Analog and Digital Transmission
1. What do you mean by data and signal?
Ans: Data refers to an entity that expresses some meaning based on rules which are agreed upon by the sender and the receiver. It can be available in various forms such as text, graphics, audio and video. Signal refers to the representation of the data in a form (such as electrical, electronic or optical) suitable for transmission over a transmission medium.
2. Define analog and digital data.
Ans: Analog data is defined as the data having continuous states. For example, sounds produced while speaking take continuous values. Digital data is defined as the data having discrete states. For example, data to be stored in the memory of computer is in the form of 0s and 1s.
3. Distinguish analog and digital signals?
Ans: Data is transmitted across a transmission medium in the form of electromagnetic signals. An electromagnetic signal can be either analog or digital. A signal that passes through and includes a wide range of varying values of intensity over a period of time is referred to as analog signal [Figure 3.1(a)]. In contrast, a signal that has only a finite range of values (generally 0 and 1) is referred to as digital signal [Figure 3.1(b)]. Either of the analog or digital signals can be used to transmit either analog or digital data.
4. What do you understand by periodic and non-periodic signals?
Ans: Both analog and digital signals can be either periodic or non-periodic. A periodic signal exhibits a specific signal pattern that repeats over time [Figures 3.2(a) and 3.2(b)]. Sine waves and square waves are the most common examples of periodic analog and digital signals, respectively. On the other hand, a non-periodic (or aperiodic) signal does not repeat any specific signal pattern over time [Figures 3.2(c) and 3.2(d)]. Usually, in data communications, periodic analog and non-periodic digital signals are used.
5. Describe the parameters that characterize the sine waves.
Ans: The sine wave is the most fundamental form of periodic analog signal. It is characterized by the following three parameters.
|Peak Amplitude: It is the highest value or strength of the signal at any point of time [Figure 3.3(a)]. The unit for amplitude depends on the type of the signal. For example, in case of electrical signals, the unit of amplitude is normally volts and amperes.|
Frequency: It refers to the number of cycles a signal completes in 1 s. It is measured in cycles per second or Hertz (Hz). Another parameter equivalent to frequency is the period which is defined as the time taken by a signal to complete one cycle [Figure 3.3(a)]. Thus, we can say that frequency is equal to the number of periods per second. Both frequency (f) and period (T) are related to each other by the following formula.
f = 1/T
For example, if a signal wave completes one period in 1 s, its frequency is 1 Hz.
|Phase: It refers to the measure of shift in the position of a signal with respect to time 0. Figure 3.3(b) shows a shift of 90° in the sine wave shown in Figure 3.3(a). Phase is measured in degree (°) or radians (rad) where 1° = 2π/360 rad.|
6. Define wavelength. What is the relation between wavelength and frequency?
Ans: Wavelength refers to the distance covered by the signal in one cycle. In other words, it is the distance between two similar points in corresponding phases of two adjacent cycles of a signal (Figure 3.4). It is usually denoted by λ.
Wavelength relates the frequency (f) of a signal with the speed of propagation of the medium through which the signal is being transmitted. The relation between frequency and wavelength is expressed using the following formula.
λ = propagation speed/f
⇒ λ = propagation speed × period
For example, in vacuum, the signal propagates with speed of light (c), which is equal to 3 × 108 m/s. Thus,
λ = c/f
7. Distinguish time-domain and frequency-domain representations.
Ans: A signal can be represented in two ways: time-domain and frequency-domain. In time-domain representation, the signal is represented as a function of time. The time-domain plot of a signal depicts the changes in the amplitude of a signal with time. Figure 3.5 represents the sine wave using time-domain plot. The horizontal axis represents the time and the vertical axis represents the amplitude.
In frequency-domain representation, a signal is represented as a function of frequency. Unlike time-domain plot, the frequency-domain plot does not depict the changes in amplitude; rather it depicts only the peak amplitude of the signal and frequency. In addition, the complete sine wave is represented just by a spike. Figure 3.6 shows the frequency domain plot for the sine wave shown in Figure 3.5.
8. What do you understand by composite signal?
Ans: In data communications, single frequency sine waves are not used to communicate data, rather composite signals are used. A composite signal is a collection of one or more signals having different frequencies, amplitude and phases (Figure 3.7). It can be periodic or non-periodic. The decomposition of a periodic composite signal produces a series of signals with discrete frequencies, whereas the decomposition of a non-periodic composite signal produces a combination of signals with continuous frequencies in the range of zero to infinity.
9. Why digital transmission is preferred over analog transmission?
Ans: Analog transmission is a means of transmitting analog signals (representing either analog or digital data) while the digital transmission is a means of transmitting digital signals (representing either analog or digital data). Digital transmission offers certain advantages due to which it is preferred over analog transmission. These advantages are as follows:
|To achieve long-distance analog transmission, amplifiers are to be used for boosting up the signal energy. However, the amplifiers boost up the noise too, thereby resulting in more distorted signal. In contrast, repeaters used in digital transmission do not contribute to noise and other impairments. As a result, the data can be transmitted to longer distances while maintaining the data integrity. The digital data as well as the digitized analog data can be easily encrypted using encryption techniques.|
|With the advent of large-scale integration (LSI) and very large-scale integration (VLSI) technologies, the size and cost of digital equipments have reduced to a greater extent as compared to analog equipments.|
|To utilize the large channel capacity effectively, multiplexing is required and it is quite easier and economical to accomplish this using digital techniques than analog techniques.|
|All data including voice, video and digital data can be integrated and passed through digital circuits conveniently.|
10. How Fourier analysis is important in data communication?
Ans: Fourier analysis is a tool that is used to convert a time domain signal to a frequency domain signal and vice versa. According to this analysis, a composite periodic signal with a period T can be decomposed into a series called Fourier series of sine and cosine functions with each function being an integral harmonic of the fundamental frequency f of the composite signal as shown here.
