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Signals : Physical Layer

Signals : Physical Layer. Shashank Srivastava Motilal Nehru National Institute Of Technology, Allahabad. ANALOG DATA. Analog data refers to information that is continuous. Example: human voice. When someone speaks, an analog wave is created in the air.

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Signals : Physical Layer

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  1. Signals : Physical Layer Shashank Srivastava Motilal Nehru National Institute Of Technology, Allahabad

  2. ANALOG DATA • Analog data refers to information that is continuous. • Example: human voice. When someone speaks, an analog wave is created in the air. • This can be captured by a microphone and converted to an analog signal or sampled and converted to a digital signal.

  3. DIGITAL DATA • Digital data refers to information that has discrete values. • Example: A digital clock in which the value will change suddenly from 8.05 to 8.06. • Data are stored in computer memory in the form of 0s and 1s . It can be converted to a digital signal when it is transferred from one position to another inside computer or modulated into an analog signal for transmission across a medium.

  4. Analog and Digital Clocks

  5. Transformation of Information to Signals

  6. Comparison Of Analog and Digital Signals • An analog signal has infinitely many levels of intensity over a period of time. • Or • An analog signal can have an infinite number of values in a range. • As the wave moves from value A to B, it passes through an infinite number of values along its path. • A digital signal, can have only limited number of defined values often as simple as 1 and 0 .

  7. Comparison Of Analog and Digital Signals • Graphical representation: • The vertical axis represents the value of strength • The horizontal axis represent the time • The curve representing the analog signal passes through an infinite number of points. • The vertical lines of the digital signals shows the sudden jump that the signals makes from value to value.

  8. Analog and Digital Signals

  9. PERIODIC AND APERIODIC (NONPERIODIC) SIGNALS • A periodic signal completes a pattern with a measurable time frame called a period, and repeats that pattern over subsequent identical periods. • The completion of one full pattern is called a cycle. • Both digital and analog data can be periodic or non-periodic. • A non periodic signal changes without exhibiting a pattern or cycle that repeats over time.

  10. Periodic Signals

  11. Aperiodic Signals

  12. SIMPLE AND COMPOSITE ANALOG SIGNALS • A simple periodic analog signal, a sine wave , cannot be decomposed into simpler signals. • A composite periodic analog signal is composed of multiple sine waves. • Sine Waves This is most fundamental form of a periodic analog signal. • Each cycle consists of a single arc above the time axis followed by single arc below it.

  13. SINE WAVE • A sine wave can be represented by three parameters: peak amplitude , the frequency and the phase. • PEAK Amplitude: • The peak amplitude of a signal is the absolute value of its highest intensity , proportional to the energy it carries. • For electric signals, peak amplitude is measured in volts

  14. Period refers to the amount of time , in seconds, a signal needs to complete one cycle . • Frequency refers to the number of periods in 1 s • Frequency and periods are the inverse of each other. • F=1/T and T =1/f • Frequency is the rate of change with respect to time. Change in sort span of time means high frequency. Change over a long span of time means low frequency. • If signal does not change at all, its frequency is zero. • If a signal changes instantaneously, its frequency is infinite.

  15. Example Suppose that power has a frequency of 60 Hz. The period of this sine wave can be determined as follows: T=1/f=1/60 = 0.0166s = 0.0166 x 10 3 ms = 16.6 ms that the period of the power is 0.0116 s or 16.6 ms .Our eyes are not sensitive enough to distinguish these rapid changes.

  16. Sine Wave

  17. PHASES • Phase describes the position of the wave form relative to time 0 • Phase is measured in degree or radians. • A phase shift of 360 0 corresponds to a shift of a complete period. • A phase shift of 180 0corresponds to a shift of one-half of a period. • A phase shift of 90 0corresponds to a shift of one-quarter of a period.

  18. Point To Be Noted 1.A sine wave with phase of 0 0 starts with a zero amplitude. The amplitude is increasing. 2. A sine wave with a phase of 90 0 starts at time 0 with a peak amplitude .The amplitude is decreasing. 3. A sine wave with a phase of 1800 starts at time 0 with a zero amplitude. The amplitude is decreasing. In terms of Shift or offset 1.A sine wave with a phase of 00 is not shifted 2.A sine wave with a phase 900 is shifted to the left by 1/4 cycle . Note that the signal does not really exist before time 0. 3. A sine wave with a phase of 1800 is shifted to the left by 1/2 cycle. Note that the signal does not really exist before time 0.

