Lab/Tutorial :
SEHH2238 : Computer Networking
Session 2 : Signal Transmission (Solution)
1) Time and Frequency Domain
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1) Draw the time-domain plot of a sine wave (for only 1s) with maximum amplitude of 15V, frequency of 5 and phase 270o.
2) What is the bandwidth of a signal that can be decomposed into four sine waves with frequencies at 0, 20, 50, 100 and 200 Hz? All maximum amplitudes are the same. Draw the frequency spectrum.
3) A periodic composite signal with a bandwidth of 2000 Hz is composed of two sine waves. The first one has a frequency of 100 Hz with a maximum amplitude of 20V; the second one has a maximum amplitude of 5 V. Draw the frequency spectrum.
SEHH2238 Computer Networking Tutorial 2 Page 1
2) Transmission Impairment
1) The loss in a cable is usually defined in decibels per kilometer (dB/km). If the signal at the beginning of a cable with −0.3 dB/km has a power of 2 mW, what is the power of the signal at 5 km?
The loss in the cable in decibels is 5 × (−0.3) = −1.5 dB. We can calculate the power as
2) If the power of a signal is 0.5W and the power of the noise is 10mW. What is the SNR? What is SNRdB?
SNR = 0.5 / ( 10 x 10-3 ) = 50 SNRdB = 10 log10 50 = 16.99
3) Data Rate
1) We need to send data at a rate 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need?
Bit Rate 265 x 103 log2L
= 2 x B x log2L (by Nyquist Bit Rate) = 2 x 20 x 103 x log2L
2) A measurement is done on a telephone line (4kHz of bandwidth). When the signal is 10W, the noise is 5mW. What is the maximum data rate supported by this telephone line?
Max. Bit Rate = B x log2(1 + SNR) (by ) = 4000 x log2(1 + 10 / ( 5 x 10-3))
= 4000 x 11 = 44Kbps
3) We have a channel with a 1-MHz bandwidth. The SNR for this channel is 63. What are the appropriate bit rate and signal level?
Max. Bit Rate = B x log2(1 + SNR) (by ) =1×106 xlog2(1+63)
= 6 x 106 = 6Mbps
Hence, 6 Mbps is the upper limit of bit rate in this noisy channel. For better performance, we choose something lower, says 4 Mbps. Since noise has been considered, we can use Nyquist formula to determine the number of signal levels.
Bit Rate = 2 x B x log2L (by Nyquist Bit Rate) 4×106 =2x1x106 xlog2L
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4) What is the bit rate for signal in the following figure? 32 ns
Duration of one signal = 32 ns / 8 = 4 ns
One signal carries log24 = 2 bits
Bit rate = No. of bits per signal / Duration of one signal
= 2 / ( 4 x 10-9) = 5 x 108 = 500 Mbps
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