Frenzel - Principles of Electronic Communication Systems 4th Ed, Part 2 PDF

chapter 2 Electronic Fundamentals for Communications T o understand communication electronics as presented in this book, you need a knowledge of certain basic principles of electronics, including the fundamentals of alternatin...

chapter 2 Electronic Fundamentals for Communications T o understand communication electronics as presented in this book, you need a knowledge of certain basic principles of electronics, including the fundamentals of alternating-current (ac) and direct-current (dc) circuits, semiconductor operation and characteristics, and basic electronic circuit operation (amplifiers, oscillators, power supplies, and digital logic circuits). Some of the basics are particularly critical to understanding the chapters that follow. These include the expression of gain and loss in decibels, LC tuned circuits, resonance and filters, and Fourier theory. The purpose of this chapter is to briefly review all these subjects. If you have studied the material before, it will simply serve as a review and reference. If, because of your own schedule or the school’s curriculum, you have not previously covered this material, use this chapter to learn the necessary information before you continue. Objectives After completing this chapter, you will be able to: Calculate voltage, current, gain, and attenuation in decibels and apply these formulas in applications involving cascaded circuits. Explain the relationship between Q, resonant frequency, and bandwidth. Describe the basic configuration of the diferent types of filters that are used in communication networks and compare and contrast active filters with passive filters. Explain how using switched capacitor filters enhances selectivity. Explain the benefits and operation of crystal, ceramic, and SAW filters. Calculate bandwidth by using Fourier analysis. 30 2-1 Gain, Attenuation, and Decibels Most electronic circuits in communication are used to process signals, i.e., to manipulate signals to produce a desired result. All signal processing circuits involve either gain or attenuation. Gain Gain means ampliication. If a signal is applied to a circuit such as the ampliier shown Gain in Fig. 2-1 and the output of the circuit has a greater amplitude than the input signal, the circuit has gain. Gain is simply the ratio of the output to the input. For input (Vin ) and output (Vout ) voltages, voltage gain AV is expressed as follows: Figure 2-1 An amplifier has output Vout gain. AV 5 5 input Vin Amplifier The number obtained by dividing the output by the input shows how much larger Vin Vout the output is than the input. For example, if the input is 150 µV and the output is Input signal Output signal 75 mV, the gain is AV 5 (75 3 1023 )y(150 3 1026 ) 5 500. Vout A ⫽ gain ⫽ The formula can be rearranged to obtain the input or the output, given the other two Vin variables: Vout 5 Vin 3 AV and Vin 5 Vout /AV. If the output is 0.6 V and the gain is 240, the input is Vin 5 0.6/240 5 2.5 3 10 23 5 2.5 mV. Example 2-1 What is the voltage gain of an ampliier that produces an output of 750 mV for a 30-µV input? Vout 750 3 1023 AV 5 5 5 25,000 Vin 30 3 1026 Since most ampliiers are also power ampliiers, the same procedure can be used to calculate power gain AP: Pout AP 5 Pin where Pin is the power input and Pout is the power output. Example 2-2 The power output of an ampliier is 6 watts (W). The power gain is 80. What is the input power? Pout Pout AP 5 therefore Pin 5 Pin AP 6 Pin 5 5 0.075 W 5 75 mW 80 Electronic