GATE Practice - Computer Networks & Communications

This is an incorrect option because in FDM, the available bandwidth is shared among signals. So, it is not possible to utilise full available bandwidth for each channel. It is only possible in TDM.

MOV A,# 85HCPL A ;ADD ,#85 H = 1000 0101
1’s = 0111 1010
2’s = + 10111 1011 = > 7BH

This is the correct answer. Asynchronous communication transfers a single byte at a time.

This is the right answer as:
Bandwidth = freq/Q; where Q is quality factor
= 110 x 10⁶/70 Hz
= 1.571 MHz

MOV A, #0F5H; A=F5 hex ADDA, #0BH; A=F5+0B=00F5 +0B => 1111 0101 + 0000 1011 = 0000 0000
After the addition register A(destination) contains 00, the flags are as follows:
CY = 1 as there is carry out from D7. PF = 1 as the number of 1’s is zero.

It works on the principle of broadcasting and switching in which each station can transmit and receive.

This is not a network-connecting device. Token ring is an IEEE standard of LAN (IEEE 802.5) in the layered architecture developed by Institution of Electrical and Electronics Engineers, under the project 802 set up in 1980.

This is the right answer as:

MAN - Metropolitan Area Network = IEEE 802.6
CSMA/CD - Carrier Sense Multiple Access/Collision Detect = IEEE 802.3

In order to make the entire arrangement substantially unidirectional, many identical antennas are spaced equally along the line.

Assuming minimum Hamming distance (d_min) = 5
Number of errors (s) that can be detected can be obtained from (s + 1) <= d_min <= 5 - 1
s <= 4
Thus, at most 4 errors can be detected.

T_f = Transmission time required to transmit a frame
T_p = Propagation time required to reach the destination for a transmitted bit
Let A be a sender.
T_f = Frame size/Bit rate = 4800/9600 = 0.5 sec
T_p = 2000/200000 = 0.01 sec
We know A = T_p/T_f
A = 0.01/0.5 = 0.02
Throughput = 1/(1 + 2A) = 1/(1 + 2 x 0.02) = 0.96, i.e. 96%

L = 100 km = 100000 m
Data rate of T1 = 1.544 Mbps
Speed (v) = 2/3 x 3 x 100 000 000 m/s
Number of bits in 1 sec = 1.544 x 1000,000 bits
Distance covered in 1 sec = 2 x 100,000,000 m
Number of bits corresponding to 100000 m cable is given by:
= ((1.544 x 1000000) x 100000)/2 x 100 000 000 => 772 bits 

When the field behind a screen with an opening is added to the field of a complementary screen, the sum is equal to the field when there is no screen.

This is the right answer.
Net token is put every 5 microseconds.
Maximum output data rate = m
Token arrival rate = 1/5 microsecond =>1000000/5 = 200 kbps
The maximum output rate can be obtained from the following expression:
C= S(m - token arrival rate), where C = bucket capacity, M = Maximum output rate, S = time for which maximum output is obtained M = C/S + token arrival rate
Assume C = 1 Mbps and S = 20 msec to get:
M = 1000000000/20 + (200 x 1000)
M = 50.2 Mbps