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ELECTRONIC COMMUNICATION SYSTEMS BY GEORGE KENNEDY PDF

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ELECTRONICS COMMUNICATION SYSTEM BY GEORGE cittadelmonte.info George, date Electronic Communication system/George Kennedy, Bernard Davis . Electronic Communication System (4th Edition) by Kennedy & cittadelmonte.info R. Stumy, Frequency-Modulated Radio,2d ed., George Newnes Ltd., London, , . ELECTRONIC COMMUNICATION SYSTEM BY GEORGE cittadelmonte.info - Ebook download as PDF File .pdf), Text File .txt) or read book online.


Electronic Communication Systems By George Kennedy Pdf

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Electronic Communication system/George Kennedy Advanced Electronic Communications Systems Wayne Tomasi Sixth Edition Advanced Electronic. Electronic Communication Systems. Fifth Edition. George Kennedy. Supervising Engineer. Overseas Telecommun/catlons Commission. Austral/a. Bernard Davis. Kennedy. George, date. Electronic Communication system/George Kennedy, Bernard Davis,. 4th ed p. cm. Includes bibliographical references and index.

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Noise 17 can become significant. We also receive noise from the center of our own galaxy the Milky Way. Random noise power is proportional to the bandwidth ove. Not very much of it below 20 MHz penetrates down through the ionosphere. Such noise is generally random.. A random voltage across the resistor definitely exists and may be both measured and calculated. Using Equation 2.

This noise voltage is caused by the random movement of electrons within the resistor. Since it is random and therefore has a finite nns value but no de component. That is correct if the measuring instrument is a direct current de voltmeter.

Assmne that RL is noiseless and is receiving the maximum noise power generated by R. Then P. It is tnte that as many electrons arrive at one end of the resistor as at the other over any long period oftirne. It must be realized that all fonn ulns referring to random noise arc applicable only to the m1s value of such noise.

At any instant of time. The rate of arrival of electrons at either end of the resistor therefore varies randomly. The resistor is a noise generatur. The paths taken are random and therefore unequal. Noise 19 and V. Note especially that the generated noise voltage is quite independent of the frequency at which it is measured. V Fig. Example 2. It is caused by rando. The most important of all the other sources is the shot effect.

Hence the name shot noise. Although the average output current of a device is governed by tlle various bias voltages. When amplified. A low voltage fed to this amplifier would be masked by the noise and lost. This stems from the fact that it is random and therefore evenly distributed over the frequency spectmm. In bipolar transistors. They all show that such noise is inversely proportional to transconductance and also directly proportional to output current.

The minute currents induced in the input of the device by random fluctuations in the output current become of great importance at such frequencies and create random noise frequency distortion. In addition. A noise figure see Sect-ion 2.

Many variables are involved in the generation of this noise in the various amplifying devices. Shot noise behaves in a similar manner to thennal agitation noise.

So far as the use of R is concerned. The most convenient method of dealing with shot noise is to find the value or fortnula for an equivalent input-noise resistor. This precedes the device. For noise only.

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Once this high-frequency noise makes its presence felt. Tbc noise current has been replaced by a resistance so that it is now easier to add shot noise to thermal noi. RF transistors are remarkably low-noise. The value of the equivalent shot-noise resistance R. The sum of two such nns voltages in series is given by the square root of the sum of their squares.

Approximate formulas for equivalent shot. The result of all this is that it is preferable to measure noise at such high frequencies. This means that the total noise voltage is less than that due to any of the individual resistors.

Tt may appear logical to combine all the noise resistances at the input. The process might then be continued. The bandwidth of the amplifier is 6 MHz.. This is. Solution V. This greatly simplifies subsequent calculations.

It is even better to go one step further and find an equivalent resistance for such an input voltage. The result JJ useless because the argument assumed that it is important to find the total output noise voltage. Now the noise resistance actually present at the input of the second stage is R2.

The same noise voltage would be present at the output if there were no R3 there.. Tnstead R. The-series resistance of the coil. As Example 2.

