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Data Communications and Computer Networks Dr. Kavita Jaiswal Assistant Professor Department of Computer Science & Engineering IIIT Naya Raipur Module II  Introduction to Physical Layer One of the major functions of the physical layer i...

Data Communications and Computer Networks Dr. Kavita Jaiswal Assistant Professor Department of Computer Science & Engineering IIIT Naya Raipur Module II  Introduction to Physical Layer One of the major functions of the physical layer is to move data in the form of electromagnetic signals across a transmission medium. How data and signals can be either analog or digital. Analog refers to an entity that is continuous; digital refers to an entity that is discrete. Periodic analog signals used in data communication. The attributes of analog signals such as period, frequency, and phase are also explained. Nonperiodic digital signals used in data communication. The attributes of a digital signal such as bit rate and bit length. Transmission impairment: How attenuation, distortion, and noise can impair a signal. The data rate limit: How many bits per second we can send with the available channel. The data rates of noiseless and noisy channels. The performance of data transmission including bandwidth, throughput, latency, and jitter. Data and Signals Data are entities that convey meaning Examples: Computer files, music on CD Sigmals are the electric or electromagnetic encoding of data. Examples: Computer networks and data/voice communication systems transmit signals Data need to be transmitted and recieved, but the media have to change data to signals. Data and signals can be analog or digital Analog and Digital data Data can be analog or digital. The term analog data refers to information that is continuous; digital data refers to information that has discrete states. Analog data take on continuous values. Digital data take on discrete values. Example: Analog data, such as the sounds made by a human voice, take on continuous values. 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. Digital data take on discrete values. For example, data are stored in computer memory in the form of Os and 1s. They can be converted to a digital signal or modulated into an analog signal for transmission across a medium. Note Data can be analog or digital. Analog data are continuous and take continuous values. Digital data have discrete states and take discrete values.  Analog and Digital Signal An analog signal has infinitely many levels of intensity over a period of time. As the wave moves from value A to value B, it passes through and includes an infinite number of values along its path. A digital signal, on the other hand, can have only a limited number of defined values. Although each value can be any number, it is often as simple as 1 and 0. Note Signals can be analog or digital. Analog signals can have an infinite number of values in a range; digital signals can have only a limited number of values. Periodic and Nonperiodic A periodic signal completes a pattern within 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. A nonperiodic signal changes without exhibiting a pattern or cycle that repeats over time. Both analog and digital signals can be periodic or nonperiodic. Note In data communications, we commonly use periodic analog signals and nonperiodic digital signals.  Periodic Analog Signals Periodic analog signals can be classified as simple or composite. 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. A sine wave can be represented by three parameters: the peak amplitude, the fre- quency, and the phase. These three parameters fully describe a sine wave. Peak Amplitude The peak amplitude of a signal is the absolute value of its highest intensity, propor-tonal to the energy it carries. For electric signals, peak amplitude is normally measured in volts. Period and Frequency Period refers to the amount of time, in seconds, a signal needs to complete 1 cycle. Frequency refers to the number of periods in 1s. Note that period and frequency are just one characteristic defined in two ways. Frequency and period are the inverse of each other. Table : Units of period and frequency Example The power we use at home has a frequency of 60 Hz. The period of this sine wave can be determined as follows: Example The period of a signal is 100 ms. What is its frequency in kilohertz? Solution First we change 100 ms to seconds, and then we calculate the frequency from the period (1 Hz = 10−3 kHz). Note Frequency is the rate of change with respect to time. Change in a short span of time means high frequency. Change over a long span of time means low frequency. Note If a signal does not change at all, its frequency is zero. If a signal changes instantaneously, its frequency is infinite.  