Fourier series converts time domain of a periodic signal to discrete frequency domain. To convert time domain of a non-periodic signal to continuous frequency domain, Fourier transform is used as expressed below:
Inverse Fourier transform is expressed by the following formula:
11. Define bit interval, bit rate and bit length of a digital signal.
|Bit Interval: It is the time required to transmit one bit.|
|Bit Rate: The number of bits transmitted per second is known as bit rate. It can also be defined as the number of bit intervals per second. Its unit is bits per second (bps). The bit rate for a bit having bit interval t will be 1/t.|
|Bit Length: The distance occupied by a single bit while transmission on a transmission medium is known as bit length. It is related to bit interval as per the formula given below:|
Bit length = bit interval × speed of propagation
12. Differentiate baseband and broadband transmissions?
Ans: Though both baseband and broadband transmissions are the approaches used to transmit digital signals, there are certain differences between the two. These differences are listed in Table 3.1.
|Baseband transmission||Broadband transmission|
|• The digital signal is sent over a channel without converting it into an analog signal.||• The digital signal is first converted into analog signal and then sent over the channel.|
|• The entire bandwidth of the cable is consumed by a single signal.||• Signals are sent on multiple frequencies, allowing multiple signals to be sent simultaneously.|
|• It requires a low-pass channel—a channel whose bandwidth starts from zero.||• It requires a bandpass channel—a channel whose bandwidth does not start from zero.|
13. Discuss different types of transmission impairments.
Ans: Sometimes, when a signal passes through a transmission medium, it gets deteriorated due to imperfections in the transmission medium. As a result, the signal received at the receiver's end differs from the one being sent. For example, while transmitting digital signals, bit errors may occur such that in the received signal, some binary 1 is transformed to binary 0 or vice versa. The types of impairments that may occur during transmission are as follows:
|Attenuation: While a signal passes through a transmission medium, it loses some of its strength (or weakens) to get over the resistance of transmission medium. This loss of strength or energy is termed as attenuation. In case the signal strength becomes very low, it cannot be detected and interpreted properly at the receiver's end. To compensate the loss of strength, amplifiers or repeaters can be used. The extent of attenuation of a signal depends greatly on its frequency; the effect of attenuation is different for different frequency components. For example, during the transmission of a composite signal, some frequency components bypass without any attenuation, some lose little bit strength while others get blocked. This dependence of attenuation upon frequency leads to a distortion what is called attenuation distortion, which is more prevalent in analog signals than digital signals. However, its effect can be reduced by using a suitable equalizer between the channel and the receiver.|
|Delay Distortion: This is another type of distortion that results because of difference in the delays experienced by different frequency components while passing through the transmission medium. Since the speed of propagation of a signal varies with frequency, different frequency components of a composite signal arrive at the receiver at different times leading to delay distortion. As a result, the shape of received signal changes or gets distorted. Like attenuation distortion, the effect of delay distortion can be neutralized with the use of equalizers. However, unlike attenuation distortion, delay distortion is predominant in digital signals.|
|Noise: When a signal transits through a transmission medium, some types of undesired signals may mix with it such as intermodulation noise and crosstalk. Intermodulation noise occurs in the cases where signals with different frequencies pass through the same transmission medium. In such cases, the frequencies of some signals may combine (add or subtract) to generate new frequency components which may interfere with the signals of same frequency sent by the transmitter. This leads to distortion in the signal what is known as intermodulation noise. Crosstalk results when the electromagnetic signals passing through one wire are induced on other wires due to the close proximity of wires.|
14. What does the signal-to-noise ratio (SNR) determine?
Ans: The SNR is used to determine the effect of noise on the signal. Formally, it is defined as the ratio of the average power of a signal to the average noise power, as shown here.
SNR = avg(Ps)/avg(PN)
More the value of SNR is, the less is the signal distorted due to noise and vice versa. SNR is usually expressed in decibel (dB) units. SNRdB is defined as:
SNRdB = 10 log10 SNR
15. Write down the theoretical formula to calculate data rate for a noiseless channel and a noisy channel.
Ans: Data rate for a noiseless channel can be determined by using the Nyquist bit rate formula as given below.
Bit rate = 2 × B × log2 L,
where B = bandwidth of channel and L = the number of signal levels used to represent data. Data rate for a noisy channel can be determined by Shannon capacity, as given below.
C = B × log2(1 + SNR),
where C = data rate and B = bandwidth of channel.
16. Describe the factors used to measure the performance of a network?
Ans: The performance of a network is one of the most important issues in networking. Some major factors that measure the performance of a network are as follows.
|Bandwidth: It is one of the main characteristics that measure the performance of a network. It can be used in two different contexts. First is the bandwidth in hertz which refers to the range of frequencies in a composite signal and second is the bandwidth in bps which refers to the number of bits that can be transmitted by a channel or network in 1 s. An increase in bandwidth in hertz implies the increase in bandwidth in bps.|
|Throughput: It is the measure of how much data can be transmitted through a network in 1 s. Though from the definitions throughput sounds similar to bandwidth, it can be less than bandwidth. For example, a network may have bandwidth of 10 Mbps but only 2 Mbps can be transmitted through it.|
|Latency (Delay): It refers to the time elapsed between the first bit transmitted from the source and the complete message arrived at the destination. It is the sum of propagation time, transmission time, queuing time and processing delay of a frame.
