Anil K. Maini, Varsha Agrawal, Satellite Communications,

Frequency Modulation 211 through the lowpass filter and error amplifier drives the control input of the VCO to keep its output frequency always in lock with the instantaneous frequency of the input FM signal. As a result, the error amplifier always represents the detected output. The double balanced mixer nature of the phase detector suppresses any carrier level changes and therefore the PLL-based FM detector requires no additional limiter circuit. A comparison of the three types of FM detector reveals that the Foster–Seeley type frequency discriminator offers excellent linearity of response, is easy to balance and the detected output depends only on the frequency deviation. However, it needs high gain RF and IF stages to ensure limiting action. The ratio detector circuit, on the other hand, requires no additional limiter circuit; detected output depends both on the frequency deviation as well as on the average carrier level. However, it is difficult to balance. The PLL-based FM detector offers excellent reproduction of the modulating signal, is easy to balance and has low cost and high reliability. Problem 5.5 An FM signal is represented by the equation v(t) = 10 sin 7.8 × 108t + 6 sin 1450t . Determine the unmodulated carrier frequency, the modulating frequency and the mod- ulation index. Solution: Comparing the given equation with standard form of equation for FM signal given by v(t) = Vm sin (ωct + m sin ωmt). Unmodulated carrier frequency, ωc = 7.8 × 108 or fc = (7.8 × 108/2 )Hz = 124.14 MHz Modulation index, m = 6 Modulating frequency, ωm = 1450 or fm = 1450/2 = 230.77 Hz Problem 5.6 An FM signal is represented by v(t) = 15 cos 108 t + 6 sin 2 × 103 t . Determine (a) Maximum phase deviation and (b) Maximum frequency deviation Solution: Comparing the given equation with the standard form of FM signal expression, we get, (ωct + ) = 108 t + 6 sin 2 × 103 t . Therefore, phase ( ) = 6 sin 2 (103)t. This gives maximum phase deviation max = 6 radian. Frequency is nothing but rate of change of phase. Therefore, maximum frequency deviation is nothing but maximum value of (d /dt). Now, (d /dt) = 6 × 2 (103) cos 2 (103)t = 12 (103) cos 2 (103)t Maximum frequency deviation = 12 (103) radian/sec = 6 kHz Problem 5.7 An FM signal is represented by v(t) = 15 cos [2 (108)t + 150 cos 2 (103)t] Determine the bandwidth of the signal. Also write the expression for the instantaneous frequency of the signal.

212 Communication Techniques Solution: Comparing the given expression with the standard expression for the FM signal, we get the value of modulation index (mf) as mf = 150 The modulating frequency fm = 2 (103)/(2 ) = 1 kHz. Therefore, the frequency deviation ı = 150 × 1 kHz = 150 kHz. The bandwidth can now be computed from: Bandwidth = 2(mf + 1)fm = (2 × 151 × 1)kHz = 302 kHz Expression for instantaneous frequency can be written by taking the derivative of the expression for instantaneous phase , where = 2 (108)t + 150 cos 2 (103)t d = 2 (108) − 150 × 2 (103) sin 2 (103)t Instantaneous frequency ω = dt = 2 (108) − 300 (103) sin 2 (103)t f = 108 − 150(103) sin 2 (103)t Problem 5.8 A 95 MHz carrier is frequency modulated by a sinusoidal signal and the modulated signal is such that maximum frequency deviation achieved is 50 kHz. Determine the modulation index and bandwidth of the modulated signal, if the modulating signal frequency is (a) 1 kHz and (b) 100 kHz. Solution: In the first case, frequency deviation = 50 kHz and modulating frequency = 1 kHz Therefore, modulation index = (50 × 103)/(1 × 103) = 50 Bandwidth = [2 × (50 + 1) × 1] kHz = 102 kHz In the second case, modulating frequency = 100 kHz Therefore, modulation index = (50 × 103)/(100 × 103) = 0.5 Bandwidth = [2 × (0.5 + 1) × 100] kHz = 300 kHz Problem 5.9 Find the bandwidth of an FM signal produced in a commercial FM broadcast with mod- ulating signal frequency being in the range of 50 Hz to 15 kHz and maximum allowable frequency deviation being 75 kHz. Solution: Maximum allowed frequency deviation = 75 kHz and highest modulating frequency = 15 KHz Therefore, deviation ratio, D = (75 × 103)/(15 × 103) = 5 Bandwidth = 2 (D + 1) fm = [2 × (5 + 1) × 15] kHz = 180 kHz

