Communication Systems

Effect of Noise on Analog Communication SystemsBy Teguh Firmansyah, S.T, M.TElectrical EngineeringOutlineEffect of Noise on Amplitude-Modulation SystemsEffect of Noise on Angle ModulationComparison of Analog-Modulation SystemsCarrier-Phase Estimation with a Phase-Locked Loop (PLL)Effects of Transmission Losses and Noise in Analog Communication Systems22Effect of Noise on Amplitude-Modulation SystemsEffect of Noise on a Baseband SystemBaseband system: there is no carrier demodulation to be performed. The receiver consists only of an ideal lowpass filter with the bandwidth W.The noise power at the output of the receiver for a white noise input:

ThenThe baseband signal-to-noise ratio (SNR) is

PR: the received power -WWN0/2fSX(f)ExampleFind the SNR in a baseband system with a bandwidth of 5 kHz and with N0/2 = 1014W/Hz. The transmitter power is one kilowatt and the channel attenuation is 1012.SolutionWe have PR = 1012 PT = 1012 103= 10-9 Watts. Therefore,

This is equivalent to 10 log10 20 = 13 dB.NoiseThe message signal m(t)modulatoru(t)channel+additive noisebandpassfilterdemodulatorr(t)=u(t)+n(t)filtered noise output signal y(t)lowpassfilter

Effect of Noise on DSB-SC AMIn DSB-SC AM, the transmitted signal is

The received signal at the output of the receiver noise-limiting filter is the sum of this signal and filtered noise.

Multiplying r(t) by a locally generated sinusoid cos(2fct + )

baseband signal Effect of Noise on DSB-SC AM

The output after the lowpass filter isIf a phase-locked loop is employed, then = 0 and the demodulator is called a coherent or synchronous demodulator.Assume that = 0. Hence,

The message signal power is

The power content of the message signalThe noise power is

since the power contents of nc(t) and n(t) are equalEffect of Noise on DSB-SC AMThe power content of n(t)

The noise power is

The output SNR is

PR: the received signal powerDSB-SC AM does not provide any SNR improvement over a simple baseband communication system.is identical to (S/N)bEffect of Noise on SSB AMIn SSB AM, the transmitted signal is

The input to the demodulator is

Assume that demodulation occurs with an ideal phase reference, the output of the lowpass filter is

Again, the signal and the noise components are additive and a meaningful SNR at the receiver output can be defined.Effect of Noise on SSB AMThe message signal power is

The power content of the message signalThe noise power is

where

The output SNR is

The SNR in an SSB system is equivalent to that of a DSB system.in SSB AM,

thus,Effect of Noise on Conventional AMIn conventional DSB AM, the modulated signal is

The received signal at the input to the demodulator is

Assume a synchronous demodulator is employed. After mixing and lowpass filtering, we have

the modulation indexnormalized message signalThe DC component in the demodulated waveform is removed by a DC block and, hence, the lowpass filter output is

In conventional DSB AM, the received signal power iswhere we have assumed that the message process is zero mean (See pp. 137 ).The output SNR is

the modulation efficiency, small than 1The SNR in conventional AM is always smaller than the SNR in a baseband systemThe reason for this loss is that a large part of the transmitter power is used to send the carrier component of the modulated signal and not the desired signal.

Effect of Noise on the envelope detectorThe input to the envelope detector is

The envelope of r(t) is given by

Assume that the signal component in r(t) is much stronger than the noise component, that is,

Therefore, we have a high probability that

After removing the DC component, we obtain

Under the assumption of high SNR at the receiver input, the performance of synchronous and envelope demodulators is the same.If the noise power is much stronger than the signal power, the signal and the noise components are no longer additive (the signal component is multiplied by noise and is no longer distinguishable).Example

Answer

Effects of Transmission Losses and Noise in Analog Communication SystemsTwo dominant factors that limit the performance of the systemAdditive noiseSignal attenuationBasically all physical channels, including wireline and radio channels, are lossy.Signal attenuation renders the communication signal more vulnerable to additive noise

Characterization of Thermal Noise SourcesThermal noise is produced by the random movement of electrons due to thermal agitation.

