Question

4. If an asynchronous frame is used to send ASCII characters in the form of bytes (8 bits), what is the shortest time it could take to send 1000 characters if each bit in a frame is 1 msec long? 5. Suppose an asynchronous frame holds 8 bits of data, a parity bit, and two stop bits (it could happen). Calculate the efficiency of the communication system. Suppose a synchronous frame has 16 bits of non-data in the front and a 16-bit BCC at the end. The frame carries 1024 bytes of actual data. Calculate the efficiency of the communication system. 6.

Answer 1

4.

In asynchronous frame along with data bits, two more bits needed for START and STOP bits.

Hence 10 bits / character is minimum requirement for asynchronous transmission.

For 1000 characters, 1000 x 10 = 10,000 bits needed.

Total time = bit time X 10,000 bits = 1ms x 10000 = 10000 ms or 10 s

5.

Efficiency = (data bits / total bits) x 100

= (8/11) x 100 = 72.72%

6.

Efficiency = (1024 bytes x 8 bits) / (1024 bytes x 8 bits + 16 bits non data + 16 bits BCC)

= 8192/8224 = 99.6%