Pulse Communication Systems 213 Problem 5.10 A carrier when frequency modulated by a certain sinusoidal signal of 1 kHz produces a modulated signal with a bandwidth of 20 kHz. If the same carrier signal is frequency mod- ulated by another modulating signal whose peak amplitude is 3 times that of the previous signal and the frequency is one-half of the previous signal, determine the bandwidth of the new modulated signal. Solution: The bandwidth = 2(mf + 1)fm. This gives 2(mf + 1) × 1 = 20 or mf = 9. Since the amplitude of the new modulating signal is 3 times that of the previous signal, it will produce a frequency deviation that is 3 times that produced by the previous signal. Also, new modulating frequency is one-half of the previous signal frequency. Therefore, the new modulation index, which is the ratio of the frequency deviation to the modulating frequency, will be 6 times that produced by the previous signal. Therefore, New modulation index = 9 × 6 = 54 New bandwidth = 2(mf + 1)fm = [2 × (54 + 1) × 0.5]kHz = 55 kHz 5.4 Pulse Communication Systems Pulse communication systems differ from continuous wave communication systems in the sense that the message signal or intelligence to be transmitted is not supplied continuously as in case of AM or FM. In this case, it is sampled at regular intervals and it is the sampled data that is transmitted. All pulse communication systems fall into either of two categories, namely ana- logue pulse communication systems and digital pulse communication systems. Analogue and digital pulse communication systems differ in the mode of transmission of sampled information. In the case of analogue pulse communication systems, representation of the sampled amplitude may be infinitely variable whereas in the case of digital pulse communication systems, a code representing the sampled amplitude to the nearest predetermined level is transmitted. 5.4.1 Analogue Pulse Communication Systems Important techniques that fall into the category of analogue pulse communication systems include: 1. Pulse amplitude modulation 2. Pulse width (or duration) modulation 3. Pulse position modulation 5.4.1.1 Pulse Amplitude Modulation In the case of pulse amplitude modulation (PAM), the signal is sampled at regular intervals and the amplitude of each sample, which is a pulse, is proportional to the amplitude of the modulating signal at the time instant of sampling. The samples, shown in Figure 5.31, can have either a positive or negative polarity. In a single polarity PAM, a fixed DC level can be added

214 Communication Techniques to the signal, as shown in Figure 5.31 (c). These samples can then be transmitted either by a cable or used to modulate a carrier for wireless transmission. Frequency modulation is usually employed for the purpose and the system is known as PAM-FM. Figure 5.31 Pulse amplitude modulation 5.4.1.2 Pulse Width Modulation In the case of pulse width modulation (PWM), as shown in Figure 5.32, the starting time of the sampled pulses and their amplitude is fixed. The width of each pulse is proportional to the amplitude of the modulating signal at the sampling time instant. Figure 5.32 Pulse width modulation 5.4.1.3 Pulse Position Modulation In the case of pulse position modulation (PPM), the amplitude and width of the sampled pulses is maintained as constant and the position of each pulse with respect to the position of a