The power spectral density of thermal noise is given as

: Plancks constant (equal to 6.6 1034 J sec)k : Boltzmanns constant (equal to 1.38 1023 J/K)T : the temperature in degrees KelvinR : the resistance ()At frequencies below 1012Hz (which includes all conventional communication systems) and at room temperature,

Consequently, the power spectral density is well approximated as

When connected to a load resistance with value RL, the noise voltage delivers the maximum power when R = RL. In such a case, the load is matched to the source and the maximum power delivered to the load is E[N2(t)]/4RL. Therefore, the power spectral density of the noise voltage across the load resistor is

kT is usually denoted by N0. Hence, the power spectral density of thermal noise is generally expressed as

For example, at room temperature (T0 = 290 K), N0 = 4 1021 W/Hz.

Effective Noise Temperature and Noise FigureWhen we employ amplifiers in communication systems to boost the level of a signal, we are also amplifying the noise corrupting the signal.we may model an amplifier as a filter with the frequency response characteristic H(f).

Noise power at the output of the network is

Recall that the noise equivalent bandwidth of the filter is defined asG = |H(f)|2max is the maximum available power gain of the amplifier.The output noise power from an ideal amplifier may be expressed as

Any practical amplifier introduces additional noise at its output due to internally generated noise. Hence, the noise power at its output may be expressed as

the power of the amplifier output due to internally generated noise

Therefore,

whereis called the effective noise temperature of the two-port network (amplifier).A signal source at the input to the amplifier with power Psi will produce an output with power

Hence, the output SNR from the two-port network is

the input SNR to the two-port networkThe SNR at the output of the amplifier is degraded (reduced) by the factor (1+Te/T). An ideal amplifier is one for which Te = 0.When T is taken as room temperature T0 (290K), the factor F= (1+Te/ T0) iscalled the noise figure of the amplifier.

Consequently,By taking the logarithm of both sides,

the loss in SNR due to the additional noise introduced by the amplifierThe overall noise figure of a cascade of K amplifiers with gains Gk and corresponding noise figures Fk, 1 k K is

Fries formulaF1 is the dominant term, which is the noise figure of the first amplifier stage. The front end of a receiver should have a low noise figure and a high gain.Transmission LossesThe amount of signal attenuation generally depends on the physical medium, the frequency of operation, and the distance between the transmitter and the receiver.Define the loss L in signal transmission as the ratio of the input(transmitted) power to the output (received) power of the channel, i.e.,

or, in decibels,

In wireline channels, the transmission loss is usually given in terms of decibels per unit length, e.g., dB/km.For example, the transmission loss in coaxial cable of 1 cm diameter is about 2 dB/km at a frequency of 1 Mhz. This loss generally increases with an increase in frequency.In line-of-sight radio systems, the transmission loss is given as

= c/f: the wavelength of the transmitted signalc: the speed of light (3 108m/s)f: the frequency of the transmitted signald: the distance between the transmitter and the receiver in metersfree-space path lossExampleDetermine the free-space path loss for a signal transmitted at f = 1 MHz over distances of 10 km and 20 km.Doubling the distance in radio transmission increases the free-space path loss by 6 dB.Solution = 300 m. The loss for the 10 km path is

The loss for the 20 km path is

Repeaters for Signal Transmission

Analog repeaters are basically amplifiers that are generally used in telephone wireline channels and microwave line-of-sight radio channels to boost the signal level and, thus, to offset the effect of signal attenuation in transmission through the channel.

The input signal power at the input to the repeater isThe output power from the repeater is

We may select the amplifier gain G to offset the transmission loss. Hence, G = L and P0 = PT .

Now, the SNR at the output of the repeater is

We may view the lossy transmission medium followed by the amplifier as a cascade of two networks: one with a noise figure L and the other with a noise figure Fa. For the cascade connection, the overall noise figure is

If we select Ga = 1/L, then

The cascade of the lossy transmission medium and the amplifier is equivalent to a single network with the noise figure LFa.Now, suppose that we transmit the signal over K segments of the channel, where each segment has its own repeater. Then, if Fi = iFai is the noise figure of the ith section, the overall noise figure for the K sections is

Therefore, the signal-to-noise ratio at the output of the repeater (amplifier) at the receiver isIn the important special case where the K segments are identical, i.e., Li= L for all i and Fai = Fa for all i, and where the amplifier gains are designed to offsetthe losses in each segment, i.e., Gai = Li for all i, then the overall noise figurebecomes