The time taken by a bit to travel from the source to destination is referred to as the propagation time. It can be calculated using the following formula.
Propagation time = distance/propagation speed
As the propagation speed increases, propagation time decreases.
The transmission time measures the time between the transmission of first bit from the sender's end and the arrival of last bit of the message at the destination.
Transmission time = message size/bandwidth
Greater is the message size is, the more will be the transmission time.
Whenever a message being transmitted arrives at some intermediate device, such as router, it is kept in a queue (if device is not free) maintained by the device. The device processes the queued messages one by one. The time a message spends in the queue of some intermediate or end device before being processed is referred to as the queuing time. It depends on the traffic load in the network. More the load of traffic is, the more is the queuing time.
|Bandwidth-Delay Product: It refers to the product of bandwidth and delay of a network which specifies the maximum number of bits that can be at any time on the link. For example, if a link has a bandwidth of 8 bps and delay is 5 s, then the maximum number of bits that can fill the link is 40 bits.|
|Jitter: It refers to a problem that occurs when the inter-arrival time between packets is not constant and the application using the data is time sensitive.|
17. Define line coding. List some common characteristics of line coding schemes.
Ans: Line Coding is defined as the process of converting digital data or sequence of bits to digital signals. Encoder is used at the sender's end to create a digital signal from digital data and decoder is used at the receiver's end to recreate digital data from digital signal (Figure 3.8).
Some common characteristics of line coding schemes are as follows:
|Data Element: It refers to the smallest entity (that is, bit) that expresses some information. It describes what is to be sent.|
|Signal Element: It refers to the shortest unit of digital signal. Each signal element carries one or more data elements. The ratio (r) of number of data elements to number of signal element indicates the number of data elements carried by a signal element.|
|Data Rate: Data rate, also called bit rate, refers to the number of data elements that can be transmitted per second. It is usually expressed in bps. The speed of transmission increases with the increase in data rate.|
|Signal Rate: Signal rate, also called baud rate or modulation rate or pulse rate, refers to the number of signal elements transmitted per second. It is expressed in baud. An increase in signal rate increases the demand of bandwidth. The relationship between data rate and signal rate can be expressed as:
S = CF × d × (1/r),
where S = the number of signal elements; CF = case factor; D = data rate and r = the number of data elements carried by a signal element.
|Bandwidth: Bandwidth refers to the maximum volume of data that can be transferred over any communication medium at a given point of time. More the data needed to be transmitted in a given period is, the more is the bandwidth required. On digital circuits, bandwidth is measured in bps. The bandwidth is related to signal rate (S) as per the formula shown below.
Similarly, if we are given the bandwidth (B) of channel, we can find out the maximum data rate (dmax) using the following formula.
dmax = B × r × (1/CF)
|DC Components: A signal may include very low frequency component in its spectrum. These components with frequencies around zero (called DC components) are undesirable, as some systems do not allow such components to pass through them. Thus, for these systems, the line coding with no DC component is required.|
|Baseline Wandering: Baseline refers to the average of the received signal power calculated by the receiver while decoding a digital signal. Baseline wandering refers to the movement or drift in the baseline. Due to baseline wandering, the process of decoding at the receiver's end becomes difficult. Thus, a good line coding scheme must prevent baseline wandering.|
|Error Detection: Line coding scheme should have built-in capability to detect some or all the errors that occur during transmission.|
|Self-Synchronization: Bit intervals of the receiver should match with bit intervals of the sender to interpret correctly the signal at the receiver's end. For this, the digital signal must include timing information (that is, self-synchronized) which means the transitions in the digital signal alert the receiver at different points of the pulse. This information is used to readjust or reset a clock if the receiver's clock is not synchronized.|
|Resistance to Noise: Line coding scheme should have ability to create a code that is unsusceptible to noise or other inferences in the transmission medium.|
|Complexity: Line coding scheme should be simple, as it is quite expensive to implement a complex coding scheme as compared to a simple scheme.|
18. Can the bit rate be less than the pulse rate? Why or why not?
Ans: Bit rate is always greater than or equal to the pulse rate because the relationship between pulse rate and bit rate is defined by the following formula.
Bit rate = pulse rate × log2L,
where L denotes the number of data levels of the signal and log2L denotes the number of bits per level.
If a pulse carries only 1 bit (that is, log2L=1), the pulse rate and the bit rate are the same. However, if the pulse carries more than 1 bit, then the bit rate is greater than the pulse rate.
19. Discuss various line coding schemes.
Ans: There are various line coding schemes that can be classified into three basic categories namely, unipolar, polar and bipolar.
This scheme uses two voltage levels of a signal and both of these voltage levels are on one side of the time axis (above or below). In this scheme, bit rate and baud rate are equal. However, the encoded signal includes DC components and there is lack of synchronization in case of long series of 0s and 1s. The only coding scheme that falls under this category is non-return-to-zero (NRZ).
|NRZ: In this scheme, bit 1 is used to define positive voltage while bit 0 is used to define zero voltage of a signal. The name NRZ comes from the fact that the signal does not return to zero during the middle of a bit rather only between two bits (Figure 3.9). The unipolar NRZ scheme is not generally used for data communication.
