Question

Consider a synchronous TDM system in which 6 input channels are multiplexed into a 125-microsecond output frame. Each output frame includes exactly four bits from each input source, plus one synch bit per frame.

a) What is the size of each output frame in bits?

b) What is the total bit rate of the output line?

c) Suppose we would like to replace one of the input sources with a new source that has a data rate of 24 Kilobits per seconds. Calculate how many extra bits must be stuffed for this source in each output frame.

Answer 1

4 bits from each input source and there are 6 inputs. Thus there will be a total of 6* 4 = 24 bits + 1 sync bit = 25 bits per output frame.

a. Size of each output frame in bits = 25 bits / frame

b. 25 bits / frame = 25 bits / 125 micro second ( since each frame size is 125micro second ) = 0.2 Mega bits / second

c. In one second, there will be 0.2 mega bits. No . of sync bits in one second = 1/ 125 micro =8K bits

Thus the no. of bits per channel in one second =( 0.2 * 10⁶ - 8 * 10³⁾/6 = 32K bits

Thus each channel should be sending 32 kilo bits / sec

Since one channel has only 24 kilo bits / sec , it should be stuffed with 32 - 24 = 8 kilo bits / sec

Each frame size is 125 micro second => in each frame, there should be 125*10^-6 * 8000 bits that should be stuffed = 1 bit / frame.

Hence for this source, 1 bit should be stufed per frame