  19. Phases

  20. Amplitude Change

  21. Frequency Change

  22. Phase Change

  23. EXAMPLE A sine wave is offset 1/6 cycle with respect to time 0. What is its phase in degrees and radians? Sol: One complete cycle is 360 0 Therefore 1/6 cycle =1/6 X 360=60 0 = 60 X 2 X 3.14/360 Rad =1.046 Rad

  24. WAVELENGTH • Wavelength : It is the distance a simple signal can travel in one period. • Wavelength binds the period or frequency of a simple sine wave to the propagation speed of the medium. • The wavelength depends on both the frequency and the medium. • Wavelength=propagation speed x period = propagation speed • Frequency • λ= c/f • λ is wavelength, c is propagation speed, and f is frequency. • It is normally measured in micrometers(microns).

  25. EXAMPLE The wavelength of red light(frequency = 4X1014) in air is =(3X108 / 4X1014 ) =0.75X10-6m =.75um Wavelength is measured in micrometers.

  26. Time and Frequency Domain • The Time domain plot shows changes in signal amplitude with respect to time.(Amplitude versus Time) • A frequency –domain is concerned with only the peak value and the frequency. • The advantage of the frequency domain is that we can immediately see the values of the frequency and peak amplitude. • A complete sine wave is represented by one spike in the frequency domain. • It is more compact and useful when we are dealing with more than one sine wave. • The phase of a signal cannot be shown in frequency domain.

  27. Time and Frequency Domain

  28. Examples

  29. COMPOSITE SIGNALS • A single frequency sine wave is not useful in data communications, we need to send a composite signal ,a signal made of many simple sine waves. • Example: A single sine wave can be used to send an alarm when a burglar opens a door in our house. • But if we send a single sine wave to convey data, we would be sending alternating 0s and 1s,which does not have any communication value. • A composite is a combination of simple sine waves with different frequencies ,amplitude , and phases. • It can be periodic or non-periodic .

  30. COMPOSITE SIGNAL • A periodic composite signal can be decomposed into a series of simple sine waves with discrete frequencies-frequencies that have integer values(1,2,3…) • A non-periodic composite signal can be decomposed into a combination of an infinite number of simple sine waves with continuous frequencies , frequencies that have real values. • In a time domain representation of the composite signal, there are an infinite number of simple sine frequencies. • A medium may pass some frequencies and may block or weaken others. So, we may not receive the same signal at the other end.

  31. Bandwidth • The range of frequencies contained in a composite signal is its bandwidth. • The bandwidth of a composite signals is the difference between the highest and the lowest frequencies contained in that signal. • The range of frequencies a medium can pass is called its bandwidth. Normally a medium cannot pass all frequencies, so bandwidth generally refers to the range of frequencies that a medium can pass without losing one-half of the power contained in that signal. • It is the difference between the highest and lowest frequencies that a medium can pass. • Example: If a composite signal contains frequencies between 1000 and 5000, its bandwidth is (5000-1000) or 4000.

  32. Bandwidth

  33. EXAMPLE If a periodic signal is decomposed into five sine waves with frequencies of 100, 400, 600,900 and 1200 Hz, what is its bandwidth? Draw the spectrum , assuming all components have a maximum amplitude of 20V. Solution: Let fh be the highest frequency and fl the lowest frequency , and B the bandwidth . Then B = fh-fl =1200-100 =1100 Hz

  34. DIGITAL SIGNALS Data can be represented by a digital signal. Example: a 1 can be encoded as a positive voltage and a 0 as zero voltage. A digital signal can have more than two levels. Example: A digital signal has eight levels. How many bits are needed per level ? We calculate the number of bits from the formula Number of bits per level=log2l=log28=3 Each signal level is represented by 3 bits.

  35. Digital Signal

  36. Amplitude, Period, and Phase for a Digital Signal

  37. Bit interval and Bit Rate Most digital signals are aperiodic, thus period or frequency is not appropriate. The bit interval or bit duration(instead of period) is the time required to send one bit. The bit rate (instead of frequency) is the number of bit intervals per second means the number of bits sent in 1s ,expressed in bits per second (bps). Bit interval=1/bit rate

  38. BIT LENGTH The bit length is the distance one bit occupies on the transmission medium. Bit Length=propagation speed X bit duration

  39. The bit rate and bandwidth are proportional to each other. B >= n/2 or n<=2B n is the bit rate, B is the bandwidth. The analog bandwidth of a medium is expressed in hertz, the digital bandwidth in bits per second. Digital bandwidth means the maximum bit rate that a medium can pass.