Phase Phase describes the position of the waveform relative to time 0. A sine wave with a phase of 0° is not shifted. A sine wave with a phase of 90° is shifted to the left by 1/4 cycle. However, note that the signal does not really exist before time 0. A sine wave with a phase of 180° is shifted to the left by 1/2 cycle. However, note that the signal does not really exist before time 0. Example A sine wave is offset 1/6 cycle with respect to time 0. What is its phase in degrees and radians? Solution We know that 1 complete cycle is 360°. Therefore, 1/6 cycle is Wavelength Wavelength is another characteristic of a signal traveling through a transmission medium. Wavelength binds the period or the frequency of a simple sine wave to the propagation speed of the medium Task: 1. what is the frequency of the signal in figure ? 2. The light of the sun takes approximately eight minutes to reach the earth. What is the distance between the sun and the earth? 3. A signal has a wavelength of 1um in air. How far can the front of the wave travel during 500 period Time-domain and Frequency-domain plots of a sine wave The time-domain plot shows changes in signal amplitude with respect to time(it is an amplitude versus time plot). To show the relationship between amplitude and frequency, we can use what is called a frequency domain plot. A frequency domain plot is concerned with only the peak value and frequency. Note A complete sine wave in the time domain can be represented by one single spike in the frequency domain. Example The frequency domain is more compact and useful when we are dealing with more than one sine wave. For example, Figure (a) shows three sine waves, each with different amplitude and frequency.  Composite Signals A composite signal is a signal made up of multiple simple sinusoidal signals. According to Fourier analysis, any composite signal is a combination of multiple simple sine waves with different amplitudes, frequencies and phases. A composite periodic signal Decomposition of a composite periodic signal in the time and frequency domains Example Figure shows a nonperiodic composite signal. It can be the signal created by a microphone or a telephone set when a word or two is pronounced. In this case, the composite signal cannot be periodic, because that implies that we are repeating the same word or words with exactly the same tone. Bandwidth The bandwidth of a composite signal is the difference between the highest and the lowest frequencies contained in that signal. Example If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is its bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V. Solution Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth. Then The spectrum has only five spikes, at 100, 300, 500, 700, and 900 Hz Example A periodic signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum if the signal contains all frequencies of the same amplitude. Solution Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth. Then Example A nonperiodic composite signal has a bandwidth of 200 kHz, with a middle frequency of 140 kHz and peak amplitude of 20 V. The two extreme frequencies have an amplitude of 0. Draw the frequency domain of the signal. Solution The lowest frequency must be at 40 kHz and the highest at 240 kHz. Figure shows the frequency domain and the bandwidth. Digital Signals In addition to being represented by an analog signal, information can also be represented by a digital signal. For 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. In this case, we can send more than 1 bit for each level.  Example A digital signal has eight levels. How many bits are needed per level? We calculate the number of bits from the formula Each signal level is represented by 3 bits. Question : A digital signal has nine levels. How many bits are needed per level? Bit Rate Bit Length Most digital signals are nonperiodic, and We discussed the concept of the thus period and frequency are not wavelength for an analog signal: the appropriate characteristics. distance one cycle occupies on the Another term bit rate (instead of transmission medium. frequency) is used to describe digital We can define something similar for a signals. digital signal: the bit length. The bit rate is the number of bits sent in 1 sec, expressed in bits per second (bps). The bit length is the distance one bit occupies on the transmission medium. Relation between bit rate and bit interval can be as follows Bit rate = 1 / Bit length Bit length = propagation speed × bit duration  Types of channels Each composite signal has a lowest possible(minimum) frequency and a highest possible (maximum) frequency. From the point of view of transmission, there are two types of channels: Low pass Channel Band pass channel This channel has the lowest frequency This channel has the lowest frequency as 0 as some non-zero frequency f1 and and highest frequency as some non-zero frequency f1. highest frequency as some non-zero frequency f2 This channel can pass all the frequencies in This channel