I required to determine the noise voltage across the capacitor. For the second stage. Noise 23 ll. To either side ofresonance the presence of the tuned circuit affects noise in just the same way as any other voltage.

This will allow us Consider Fig.. In the preceding sections dealing with noise calculations. Calculate the equivalent input-noise 1'esistance of this two-stage amplifier.. The more interesting case is a tuned circuit which is not ideal. Thus i. Equation 2. The second 1s companson of n01se and signal at the same point to ensure that the n01se 1s.. The noise current in the circuit will be. It is fed from a source antenna of internal impedance R.

For example. Instead of equivalent noise resistance. Noise 25 not excessive. In the second instance. Therefore 2. In addi- tion. The noise figure of practical receivers can be kept to below a couple of decibels up to frequencies in the lower gigahertz range by a suitable choice of the first transistor. At frequencies higher than that.

The noise figure Fis defined as the ratio of the signal-to-noise power supplied to the input tenninals of a receiver or amplifier to the signal-to-noise power supplied to the output or load resistor. As a matter offact. Each is treated as a four-tenninal network having an input impedance R1. An effort is naturally made to keep the signal. It is defined as the ratio of signal power to noise power at the same point. Calculate the generalized form of noise figure from steps 3 and 6 2.

Calculate Pno from Rcq if possible 2. Each is now shown. The calculation procedure may be broken down into a number of general steps. Rt gain"' A Fig. It is seen from Fig. Write P for the noise output power to be determined later 2.

Determine the signal output power P. An actual fonnula for F may now be obtained by substitution for the output noise power Putting it another way. All this applies here 1 with the minor exception that these noise resistances must now be added to the parallel combination of R0 and R. Note that Equation 2. Noise 27 2.

For the ti.

Electronic Communication system/George Kennedy

It is convenient to define R: This may be seen from reexamining Equation 2. Not the least reason for its use is convenience. Note that this constitutes n lnrgc enough mismatch. This is a situation exploited very often in prac6ce. Under matched conditions R. Controversy exists regarding which is the better all-around measurement.

Jf all the noise of the receiver were generated by R0. Noise 29 P. T0F -. It is then possible to use Equation' If this is to lead to the conect value of noise output - power. It must be repeated that the equiva- lent noise temperature is just a convenient fiction.

In defining the equivalent noise temperature of a receiver or amplifier. Tcq' the equivalent noise temperature. Finally we have.. It will be recalled that the equivalent noise resistance introduced in Section 2.

It is the d. When dealing with random noise calculations it comparing the noise performance of receivers: Circle the letter preceding the a.

Thermal noise is independent of the frequency a. Equivalent noise resistance vaJues. Industrial noise is usually o the impulse c. Galactic noise a. Cosmic noise L. Space noise generally covers a wide frequency doubled. The square of the b. One of the following types of noise becomes of c. One of the following is not a useful quantity for 9. Noise figure c. The value of a resistor creating thermal noise is 8. Solar noise line that correctly completes each sentence. Atmospheric noise great importance at high frequencies.

The noise power generated is therefore spectrum. Input noise voltage a. Indicate the noise whose source is in a category consists ofan incomplete statement followed by four different from that oftbe other three. Boltzmann's constant fiers. Which two broad classifications of noise are the width.

HF mixers are generally noisier than HF ampli. Noise temperature b. Random noise power is inversely proportional surement for comparing amplifier noise charac. A random voltage across a resistance cannot The noise output of a resistor is amplified by a noiseless amplifier having a gain of 60 and a bandwidth of20 kHz.

Noise 31 JO. When might the latter. Review Problems I. If this receiver is connected to an antenna with an impedance of 75 fl. How can some of them be avoided or minimized? What is the strongest source of extraterrestrial noise? The front end of a television receiver. Define signal-to-noise ratio and noise figure ofa receiver.