Hence,

The overall noise figure for the cascade of the K identical segments is simply K times the noise figure of one segment.ExampleA signal with the bandwidth 4 kHz is to be transmitted a distance of 200 km overa wireline channel that has an attenuation of 2 dB/km.Determine the transmitter power PT required to achieve an SNR of (S/N)o = 30 dB at the output of the receiver amplifier that has a noise figure FadB = 5 dB.Repeat the calculation when a repeater is inserted every 10 km in the wireline channel, where the repeater has a gain of 20 dB and a noise figure of Fa = 5 dB. Assume that the noise equivalent bandwidth of each repeater is Bneq = 4 kHz and that No = 4 1021 W/Hz.(a). The total loss in the 200 km wireline is 400 dB. From Equation (6.5.35), with K = 1, we have

Hence,

But

where dBW denotes the power level relative to one Watt. Therefore,

which is an astronomical figure.(b). The use of a repeater every 10 km reduces the per segment loss to LdB = 20 dB. There are 20 repeaters and each repeater has a noise figure of 5 dB. Hence,Equation (6.5.35) yields

and

Therefore,

However, analog repeaters add noise to the signal and, consequently, degrade theoutput SNR.the advantage of using analog repeaters in communication channels that span large distancesThe transmitted power PT must be increased linearly with the number K of repeaters in order to maintain the same (S/N)o as K increases. For every factor of two increase in K, the transmitted power PT must be increased by 3 dB.

ExampleWe assume that the message is a wide-sense stationary random process M(t) with the autocorrelation functionRM() = 16 sinc2(10000).We also know that all the realizations of the message process satisfy the conditionmax |m(t)| = 6. We want to transmit this message to a destination via a channelwith a 50 dB attenuation and additive white noise with the power spectral densitySn(f) = N0/2 = 1012 W/Hz. We also want to achieve an SNR at the modulator outputof at least 50 dB. What is the required transmitter power and channel bandwidth if we employ the following modulation schemes?1. DSB AM.2. SSB AM.3. Conventional AM with a modulation index equal to 0.8.SolutionThe power spectral density of the message process is

Therefore, the bandwidth W = 10,000 Hz.

The baseband SNR isThe channel attenuation is 50 dB. That is,

Hence,

1. For DSB-SC AM, we have

Therefore, PT = 200 WattandBW = 2W = 2 10000Hz = 20 kHz.2. For SSB AM, we have

Therefore, PT = 200 WattandBW = W = 10000Hz = 10 kHz.3. For conventional AM, with a = 0.8,

where is the modulation efficiency given by

Since max |m(t)| = 6, we have

therefore,

Hence,

Therefore,PT 909 Watt.The bandwidth of conventional AM is equal to the bandwidth of DSB AM, i.e., BW = 2W = 20 kHz.OutlineEffect of Noise on Amplitude-Modulation SystemsEffect of Noise on Angle ModulationComparison of Analog-Modulation SystemsCarrier-Phase Estimation with a Phase-Locked Loop (PLL)Effects of Transmission Losses and Noise in Analog Communication Systems3535Effect of Noise on Angle ModulationSince noise is additive, the noise is directly added to the amplitude-modulated signals.In a frequency-modulated signal, the noise is added to the amplitude and the message information is contained in the frequency of the modulated signal.Therefore, the message is contaminated by the noise to the extent that the added noise changes the frequency of the modulated signal.

The frequency of a signal can be described by its zero crossings. Therefore, the effect of additive noise on the demodulated FM signal can be described by the changes that it produces in the zero crossings of the modulated FM signal.

The effect of noise in a low-power FM system is more severe than in a high-power FM system.In a low power signal, noise causes more changes in the zero crossings.In angle modulations, the higher the signal level, the lower the noise level.

The angle-modulated signal is represented asThe output of this filter is

Assume that the signal power is much higher than the noise power. Then, the bandpass noise is represented as

Vn(t): the envelope of the bandpass noise processn(t): the phase of the bandpass noise processThe assumption that the signal is much larger than the noise means that

Note:As with conventional-AM noise-performance analysis, a precise analysis is quite involved due to the nonlinearity of the demodulation process.

noise component Noting that

we see that the output of the demodulator is given by

where we have definedThe noise component is inversely proportional to the signal amplitude Ac. The higher the signal level, the lower the noise level.In AM systems, the noise component is independent of the signal component, and a scaling of the signal power does not affect the received noise power.

(t) is either proportional to the message signal or proportional to its integral. In both cases, it is a slowly varying signal compared to nc(t) and ns(t), i.e., (t) . The bandwidth of the filtered noise at the demodulator input is half of the bandwidth of the modulated signal, which is many times the bandwidth of the message signal.

