Tuesday, 10 January 2017

Computer Networks problems and solutions for UGC NET/GATE exam - Set 5

21.       A movie DVD holds 120 minutes' movie and takes 5 GB (G is specified as 109 in this case). What is the bandwidth to play this movie DVD?
Solution
5 x 109 x 8 / (120 x 60) = 5.556 Mbps (1 GB = 109 B)
or
5 x 230 x 8 / (120 x 60) = 5.965 Mbps (1 GB = 230 B)
22.       It is desired to send a sequence of computer screen images over an optical fiber. The screen is 480 x 640 pixels, each pixel being 24 bits. There are 60 screen images per second. How much bandwidth is needed, and how many microns of wavelength are needed for this band at 1.30 microns?
Solution
The data rate Δf is 480 x 640 x 24 x 60 bps, which is 442 Mbps. For simplicity, let us assume 1 bps/Hz.
Δf = c Δλ/λ2
Δλ = λ2 Δf/c
where Δf is the bandwidth range, c is the light speed, Δλ is the wavelength range, and lambda is a certain wavelength.
We have Δf = 4.42 x 108, so
          Δλ = 2.5 x 10-6 microns.
The range of wavelengths used is very short.
or If we assume 8 bps/Hz, Δλ = 0.17 x 10-6 microns.
23.       An ADSL system using DMT allocates 3/4 of the available data channels to the downstream link. It uses QAM-64 modulation on each channel. What is the capacity of the downstream link?
Solution
There are 256 channels in all, minus 6 for POTS and 2 for control, leaving 248 for data. If 3/4 of these are for downstream, that gives 186 channels for downstream. ADSL modulation is at 4000 baud, so with QAM-64 (6 bits/baud) we have 24,000 bps in each of the 186 channels.
The total bandwidth is then 4.464 Mbps downstream.
Note:
Discrete multitone (DMT) is a method of separating a Digital Subscriber Line (DSL) signal so that the usable frequency range is separated into 256 frequency bands (or channels) of 4.3125 KHz each. DMT uses the fast Fourier transform (FFT) algorithm for modulation and demodulation. Dividing the frequency spectrum into multiple channels allows DMT to work better when AM radio transmitters are present. Within each channel, modulation uses quadrature amplitude modulation (QAM).
QAM (quadrature amplitude modulation) is a method of combining two amplitude-modulated (AM) signals into a single channel, thereby doubling the effective bandwidth. QAM is used with pulse amplitude modulation (PAM) in digital systems, especially in wireless applications.
POTS, Short for plain old telephone service, which refers to the standard telephone service that most homes use. In contrast, telephone services based on high-speed, digital communications lines, such as ISDN and FDDI, are not POTS. The main distinctions between POTS and non-POTS services are speed and bandwidth.


24.       In a typical mobile phone system with hexagonal cells, it is forbidden to reuse a frequency band in an adjacent cell. If 840 frequencies are available, how many can be used in a given cell?
Solution
Each cell has six other adjacent cells, in a hexagonal grid. If the central cell uses frequency group A, its six adjacent cells can use B, C, B, C, B, and C respectively. In other words, only 3 unique cells are needed. So the answer is 840/3 = 280 frequencies.
25.       If the bit string 0111110111110111110 is subjected to bit stuffing, what output string will be transmitted?
Solution
The output is 0111110011111001111100.
Note:
In data transmission and telecommunication, bit stuffing is the insertion of non information bits into data. Stuffed bits should not be confused with overhead bits (nondata bits).
- Call a flag six consecutive 1s
- On transmit, after five 1s in the data, insert a 0
- On receive, a 0 after five 1s is deleted

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