Noise in mixers is caused by inadequate image c. Which of lhe following statements is tme? What does the meter read now? Which of the following is the most reliable mea. The RF amplifier of a receiver has an input resistance of l n. Flicker is sometimes called demodulation a. Discuss the types. The amplifier has a input resistor and a shot-noise equivalent resistance of fl. A meter connected to the output of the amplifier reads I mV rms. Calculate the minimum signal voltage that the receiver of Problem 2.

Review Questions I. Describe briefly the forms of noise to which a transistor is prone. Given that the bandwidth is 1. A parallel-tuned circuit. An amplifier operating over the frequency range of to kHz bas a kfl input resistor. A receiver has an overall gain A. What is transit-time effect? How it is generated? Describe briefly how this can be measured using the diode generator. Write the relation for maximum noise power output of a resistor.

What is ideal and practical values of noise figure? Why they arc so explain. One of the terms of this formula will be the noise output power. What is noise temperature'? How is it related to noise figure? Write the expression for therms noise voltage. Derive the relation between noise figure and temperature. The students will also be able to calculate the frequencies present.

This clas- sification is mainly based on the nature of message or modulating signal. This is based on how many components of the basic amplitude modulated signal are chosen for transmission. If the message to be transmitted is continuous or analog in nature. The amplitude of the carrier signal is varied in accordance with the message to obtain modulated signal in case of amplitude modulation.

Upon studying this chapter. Based on this observation. In analog communication. Understand the differences between AM and its variants. Tbis chapter deals with the amplitude modulation techniques employed in analog communication. After studying the theory of amplitude modulation techniques. This chapter deals with amplitude modulation techniques in detail.

Objectives Upon completing the material in Chapter The modulation techniques in analog communicatiot1 can be classified into amplitude modulation AM and angle modulation techniques. The angle modulation employs variation of angle of the carrier signal in proportion to the message. SSB and VSB signals Anulyze and detem1ine through computation the carrier power and sideband power in AM and its variants Solve problems involving frequency components. This is followed by a description of different methods for the generation of AM.

The next chapter deals with angle modu. Explain different approaches for the generation of AM. The communication process can be broadly divided into two types. AM is defined as a system of modulation in which the amplitude of the carrier is made proportional to the instantaneous amplitude of the modulating voltage.

In practice. Hence in this section and later. The analog message is passed onto the final destination. Let the carrier voltage and the modulating voltage. This block diagram is drawn by referring to the communication system block diagram given in Fig.

This results in the modulated signal which is also analog in nature. The continuous message signal is subjected to analog modulation with the help of a sine wave carrier at the transmitter.

Its inclusion here would merely complicate the proceedings. This chapter deals with various amplitude modulation techniques employed in analog modulation block shown in Fig.

At the receiver the incoming modulated signal is passed through an analog demodulation process which extracts out the analog message signal. In amplitude modulation.. The analog modulated signal is transmitted via the cornmuication channel towards the receiver. Figure 3. The modulation index is a number lying between O and I. Thus when there is temporarily no modulation. COS w e. We have A "' V.

Amplitude Modulation 35 From the definition of AM. From Fig. Freqttettcy Spectnmi of the AM Wave We shall show mathematically that the frequencies present in the AM wave are the carrier frequency and the first pair of sideband frequencies. The situation is illustrated in Fig. When a carrier is amplitude modulated.

ELECTRONIC COMMUNICATION SYSTEM BY GEORGE KENNEDY.pdf

When modula- tion is present. Equation 3. As illustrated Solution I I J.. The very important conclusion to be made at this stage is that the bandwidth required for amplitude modulation is twice the frequency of the modulating signal. The two additional terms produced are the two sidebands outlined. The frequency spectrum ofAM wave is shown in Fig.

If the oscillator output is modulated by audio frequencies up to 10 kH. Example 3. The first tenn is identical to Equation 3. AM consists of three discrete frequencies.

Of these. It is apparent that the process of amplihtde modulation has the effect of adding to the unmodulated wave.