Therefore,Now notice that in this case, fc is the axis of symmetry of the bandpass noiseprocess. Therefore, the conditions leading to the result of Exercise 5.3.3 are validwith a = cos /Ac and b = sin /Ac. By using the result of Exercise 5.3.3, we have

Note: the bandwidth of the filtered noise process extends from fc Bc/2 to fc + Bc/2; hence, the spectrum of nc(t) extends from - Bc/2 to Bc/2. Therefore,

and

This equation provides an expression for the power spectral density of the filtered noise at the front end of the receiver.In the case of FM modulation, the process Yn(t) is differentiated and scaled by 1/2.The power spectral density of the process is given by

The demodulated-noise power spectral density is given

PMFMThe effect of noise in FM for higher frequency components is much higher than the effect of noise on lower frequency components.The noise power at the output of the lowpass filter is

The output signal power isThen the signal-to-noise ratio is

Noting that is the received signal power, denoted by PR, and

we may express the output SNR as

where is denoted by

The output SNR is proportional to the square of the modulation index . Therefore, increasing increases the output SNR.The increase in the received signal-to-noise ratio is obtained by increasing the bandwidth. Therefore, angle modulation provides a way to trade off bandwidth for transmitted power.Note that in the preceding, expression, PM/(max|m(t)|)2 is the average-to-peak-power-ratio of the message signal (or equivalently, the power content of the normalized message, PMn). Therefore,

Now using Carsons rule Bc = 2( + 1)W, we can express the output SNR interms of the bandwidth expansion factor, which is defined as the ratio of the channel bandwidth to the message bandwidth and is denoted by :

Therefore,

Although we can increase the output signal-to-noise ratio by increasing , having a large means having a large Bc (by Carsons rule). Having a large Bc means having a large noise power at the input of the demodulator.

will no longer apply and that the preceding analysis will not hold.The approximationHowever, in angle modulation, the message is in the phase of the modulated signal and, consequently, increasing the transmitter power does not increase the demodulated message power.In both amplitude modulation and angle modulation, increasing the transmitter power (and consequently the received power) will increase the output signal-to-noise ratio, but the mechanisms are totally different.In AM, any increase in the received power directly increases the signal power at the output of the demodulator.In angle modulation, the output signal-to-noise ratio is increased by a decrease in the received noise power.

In FM, the effect of noise is higher at higher frequencies.Signal components at higher frequencies will suffer more from noise than signal components at lower frequencies.In some applications where FM is used to transmit SSB-FDM signals, those channels that are modulated on higher frequency carriers suffer from more noise.To compensate for this effect, such channels must have a higher signal level.The quadratic characteristics of the demodulated noise spectrum in FM is the basis of preemphasis and deemphasis filtering.

ExampleWhat is the required received power in an FM system with = 5 if W = 15 kHz and N0 = 1014 W/Hz? The power of the normalized message signal is assumed to be 0.1 Watt and the required SNR after demodulation is 60 dB.SolutionWe use the relationwith= 106, = 5, PMn = 0.1, N0 = 1014, and W = 15,000, to obtain PR = 2 105 or 20 microwatts.

Threshold Effect in Angle ModulationThe noise analysis of angle-demodulation schemes is based on the assumption that the signal-to-noise ratio at the demodulator input is high.With this crucial assumption, we observe that the signal and noise components at the demodulator output are additive and we are able to carry out the analysis.Particularly at low signal-to-noise ratios, signal and noise components are so intermingled that we cannot recognize the signal from the noise.Therefore, no meaningful signal-to-noise ratio as a measure of performance can be defined.Threshold Effect in Angle Modulation (cont.)Threshold Effect - there exists a specific signal-to-noise ratio at the input of the demodulator (known as the threshold SNR) below which signal mutilation occurs.The existence of the threshold effect places an upper limit on the tradeoff between bandwidth and power in an FM system.This limit is a practical limit in the value of the modulation index f. (f , Bc , noise power at the input of the demodulator )Threshold Effect analysis in FMAt threshold, the following approximate relation between and f holds in an FM system:

Given a received power PR, we can calculate the maximum allowed to make sure that the system works above threshold.Also, given a bandwidth allocation of Bc, we can find an appropriate using Carsons rule Bc = 2( + 1)W.Then, using the preceding threshold relation, we determine the required minimum received power to make the whole allocated bandwidth usable.Two factors that limit the value of the modulation index The limitations on channel bandwidth (which affect through Carsons rule)

The limitation on the received power (which limits the value of )