This scheme uses two voltage levels of a signal, positive and negative, that can be on both sides of the time axis. The positive voltage may represent 0 while negative voltage may represent 1 or vice versa. Four different schemes fall under this category, which are discussed as follows.
|NRZ: It is the most common type of polar coding scheme. In this scheme, positive voltage is used to represent one binary value and negative voltage is used to represent another. There are two types of polar NRZ schemes namely, NRZ-level (NRZ-L) and NRZ-invert (NRZ-I). In NRZ-L, the value of bit is determined by the signal level which remains constant during bit duration (Figure 3.10). In NRZ-I, the value of bit is determined by inversion or lack of inversion in the signal level. If there is change in the signal level, the value of bit will be 0, whereas if there is lack of change in the signal level, the value of bit will be 1 (Figure 3.11). Both NRZ-L and NRZ-I suffer from synchronization problem in case of long sequence of 0s. However, NRZ-I suffers in case of long sequence of 1s also. In addition, both NRZ-L and NRZ-I have DC component problem.|
|Return-to-Zero (RZ): This scheme solves the synchronization problem of NRZ scheme. It uses three values of voltage namely, zero, positive and negative. Unlike NRZ, the signal changes during the middle of a bit but not between the bits. Once it changes during the bit, it remains there until the next bit starts. Thus, to encode a single bit, two signal changes are required (Figure 3.12). There is no DC component problem in RZ coding scheme. However, it is complex as it requires three levels of voltage.|
|Manchester Scheme: This is a biphase scheme that combines the idea of RZ and NRZ-L schemes. In this scheme, the level of voltage changes at the middle of bit to provide synchronization (Figure 3.13). The low-to-high transition in the middle of a bit indicates 1 while the high-to-low indicates 0. This scheme overcomes the disadvantages of NRZ-L scheme.|
|Differential Manchester: This is also a biphase scheme. It combines the idea of RZ and NRZ-I schemes. It is similar to the manchester scheme in the sense that it changes during the middle of a bit but the only difference is that the bit values are determined at the starting of bit. Transition occurs when the next bit is 0; otherwise, no transition occurs (Figure 3.14). This scheme overcomes all the disadvantages of NRZ-I scheme.
This scheme is similar to RZ encoding scheme and has three levels of voltage. The only difference is that bit 0 is represented by zero level and bit 1 is represented by positive and negative levels of voltage. In bipolar encoding, a stream of bits containing long sequence of binary 0s creates a constant zero voltage. Therefore, bipolar encoding schemes does not have DC component. This scheme is available in two forms, which are as follows:
|Alternate Mark Inversion (AMI): This is the most commonly used bipolar coding scheme. It is so called because it involves representation of binary 1s by alternate positive and negative levels of voltages. Here, bit 0 is represented using zero level of voltage (Figure 3.15). This scheme is used for communication between devices placed at a large distance from each other. However, it is difficult to achieve synchronization in this scheme when a continuous stream of bit 0 is present in the data.|
|Pseudoternary: This scheme is a modification of AMI encoding scheme. In this scheme, binary 1 is represented using zero level of voltage while binary 0s are represented by alternate positive and negative voltages (Figure 3.16).|
20. Explain the concept of block coding.
Ans: Block Coding is an alternative to line coding scheme that is also used to convert digital data to digital signals. However, it is much better than line coding, as it ensures synchronization and has built-in error detecting capability, which results in better performance than line coding. In this scheme, a three-step process (Figure 3.17) is used to code the digital data.
|1.||Division: The original sequence of bits is divided into blocks or groups of n bits each.|
|2.||Substitution: Each n-bit group is replaced with m bits where m >n.|
|3.||Line Coding: An appropriate line coding scheme is used to combine the m-bit groups to form a stream.|
The block coding is usually represented as nB / mB (n binary/m binary) such as 4B/5B and 5B/6B. For example, in 4B/5B coding, original bit sequence is divided into 4–bit codes and each 4–bit code is replaced with a 5–bit block and then NRZ-I line coding is used to convert 5–bit groups into digital signal.
21. Explain pulse code modulation (PCM) and delta modulation (DM).
Ans: Both PCM and DM are the techniques used to convert analog signal to digital data.
Pulse Code Modulation
This technique involves PCM encoder which encodes analog signal using three steps, namely, sampling, quantization and encoding (Figure 3.18).
|1.||Sampling: In this process, the numbers of samples of the original analog signal are taken at regular intervals of time (called sampling period). The inverse of sampling period is referred to as sampling rate. According to Nyquist theorem, the sampling rate should be at least two times of the highest frequency contained in the signal to regenerate the original analog signal at the receiver's end. For example, during the sampling of voice data, with frequency in the range of 400–5,000 Hz, 10,000 samples per second are sufficient for the coding. The sampling process is also termed as pulse amplitude modulation (PAM) because it produces a PAM signal—a series of pulses having amplitude between the highest and the lowest amplitudes of the original analog signal. Sample and hold is the most common sampling method used today for creating flat-top samples.|
|2.||Quantization: The sampled PAM pulses may have non-integral values of amplitude which cannot be encoded directly. Thus, the sampled pulses need to be quantized and approximated to integral values using analog-to-digital converter. Considering the original analog signal has amplitudes between Vmaximum and Vminimum, the following steps are used for quantizing the signal.|
|i)||The range of amplitudes of analog signal is partitioned into L levels or zones each of height h, where
h = (Vmaximum – Vminimum)/L.
The value of L is chosen depending on the range of amplitudes of the original analog signal as well as the extent of accuracy required in the recovered signal. The signal having more amplitude values require more number of quantization levels.
|ii)||The quantized values of 0 to L–1 are assigned at the midpoint of each zone.|
|iii)||The value of the sample amplitude is approximated to the quantized value.