can pass all the the range 0 to f1. frequencies in the range f1 to f2. Bandwidth of the channel A channel is the medium through which the signal carrying information will be passed. In terms of analog signal, bandwidth of the channel is the range of frequencies that the channel can carry. In terms of digital signal, bandwidth of the channel is the maximum bit rate supported by the channel. i.e. the maximum amount of data that the channel can carry per second. The bandwidth of the medium should always be greater than the bandwidth of the signal to be transmitted else the transmitted signal will be either attenuated or distorted or both leading in loss of information. The channel bandwidth determines the type of signal to be transmitted i.e. analog or digital. Transmission of Digital signal Digital signal can be transmitted in the following two ways: Baseband Transmission Broadband transmission Baseband Transimssion The signal is transmitted without making any change to it (ie. Without modulation) In baseband transmission, the bandwidth of the signal to be transmitted has to be less than the bandwidth of the channel. Ex. Consider a Baseband channel with lower frequency 0Hz and higher frequency 100Hz, hence its bandwidth is 100 (Bandwidth is calculated by getting the difference between the highest and lowest frequency). We can easily transmit a signal with frequency below 100Hz, such a channel whose bandwidth is more than the bandwidth of the signal is called Wideband channel. Logically a signal with frequency say 120Hz will be blocked resulting in loss of information, such a channel whose bandwidth is less than the bandwidth of the signal is called Narrowband channel. Broadband Transimssion Given a bandpass channel, a digital signal cannot be transmitted directly through it In broadband transmission we use modulation, i.e we change the signal to analog signal before transmitting it. The digital signal is first converted to an analog signal, since we have a bandpass channel we cannot directly send this signal through the available channel. Ex. Consider the bandpass channel with lower frequency 50Hz and higher frequency 80Hz, and the signal to be transmitted has frequency 10Hz. To pass the analog signal through the bandpass channel, the signal is modulated using a carrier frequency. Ex. The analog signal (10Hz) is modulated by a carrier frequency of 50Hz resulting in an signal of frequency 60Hz which can pass through our bandpass channel. Modulation of a digital signal for transmission on a bandpass channel Task: 1. What is the length of a bit in a channel with a propagation speed of 2 x 108 m/s if the channel bandwidth is a. 10 Mbps? b. 100 Mbps? c. 1 Gbps? 2. What is the bit rate for each of the following signals? a. A signal in which 1 bit lasts 0.001 s b. A signal in which 1 bit lasts 2 ms c. A signal in which 10 bits last 20 us 3. If the bandwidth of the channel is 5 Kbps, how long does it take to send a frame of 1,000,000 bits out of this device? 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. Attenuation It means loss of energy. The strength of signal decreases with increasing distance which causes loss of energy in overcoming resistance of medium. This is also known as attenuated signal. Amplifiers are used to amplify the attenuated signal which gives the original signal back and compensate for this loss. Attenuation is measured in decibels(dB). It measures the relative strengths of two signals or one signal at two different point.  Example Suppose a signal travels through a transmission medium and its power is reduced to one-half. This means that P2 is (1/2)P1. In this case, the attenuation (loss of power) can be calculated as A loss of 3 dB (–3 dB) is equivalent to losing one-half the power. Note that the decibel is negative if a signal is attenuated and positive if a signal is amplified. Example A signal travels through an amplifier, and its power is increased 10 times. This means that P2 = 10P1. In this case, the amplification (gain of power) can be calculated as Example One reason that engineers use the decibel to measure the changes in the strength of a signal is that decibel numbers can be added (or subtracted) when we are measuring several points (cascading) instead of just two. In Figure a signal travels from point 1 to point 4. In this case, the decibel value can be calculated as Example Sometimes the decibel is used to measure signal power in milliwatts. In this case, it is referred to as dBm and is calculated as dBm = 10 log10 Pm , where Pm is the power in milliwatts. Calculate the power of a signal with dBm = −30. Solution We can calculate the power in the signal as  Example 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? Solution The loss in the cable in decibels is 5 × (−0.3) = −1.5 dB. We can calculate the power as Distortion It means changes in the form or shape of the signal. This is generally seen in composite signals