ThresholdThese plots are drawn for a sinusoidal message for which

In such a case,As an example, for = 5, this relation yields

and

ThresholdFor = 5, we have

and

For example,= 20 dB. Then regardless of the available bandwidth, we cannot use = 5 for such a system because the system will operate below threshold.

we can use = 2. This yields an SNR equal to 27.8 dB at the output of the receiver.This is an improvement of 7.8 dB over a baseband system.In general, if we want to employ the maximum available bandwidth, we must choose the largest possible that guarantees that the system will operate above threshold. This is the value of that satisfies

By substituting this value into Equation (6.2.22), we obtain

which relates a desired output SNR to the highest possible that achieves that SNR(for FM system)ExampleDesign an FM system that achieves an SNR at the receiver equal to 40 dB and requires the minimum amount of transmitter power. The bandwidth of the channel is 120 kHz; the message bandwidth is 10 kHz; the average-to-peak power ratio for the message, is 0.5; and the (one-sided) noise power spectral density is N0 = 108 W/Hz. What is the required transmitter power if the signal is attenuated by 40 dB in transmission through the channel?

SolutionFirst, we have to see whether the threshold or the bandwidth impose a more restrictive bound on the modulation index. By Carsons rule,

from which we obtain = 5. Using the relation

with = 104, we obtain 6.6.

Since the value of given by the bandwidth constraint is less than the value of given by the power constraint, we are limited in bandwidth (as opposed to being limited in power).Therefore, we choose = 5, which, when substituted in the expansion for the output SNR,

yields

Since with W = 10000 and N0 = 108, we obtain

and

Had there been no bandwidth constraint, we could have chosen = 6.6, which would result in . In turn, we would have PR 0.0153 and PT 153 Watts.

Preemphasis and Deemphasis FilteringThe noise power spectral density at the output of the demodulator in PM is flat within the message bandwidth; however, for FM, the noise power spectrum has a parabolic shape.FM performs better in low-frequency components of the message signal and PM performs better in high-frequency components.The objective in preemphasis and deemphasis filteringTo design a system which behaves like an ordinary frequency modulatordemodulator pair in the low-frequency band of the message signal, and like a phase modulatordemodulator pair in the high-frequency band of the message signal.A phase modulator is simply the cascade connection of a differentiator and a frequency modulatorwe need a filter in cascade with the modulator that does not affect the signal at low frequencies and acts as a differentiator at high frequencies. A simple highpass filter is a very good approximation to such a system.

The filter has a constant gain for low frequencies; at higher frequencies, it has frequency characteristics approximated by K|f|, which is the frequency characteristic of a differentiator.At the demodulator side, low frequencies have a simple FM demodulator and high frequency components have a phase demodulator, which is the cascade of a simple FM demodulator and an integrator.Therefore, the demodulator needs a filter that has a constant gain at low frequencies and behaves as an integrator at high frequencies. A good approximation to such a filter is a simple lowpass filter.

The modulator filter, which emphasizes high frequencies, is called the preemphasis filter; the demodulator filter, which is the inverse of the modulator filter, is called the deemphasis filter.

In commercial FM broadcasting of music and voice, first-order lowpass and highpass RC filters with a time constant of 75 s are employed.The frequency response of the receiver (deemphasis) filter is given by

where f0 = 1/(275106) 2100 Hz is the 3 dB frequency of the filter.The transmitter and the receiver filters cancel the effect of each other, the received power in the message signal remains unchanged. We only have to consider the effect of filtering on the received noise.The noise component after the deemphasis filter has a power spectral density given by

The noise power at the output of the demodulator can be obtained as

The ratio of the output SNRs in a simple FM system and a FM system with preemphasis and deemphasis filtering is inversely proportional to the noise power ratios, i.e.,

The improvement obtained by employing preemphasis and deemphasis filteringExampleIn commercial FM broadcasting, W = 15 kHz, f0 = 2100 Hz, and = 5. Assuming that the average-to-peak power ratio of the message signal is 0.5, find the improvement in the output SNR of FM when we use preemphasis and deemphasis filtering rather than a baseband system.SolutionIn a regular FM system