Since the output values of the quantization process are only the approximated values, an error known as quantization error may occur due to which it may not be possible to recover the original signal exactly. The quantization error also affects the signal-to-noise ratio (SNRdB) of the signal and the amount of effect depends on the value of L or the number of bits per sample (nb) as shown in the following formula.
SNRdB = 6.02 nb + 1.76 dB.
The effect of quantization error can be minimized by using a process called companding and expanding. This process uses a compressor before encoding and uses an expander after decoding. Companding refers to decreasing the instantaneous voltage amplitude for larger values while expanding refers to increasing the instantaneous voltage amplitude for smaller values. This helps to improve the SNRdB of the signal.
|3.||Encoding: After quantization, encoding is done in which each sample is converted to m-bit codeword where m is equal to number of bits per sample (nb). The value of nb depends on the value of L as shown in the following formula.
m = nb = log2 L
The relationship between bit rate and the number of bits per sample (nb) can be expressed as:
Bit rate = nb × sampling rate
At the receiver's side, original signal is recovered using PCM decoder which converts the codeword into a staircase signal. The staircase signal is formed by changing the codeword into a pulse that maintains the amplitude till the next pulse. Then, a low-pass filter is used to smoothen the staircase signal into an analog signal. Figure 3.19 depicts this process.
This is an alternative to PCM technique with much reduced quantization error. In this technique, a modulator is used at the sender's side that produces the bits from an analog signal and these bits are sent one after another; only one bit is sent per sample. The modulator generates a staircase signal against which analog signal is compared. At each sampling interval, the amplitude value of analog signal is compared with last amplitude of staircase signal to determine the bit in the digital data. If amplitude of analog signal is smaller, the next bit will be 0. However, if the amplitude of analog signal is greater, the next bit will be 1. Figure 3.20 shows the components of DM.
To reproduce analog signal from digital data, demodulator is used at the receiver's end. The demodulator uses staircase maker and delay unit to generate analog signal which is then passed through a low-pass filter for smoothing. Figure 3.21 depicts the delta demodulation process.
22. What are the differences between serial and parallel transmissions?
Ans: The transmission of binary data across a link connecting two or more digital devices can be performed in either of two modes: serial and parallel (Figure 3.22). There are certain differences between these modes which are listed in Table 3.2.
|Serial transmission||Parallel transmission|
|• The data is transmitted by sending one bit per each clock pulse [see Figure 3.21(a)].||• The stream of bits is divided into groups and one group is sent per each clock pulse [see Figure 3.21(b)].|
|• It is a slower mode of transmission, as only one bit can be transmitted at a time.||• It is a faster mode of transmission, as several bits can be transmitted at a time.|
|• It requires only one communication channel between communicating devices; thus, it is a cheaper mode of transmission.||• To send n bits at a time, it requires n communication channel between communicating devices. As a result, cost is increased by a factor of n as compared to serial transmission.|
|• As communication within devices is parallel, both sender and receiver require converter at the interface between the device and the communication channel. The converter at the interface between sender device and communication channel converts parallel transmission to serial transmission while the converter at the interface between communication channel and receiver device converts serial transmission to parallel transmission.||• No such converters are required.|
23. Discuss the three types of serial mode transmissions.
Ans: Serial mode transmission can be classified into three types, namely, synchronous, asynchronous and isochronous.
|Asynchronous Transmission: In this transmission, the entire bit stream is divided into groups of 8 bits (that is, one byte) each (Figure 3.23). Each byte is treated independently and transmitted whenever ready regardless of the timer. To let the receiver know about the arrival of a byte, a start bit (usually represented by 0) is added to the starting of each byte. Similarly, one or more stop bits (usually represented by 1) are added to indicate the end of each byte. Though the transmission is asynchronous at the byte level, still some synchronization is needed during the transmission of bits within a byte. To achieve this synchronization, the receiver starts the timer after finding the start bit and counts the number of bits until it finds the stop bit. There may be gap between the transmission of two bytes which is filled with a series of stop bits or idle channel. Asynchronous transmission is slow as it adds control information such as start bits, stop bits, and gap between bytes. However, it is a cheaper and effective mode of transmission and thus is suitable for communication between devices which do not demand fast speed.|
|Synchronous Transmission: In this transmission, timing source is used for synchronization, so that the receiver can receive the information in the order in which it is sent. Multiple bytes are combined together to form frames. Data is transmitted as a continuous sequence of 1s and 0s with no gap in between. However, if there is any gap, that gap is filled with a special sequence of 0s and 1s called idle (Figure 3.24). At the receiver's end, the receiver keeps counting the bit and separates them into byte groups for decoding. Synchronous transmission is fast as compared to asynchronous transmission, as it does not use any extra bits such as start bits and stop bits. Thus, it is best suitable for applications requiring high speed. However, this transmission cannot be used for real-time applications such as television broadcasting, as there can be uneven delays between the arrival of adjacent frames at the receiver's end which in turn results in poor quality of video.
|Isochronous Transmission: This transmission provides synchronization in the entire stream of bits and not only at the byte level thereby ensuring the fixed arrival rate of data. This characteristic makes this transmission suitable for real-time applications for which synchronous transmission is not suitable.|
24. What do you mean by carrier signal?
Ans: Carrier Signal, also called carrier frequency, refers to the high-frequency signal produced by the sender device to which receiver device is tuned and is used as a base for the information signal.