Jt is derived from Fig. Dividing the equation of Vmby the equation of Ve. II is the standard method of evaluating the modulation index when calculating from a wavefonn such as may be seen on an oscilloscope. When only the root mean square nns values oftbe carrier and the modulated voltage or current are known..

The modulated wave extends between these two limiting envelopes and has a repetition rate equal to the unmodulated carrier frequency. It may not be used in any other siniation.

Vmln 3.

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The maximum negative amplitude. Vm a Vmn Power Relations in the AM Wave It has been shown that the carrier component of the modulated wave has the same amplitude as the unmodulated carrier. This is important. This relation may now be derived. Since amplitude of the sidebands depends on the modulation index V IV. The first tenn of Equation 3. Solution P AM'!!!! It is interesting to know from Equation 3. The modulated wave contains extra energy in the two sideband components.

Calculate the total power in the modulated wave. C it is anticipated that the total power Ill in the modulated wave will depend on the modulation index also..

How much of this is carrier power? If R is the resistance in which these currents flow. Find the percentage modulation. Solution i. This occurs when the antenna current of the transmitter is metered. Determine the antenna current when the percent of modulation clumges to 0. There are two methods of calculating the total modulation index. Let V1. Equation Then the total modulating voltage V. The procedure consists of calculating the total modulation index and then substituting it into Equation 3.

To calculate the total modulation index. I sideband power will now be the sum of individual sideband powers. It is seen that there are two approaches It increases to 12 A as a result of simultaneous modulation.

Calculate the modulation index. If another sine wave is simultaneously transmitted wW1 modulation index 0. What is the modulation index due to this second wave? Solution From Equation. Note also that th.. DSBSC technique accordingly adds complexity at the receiving point to recover the message.. II c 2 Therefore. Suppose your application requirement is cost ofreceiver needs to be significantly low Thus depending on the application.

A significant saving in power requirement can be achieved by supressing the carrier before transmission. PAM 3. The top envelope crosses below ilie zero reference amplitude value and similarly. For better distinction. It is derived from Fig.

As illustrated. The instantaneous voltage of the resulting amplitude modulated wave is v.. At the most.. This is tbe price we pay by suppressing the carrier. V sin ro. Thus the information from AM can be recovered uniquely either from top or bottom envelope by a simple envelope detector circuit assume it as diode rectifier for time being.

Samething is true with respect of bottom envelope also. Example Since amplitude of the sidebands depends on the modulation index V. The modulated wave contains energy only due to the two sideband components. SR 4 2R Substituting Equation How much power saving in W is ac. This is correct also. This thought process led to the development of another variant of AM..

B SSII. COS W. Both the sidebands. Since only one of the sidebands is selected for transmis- sion. SSB needs a bandwidth equal to that of message.. In this bookj unless specified. We require 5. It will be maximum. I case. Hence saving in bandwidth can be achieved by suppressing one of the sidebands..

Case 2 Given. Pososc "' P. Carrier power. Total power in AM. How much of carrier power ill kW is required if we want to transmit the same message by an AM transmitter? Solution Given. The mathematical treatment here follows this assumption.

That is..

From Fig.. SSB signficantly saves power. The SSB technique further complicates the receiver structure to recover message. SSB consists ofone discrete frequency either atf. The bandwidth required for SSB is the frequency of the modulating signal..

Therefore still all the three versions ofAM. If the cut-off frequencies are if. Then the next question is why not use only SSB? The answer is same as in the case of existence of AM. As will be explained later. The envelope of SSB does not contain message and hence a simple envelope detector circuit is not useful for recovering the message. This is the price we pay by suppressing the carrier and one of the sidebands.

Its frequency will be either lower or more than carrier frequency by au amount of modulating signal frequency. LSB I fc. The modulated wave contains energy only due to one sideband compcment. The modulated wave will have only one sine wave. USB fo-fm fr. The only wave to distinguish is to compare with carrier signal.