For FM with preemphasis and deemphasis filtering, we have

OutlineEffect of Noise on Amplitude-Modulation SystemsEffect of Noise on Angle ModulationComparison of Analog-Modulation SystemsCarrier-Phase Estimation with a Phase-Locked Loop (PLL)Effects of Transmission Losses and Noise in Analog Communication Systems7070Comparison of Analog-Modulation SystemsThree important practical criteriaBandwidth efficiencyPower efficiency (as reflected in its performance in the presence of noise)The ease of implementationBandwidth efficiencyThe most bandwidth efficient analog communication system is the SSB-SC system with a transmission bandwidth equal to the signal bandwidth.Widely used in bandwidth critical applications (such as voice transmission over microwave and satellite links and some point-to-point communication systems in congested areas).Cannot be used for the transmission of signals that have a significant DC component (e.g., image signals).The VSB system is a good compromise, which has a bandwidth slightly larger than SSB and is capable of transmitting DC values.PM (and particularly FM) are the least favorable systems when bandwidth is the major concern, and their use is only justified by their high level of noise immunity.Power EfficiencyAngle-modulation schemes (particularly FM) provide a high level of noise immunity and, therefore, power efficiency.FM is widely used on power-critical communication links, such as point-to-point communication systems and high-fidelity radio broadcasting.Conventional AM and VSB+C are the least power-efficient systems and are not used when the transmitter power is a major concern.Their use is justified by the simplicity of the receiver structure.Ease of ImplementationThe receiver for conventional AM is the simplest receiver structure The structure of the receiver for VSB+C system is only slightly more complicated.FM receivers are also easy to implement.These three systems are widely used for AM, TV, and high-fidelity FM broadcasting (including FM stereo).The power inefficiency of the AM transmitter is compensated by the extremely simple structure of literally hundreds of millions of receivers.DSB-SC and SSB-SC require synchronous demodulation and, therefore, their receiver structure is much more complicated (never used for broadcasting purposes).DSB-SC is hardly used in analog signal transmission, due to its relative bandwidth inefficiency.

OutlineEffect of Noise on Amplitude-Modulation SystemsEffect of Noise on Angle ModulationComparison of Analog-Modulation SystemsCarrier-Phase Estimation with a Phase-Locked Loop (PLL)Effects of Transmission Losses and Noise in Analog Communication Systems7575Carrier-Phase Estimation with a Phase-Locked Loop (PLL)For DSB-SC AM, the received noise-corrupted signal at the input to the demodulator is given by

where m(t) is the message signal, which is assumed to be a sample function of a zero-mean random process M(t).r(t) has a zero mean, since the message signal m(t) is zero mean, i.e., m(t) contains no DC component.Note:The average power at the output of a narrowband filter tuned to the carrier frequency fc is zero. we cannot extract a carrier-signal component directly from r(t).If we square r(t), the squared signal contains a spectral component at twice the carrier frequency. That is,

E[M2(t)] = RM(0) > 0 there is signal power at the frequency 2fc can be used to drive a phase-locked loop (PLL).

A sinusoidal component at 2fc

see pp. 186lowpass filter

This filter is usually selected to have the relatively simple transfer function

The time constants 1 and 2 are design parameters (1 >> 2) that control the bandwidth of the loop.The output of the loop provides the control voltage for the VCO.A sinusoidal signal with an instantaneous phase given by

In a steady-state operation when the loop is tracking the phase of the received carrier, the phase error is small; hence,

With this approximation, the PLL is represented by the linear model which has a closed-loop transfer function

By substituting from Equation (6.4.4) for G(s), we obtainThe denominator of H(s) may be expressed in the standard form

:the loop-damping factor:the natural frequency of the loop

In terms of the loop parameters,

andthe closed-loop transfer function becomes

= 1: critically damped loop response < 1: underdamped loop response > 1: overdamped loop responseIn practice, the selection of the bandwidth of the PLL involves a trade-off between the speed of response and the noise in the phase estimate.On the one hand, we want to select the bandwidth of the loop to be sufficiently wide in order to track any time variations in the phase of the received carrier.On the other hand, a wideband PLL allows more noise to pass into the loop, which corrupts the phase estimate.