25. Define digital-to-analog conversion. Discuss various techniques for digital-to-analog conversion.
Ans: The process of changing one or more of the attributes of analog signal based on information in digital data is referred to as digital-to-analog conversion. It is also called the modulation of a digital signal. Depending on whether the amplitude, frequency or phase of the carrier signal is modified, there are three basic techniques for digital-to-analog conversion, which include amplitude shift keying (ASK), frequency shift keying (FSK) and phase shift keying (PSK). In addition, one more efficient technique for digital-to-analog conversion is quadrature amplitude modulation (QAM) (explained in Q27) that involves changing of both amplitude and phase of the carrier signal.
Amplitude Shift Keying
It involves changing the amplitude of the carrier signal without changing its frequency and phase. The amplitude of a carrier signal is multiplied by binary 0 or 1. A special case of ASK is binary ASK (BASK), also known as on-off keying (OOK), where peak amplitude for one binary digit is taken as 0 while the other binary digit has amplitude equal to the peak amplitude of carrier frequency (Figure 3.25). Another variation of ASK is the multilevel ASK (MFSK) which involves more than two levels of amplitude.
The bandwidth (B) for ASK is directly proportional to signal rate (S) as shown in the formula given below:
B = (1 + fc) × S
Here, fc is the factor whose value lies between 0 and 1 and it depends on the modulation and filtering process.
It is clear from the above formula that Bmin = S (when fc = 0) and Bmax = 2 × S (when fc = 1).
Though ASK is the simplest technique, it is highly susceptible to noise and thus is an inefficient modulation technique.
Frequency Shift Keying
In this technique, the frequency of the carrier signal is changed without changing its amplitude and phase. The simplest form of FSK is binary FSK (BFSK) in which two different frequencies (say f1 and f2) close to the carrier frequency are taken to represent two binary values in digital data (Figure 3.26).
The bandwidth for BFSK is calculated by the following given formula:
B = (l + fc) × S + (f1 – f2),
where, f1 – f2 = 2f
Another variation of FSK is multilevel FSK (MFSK) which involves using more than two carrier frequencies. For example, four frequencies can be used to transmit two bits, eight frequencies to transmit three bits and so on. The only requirement is that the distance between any two carrier frequencies should be 2f. FSK is less susceptible to noise than ASK and thus is suitable for high-frequency radio transmission.
Phase Shift Keying
In this technique, the phase of the carrier signal is changed without changing its amplitude and frequency. There are two forms of PSK techniques; namely, binary PSK (BPSK) and quadrature PSK (QPSK). BPSK, also called as two-level PSK, uses two phase values. To represent one binary value, the signal is sent with the same phase as that of its predecessor and to represent another binary value; the signal is sent with opposite phase from its predecessor (Figure 3.27). The bandwidth of BPSK is equal to BASK, but less than that for BFSK.
QPSK, also called as four-level PSK, offers an efficient use of bandwidth by using two bits at a time in each signal element. It is so called, because it involves two BPSK modulations of which one is in-phase and another is out-of-phase (quadrature). In this technique, the stream of bits is first passed through a converter which converts serial transmission into parallel transmission and then, one bit is transmitted to in-phase modulator and other bit is transmitted to out-of-phase modulator. Both modulators produce composite signals which have the same frequency but different phases. When these signals are combined, the resultant signal may involve one of four different phases: 45°, –45°, 135° and –135°.
PSK is more commonly used than ASK and FSK. It is less susceptible to noise than ASK as phase is not likely to be affected easily by the noise. Moreover, it does not require two carriers as required in FSK.
26. What is constellation diagram? Explain with a suitable example.
Ans: Constellation Diagram is a 2D diagram used to define the amplitude and phase of a signal element especially in the cases where two carrier signals (one in-phase and other quadrature) are being used. The horizontal axis of the diagram represents the in-phase carrier while the vertical axis represents the quadrature carrier signal (Figure 3.28). Each signal element is represented by a point (black dot) with bit or combinations of bits written next to it. The projection of the point on the horizontal axis defines the peak amplitude of in-phase component while the projection of point on the vertical axis defines the peak amplitude of the quadrature component. The line joining the point to origin indicates the peak amplitude of signal element and the angle that this line makes with the horizontal axis indicates the phase of the signal element.
For example, the constellation diagram for QPSK as discussed in previous question may involve four phases as shown in Figure 3.29.
27. Write a short note on quadrature amplitude modulation (QAM).
Ans: QAM technique combines the idea of ASK and PSK. That is, instead of changing only one attribute of a carrier, two attributes including amplitude and phase of the carrier signal are changed. Like QPSK, it uses two carriers (one is in-phase and the other is quadrature) with different amplitude levels. Figure 3.30 shows the constellation diagram of a QAM signal with two amplitude levels and four phases.
The minimum bandwidth required by QAM is to that for ASK and PSK. It is less susceptible to noise than ASK as well as does not require two carriers as required in FSK.
28. Define analog-to-analog conversion. Discuss various techniques for analog-to-analog conversion.
Ans: The process of representing analog data by analog signal is known as analog-to-analog conversion (also called analog modulation). The techniques used for analog-to-analog conversion include amplitude modulation (AM), frequency modulation (FM) and phase modulation (PM).
In this modulation, the amplitude of a carrier wave is varied in accordance with the characteristic of the modulating signal. The frequency of the carrier remains the same, only the amplitude changes to follow variations in the signal. In simpler words, the two discrete binary digits (0 and 1) are represented by two different amplitudes of the carrier signal. Figure 3.31 depicts how the modulating signal is superimposed over the carrier signal that results in an amplitude-modulated signal.
The bandwidth of the amplitude-modulated signal is twice of that of modulating signal. That is,
BAM = 2B,
where B = bandwidth of modulating signal.