SO in 3. Pc"" W and m. Calculate the total power in case of SSB tech- nique. Total power in SSB. Compare the powers required for SSB in both the cases and comment on the reason for change in tile power levels. This infers that the total power in SSB also depends on the depth of modulation. The total power in the SSB modulated wave will be P. Solution Case 1 Given. In total 6. This is because.. How much of carrier power in kW is required if we want to transmit the same message by an AM transmitter?

Solution Given.. Ill2 o. It was observed in practice that such a process results in eliminating even some por- tion of the wanted sideband.

Carrier power.. One way to compensate for this loss is to allow a vestige or trace or fraction-of unwanted sideband along with the wanted sideband. Therefore an attempt to attenuate unwanted component will in tum leads to attenuation of wanted component.

If the cut-off frequencies arc f. COS W.. VSB consists of two discrete frequencies either at ifc. Thif book also follows the same convention. The bandwidth required for VSB is the frequency of the modulating signal plus vestiage. COS W r. Since amplitude of the sidebands depends on the modulation index VJ Vr.. Amplitude Mod11latio11 51 t j a j b Fig. The shape of the signal in the time cloinain depends on the value of vestige frequency.

BR 4 2R Substituling these equations in the total power equation. The modulated wave contains energy due to these two components. Now adding the unmodulated carrier component to thtS.. Solution G iven. We require Calculate tlze total power i11 case oJVSB technique. The output of the analog multiplier is given by V. How much power saving in W is. In-a nonlinear resistance.

If the above equation refers to a resistor. The previous linear relation seems to apply to ce1iain point. Current now becomes proportional not only to voltage but also to the square. Whether the increase is more or less rapid depends on whether the device begins to saturate. Only the square term is large enough to be taken into consideration for most applications.. The reason that the initial portion of the graph is linear is simply that the coefficient c is much smaller than b.

Vme c. Since Equation 3. The diode is biased such that it exhibits the negative resistance property. The requisite AM components can be selected by using the tuning circuit that resonates at the carrier frequency with a bandwidth equal to twice the me. The devices like diodes. The output of the diode is collected via a tuned ci. Under this condition. The output of the balanced modulator contait1s the two sidebands and some of the miscellaneous components which are taken care of by tuning the output tranfom1er's secondary winding.

Vo D2 Fig. No system can of course be perfectly symmetrical in practice. D1 and vf'. Analog ll "Vn1 Ve multiplier l Ve Fig. Cb RF. The diodes use the nonlinear resistance property for generating modulated signals.

The modulated output currents of the two diodes arc combined in the center-tapped prirnary of the output trilnsfonner. Both the diodes receive tbe carrier volt- age in phase. The final output consists only of sidebands. As indicated. The balanced modulator using the diodes is given in Fig. The above equation has the standard AM signal components.. In this way we can generate the AM signal with the help of device that exibblts nonlinear resistance property.

If this system is made cornpletely symmetrical. They therefore subtract. This signal is passed through a. The output of the analog multiplier is given by. The two diode output currents will be i. The tuning of the output transformer will remove the modulating frequencies from the output. If the lower sideband is passed out then the output of the bandpass filter will be mVc v a Using the Filter Method The basis for the filter method is that after the balanced modulator the unwanted sideband is removed by a filter.

One of the sidebands. Thus by suitable polarity for M1 output and addiJ. Vc cos co. Usittg the Phase Shift Method The phase shift method avoids filters and some of tbeir inherent disad- vantages.. Both the modulators produce the two-sidebands.

CVmVt cos w. One of the balanced modulators. The balanced modulator generates the DSBSC signal and the sideband suppression filter suprcsses-the unwanted sideband and al lows the wanted sideband.. Depending on the cut-off frequency values we can-represent the output of tbe fi Iler as v "" 2acV. In this way SSB is generated in case of filter method. As derived in the previous section.

Let v. The output of the balanced modulator M1 is given by. We can see that the later part of this circuit is identical to that of the phase sbift method.. A phase shift is then appl ied to this frequency only. The block diagram oftbe third method is shown in Fig. The output of M. I A ltematively. The output oftbe analog multiplier is given by v1.

