Computer Networks problems and solutions for UGC NET/GATE exam - Set 1

1.       The power we use at home has a frequency of 60 Hz. The period of this sine wave can be determined as follows:
T = 1/f = 1/60 = 0.0166 s
= 0.0166 x 103 ms = 16.6 ms
2.       The period of a signal is 100 ms. What is its frequency in kilohertz?
First we change 100 ms to seconds, and then we calculate the frequency from the period (1 Hz = 10−3 kHz).
100 ms = 100 x 10-3 s = 10-1 s
f = 1/T = 1/10-1 Hz = 10 Hz
= 10 x 10-3 kHz = 10-2 kHz
Frequency is the rate of change with respect to time.
Change in a short span of time means high frequency.
Change over a long span of time means low frequency.

3.       A sine wave is offset 1/6 cycle with respect to time 0. What is its phase in degrees and radians?
We know that 1 complete cycle is 360°. Therefore, 1/6 cycle is
1/6 x 360 = 60o
= 60 x 2π/360 rad = π/3 rad = 1.046 rad
Phase describes the position of the waveform relative to time 0.
Three sine waves with the same amplitude and frequency, but different phases
4.       If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is its bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V.
Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth. Then
B = fh - fl = 900 – 100 = 800 Hz
The spectrum has only five spikes, at 100, 300, 500, 700, and 900 Hz (see Figure).
The bandwidth of a composite signal is the difference between the highest and the lowest frequencies contained in that signal.
5.       A periodic signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum if the signal contains all frequencies of the same amplitude.
Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth. Then
B = fh - fl => 20 = 60 - fl => fl = 60 – 20 = 40 Hz

The spectrum contains all integer frequencies. We show this by a series of spikes (see Figure).

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