Tuesday, 27 December 2016

Computer Networks problems and solutions for UGC NET/GATE exam - Set 2

6.       A nonperiodic composite signal has a bandwidth of 200 kHz, with a middle frequency of 140 kHz and peak amplitude of 20 V. The two extreme frequencies have an amplitude of 0. Draw the frequency domain of the signal.
Solution
The lowest frequency must be at 40 kHz and the highest at 240 kHz. Figure below shows the frequency domain and the bandwidth.
7.       A digital signal has eight levels. How many bits are needed per level?
Solution
We calculate the number of bits from the formula,
Number of bits per level = log28 = 3
Each signal level is represented by 3 bits.
Note:
Two digital signals: one with two signal levels and the other with four signal levels.
8.       A digital signal has nine levels. How many bits are needed per level?
Solution
We calculate the number of bits by using the formula.
Number of bits per level = log29 = 3.17
Each signal level is represented by 3.17 bits. However, this answer is not realistic. The number of bits sent per level needs to be an integer as well as a power of 2. For this example, 4 bits can represent one level.

9.       Assume we need to download text documents at the rate of 100 pages per sec. What is the required bit rate of the channel?
Solution
A page is an average of 24 lines with 80 characters in each line. If we assume that one character requires 8 bits (ascii), the bit rate is
100 x 24 x 80 x 8 = 1,536,000 bps = 1.536 Mbps
10.       A digitized voice channel, is made by digitizing a 4-kHz bandwidth analog voice signal. We need to sample the signal at twice the highest frequency (two samples per hertz). We assume that each sample requires 8 bits. What is the required bit rate?
Solution
The bit rate can be calculated as
2 x 4000 x 8 = 64,000 bps = 64 kbps

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