UGC NET CSE | Junet 2015 | Part 3 | Question: 2

Question

Consider a 32-bit microprocessor, with a 16-bit external data bus, driven by an 8 MHz input clock. Assume that this microprocessor has a bus cycle whose minimum duration equals four input clock cycles. What is the maximum data transfer rate for this microprocessor?

• A. $8 \times 10^6$ bytes/sec
• B. $4 \times 10^6$ bytes/sec
• C. $16 \times 10^6$ bytes/sec
• D. $4 \times 10^9$ bytes/sec

Comments

Correct the tag
Its from ugcnetjune2015iii ie paper 3

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fixed

Answer 1

Since the microprocessor has a 16-bit external data-bus, this means that 16 bits are transferred in 1 bus cycle.

now input clock= 8MHz

This means 8x(10^6) cycles in 1 second.

therefore, 1 cycle takes 1/(8x(10^6)) seconds

 

since 1 bus cycle= 4 clock cycles= 4/(8x(10^6)) seconds

now 1 bus cycle -------> 16 bits

4/(8x(10^6)) seconds ---------> (16x8x(10^6))/4 bits = (16x8x(10^6))/(4x8) BYTES

therefore answer= 4x(10^6) Bytes/sec

Answer 2

Since minimum bus cycle duration = 4 clock cycles

Bus clock = 8 MHz

Then, maximum bus cycle rate = 8 M / 4 = 2 M/sec

Data transferred per bus cycle = 16 bit = 2B

Data transfer rate per second = bus cycle rate * data per bus cycle = 2 M * 2 = 4 * 10⁶ B/sec

B is answer

Comments

This can also be visualized as :

Clock cycle=$\frac{1}{8MHz}=\frac{1}{8*10^6}=125ns$

Bus Cycle=4*125ns=500ns

2 Bytes are transferred (16-bit data) for every 500ns.

 

Thus transfer rate=$\frac{2}{500ns}=\frac{2}{5*10^-9}=4MBps$