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# CS601 Fall 2011 Final Term Feb 2012 – VU Current Paper – 08 Feb 2012

Question No: 41 ( Marks: 2 )
What are the conditions for the polynomial used by the CRC generator?
CRC generator:
CRC generator (the divisor) is most often represented not as a1’s and 0’s but as an algebraic polynomial.
Conditions for the polynomial:
it should have following properties:
It should not be divisible by “x”.
It should not be divisible by “x+1”.
The first condition guarantees that all burst error of a length equal to degree of the polynomial is detected.
The 2nd condition guarantees that all burst error affecting an odd number of bits are detected.

Question No: 42 ( Marks: 2 )
What are intelligent modems?
Intelligent modems:
A modem that responds to commands and can accept new instructions during online transmission. It was originally developed by Hayes.
Example:
> Automatic answering,
> Dialing etc.

Question No: 43 ( Marks: 2 )
What is the basic purpose of Router?
Basic purpose of Router:
“A router is a device that extracts the destination of a packet it receives, selects the best path to that destination, and forwards data packets to the next

device along this path. They connect networks together; a LAN to a WAN for example, to access the Internet.
“A more precise definition of a router is a computer networking device that interconnects separate logical subnets.”

Question No: 44 ( Marks: 3 )
What are the fractional T Lines?
The fractional T Lines:
Many subscribers don’t need the entire capacity of the T-line.
For example,
A small business may need only one-fourth of the capacity of T-line. if four business of same size lie in the same building, they can share T-line.DSU/CSU

allow the capacity of T-line to be interleaved in to four channels

Question No: 45 ( Marks: 3 )
What are the light sources used for optic fiber?
> light sources used for optic fiber:
> The light source can weather be an LED or ILD
> LED (Light emitting diode) cheaper but provide unfocused light that strikes the boundaries of channel at uncontrollable angles.
> Limited to short distance use.
> LASSER
> Can be focused to a narrow range allowing control over angle of incidence.

Question No: 46 ( Marks: 3 )
What is Multi Access Unit (MAU) in Token Ring?
Multi Access Unit (MAU) in Token Ring:
> Individual automatic switches are combined in to a hub
> One MAU can support up to 8 stations.
> Although it looks like a star, it is in fact a ring.

Question No: 47 ( Marks: 5 )
Give characteristics of Dual Ring, if necessary then draw the diagram. [5]
Characteristics of Dual Ring:
A network topology in which two concentric rings connect each node on a network instead of one network ring that is used in a ring topology. Typically, the

secondary ring in a dual-ring topology is redundant. It is used as a backup in case the primary ring fails. In these configurations, data moves in opposite

directions around the rings. Each ring is independent of the other until the primary ring fails and the two rings are connected to continue the flow of data

traffic.

Question No: 48 ( Marks: 5 )
What the receiver will receive if the checksum method is applied to the following bit. 10101001 00111001
Ans:
the receiver will receive the checksum method is applied to the following bit.
10101001 00111001
10101001 00111001
Sum of 2 bits are
10101001
00111001
——————-
11100010
00011101 1′s complement
1
————–
00011110 2′s complement
———–
10101001 00111001 ==> 00011110
So the data transmitted which will receiver get:
10101001 00111001 00011110

Question No: 49 ( Marks: 5 )
What is refraction in terms of optic fiber? Give one example.
Refraction:
Light travels in a straight line as long as it is moving through a single uniform structure If a ray of light traveling through one substance enters another

(more or less dense) substance, its speed changes abruptly causing the ray to change direction. This phenomenon is called Refraction.
Refraction in terms of optic fiber:
The propagation of light in an optical fiber which in its simplest form consists of a circular core of uniform refractive
index surrounded by a cladding of slightly lower refractive index. The light is launched into the entrance face of the fiber.
The light is propagated by the total internal reflection at the interface between core and cladding. However the rays incident at angles larger than a

certain angle, called the cut-off angle, suffer both refraction and reflection at the interface between the core and the cladding.
They, therefore, are not guided. Due to this the optical fiber has a numerical aperture. The numerical aperture is given by the square root of (n12-n22).

Typical values of numerical aperture lie between 0.1 and 0.3.

The refractive indices of the core and the cladding are n1 and n 2 respectively. The fiber is normally in air (n0=1)
but could also be in a medium of refractive index n0.

Question No: 50 ( Marks: 10 )
What are the asynchronous protocols in data link layer? Discuss in detail with examples. [10 marks]
Asynchronous protocols in data link layer:
Asynchronous communication at the data link layer or higher protocol layers is known as statistical multiplexing or packet mode communication, For example :
Asynchronous transfer mode (ATM). In this case the asynchronously transferred blocks are called data packets, Async protocols in Data link layer is called

statistical multiplexing. for example ATM cells.
The opposite is circuit switched communication, which provides constant bit rate, for example ISDN and SONET/SDH.
The packets may be encapsulated in a data frame, with a frame synchronization bit sequence indicating the start of the frame, and sometimes also a bit

synchronization bit sequence, typically 01010101, for identification of the bit transition times. Note that at the physical layer, this is considered as

synchronous serial communication.
Examples of packet mode data link protocols that can be/are transferred using synchronous serial communication are the
> HDLC,
> Ethernet,
> PPP and
> USB Protocol

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