 * 1.2 microseconds = 4.8 microseconds + 2.4 microseconds = 7.2 microseconds
3. Propagation time of one Full Adder = Propagation time of two Half Adders + Propagation time of one OR gate = 2 * 7.2 microseconds + 1.2 microseconds = 14.4 microseconds + 1.2 microseconds = 15.6 microseconds
4. Total Propagation time of the 4-bit binary adder = Propagation time of one Full Adder * Number of Full Adders = 15.6 microseconds * 4 = 62.4 microseconds
Since the given answer is 12 microseconds, it seems there might be a mistake in the given information or the answer itself. Please double-check the problem statement or provide further clarification to resolve the discrepancy.
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A half adder is implemented with XOR and AND gates. A full adder is implemented with two half adders and one OR gate.The propagation delay of an XOR gate is twice that of an AND/OR gate. The propagation delay of an AND/OR gate is 1.2 microseconds. A 4-bit-ripple-carry binary adder is implemented by using four full adders. The total propagation time of this 4-bit binary adder in microseconds is ______.Correct answer is '12'. Can you explain this answer?
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A half adder is implemented with XOR and AND gates. A full adder is implemented with two half adders and one OR gate.The propagation delay of an XOR gate is twice that of an AND/OR gate. The propagation delay of an AND/OR gate is 1.2 microseconds. A 4-bit-ripple-carry binary adder is implemented by using four full adders. The total propagation time of this 4-bit binary adder in microseconds is ______.Correct answer is '12'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about A half adder is implemented with XOR and AND gates. A full adder is implemented with two half adders and one OR gate.The propagation delay of an XOR gate is twice that of an AND/OR gate. The propagation delay of an AND/OR gate is 1.2 microseconds. A 4-bit-ripple-carry binary adder is implemented by using four full adders. The total propagation time of this 4-bit binary adder in microseconds is ______.Correct answer is '12'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A half adder is implemented with XOR and AND gates. A full adder is implemented with two half adders and one OR gate.The propagation delay of an XOR gate is twice that of an AND/OR gate. The propagation delay of an AND/OR gate is 1.2 microseconds. A 4-bit-ripple-carry binary adder is implemented by using four full adders. The total propagation time of this 4-bit binary adder in microseconds is ______.Correct answer is '12'. Can you explain this answer?.
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