Fundamentals for Communications 31  Figure 2-2 Total gain of cascaded circuits is the product of individual stage gains. Vin ⫽ 1 mV 5 mV 15 mV Vout ⫽ 60 mV A1 ⫽ 5 A2 ⫽ 3 A3 ⫽ 4 AT ⫽ A1 ⫻ A2 ⫻ A3 ⫽ 5 ⫻ 3 ⫻ 4 ⫽ 60 When two or more stages of ampliication or other forms of signal processing are cascaded, the overall gain of the combination is the product of the individual circuit gains. Fig. 2-2 shows three ampliiers connected one after the other so that the output of one is the input to the next. The voltage gains of the individual circuits are marked. To ind the total gain of this circuit, simply multiply the individual circuit gains: AT 5 A1 3 A2 3 A3 5 5 3 3 3 4 5 60. If an input signal of 1 mV is applied to the irst ampliier, the output of the third ampliier will be 60 mV. The outputs of the individual ampliiers depend upon their individual gains. The output voltage from each ampliier is shown in Fig. 2-2. Example 2-3 Three cascaded ampliiers have power gains of 5, 2, and 17. The input power is 40 mW. What is the output power? AP 5 A1 3 A2 3 A3 5 5 3 2 3 17 5 170 Pout AP 5 therefore Pout 5 APPin Pin Pout 5 170(40 3 1023 ) 5 6.8 W Example 2-4 A two-stage ampliier has an input power of 25 µW and an output power of 1.5 mW. One stage has a gain of 3. What is the gain of the second stage? Pout 1.5 3 1023 AP 5 5 5 60 Pin 25 3 1026 AP 5 A1 3 A2 If A1 5 3, then 60 5 3 3 A2 and A2 5 60y3 5 20. 32 Chapter 2 Figure 2-3 A voltage divider introduces attenuation. Vin R1 ⫽ 200 ⍀ R2 Vout ⫽ Vin R1 ⫹ R2 R2 ⫽ 100 ⍀ R2 100 A⫽ ⫽ ⫽ 0.3333 R1 ⫹ R2 300 Attenuation Attenuation refers to a loss introduced by a circuit or component. Many electronic circuits, Attenuation sometimes called stages, reduce the amplitude of a signal rather than increase it. If the output signal is lower in amplitude than the input, the circuit has loss, or attenuation. Like gain, attenuation is simply the ratio of the output to the input. The letter A is used to represent attenuation as well as gain: output Vout Attenuation A 5 5 input Vin Circuits that introduce attenuation have a gain that is less than 1. In other words, the output is some fraction of the input. An example of a simple circuit with attenuation is a voltage divider such as that shown in Fig. 2-3. The output voltage is the input voltage multiplied by a ratio based on the resistor values. With the resistor values shown, the gain or attenuation factor of the circuit is A 5 R2/(R1 1 R2 ) 5 100/(200 1 100) 5 100/300 5 0.3333. If a signal of 10 V is applied to the attenuator, the output is Vout 5 Vin A 5 10(0.3333) 5 3.333 V. When several circuits with attenuation are cascaded, the total attenuation is, again, the product of the individual attenuations. The circuit in Fig. 2-4 is an example. The attenuation factors for each circuit are shown. The overall attenuation is AT 5 A1 3 A2 3 A3 With the values shown in Fig. 2-4, the overall attenuation is AT 5 0.2 3 0.9 3 0.06 5 0.0108 Given an input of 3 V, the output voltage is Vout 5 ATVin 5 0.0108(3) 5 0.0324 5 32.4 mV Figure 2-4 Total attenuation is the product of individual attenuations of each cascaded circuit. Vin ⫽ 3 V Vout Loss Loss Loss stage circuit component A1 ⫽ 0.2 A2 ⫽ 0.9 A3 ⫽ 0.06 AT ⫽ A1 ⫻ A2 ⫻ A3 ⫽ 0.2 ⫻ 0.9 ⫻ 0.06 ⫽ 0.0108 Vout ⫽ ATVin ⫽ 0.0324 ⫽ 32.4 mV Electronic Fundamentals for Communications 33  Figure 2-5 Gain exactly ofsets the attenuation. Vin 750 ⍀ Vout ⫽ Vin A2 ⫽ 4 250 ⍀ 250 A1 ⫽ AT ⫽ A1A2 ⫽ 0.25(4) ⫽ 1 750 ⫹ 250 250 A1 ⫽ ⫽ 0.25 1000 It is common in communication systems and equipment to cascade circuits and components