Pulse Communication Systems 215 recurrent reference pulse varies as a function of the instantaneous sampled amplitude of the modulating signal. In this case, the transmitter sends synchronizing pulses to operate timing circuits in the receiver. A pulse position modulated signal can be generated from a pulse width modulated signal. In a PWM signal, the position of the leading edges is fixed whereas that of the trailing edges depends upon the width of the pulse, which in turn is proportional to the amplitude of the modulating signal at the time instant of sampling. Quite obviously, the trailing edges constitute the pulse position modulated signal. The sequence of trailing edges can be obtained by differentiating the PWM signal and then clipping the leading edges as shown in Figure 5.33. Pulse width modula- tion and pulse position modulation both fall into the category of pulse time modulation (PTM). Figure 5.33 Pulse position modulation 5.4.2 Digital Pulse Communication Systems Digital pulse communication techniques differ from analogue pulse communication techniques described in the previous paragraphs in the sense that in the case of analogue pulse modulation the sampling process transforms the modulating signal into a train of pulses, with each pulse in the pulse train representing the sampled amplitude at that instant of time. It is one of

216 Communication Techniques the characteristic features of the pulse, such as amplitude in the case of PAM, width in the case of PWM and position of leading or trailing edges in the case of PPM, that is varied in accordance with the amplitude of the modulating signal. What is important to note here is that the characteristic parameter of the pulse, which is amplitude or width or position, is infinitely variable. As an illustration, if in the case of pulse width modulation, every volt of modulating signal amplitude corresponded to 1 μs of pulse width, then 5.23 volt and 5.24 volt amplitudes would be represented by 5.23μs and 5.24 μs respectively. Further, there could be any number of amplitudes between 5.23 volts and 5.24 volts. It is not the same in the case of digital pulse communication techniques, to be discussed in the paragraphs to follow, where each sampled amplitude is transmitted by a digital code representing the nearest predetermined level. Important techniques that fall into the category of digital pulse communication systems include: 1. Pulse code modulation (PCM) 2. Differential PCM 3. Delta modulation 4. Adaptive delta modulation 5.4.2.1 Pulse Code Modulation In pulse code modulation (PCM), the peak-to-peak amplitude range of the modulating signal is divided into a number of standard levels, which in the case of a binary system is an integral power of 2. The amplitude of the signal to be sent at any sampling instant is the nearest standard level. For example, if at a particular sampling instant, the signal amplitude is 3.2 volts, it will not be sent as a 3.2 volt pulse, as might have been the case with PAM, or a 3.2 μs wide pulse, as for the case with PWM; instead it will be sent as the digit 3, if 3 volts is the nearest standard amplitude. If the signal range has been divided into 128 levels, it will be transmitted as 0000011. The coded waveform would be like that shown in Figure 5.34 (a). This process is known as quantizing. In fact, a supervisory pulse is also added with each code group to facilitate reception. Thus, the number of bits for 2n chosen standard levels per code group is n + 1. Figure 5.34 (b) illustrates the quantizing process in PCM. It is evident from Figure 5.34 (b) that the quantizing process distorts the signal. This distortion is referred to as quantization noise, which is random in nature as the error in the signal’s amplitude and that actually sent after quantization is random. The maximum error can be as high as half of the sampling interval, which means that if the number of levels used were 16, it would be 1/32 of the total signal amplitude range. It should be mentioned here that it would be unfair to say that a PCM system with 16 standard levels will necessarily have a signal- to-quantizing noise ratio of 32:1, as neither the signal nor the quantizing noise will always have its maximum value. The signal-to-noise ratio also depends upon many other factors and also its dependence on the number of quantizing levels is statistical in nature. Nevertheless, an increase in the number of standard levels leads to an increase in the signal-to-noise ratio. In practice, for speech signals, 128 levels are considered as adequate. In addition, the more the number of levels, the larger is the number of bits to be transmitted and therefore higher is the required bandwidth.

