Question

1. (ch.7) A PCM/TDM system multiplexes 24 voice band channels. Each sample is encoded into 7 bits and a frame consists of one sample from each channel. A framing bit is added to each frame. The sampling rate is 9000 samples/sec. What is the line speed in bits per second? 2. (ch.9) An asynchronous protocol uses 1 start bit, 7 data bits, 2 stop bits and 1 even parity bit for each character. a. How many bits are in a character? b. What is the efficiency of this system? c. If the data rate is 121 kbps, how many characters can be sent in 1 minute? 3. (ch.10) Assume an Ethernet cable connecting 2 nodes has a velocity factor of 0.66 and a length of 300m between nodes and supports a data rate of 100Mb/s. a. Calculate the length of time for a signal to pass 300m from one node A to node B. b. If two node transmit at the same time how long will they transmit before a collision is detected? c. How may bits will be transmitted in that time? d. If node A transmits 1200 bits what is the length of time from when node A begins to transmit and when node B gets the last bit?

Answer 1

1. each frame has 7 encoded bits for each sample and 1 framing bit so there are total 8 bits in a frame.

TDM/PCM multiplexes 24 channel, In one TDM sample there are 24 * 8 = 192 bits/sample.

total Data rate = 192 bits/sample * 9000 samples/sec = 17,28,000 = 1.76 Mbps

2. In asynchroronous protocol, 1 start bit, 7 data bits, 2 stops bits and 1 even parity bit for each character

a. total bits in character N = 1 (start bit) + 7 (data bits) + 2 (stops bits) + 1 (even parity) = 7 bits

b. efficienency = data bits / total bits = 7/11 = 63.63 %

c. for data rate 121 kbps, total characters sent in a minute = total bits in a minute / bits in a one character

$total \ character \ in \ a \ minute = \frac{121*1000*60}{11}=6,60,000$

3. Given, velocity factor = .66, data rate = 100 Mbps, distance between nodes = 300 m.

a) time taken by a signal to travel form node A to node $T=\frac{300}{3*10^{8}*.66}=1.52 \ \mu sec$

b) if two node transmits at same time then it takes T time to detect a collision hence they will transmit uptill T = 1.52 micro sec.

c) At 100 Mbps date rate, bits transmitted in this time  $1.52 \ \mu sec * 100 \ Mbps = 152 \ bits$

d) if 1200 bits are transmitted by note A so time t taken by the node B to receive all 1200 bits.

t = time taken by Node A to transmit the data (1200 bits) + time taken by signal to travel from node A to node B

t = 1200 / 100 Mbps + T = 20 micro sec + 1.52 micro sec = 21.52 micro sec