that have gain and attenuation. For example, loss introduced by a circuit can be compensated for by adding a stage of ampliication that offsets it. An example of this is shown in Fig. 2-5. Here the voltage divider introduces a 4-to-1 voltage loss, or an attenuation of 0.25. To offset this, it is followed with an ampliier whose gain is 4. The overall gain or attenuation of the circuit is simply the product of the attenuation and gain factors. In this case, the overall gain is AT 5 A1A2 5 0.25(4) 5 1. Another example is shown in Fig. 2-6, which shows two attenuation circuits and two ampliier circuits. The individual gain and attenuation factors are given. The overall circuit gain is AT 5 A1 A2 A3 A4 5 (0.1)(10)(0.3)(15) 5 4.5. For an input voltage of 1.5 V, the output voltage at each circuit is shown in Fig. 2-6. In this example, the overall circuit has a net gain. But in some instances, the overall circuit or system may have a net loss. In any case, the overall gain or loss is obtained by multiplying the individual gain and attenuation factors. Example 2-5 A voltage divider such as that shown in Fig. 2-5 has values of R1 5 10 kV and R2 5 470 V. a. What is the attenuation? R2 470 A1 5 5 A1 5 0.045 R1 1 R2 10,470 b. What amplifier gain would you need to offset the loss for an overall gain of 1? AT 5 A1A2 where A1 is the attenuation and A2 is the amplifier gain. 1 1 5 0.045A2 A2 5 5 22.3 0.045 Note: To ind the gain that will offset the loss for unity gain, just take the reciprocal of attenuation: A2 5 1yA1. 34 Chapter 2 Figure 2-6 The total gain is the product of the individual stage gains and attenuations. Vin ⫽ 1.5 V Loss 0.15 V 1.5 V Loss 0.45 V Vout ⫽ 6.75 V stage stage A1 ⫽ 0.1 A2 ⫽ 10 A3 ⫽ 0.3 A4 ⫽ 15 AT ⫽ A1A2A3A4 ⫽ (0.1)(10)(0.3)(15) ⫽ 4.5 Example 2-6 An ampliier has a gain of 45,000, which is too much for the application. With an input voltage of 20 µV, what attenuation factor is needed to keep the output voltage from exceeding 100 mV? Let A1 5 ampliier gain 5 45,000; A2 5 attenuation factor; AT 5 total gain. Vout 100 3 1023 AT 5 5 5 5000 Vin 20 3 1026 AT 5000 AT 5 A1A2 therefore A2 5 5 5 0.1111 A1 45,000 Decibels The gain or loss of a circuit is usually expressed in decibels (dB), a unit of measurement Decibel (dB) that was originally created as a way of expressing the hearing response of the human ear to various sound levels. A decibel is one-tenth of a bel. When gain and attenuation are both converted to decibels, the overall gain or atten- uation of an electronic circuit can be computed by simply adding the individual gains or attenuations, expressed in decibels. It is common for electronic circuits and systems to have extremely high gains or attenuations, often in excess of 1 million. Converting these factors to decibels and using logarithms result in smaller gain and attenuation igures, which are easier to use. Decibel Calculations. The formulas for computing the decibel gain or loss of a circuit are Vout dB 5 20 log (1) Vin Iout dB 5 20 log (2) Iin Pout dB 5 10 log (3) Pin Formula (1) is used for expressing the voltage gain or attenuation of a circuit; for- mula (2), for current gain or attenuation. The ratio of the output voltage or current to the input voltage or current is determined as usual. The base-10 or common log of the input/output ratio is then obtained and multiplied by 20. The resulting number is the gain or attenuation in decibels. Electronic Fundamentals for Communications 35  Formula (3) is used to compute power gain or attenuation. The ratio of the power output to the power input is computed, and then its logarithm is multiplied by 10. Example 2-7 a. An amplifier has an input of 3 mV and an output of 5 V. What is the gain in decibels? 