In this modulation, the instantaneous frequency of carrier wave is caused to depart from the centre frequency by an amount proportional to the instantaneous value of the modulating signal. In simple words, FM is the method of impressing modulating signal onto a carrier signal wave by varying its instantaneous frequency rather than its amplitude (Figure 3.32).
The total bandwidth that is needed for the frequency-modulated signal can be calculated from the below-given formula.
BFM = 2 (1 + ß) B,
where ß is a factor whose value depends on the modulation technique with a common value of 4 and ß is the bandwidth of modulating signal.
This is the encoding of information into a carrier wave by variation of its phase in accordance with an input signal. In this modulation technique, the phase of sine wave carrier is modified according to the amplitude of the message to be transmitted (Figure 3.33).
The total bandwidth that is needed by a PM signal can be calculated from the bandwidth and maximum amplitude of the modulating signal as shown here.
BPM = 2 (1 + ß) B,
where ß is a factor whose value is lower in PM than FM. For narrowband, its value is 1 and for wideband, its value is 3. B is the bandwidth of modulating signal.
29. What does a decibel (dB) measure? Give example.
Ans: Decibel (dB) is a measure of comparative strength of two signals or one signal at two different points. It is used by engineers to determine whether the signal has lost or gained strength.
The positive value of a signal indicates gain in strength while negative value indicates that the signal is attenuated.
dB = 10 log10(P2/P1),
where P1 and P2 are the powers of signal at two different points or the powers of two different signals.
30. For the following frequencies, calculate the corresponding periods. Write the result in seconds (s), milliseconds (ms) and microseconds (ms): 24 Hz; 8 MHz.
Ans: As frequency = 1/period. Thus, for frequency 24 Hz, period = 1/24 s = 0.041 s
As 1 s = 103 ms ⇒ 0.041 s = 0.041 × 103 = 41 ms
As 1 s = 106 us ⇒ 0.041 s = 0.041 × 106 = 41000 ms.
Similarly, for frequency 8 MHz, period = 1/8000000 s = 0.000000125 s
0.000000125 × 103 ms = 0.000125 ms
0.000000125 × 106 μs = 0.125 ms.
31. A sine wave completes one cycle in 25 microseconds. What is its frequency? Express the frequency in hertz.
Ans: Frequency = 1/period. Here, period = 25 μs = 25 × 10−6 s. Therefore, frequency = 106/25 = 40000 Hz.
32. A digital signal has a bit interval of 40 microseconds. What is the bit rate? Express the bit rate in kbps.
Ans: Bit rate = number of bit intervals per second. Therefore, bit rate = 1/(40 × 10−6) = 25000 bps.
⇒ Bit rate = 25 kbps.
33. A signal has a bandwidth of 30 Hz. The highest frequency is 80 Hz. What is its lowest frequency? Draw the spectrum if the signal contains all the frequency of same amplitude.
Ans: Bandwidth = the highest frequency – the lowest frequency. Bandwidth = 30 Hz (given). The highest frequency = 80 Hz. Thus, the lowest frequency = 80–30 = 50 Hz. The spectrum of the signal containing all the frequencies of same amplitude is shown in Figure 3.34.
34. An image has the size of 1024 x 786 pixel with 256 colours. Assume the image is uncompressed. How does it take over a 56–kbps modem channel?
Ans: To represent 256 colours, we need log2256 = 8 bits. Therefore, the total number of bits that are to be transmitted will be equal to 1024 × 786 × 8 = 6438912 bits. Thus, bit rate = 6438912/ (56 × 1000) = 115 bps.
35. An analog signal carries four bits in each signal element. If 1000 signal elements are sent per second, find the baud rate and bit rate.
Ans: Baud rate = the number of signal elements per second = 1000. As we know that baud rate = bit rate × (1/number of data elements carried in one signal element). Bit rate = 1000 × 4 = 4000 bps.
36. The power of a signal is 10 mW and the power of the noise is 1 μW. What is the value of SNR in dB?
Ans: SNR = signal power/noise power. Thus, SNR = (10 × 10−3)/(1 × 10−6) = 10,000. SNRdB = 10 log10 SNR = 10 log10 10000 =10 × 4 = 40.
37. Calculate the highest bit rate for a telephone channel, given the bandwidth of a line to be 3000 Hz and the SNR being 35 dB.
Ans: As SNRdB = 10 log10 SNR. SNR = 10 (SNIW10) = 103.5 = 3162 (rounded off). According to Shannon's capacity, The highest bit rate = bandwidth × log2 (1 + SNR). Therefore, the highest bit rate = 3000 log2 (1 + 3162) = 3000 × log2 3163 = 3000 × 11.6 = 34800 bps.
38. What is the Nyquist minimum sampling rate for the following?
(a) A complex low-pass signal with a bandwidth of 300 kHz.
Ans: According to Nyquist theorem, the sampling rate is two times of the highest frequency in the signal. As the frequency of a low-pass signal starts from zero, the highest frequency in the signal = bandwidth; that is, 300 kHz. Thus, sampling rate = 2 × 300 kHz = 600000 samples per second.
(b) A complex bandpass signal with a bandwidth of 300 kHz.
Ans: The sampling rate cannot be found in this case because we do not know the maximum frequency of the signal.
(c) A complex bandpass signal with bandwidth of 300 kHz and the lowest frequency of 100 Hz.
Ans: The highest frequency = 100 +300 = 400 kHz. Thus, sampling rate = 2 × 400 kHz = 800000 samples per second.
39. What is the maximum bit rate for a noiseless channel with a bandwidth of 6000 Hz transmitting a signal with four signal levels?