made up with different frequencies. Each frequency component has its own propagation speed travelling through a medium. And thats why it delay in arriving at the final destination. Every component arrive at different time which leads to distortion. Therefore, they have different phases at receiver end from what they had at senders end. Noise The random or unwanted signal that mixes up with the original signal is called noise. There are several types of noise such as induced noise, crosstalk noise, thermal noise and impulse noise which may corrupt the signal. Induced noise comes from sources such as motors and appliances. These devices act as sending antenna and transmission medium act as receiving antenna. Thermal noise is movement of electrons in wire which creates an extra signal. Crosstalk noise is when one wire affects the other wire. Impulse noise is a signal with high energy that comes from lightning or power lines Signal-to-noise ratio To find the theoretical bit rate limit, we need to know the ration. The signal-to-noise ratio is defined as SNR = AVG SIGNAL POWER / AVG NOISE POWER SNRdB= 10Log10SNR Figure Two cases of SNR: a high SNR and a low SNR Example 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: Data Rate Limits A very important consideration in data communications is how fast we can send data, in bits per second, over a channel. Data rate depends on three factors: 1. The bandwidth available 2. The level of the signals we use 3. The quality of the channel (the level of noise) Two theoretical formulas were developed to calculate the data rate: Nyquist for Noiseless Channel Shannon for noisy channel Noiseless Channel: Nquist Bit Rate The maximum data rate for noiseless channels is calculated using the Nyquist Theorem, which states: Maximum Data Rate = 2 * Bandwidth * log2(L) where L is the number of signal levels.  Example 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 We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need? Solution We can use the Nyquist formula as shown: Since this result is not a power of 2, we need to either increase the number of levels or reduce the bit rate. If we have 128 levels, the bit rate is 280 kbps. If we have 64 levels, the bit rate is 240 kbps. Noisy Channel: Shannon Capacity Shannon Theorem determines the maximum data rate for noisy channels, with the formula: Maximum Data Rate = Bandwidth * log2 (1 + SNR).  Example 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 Question ? 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? Note The Shannon capacity gives us the upper limit; the Nyquist formula tells us how many signal levels we need. Performance Up to now, we have discussed the tools of transmitting data (signals) over a network and how the data behave. One important issue in networking is the performance of the network--how good is it? Bandwidth Throughput Latency (Delay) Bandwidth-Delay Product Bandwidth In networking, we use the term bandwidth in two contexts. The first, bandwidth in hertz, refers to the range of frequencies in a composite signal or the range of frequencies that a channel can pass. The second, bandwidth in bits per second, refers to the speed of bit transmission in a channel or link. Throughput The throughput is a measure of how fast we can actually send data through a network. Although, at first glance, bandwidth in bits per second and throughput seem the same, they are different. A link may have a bandwidth of B bps, but we can only send T bps through this link with T always less than B. In other words, the bandwidth is a potential measurement of a link; the throughput is an actual measurement of how fast we can send data. Example A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network? Solution We can calculate the throughput as The throughput is almost one-fifth of the bandwidth in this case. Latency(Delay) The latency or delay defines how long it takes for an entire message to completely arrive at the destination from the time the first bit is sent out from the source. Latency = propagation time + transmission time + queuing time + processing delay Propagation time = Distance / (Propagation Speed) Transmission time = (Message size) / Bandwidth Example 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 We can calculate the propagation and transmission time as Figure Filling the link with bits for case 1 Figure Filling the link with bits in case 2 Note The bandwidth-delay product defines the number of bits that can fill the link. Task: 1. What are the propagation time and the transmission time for a 5-Mbyte message (an image) if the bandwidth of the network is 1 Mbps? Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 108 m/s. 2. What is the total delay (latency) for a frame of size 5 million bits that is being sent on a link with 10 routers each having a queuing time of 2 us and a processing time of 1 us. The length of the link is 2000 Km. The speed of light inside the link is 2 × 108 m/s. The link has a bandwidth of 5 Mbps. Which component of the total delay is dominant? Which one is negligible? 