Electronics and Communication Engineering - Digital Electronics - Discussion
Discussion Forum : Digital Electronics - Section 8 (Q.No. 8)
8.
The multiplication time for 10-bit numbers with 1 MHz clock will be
Discussion:
5 comments Page 1 of 1.
MOHIT SEHRAWAT said:
9 years ago
Please give explanation for this question.
Tarun sah said:
5 years ago
How to solve this? Please explain in detail.
Gopal Samy said:
4 years ago
Explain the concept.
Mahi said:
4 years ago
Please explain it in detail.
Shadik said:
10 months ago
To calculate the multiplication time for 10-bit numbers using a 1 MHz clock, we need to understand how many clock cycles the multiplication process takes and the time per clock cycle.
1. Clock Frequency: 1 MHz means the clock has a period of 1 μs (microsecond).
2. 10-bit Numbers: Multiplying two 10-bit numbers typically requires 10 cycles for a straightforward, bit-by-bit multiplication algorithm. However, the actual number of cycles depends on the specific implementation of the multiplier (e.g., array multiplier, shift-and-add, or more complex methods).
Assuming a basic 10-cycle multiplication:
{Time per cycle} = \frac{1}{1\,\{MHz}} = 1\,\mu\{s}.
{Multiplication time} = 10 , \{cycles} \times 1 , \mu\{s/cycle} = 10 , \mu\{s}].
However, the options given are all above 20 μs. This could imply a more complex multiplication method or a serial processing approach, where each bit is processed in multiple cycles.
For example, if a more complex algorithm like a shift-and-add method is used, it might take around 21 to 22 cycles to complete.
Correct me if I am wrong.
1. Clock Frequency: 1 MHz means the clock has a period of 1 μs (microsecond).
2. 10-bit Numbers: Multiplying two 10-bit numbers typically requires 10 cycles for a straightforward, bit-by-bit multiplication algorithm. However, the actual number of cycles depends on the specific implementation of the multiplier (e.g., array multiplier, shift-and-add, or more complex methods).
Assuming a basic 10-cycle multiplication:
{Time per cycle} = \frac{1}{1\,\{MHz}} = 1\,\mu\{s}.
{Multiplication time} = 10 , \{cycles} \times 1 , \mu\{s/cycle} = 10 , \mu\{s}].
However, the options given are all above 20 μs. This could imply a more complex multiplication method or a serial processing approach, where each bit is processed in multiple cycles.
For example, if a more complex algorithm like a shift-and-add method is used, it might take around 21 to 22 cycles to complete.
Correct me if I am wrong.
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