Ans: Bit rate = 2 × channel bandwidth × log2 (number of signal levels). Therefore, bit rate = 2 × 6000 × log2 4 = 24000 bps.
40. What is the total delay (latency) for a frame size of 10 million bits that is being set up on a link with 15 routers, each having a queuing time of 2 μs and a processing time of 1 μs? The length of link is 3000 km. The speed of light inside the link is 2 × 108 m/s. The link has bandwidth of 6 Mbps.
Ans: Here, propagation time = distance / propagation speed
⇒ (3000 × 1000)/(2 × 108) = 1.5 × 10−2 s.
Transmission time = message size / bandwidth
⇒ (10 × 106)/(6 × 106) = 1.7 s (approx).
As there are 15 routers, total queuing time = 15 × 2 × 10−6 = 30 × 10−6 s.
Processing time = 15 × 1 × 10−6 = 15 × 10−6 s. Now, latency = propagation time + transmission time + queuing time + processing time
⇒ 1.5 × 10−2 + 1.7 + 30 × 10−6 s + 15 × 10−6 s = 1.715045 s.
41. A data stream is made of 10 alternating 0s and 1s. Encode this stream using the following encoding schemes.
(b) Polar NRZ-L
(e) Manchester Ans:
(f) Differential Manchester
42. A signal is quantized using 10–bit PCM. Find the SNR in dB.
Ans: SNRdB = 6.02 nb + 1.76. Here, nb is the number of bits used for quantization = 10. Thus, SNRdB = 6.02 × 10 + 1.76 = 61.96 dB.
43. A system is designed to sample analog signals, convert them to digital form with a 4–bit converter, and transmit them. What bit rate is required if the analog signal consists of frequencies between 400 Hz and 3400 Hz.
Ans: Bandwidth = 3400 – 400 = 3000 Hz. Bit rate = 2 × bandwidth × log2 L = 2 × 3000 × log2 4 ⇒ 2 × 2 × 3000 = 12000 bps = 12 kbps.
44. Given the bit pattern 01100, encode this data using ASK, BFSK, and BPSK
Ans: Bit pattern 01100 can be encoded using ASK, BFSK and BPSK as shown in Figure 3.35(a), 3.35 (b) and 3.35 (c), respectively.
45. Find the maximum bit rate for an FSK signal if the bandwidth of the medium is 12,000 Hz and the difference between the two carriers must be at 2000 Hz. Transmission is in full-duplex mode.
Ans: As we know that in FSK, bandwidth = (1 + fc) × signal rate + difference between carrier frequencies.
⇒ 12000 = (1 + 1) × signal rate + 2000
⇒ Signal rate = 5000 baud.
Now, as we know that signal rate = bit rate × (1/r). For FSK, one data element is carried by one signal element, that is, r = 1
⇒ Bit rate = 4000 kbps.
46. Find the bandwidth for the following situations if we need to modulate a 5–kHz voice.
Ans: BAM = 2 × B = 2 × 5 = 10 kHz.
(b) PM (set ß = 5)
Ans: BFM = 2 × (1 + ß) × B = 2 × (1 + 5) × 5 = 60 kHz.
(c) PM (set ß =1)
Ans: BPM = 2 × (1 + ß) × B = 2 × (1 + 1) × 5 = 20 kHz.
Multiple Choice Questions
|1.||Which of the following is a characteristic of an analog quantity?
(a) Changes in quantity are represented by discrete values.
(b) Its values follow a logarithmic response curve.
(c) It can be described with a finite number of steps.
(d) It has a continuous set of values over a given range.
|2.||In which of the following form data has to be transformed so that it can be transmitted.
(a) Electromagnetic signals
(b) Aperiodic signals
(c) Low-frequency sine waves
(d) None of these
|3.||A sine wave is
(a) Periodic and continuous
(b) Aperiodic and continuous
(c) Periodic and discrete
(d) Aperiodic and discrete
|4.||What is the bandwidth of a signal that ranges from 200 MHz to 500 MHz?
(a) 400 MHz
(b) 100 MHz
(c) 300 MHz
(d) None of these
|5.||A sine wave has a frequency of 8 kHz. What is its period?
(a) 124 microseconds
(b) 150 microseconds
(c) 125 microseconds
(d) 156 microseconds
|6.||Shannon's noisy channel capacity theorem states that C = B log2 (1+ SNR) bps. In this equation, B stands for
(a) Bandwidth of the signal
(b) Bandwidth of the channel
(c) Strength of the signal
(d) None of these
|7.||In which of the following encoding schemes three levels of voltage are used?
(d) None of these
|8.||In differential manchester encoding, transition at the middle of the bit is used for
(a) Bit transfer
(c) Baud transfer
(d) None of these
|9.||In which of the following schemes change or lack of change in the value of voltage determines the value of bit
(c) Both (a) and (b)
(d) None of these
|10.||A 6–MHz analog video signal can be sampled and later reconstructed if it was sampled at a rate of at least (in M samples/s)?
|11.||One factor in the accuracy of a reconstructed PCM signal is the
(a) Signal bandwidth
(b) Carrier frequency
(c) Number of bits used for quantization
(d) Baud rate
|12.||Amplitude modulation is a technique used for
(a) Analog-to-digital conversion
(b) Digital-to-analog conversion
(c) Analog-to-analog conversion
(d) Digital-to-digital conversion
|13.||Which of the following factors of a carrier frequency is varied in QAM?
(d) Both (b) and (c)
|14.||How many carrier frequencies are used in BFSK?
(d) None of these
|15.||How many dots are present in the constellation diagram of 8–QAM?
(d) None of these