3. If the peak voltage value of a signal is 20 times the peak voltage value of the noise, what is the SNR? What is the SNRdB? 4. We measure the performance of a telephone line (4 KHz of bandwidth). When the signal is 10 V, the noise is 5 mV. What is the maximum data rate supported by this telephone line? Digital to Digital Conversion We see how we can represent digital data by using digital signals. The conversion involves three techniques: line coding, block coding, and scrambling. Line coding is always needed; block coding and scrambling may or may not be needed. Line Coding Line coding is the process of converting digital data to digital signals. At the sender, digital data are encoded into a digital signal; at the receiver, the digital data are recreated by decoding the digital signal. Characteristics of Line coding Signal element versus data elements Data Rate versus Signal Rate Bandwidth Baseline Wandering DC Components Self-synchronization Built-in Error Detection Immunity to Noise and Interference Complexity Figure Signal element versus data element Data Rate versus Signal Rate: The data rate defines the number of data elements (bits) sent in 1s. The unit is bits per second (bps). The signal rate is the number of signal elements sent in 1s. One goal in data communications is to increase the data rate while decreasing the signal rate. Increasing the data rate increases the speed of transmission; decreasing the signal rate decreases the bandwidth requirement. Example A signal is carrying data in which one data element is encoded as one signal element ( r = 1). If the bit rate is 100 kbps, what is the average value of the baud rate if c is between 0 and 1? Solution We assume that the average value of c is 1/2. The baud rate is then Baseline Wandering: In decoding a digital signal, the receiver calculates a running average of the received signal power. This average is called the baseline. The incoming signal power is evaluated against this baseline to determine the value of the data element. A long string of Os or 1s can cause a drift in the baseline (baseline wandering) and make it difficult for the receiver to decode correctly. A good line coding scheme needs to prevent baseline wandering. DC Components: When the voltage level in a digital signal is constant for a while, the spectrum creates very low frequencies (results of Fourier analysis). These frequencies around zero, called DC (direct-current) components, present problems for a system that cannot pass low frequencies or a system that uses electrical coupling (via a transformer). Self-synchronization: To correctly interpret the signals received from the sender, the receiver's bit intervals must correspond exactly to the sender's bit intervals. If the receiver clock is faster or slower, the bit intervals are not matched and the receiver might misinterpret the signals. Figure Effect of lack of synchronization  Built-in Error Detection: It is desirable to have a built-in error-detecting capability in the generated code to detect some of or all the errors that occurred during transmission. Immunity to Noise and Interference: Another desirable code characteristic is a code that is immune to noise and other interferences. Complexity:A complex scheme is more costly to implement than a simple one. For example, a scheme that uses four signal levels is more difficult to interpret than one that uses only two levels. Figure Line coding schemes Figure Unipolar NRZ scheme Figure Polar NRZ-L and NRZ-I schemes In polar schemes, the voltages are on both sides of the time axis. For example, the voltage level for O can be positive and the voltage level for 1 can be negative. In NRZ- L the level of the voltage determines the value of the bit. In NRZ- I the inversion or the lack of inversion determines the value of the bit. Figure Polar RZ scheme The main problem with NRZ encoding occurs when the sender and receiver clocks are not synchronized. One solution is the return-to-zero (RZ) scheme, which uses three values: positive, negative, and zero. In RZ, the signal changes not between bits but during the bit. In Figure we see that the signal goes to 0 in the middle of each bit. It remains there until the beginning of the next bit.  Figure Polar biphase: Manchester and differential Manchester schemes The idea of RZ (transition at the middle of the bit) and the idea of NRZ-L are combined into the Manchester scheme. In Manchester encoding, the duration of the bit is divided into two halves. The voltage remains at one level during the first half and moves to the other level in the second half. Differential Manchester, on the other hand, combines the ideas of RZ and NRZ-I.  