examples of physical layer
DESCRIPTION
Channel Capacity, BandwidthTRANSCRIPT
Examples of Physical Layer
Prof. Hemang KothariAssistant Professor
Computer Engineering Department MEFGI, Rajkot.
Email: [email protected]
Signal & Data Level
Example on Signal & Data Level• A digital signal has eight levels. How many bits are
needed per level? Number of bits per level = log28 = 3
Example on Signal & Data Level• What about a digital signal with 16 levels, How
many bits are needed per level?• What about 32 levels? 64 levels? 128 levels?
Examples on Bit Rate • Assume we need to download text documents at the rate of
100 pages per second. 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, the bit rate is
100 X 24 X 80 X 8 = 16,36,000 bps = 1.636 Mbps
Examples on Bit Rate • What is the bit rate for high-definition TV (HDTV)?Solution• HDTV uses digital signals to broadcast high quality video
signals. The HDTV screen is normally a ratio of 16 : 9. There are 1920 by 1080 pixels per screen, and the screen is renewed 30 times per second. Twenty-four bits represents one color pixel.
1920 X 1080 X 30 X 24 = 1492992000 = 1.5 Gbps Approx
Channel Capacity• The maximum rate at which data can be correctly
communicated over a channel in presence of noise and distortion is known as its channel capacity.
• Noise free channel (Theory)• Noisy Channel (Practical)
Noiseless - Nyquist Theorem• Nyquist gives the upper bound for the bit rate of a
transmission system by calculating the bit rate directly from the number of bits in a symbol (or signal levels) and the bandwidth of the system.
• Nyquist theorem states that for a noiseless channel:C = 2 B log22n
C= capacity in bpsB = bandwidth in Hz
Examples on Nyquist Theorem• Consider a noiseless channel with a bandwidth of
3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated as
Answer = 6000bps
Examples on Nyquist Theorem• Consider the same noiseless channel transmitting
a signal with four signal levels (for each level, we send 2 bits). The maximum bit rate can be calculated as
Examples on Nyquist Theorem• Television channels are 6 MHz wide. How many
bits/sec can be sent if four-level digital signals are used? Assume a noiseless channel.
• Answer = Data rate of 24 Mbps.
Review • Let us consider the telephone channel having
bandwidth B = 4 kHz. Assuming there is no noise, determine channel capacity for the following encoding levels: (i) 2, and (ii) 128.
(i) C = 2B = 2×4000 = 8 Kbits/s (ii) C = 2×4000×log2128 = 8000×7 = 56 Kbits/s
Note • If the available channel is a bandpass channel, we
cannot send the digital signal directly to the channel; We need to convert the digital signal to an analog signal before transmission.
Band Pass Channel
Transmission Impairment• Signals travel through transmission media, which are
not perfect. The imperfection causes signal impairment. • This means that the signal at the beginning of the
medium is not the same as the signal at the end of the medium. What is sent is not what is received.
• Three causes of impairment are attenuation, distortion, and noise.
Signal to Noise Ratio (SNR or S/N)• Signal to noise ratio shows the ratio of signal
power to noise power.• Power often expressed in watts.
S/N = signal power/noise power
SNR & SNRdb
• The power of a signal is 10 mW and the power of the noise is 1 μW; what are the values of SNR and SNRdB ?
Solution• The values of SNR and SNRdB can be calculated as
follows:• SNR = 10000 μW / 1 μW = 10000 (Decimal Value)• SNRdb = 10 log10 10000 = 10 log10 104 = 40
Noisy Channel - Shannon’s Theorem • Shannon’s theorem gives the capacity of a system
in the presence of noise.
C = B log2(1 + SNR)
Extremely Noisy Channel • Consider an extremely noisy channel in which the
value of the signal-to-noise ratio is almost zero. In other words, the noise is so strong that the signal is faint. For this channel the capacity C is calculated as
C = B log2(1 + SNR) = B log2(1 + 0) = B log21 = B X 0 = 0
Shannon’s Theorem Examples• A telephone line normally has a bandwidth of
3000. The signal-to-noise ratio is usually 3162. For this channel the capacity is calculated as
• Answer = 34860 bps
Shannon’s Theorem Examples• A channel has B = 4 KHz. Determine the channel
capacity for each of the following signal-to-noise ratios: (a) 20 dB, (b) 30 dB, (c) 40 dB.
Suppose that the spectrum of a channel is between 10 MHz and 12 Mhz, and an intended capacity of 8 Mbps.
(1) What should be the SNR in order to obtain this capacity?
(2) How many signaling levels are required to obtain this capacity?
(3) What would be the capacity if the environment starts suffering lesser noise and the SNR goes up to 27 dB.
(4) Same question as (2) but for the capacity in (3)
(1) What should be the SNR in order to obtain this capacity?Shannon's Theorem: C=B*log2(1+SNR) <=> 2^(C/B)-1=SNR<=> SNR=15
(2) How many signaling levels are required to obtain this capacity?Nyquist Theorem: C=2B*log2(M) <=> 2^(C/2B)=M <=> M=4
(3) What would be the capacity if the environment starts suffering lesser noise and the SNR goes up to 27 dB.SNR(dB)=10*log10(SNR)<=>SNR=10^2.7<=>SNR=501 (approximately) Shannon's Theorem: C=B*log2(1+501)<=>C=18 Mbps (approximately)
4) Same question as (2) but for the capacity in (3)C=18Mbps=18*10^6 bpsNyquist Theorem: C=2B*log2(M)<=>M=2^(C/2B)<=>M=22.6 ** in order to reach the desired capacity we need to round up M, so M=23. However, typically, M=2^N, so M=32 was the more “realistic” solution.