The (one-sided) noise-equivalent bandwidth of the loop is

Effect of Additive Noise on Phase EstimationAssume that the PLL is tracking a sinusoid signal of the form

which is corrupted by the additive narrowband noiseThe in-phase and quadrature components of the noise are assumed to be statistically independent, stationary Gaussian noise processes with (two-sided) power spectral density N0/2 W/Hz.n(t) can be expressed as

whereand

note that

if s(t) + n(t) is multiplied by the output of the VCO and the double frequency terms are neglected, the input to the loop filter is the noise-corrupted signal

When the power Pc = Ac2/2 of the incoming signal is much larger than thenoise power, the phase estimate . Then, we may linearize the PLL

The gain parameter Ac may be normalized to unity, provided that the noise term is scaled by 1/Ac. Thus, the noise term becomes

Since the noise n1(t) is additive at the input to the loop, the variance of the phase error , which is also the variance of the VCO output phase, is

is simply the ratio of the total noise power within the bandwidth of the PLL divided by the input signal powerthe (one-sided) equivalent-noise bandwidth of the loop

where L is defined as the signal-to-noise ratio

An exact analysis based on the nonlinear PLL is mathematically tractable when G(s) = 1, which results in a first-order loop. In this case, the probability density function for the phase error has the form

I0() is the modified Bessel function of order zero.1/LThe variance for the linear model is close to the exact variance for L > 3When the SNR at the input to the PLL drops below a certain value, the loop begins to lose lock and an impulsive-type of noise, characterized as clicks, is generated; this degrades the performance of the loop.The squaring PLL

Let the input to the squarer be u(t) + n(t). The output is y(t) = u2(t) +2 u(t)n(t)+ n2(t). the noise termsBoth components have spectral power in the frequency band centered at 2fc. The bandpass filter with bandwidth Bneq centered at 2fc, which produces the desired sinusoidal signal component that drives the PLL, also passes noise due to these two noise terms.The squaring PLL and the Costas PLL are two practical methods for deriving a carrier-phase estimate for the synchronous demodulation of a DSB-SC AM signal.Let us select the bandwidth of the loop to be significantly smaller than the bandwidth Bbp of the bandpass filter, so that the total noise spectrum at the input to the PLL may be approximated by a constant within the loop bandwidth.This approximation allows us to obtain a simple expression for the variance of the phase error as

where SL is called the squaring loss and is given as

Since SL < 1, we have an increase in the variance of the phase error; this is caused by the added noise power that results from the squaring operation. For example, when L = Bbp /2 Bneq, the loss is 3 dB or, equivalently, the variance in the estimate increases by a factor of two.Costas Loop

A second method for generating a properly phased carrier for a double sideband-suppressed carrier AM signal is Costas loopThe received signal

is multiplied by cos(2fct + ) and sin(2fct + ), which are outputs from the VCO.The two products are

An error signal is generated by multiplying the two outputs yc(t) and ys(t) ofthe lowpass filters. Thus,

the desired termsignal noisesignal noisenoise noiseOutlineEffect of Noise on Amplitude-Modulation SystemsEffect of Noise on Angle ModulationComparison of Analog-Modulation SystemsCarrier-Phase Estimation with a Phase-Locked Loop (PLL)Effects of Transmission Losses and Noise in Analog Communication Systems9191Effects of Transmission Losses and Noise in Analog Communication SystemsTwo dominant factors that limit the performance of the systemAdditive noiseSignal attenuationBasically all physical channels, including wireline and radio channels, are lossy.Signal attenuation renders the communication signal more vulnerable to additive noise

Characterization of Thermal Noise SourcesThermal noise is produced by the random movement of electrons due to thermal agitation.

The power spectral density of thermal noise is given as

: Plancks constant (equal to 6.6 1034 J sec)k : Boltzmanns constant (equal to 1.38 1023 J/K)T : the temperature in degrees KelvinR : the resistance ()At frequencies below 1012Hz (which includes all conventional communication systems) and at room temperature,

Consequently, the power spectral density is well approximated as

When connected to a load resistance with value RL, the noise voltage delivers the maximum power when R = RL. In such a case, the load is matched to the source and the maximum power delivered to the load is E[N2(t)]/4RL. Therefore, the power spectral density of the noise voltage across the load resistor is

kT is usually denoted by N0. Hence, the power spectral density of thermal noise is generally expressed as

For example, at room temperature (T0 = 290 K), N0 = 4 1021 W/Hz.

Effective Noise Temperature and Noise FigureWhen we employ amplifiers in communication systems to boost the level of a signal, we are also amplifying the noise corrupting the signal.we may model an amplifier as a filter with the frequency response characteristic H(f).

Noise power at the output of the network is

Recall that the noise equivalent bandwidth of the filter is defined asG = |H(f)|2max is the maximum available power gain of the amplifier.The output noise power from an ideal amplifier may be expressed as

Any practical amplifier introduces additional noise at its output due to internally generated noise. Hence, the noise power at its output may be expressed as

the power of the amplifier output due to internally generated noise

Therefore,

whereis called the effective noise temperature of the two-port network (amplifier).A signal source at the input to the amplifier with power Psi will produce an output with power

Hence, the output SNR from the two-port network is

the input SNR to the two-port networkThe SNR at the output of the amplifier is degraded (reduced) by the factor (1+Te/T). An ideal amplifier is one for which Te = 0.When T is taken as room temperature T0 (290K), the factor F= (1+Te/ T0) iscalled the noise figure of the amplifier.