This signal is passed through a bandpass filter which.. The study of all the amplitude modulation techniques gives 1 better understanding about their nature in time and frequency domains. Depnding on the cut-off frequency values we can represent the output of the filter as1. The balanced modulator generates the DSBSC signal and the sideband suppression filter supresses most oftbe unwanted sideband and allows a vestige ofit along with the other sideband.

Thus the basic blocks remain same as in the case of SS B generation and the only difference is in the cut-off frequency values of the bandpass filter. F 2acV. The basic technique.. If the lower sideband and vestige of upper sideband are passed out. Other methods are relatively different. The bandwidth of AM wave is given by c. Amplitude modulation is defined as the system 8. The method using analog multiplier is concephJally simple to understand..

The AM wave will have 1. Amplit11tle Mod11lalio11 61 namely.. AM needs maxi mum power and bandwidth among all its variants. The SSB technique needs mini- mum power and bandwidth. THc modulation index of AM is given by C. VI" sincoCt tionless condition is c.

The expression for total power in AM wave is Ill a. Multiple-Choice Questions Each of the followi11g multiple-choice questions b. The instantaneous voltage of the AM wave is d. The peak. Circle the letter preceding the d. This was followed by the study of different methods for the generation of AM and its variants.

IfvSll is the instantaneous voltage ofone sideband. VDS8SC '. The SSB wave wiU have a. The expression for total modulation index in case Ill b. Ill a. DSBSC b. The expression for total power in VSB wave is b. The maximum power ofSSB wave under distor. F vso b. SSB b.. The outJ ut current of a nonlinear resistor caa. VSB c. The instantaneous voltage of the VSB wave related to its input voltage by having f: JSB as wanted sideband is n. The expression for total power in SSB wave is d.

The bandwidth of SSB wave is given by The balanced modulator can be used for the gen- VSB d. The VSB wave will have lator is to a. The instantaneous voltage of the VSB wave hav. The basic working principle of a balanced modu- SSB c. DSBSC a.: To what value will this current rise if the generator is modulated additionally by another audio wave. When the modulation percentage is What is the total sideband power radiated? Hz audio sine waves.

A broadcast AM transmitter radiates 50 kW of carrier power. When a broadcast AM transmitter is 50 percent modulated. A IkHz carrier is simultaneously modulated with H: A transistor class C amplifier has maximum permissible collector dissipation of 20 W and a collector efficiency of 75 percent. The output current of a 60 percent modulated AM generator is 1.

The basic working principle of third method for A W carrier is simultaneously modulated by two audio waves with modulation percentages of55 and What will be the percentage power saving if the carrier and one of the sidebands arc now suppressed? What will be the radiated powerat 85 percent modulation?

What will be the frequencies present in the output? What will the current be when the modulation depth is increased to 0. Mention the different components of VSB wave? Derive the expression for the total power in case ofSSB wave? Amplitude Morlttlatio11 65 Review Questions I.

Derive the expression for the instantaneous voltage of VSB wave? Derive the expression for the instantaneous voltage of AM wave? How much is the bandwidth of YSB wave?

How much is the bandwidth of AM wave? Define modulation index of amplitude modulation? Mention the different components of AM wave? How much is the bandwidth ofSSB wave? Define amplitude modulation? Mention the different components of SSB wave? How do you distinguish between analog and digital communication? Write the expression for the instantaneous voltage of AM wave? George Kennedy and Bernard Davis, Electronics. George Kennedy, Bernard Davis, , Sept.

Electronics and communication System by George kennedy. Aircraft Materials and Processes by George F. Mechanics of Flight By. Electronic Communication systems: George Kennedy- Mc Graw Hill. Classify different types of communication system. Electronic communication systems by George Kennedy. May 15, Oct 18, Electronic Communication Systems by George Kennedy Dec systems through various techniques and the performance of communication system inpresence of noise. Flag for inappropriate content. Related titles.

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