Pulse Communication Systems 217 Figure 5.34 Quantizing process In binary PCM, where the binary system of representation is used for encoding various sampled amplitudes, the number of bits to be transmitted per second would be given by nf s, where n = log2L L = number of standard levels and fs ≥ fm where fm = message signal bandwidth Assuming that the PCM signal is a lowpass signal of bandwidth fPCM, then the required minimum sampling rate would be 2fPCM. Therefore, 2fPCM = nfs or fPCM = n 2 fs Generating a PCM signal is a complex process. The message signal is usually sampled and first converted into a PAM signal, which is then quantized and encoded. The encoded signal can then be transmitted either directly via a cable or used to modulate a carrier using analogue or digital modulation techniques. PCM-AM is quite common.

218 Communication Techniques 5.4.2.2 Differential PCM Differential PCM is similar to conventional PCM. The difference between the two lies in the fact that in differential PCM, each word or code group indicates a difference in amplitude (positive or negative) between the current sample and the immediately preceding one. Thus, it is not the absolute but the relative value that is indicated. As a consequence, the bandwidth required is less as compared to the one required in case of normal PCM. 5.4.2.3 Delta Modulation Delta modulation (DM) has various forms. In one of the simplest forms, only one bit is trans- mitted per sample just to indicate whether the amplitude of the current sample is greater or smaller than the amplitude of the immediately preceding sample. It has extremely simple en- coding and decoding processes but then it may result in tremendous quantizing noise in case of rapidly varying signals. Figure 5.35 (a) shows a simple delta modulator system. The message signal m(t) is added to a reference signal with the polarity shown. The reference signal is an integral part of the delta modulated signal. The error signal e(t) so produced is fed to a comparator. The output of Figure 5.35 (a) Delta modulator system. (b) Output waveform of a delta modulator system

Sampling Theorem 219 the comparator is (+ ) for e(t) > 0 and (− ) for e(t) < 0. The output of the delta modulator is a series of impulses with the polarity of each impulse depending upon the sign of e(t) at the sampling instants of time. Integration of the delta modulated output xDM(t) is a staircase approximation of the message signal m(t), as shown in Figure 5.35 (b). A delta modulated signal can be demodulated by integrating the modulated signal to obtain the staircase approximation and then passing it through a lowpass filter. The smaller the step size ( ), the better is the reproduction of the message signal. However, a small step size must be accompanied by a higher sampling rate if the slope overload phenomenon is to be avoided. In fact, to avoid slope overload and associated signal distortion, the following condition should be satisfied: dm(t) (5.29) ≥ Ts dt max where Ts = time between successive sampling time instants. 5.4.2.4 Adaptive Delta Modulator This is a type of delta modulator. In delta modulation, the dynamic range of amplitude of the message signal m(t) is very small due to threshold and overload effects. This problem is overcome in an adaptive delta modulator. In adaptive delta modulation, the step size ( ) is varied according to the level of the message signal. The step size is increased as the slope of the message signal increases to avoid overload. The step size is reduced to reduce the threshold level and hence the quantizing noise when the message signal slope is small. In the case of adaptive delta modulation, however, the receiver also needs to be adaptive. The step size at the receiver should also be made to change to match the changes in step size at the transmitter. 5.5 Sampling Theorem During the discussion on digital pulse communication techniques such as pulse code modula- tion, delta modulation, etc., it was noted that the three essential processes of such a system are sampling, quantizing and encoding. Sampling is the process in which a continuous time signal is sampled at discrete instants of time and its amplitudes at those discrete instants of time are measured. Quantization is the process by which the sampled amplitudes are represented in the form of a finite set of levels. The encoding process designates each quantized level by a code. Digital transmission of analogue signals has been made possible by sampling the continuous time signal at a certain minimum rate, which is dictated by what is called the sampling theorem. The sampling theorem states that a band limited signal with the highest frequency component as fM Hz, can be recovered completely from a set of samples taken at a rate of fs samples per second, provided that fs ≥ 2fM. This theorem is also known as the uniform sampling theorem for base band or lowpass signals. The minimum sampling rate of 2fM samples per second is called the Nyquist rate and its reciprocal the Nyquist interval. For sampling bandpass signals, lower sampling rates can sometimes be used. The sampling theorem for bandpass signals states that if a bandpass message signal has a bandwidth of fB and an upper frequency limit of fu, then the signal can be recovered from the sampled signal by bandpass filtering if fs = 2fu/k, where fs is the sampling rate and k is the largest integer not exceeding fu/fB.