Consequently,By taking the logarithm of both sides,

the loss in SNR due to the additional noise introduced by the amplifierThe overall noise figure of a cascade of K amplifiers with gains Gk and corresponding noise figures Fk, 1 k K is

Fries formulaF1 is the dominant term, which is the noise figure of the first amplifier stage. The front end of a receiver should have a low noise figure and a high gain.Transmission LossesThe amount of signal attenuation generally depends on the physical medium, the frequency of operation, and the distance between the transmitter and the receiver.Define the loss L in signal transmission as the ratio of the input(transmitted) power to the output (received) power of the channel, i.e.,

or, in decibels,

In wireline channels, the transmission loss is usually given in terms of decibels per unit length, e.g., dB/km.For example, the transmission loss in coaxial cable of 1 cm diameter is about 2 dB/km at a frequency of 1 Mhz. This loss generally increases with an increase in frequency.In line-of-sight radio systems, the transmission loss is given as

= c/f: the wavelength of the transmitted signalc: the speed of light (3 108m/s)f: the frequency of the transmitted signald: the distance between the transmitter and the receiver in metersfree-space path lossExampleDetermine the free-space path loss for a signal transmitted at f = 1 MHz over distances of 10 km and 20 km.Doubling the distance in radio transmission increases the free-space path loss by 6 dB.Solution = 300 m. The loss for the 10 km path is

The loss for the 20 km path is

Repeaters for Signal Transmission

Analog repeaters are basically amplifiers that are generally used in telephone wireline channels and microwave line-of-sight radio channels to boost the signal level and, thus, to offset the effect of signal attenuation in transmission through the channel.

The input signal power at the input to the repeater isThe output power from the repeater is

We may select the amplifier gain G to offset the transmission loss. Hence, G = L and P0 = PT .

Now, the SNR at the output of the repeater is

We may view the lossy transmission medium followed by the amplifier as a cascade of two networks: one with a noise figure L and the other with a noise figure Fa. For the cascade connection, the overall noise figure is

If we select Ga = 1/L, then

The cascade of the lossy transmission medium and the amplifier is equivalent to a single network with the noise figure LFa.Now, suppose that we transmit the signal over K segments of the channel, where each segment has its own repeater. Then, if Fi = iFai is the noise figure of the ith section, the overall noise figure for the K sections is

Therefore, the signal-to-noise ratio at the output of the repeater (amplifier) at the receiver isIn the important special case where the K segments are identical, i.e., Li= L for all i and Fai = Fa for all i, and where the amplifier gains are designed to offsetthe losses in each segment, i.e., Gai = Li for all i, then the overall noise figurebecomes

Hence,

The overall noise figure for the cascade of the K identical segments is simply K times the noise figure of one segment.ExampleA signal with the bandwidth 4 kHz is to be transmitted a distance of 200 km overa wireline channel that has an attenuation of 2 dB/km.Determine the transmitter power PT required to achieve an SNR of (S/N)o = 30 dB at the output of the receiver amplifier that has a noise figure FadB = 5 dB.Repeat the calculation when a repeater is inserted every 10 km in the wireline channel, where the repeater has a gain of 20 dB and a noise figure of Fa = 5 dB. Assume that the noise equivalent bandwidth of each repeater is Bneq = 4 kHz and that No = 4 1021 W/Hz.(a). The total loss in the 200 km wireline is 400 dB. From Equation (6.5.35), with K = 1, we have

Hence,

But

where dBW denotes the power level relative to one Watt. Therefore,

which is an astronomical figure.(b). The use of a repeater every 10 km reduces the per segment loss to LdB = 20 dB. There are 20 repeaters and each repeater has a noise figure of 5 dB. Hence,Equation (6.5.35) yields

and

Therefore,

However, analog repeaters add noise to the signal and, consequently, degrade theoutput SNR.the advantage of using analog repeaters in communication channels that span large distancesThe transmitted power PT must be increased linearly with the number K of repeaters in order to maintain the same (S/N)o as K increases. For every factor of two increase in K, the transmitted power PT must be increased by 3 dB.