5 dB 5 20 log 5 20 log 1666.67 5 20(3.22) 5 64.4 0.003 b. A filter has a power input of 50 mW and an output of 2 mW. What is the gain or attenuation? 2 dB 5 10 log 5 10 log 0.04 5 10(21.398) 5 213.98 50 Note that when the circuit has gain, the decibel igure is positive. If the gain is less than 1, which means that there is an attenuation, the decibel igure is negative. Now, to calculate the overall gain or attenuation of a circuit or system, you simply add the decibel gain and attenuation factors of each circuit. An example is shown in Fig. 2-7, where there are two gain stages and an attenuation block. The overall gain of this circuit is AT 5 A1 1 A2 1 A3 5 15 2 20 1 35 5 30 dB Decibels are widely used in the expression of gain and attenuation in communication circuits. The table on the next page shows some common gain and attenuation factors and their corresponding decibel igures. Ratios less than 1 give negative decibel values, indicating attenuation. Note that a 2:1 ratio represents a 3-dB power gain or a 6-dB voltage gain. Antilogs. To calculate the input or output voltage or power, given the decibel gain Antilog or attenuation and the output or input, the antilog is used. The antilog is the number obtained when the base is raised to the logarithm, which is the exponent: Pout dB Pout dB 5 10 log and 5 log Pin 10 Pin and Pout dB dB 5 antilog 5 log21 Pin 10 10 The antilog is simply the base 10 raised to the dB/10 power. Figure 2-7 Total gain or attenuation is the algebraic sum of the individual stage gains in decibels. A2 ⫽ ⫺20 dB A1 ⫽ 15 dB A3 ⫽ 35 dB Loss stage AT ⫽ A1 ⫹ A2 ⫹ A3 AT ⫽ 15 ⫺ 20 ⫹ 35 ⫽ 30 dB 36 Chapter 2  d B G A I N O R AT T E N U AT I O N Ratio (Power or Voltage) Power Voltage 0.000001 260 2120 0.00001 250 2100 0.0001 240 280 0.001 230 260 0.01 220 240 0.1 210 220 0.5 23 26 1 0 0 2 3 6 10 10 20 100 20 40 1000 30 60 10,000 40 80 100,000 50 100 Remember that the logarithm y of a number N is the power to which the base 10 must be raised to get the number. N 5 10 y and y 5 log N Since Pout dB 5 10 log Pin dB Pout 5 log 10 Pin Therefore Pout dB 5 10dB/10 5 log21 Pin 10 The antilog is readily calculated on a scientiic calculator. To ind the antilog for a common or base-10 logarithm, you normally press the Inv or 2nd function key on the calculator and then the log key. Sometimes the log key is marked with 10 x, which is the antilog. The antilog with base e is found in a similar way, by using the Inv or 2nd function on the In key. It is sometimes marked e x, which is the same as the antilog. Electronic Fundamentals for Communications 37  Example 2-8 A power ampliier with a 40-dB gain has an output power of 100 W. What is the input power? Pout dB 5 10 log antilog 5 log21 Pin dB Pout 5 log 10 Pin 40 Pout 5 log 10 Pin Pout 4 5 log Pin antilog 4 5 antilog alog b Pout Pin Pout log21 4 5 Pin Pout 5 104 5 10,000 Pin Pout 100 Pin 5 5 5 0.01 W 5 10 mW 10,000 10,000 Example 2-9 An ampliier has a gain of 60 dB. If the input voltage is 50 µV, what is the output voltage? Since Vout dB 5 20 log Vin dB Vout 5 log 20 Vin Therefore Vout dB 5 log21 510dB/20 Vin 20 Vout 5 1060/20 5 103 Vin Vout 5 103 5 1000 Vin Vout 5 1000Vin 5 1000 (50 3 1026 ) 5 0.05 V 5 50 mV 38 Chapter 2 dBm. When the gain or attenuation of a circuit is expressed in decibels, implicit is a comparison between two values, the output and the input. When the ratio is computed, the units of voltage