Figure Bipolar schemes: AMI and pseudoternary In bipolar encoding, we use three levels: positive, zero, and negative. AMI means alternate 1 inversion. A neutral zero voltage represents binary O. Binary 1s are represented by alternating positive and negative voltages. A variation of AMI encoding is called pseudoternary in which the 1 bit is encoded as a zero voltage and the 0 bit is encoded as alternating positive and negative voltages.  Multi Level Scheme A Multi level coding scheme is known as mBnL where m is the length of the binary pattern. B means binary data, n is the length of the signal pattern and L is the number of levels in the signaling. A letter is often used in place of L: B (binary) for L = 2, T (ternary) for L = 3, and Q (quaternary) for L = 4. Note that the first two letters define the data pattern, and the second two define the signal pattern. Note In mBnL schemes, a pattern of m data elements is encoded as a pattern of n signal elements in which 2m ≤ Ln. Figure Multilevel: 2B1Q scheme Multitransition MLT-3 MLT-3 technique uses more than two transition rules. The multiline transmission, three-level (MLT-3) scheme uses three levels (+V, 0, and –V) and three transition rules to move between the levels. If the next bit is 0, there is no transition. If the next bit is 1 and the current level is 0, the next level is the opposite of the last nonzero level. If the next bit is 1 and the current level is not 0, the next level is 0. Figure Multitransition: MLT-3 scheme  Table Summary of line coding schemes 4.25  Task: Q. Find the 8-bit data stream for each case depicted in the following figure. Block Coding We need re d u n d a n cy to ensure synchronization and to provide some kind of inherent error detecting. Block coding can give us this redundancy and improve the performance of line coding. In general, block coding changes a block of m bits into a block of n bits, where n is larger than m. Block coding is referred to as an mBInB encoding technique. Types of Block Coding I. 4B/5B It replaces each blocks of 4-bits with block of 5-bits. A 4 bit data word can have 24=16 combinations. 5 bit word can have 25=32 combinations. We therefore have 32 - 16 = 16 extra words. Some of the extra words are used for control/signalling purposes.  Figure Using block coding 4B/5B with NRZ-I line coding scheme 4.29  Table 4B/5B mapping codes 4.30 Types of Block Coding II. 8B/10B It replaces each blocks of 8-bits with block of 10-bits. It provides great error detection capability than 4B/5B. The 8B/1OB block coding is actually a combine tion of 5B/6B and 3B/4B encoding, Scrambling The function of the scrambling is to remove long string of ones (1s) and zeros (0s) from the digital binary data. Scrambler is used in physical layer transmitter where as descrambler is used at receiver. Advantages of Scrambling: It does not increase data rate unlike block coding technique. It eliminates long string of 0s to provide more transitions in the data. This helps receiver for synchronization to recover the original bit pattern. It does not have any DC components as it creates balance between positive voltage levels and negative levels. It offers error detection capability. Figure AMI used with scrambling Figure Two cases of B8ZS scrambling technique Note B8ZS substitutes eight consecutive zeros with 000VB0VB. Figure Different situations in HDB3 scrambling technique Note HDB3 substitutes four consecutive zeros with 000V or B00V depending on the number of nonzero pulses after the last substitution. Transmission Modes The transmission of binary data across a link can be accomplished in either parallel or serial mode. In parallel mode, multiple bits are sent with each clock tick. In serial mode, 1 bit is sent with each clock tick. Parallel Trasmission Serial Trasmission Figure Asynchronous transmission In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 or more stop bits (1s) at the end of each byte. There may be a gap between each byte. Figure Synchronous transmission In synchronous transmission, we send bits one after another without start or stop bits or gaps. It is the responsibility of the receiver to group the bits. Isochronous transmission In real-time audio and video, in which uneven delays between frames are not accept-able, synchronous transmission fails. For example, TV images are broadcast at the rate pres of 30 images per second; they must be viewed at the same rate. If each image is sent by using one or more frames, there should be no delays between frames. For this type of application, synchronization between characters is not enough; the entire stream of bits must be synchronized. The isochronous transmission guarantees that the data arrive at a fixed rate.. Task: Q.Draw the graph of the NRZ-I scheme using each of the following data streams, assuming that the last signal level has been positive. From the graphs, guess the bandwidth for this scheme using the average number of changes in the signal level. a. 00000000 b.11111111 c. 01010101 d. 00110011.