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Channel Capacity, Bandwidth

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- 1. Examples of Physical Layer Prof. Hemang Kothari Assistant Professor Computer Engineering Department MEFGI, Rajkot. Email: hemang.kothari@marwadieducation.edu.in
- 2. Signal & Data Level
- 3. 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
- 4. 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?
- 5. 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
- 6. 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
- 7. 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)
- 8. 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 bps B = bandwidth in Hz
- 9. 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
- 10. 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
- 11. 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.
- 12. 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
- 13. 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.
- 14. Band Pass Channel
- 15. 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.
- 16. 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
- 17. 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
- 18. Noisy Channel - Shannon’s Theorem • Shannon’s theorem gives the capacity of a system in the presence of noise. C = B log2(1 + SNR)
- 19. 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
- 20. 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
- 21. 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.
- 22. 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)
- 23. (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 bps Nyquist 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.

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