220 Communication Techniques 5.6 Shannon–Hartley Theorem The Shannon–Hartley theorem describes the capacity of a noisy channel (assuming that the noise is random). According to this theorem, C = B log2[1 + (S/N)] (5.30) where C = channel capacity in bps B = channel bandwidth in Hz S/N = signal-to-noise ratio at the channel output or receiver input The Shannon–Hartley theorem underlines the fundamental importance of bandwidth and signal-to-noise ratio in communication. It also shows that for a given channel capacity, in- creased bandwidth can be exchanged for decreased signal power. It should be mentioned that increasing the channel bandwidth by a certain factor does not increase the channel capacity by the same factor in a noisy channel, as would apparently be suggested by the Shannon–Hartley theorem. This is because increasing the bandwidth also increases noise, thus decreasing the signal-to-noise ratio. However, channel capacity does increase with an increase in bandwidth; the increase will not be in the same proportion. Problem 5.11 A binary channel with a capacity of 48 kb/second is used for PCM voice transmission. If the highest frequency component in the message signal is taken as 3.2 kHz, find appropriate values for sampling rate (fs), number of quantizing levels (L) and number of bits (n) used per sample. Solution: As per Nyquist criterion, fs ≥ 2fM . This gives fs ≥ 6400 samples/sec Also, nfs ≤ Bit transmission rate; this gives nfs ≤ 48000 or n ≤ (48000/6400) = 7.5 This gives n = 7 and L = 27 = 128 and fs = 48000/7 Hz = 6.857 kHz Problem 5.12 An analog signal is sampled at 10 kHz. If the number of quantizing levels is 128, find the time duration of one bit of binary encoded signal. Solution: Number of bits per sample (n) = log2L = log2128 = 7 Where L = Number of quantizing levels Now fs = 10 kHz. Therefore time duration of one bit of binary encoded signal = 1/(nfs) = 1/(7 × 10000) = 14.286 s Problem 5.13 Find the Nyquist rate for the message signal represented by m(t) = 10 (cos 1000 t) (cos 4000 t)

Digital Modulation Techniques 221 Solution: 10(cos 1000 t)(cos 4000 t) = 5 × [2(cos 1000 t)(cos 4000 t)] = 5[(cos 5000 t) + (cos 3000 t)] This is a band limited signal with the highest frequency component equal to (5000 ) radians/second, or 2500 Hz. Therefore, Nyquist rate = 2 × 2500 = 5000 Hz = 5 kHz Problem 5.14 A message signal given by m(t) = A sin ωmt is applied to a delta modulator having a step size of . Show that slope overload distortion will occur if A > fs where fs is 2 fm the sampling frequency. Solution: m(t) = A sin ωmt dm(t ) dt = A ωm cos ωmt The condition for avoiding slope overload is given by ≥ dm(t ) = A␻m Ts dt max or A ≤ = fs = fs (Ts␻m) 2 fm 2 fm This is the condition for avoiding slope overload. Thus slope overload will occur when A> fs (5.31) 2 fm 5.7 Digital Modulation Techniques Base band digital signals have significant power content in the lower part of the frequency spectrum. Because of this, these signals can be conveniently transmitted over a pair of wires or coaxial cables. At the same time, for the same reason, it is not possible to have efficient wireless transmission of base band signals as it would require prohibitively large antennas, which would not be a practical or a feasible proposition. Therefore, if base band digital signals are to be transmitted over a wireless communication link, they should first modulate a continuous wave (CW) high frequency carrier. Three well-known techniques available for the purpose include: 1. Amplitude shift keying (ASK) 2. Frequency shift keying (FSK) 3. Phase shift keying (PSK)