  40. Low-pass versus Band-pass A low-pass channel has a bandwidth with frequencies between 0 and f. A band-pass channel has a bandwidth with frequencies between f1 and f2. Digital transmission needs a low-pass channel. Analog signal uses a band-pass channel. In addition, bandwidth of analog signal can be shifted as long as width of bandwidth remains same. A band-pass channel is more available than a low-pass channel.

  41. Data rate limits Data rate depends on three factors: The bandwidth available. The levels of signals we can use. The quality of channel (the level of noise). Nyquist Bit Rate (for noiseless channel): Defines the maximum theoretical bit rate. BitRate = 2 * Bandwidth * - Bandwidth is the bandwidth of the channel, - L is the no. of signal levels used to represent data(example: if four signal levels are used, means for each level we send two bits), - BitRate is the bit rate in bits per second.

  42. Shannon Capacity (for noisy channel): Theoretical highest data rate for a noisy channel Capacity = Bandwidth * Bandwidth is the bandwidth of the channel, SNR is the signal to noise ratio = power of signal / power of noise, Capacity is the capacity of the channel in bits per second. This formula defines the characteristics of the channel, not the method of transmission. For increasing data rate, either increase bandwidth or improve signal-to-noise ratio. Example: if SNR is zero, capacity= B*0=0 means we cannot receive any data through this channel.

  43. Using both limits: We need to use both methods in practice. Example: we have a channel with 1 MHz bandwidth. The SNR of this channel is 63, what is the approximate bit rate and signal level? Solution: First we use Shannon formula to find upper limit. C= B * = 10^6 * = 10^6 * = 6 Mbps 6 Mbps is the upper limit. For better performance, we choose something lower, for example 4 Mbps. Now use Nyquist formula to find the number of signal levels: 4 Mbps= 2 * 1 MHz* gives L = 4.

  44. TRANSMISSION IMPAIRMENTS Attenuation –Attenuation means loss of energy. When a signal travels through a medium, it loses some of its energy in overcoming the resistance of the medium. Some of the electrical energy is converted into heat. To compensate this loss, amplifiers are used to amplify the signal. Decibel measures the relative strength of two signals or a signal at two different points. Decibel is negative if a signal is attenuated and positive if the signal is amplified. dB = 10 log10(P1/P2) where p1,p2 are powers of signal at points 1 and 2 respectively. Decibel numbers can be added(or subtracted) when we are talking about several points.

  45. 2. Distortion – Distortion means that signal changes its shape or form. Distortion occurs in a composite signal made of different frequencies. Each signal has its propagation speed and therefore its own delay in arriving at the final destination. Difference in delay may create a phase difference. So signal components at the receiver have phases different from what they had at the sender.

  46. 3. Noise –Noise like thermal noise, induced noise, cross talk, impulse noise may corrupt the signal. Thermal noise is the random motion of electrons in a wire which creates an extra signal not originally sent by the transmitter. Induced noise comes from sources and appliances. These devices act as sending antenna and the transmission medium act as receiving antenna. Cross talk is the effect of one wire on the other. One wire acts as sending antenna and other acts as receiving antenna. Impulse noise is a spike (a signal with high energy in a very short period of time) that comes from power lines, lightning and so on.

  47. PERFORMANCE- Bandwidth Network Throughput – Throughput is the measurement of how fast data can pass through an entity (such as a point on the network). Network throughput is measured in bits per second(bps). It is an actual measure of the how much data we can send.

  48. 3. Latency (Delay) – It is the time taken by an entire message to completely arrive at the destination from the time the first bit is sent out from the source machine. It is made up of four components • Propagation time • Transmission time • Queuing time • Processing time • Latency time = Propagation time + Transmission time + • Queuing time + Processing time

  49. Propagation time – Time taken by a bit to travel from source to the destination. It is given by Distance/Propagation speed Propagation speed is the distance a bit can travel through a medium in one second. Speed depends upon the medium and on the frequency of the signal. Transmission time – It is the time from the first bit until the last bit of a message has left the transmitting node. The packet transmission time in seconds can be obtained from the packet size in bit and the bit rate in bit/s as: Packet transmission time = Packet size / Bit rate or Transmission time = Message size/ bandwidth

  50. What are the propagation time and the transmission time for a 2.5-kbyte message (an e-mail) if the bandwidth of the network is 1 Gbps? Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 108 m/s. Solution

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