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Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
Lecture 2   physical layer
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Lecture 2 physical layer

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  • 1. Physical Layer Hemang Kothari Assistant Professor Computer Engineering Department MEFGI, Rajkot. Email: hemang.kothari@marwadieducation.edu.in Computer Engineering Department - MEFGI 1
  • 2. Module 2 - Content • Bandwidth • Maximum data rate of a signal • Guided and unguided transmission media. Computer Engineering Department - MEFGI 2
  • 3. Signals • Analog and digital • Aperiodic and periodic signals
  • 4. Transformation of Information to Signals
  • 5. Analog and Digital Clocks
  • 6. Analog and Digital Signals
  • 7. Periodic Signals
  • 8. Aperiodic Signals
  • 9. Phases
  • 10. Amplitude Change
  • 11. Frequency Change
  • 12. Phase Change
  • 13. Time and Frequency Domain
  • 14. Examples
  • 15. Bandwidth
  • 16. Digital Signal
  • 17. Bit Rate and Bit Interval
  • 18. Corruption Due to Insufficient Bandwidth
  • 19. Two digital signals: one with two signal levels and the other with four signal levels
  • 20. A digital signal has eight levels. How many bits are needed per level? We calculate the number of bits from the formula Example Each signal level is represented by 3 bits.
  • 21. What about a digital signal with 16 levels? How many bits are needed per level? What about 32 levels? 64 levels? 128 levels? What about 9 levels?? 2? = 9? 3.17 bits However, this answer is not realistic. Example
  • 22. 3.24 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 Example
  • 23. 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. Example The TV stations reduce this rate to 20 to 40 Mbps through compression.
  • 24. 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. • Consider first a noise-free channel of Bandwidth B.
  • 25. 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 (assuming 2 symbols/per cycle and first harmonic).  Nyquist theorem states that for a noiseless channel: C = 2 B log22n C= capacity in bps B = bandwidth in Hz
  • 26. 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 Example
  • 27. 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 Example
  • 28. 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. Note
  • 29. Modulation of a digital signal for transmission on a bandpass channel
  • 30. 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.
  • 31. Attenuation
  • 32.  Signal to noise ratio shows the ratio of signal power to noise power  Power often expressed in watts  S/N = signal power/noise power 34 Signal to Noise Ratio (SNR or S/N)
  • 33. 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: Example
  • 34. Shannon’s Theorem  Shannon’s theorem gives the capacity of a system in the presence of noise. C = B log2(1 + SNR)
  • 35. 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 Example This means that the capacity of this channel is zero regardless of the bandwidth. In other words, we cannot receive any data through this channel.
  • 36. We can calculate the theoretical highest bit rate of a regular telephone line. 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 Example This means that the highest bit rate for a telephone line is 34.860 kbps. If we want to send data faster than this, we can either increase the bandwidth of the line or improve the signal-to-noise ratio.
  • 37. The signal-to-noise ratio is often given in decibels. Assume that SNRdB = 36 and the channel bandwidth is 2 MHz. The theoretical channel capacity can be calculated as Example
  • 38. Review  Example: 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
  • 39. Review  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.
  • 40. According to the Shannon’s Channel capacity theorem: Channel capacity C = B*log (1 + S/N) Where B = Bandwidth and S/N = Signal to Noise Ratio. Answer the following questions with respect to the Channel capacity theorem: (1) When the Bandwidth increases, what happens? (2) When the Signal to Noise Ratio increases, what happens? (3) When the Signal to Noise Ration S/N tends to infinity what happens? (4) When the Bandwidth tends to infinity what happens?
  • 41. Solution: (1) When the Bandwidth increases, what happens? As the bandwidth of the channel increases, it is possible to make faster changes in the information signal, thereby increasing the information rate and hence the channel capacity. (2) When the Signal to Noise Ratio increases, what happens? As S/N increases, one can increase the information rate while still preventing errors due to noise. (3) When the Signal to Noise Ration S/N tends to infinity what happens? As S/N -> infinity, an infinite information rate is theoretically possible irrespective of the bandwidth of the channel. (4) When the Bandwidth tends to infinity what happens? As Bandwidth -> infinity, the channel capacity does not approach infinity since with an increase in Bandwidth, the noise power also increases.
  • 42. 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)
  • 43. (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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