or power are canceled, making the ratio a dimensionless, or relative, figure. When you see a decibel value, you really do not know the actual voltage or power values. In some cases, this is not a problem; in others, it is useful or necessary to know the actual values involved. When an absolute value is needed, you can use Reference value a reference value to compare any other value. dBm An often used reference level in communication is 1 mW. When a decibel value is computed by comparing a power value to 1 mW, the result is a value called the dBm. It is computed with the standard power decibel formula with 1 mW as the denominator of the ratio: GOOD TO KNOW Pout (W ) dBm 5 10 log From the standpoint of sound 0.001(W ) measurement, 0 dB is the least Here Pout is the output power, or some power value you want to compare to 1 mW, and perceptible sound (hearing 0.001 is 1 mW expressed in watts. threshold), and 120 dB equals the The output of a 1-W ampliier expressed in dBm is, e.g., pain threshold of sound. This list 1 shows intensity levels for com- dBm 5 10 log 5 10 log 1000 5 10(3) 5 30 dBm 0.001 mon sounds. (Tippens, Physics, Sometimes the output of a circuit or device is given in dBm. For example, if a micro- 6th ed., Glencoe/McGraw-Hill, phone has an output of 250 dBm, the actual output power can be computed as follows: 2001, p. 497) Pout Intensity 250 dBm 5 10 log 0.001 Sound level, dB 250 dBm Pout Hearing threshold 0 5 log 10 0.001 Rustling leaves 10 Therefore Whisper 20 Quiet radio 40 Pout 5 10250 dBm/10 5 1025 5 0.00001 Normal conversation 65 0.001 Busy street corner 80 Pout 5 0.001 3 0.00001 5 1023 3 1025 5 1028 W 5 10 3 1029 5 10 nW Subway car 100 Pain threshold 120 Jet engine 140–160 Example 2-10 A power ampliier has an input of 90 mV across 10 kV. The output is 7.8 V across an 8-V speaker. What is the power gain, in decibels? You must compute the input and output power levels irst. V2 P5 R (90 3 1023 ) 2 Pin 5 5 8.1 3 1027 W 104 (7.8) 2 Pout 5 5 7.605 W 8 Pout 7.605 AP 5 5 5 9.39 3 106 Pin 8.1 3 1027 AP (dB) 5 10 log AP 5 10 log 9.39 3 106 5 69.7 dB Electronic Fundamentals for Communications 39  dBc. This is a decibel gain attenuation figure where the reference is the carrier. The carrier is the base communication signal, a sine wave that is modulated. Often the ampli- tude’s sidebands, spurious or interfering signals, are referenced to the carrier. For exam- ple, if the spurious signal is 1 mW compared to the 10-W carrier, the dBc is Psignal dBc 5 10 log Pcarrier 0.001 dBc 5 10 log 5 10(24) 5 240 10 Example 2-11 An ampliier has a power gain of 28 dB. The input power is 36 mW. What is the output power? Pout 5 10dBy10 5 102.8 5 630.96 Pin Pout 5 630.96 Pin 5 630.96(36 3 1023 ) 5 22.71 W Example 2-12 A circuit consists of two ampliiers with gains of 6.8 and 14.3 dB and two ilters with attenuations of 216.4 and 22.9 dB. If the output voltage is 800 mV, what is the input voltage? AT 5 A1 1 A2 1 A3 1 A4 5 6.8 1 14.3 2 16.4 2 2.9 5 1.8 dB Vout AT 5 5 10dBy20 5 101.8y20 5 100.09 Vin Vout 5 100.09 5 1.23 Vin Vout 800 Vin 5 5 5 650.4 mV 1.23 1.23 Example 2-13 Express Pout 5 12.3 dBm in watts. Pout 5 10dBmy10 5 1012.3y10 5 101.23 5 17 0.001 Pout 5 0.001 3 17 5 17 mW 40 Chapter 2 2-2 Tuned Circuits Virtually all communication equipment contains tuned circuits, circuits made up of Tuned circuit inductors and capacitors that resonate at speciic frequencies. In this section, you will review how to calculate the reactance, resonant frequency, impedance, Q , and bandwidth of series and parallel resonance circuits. Reactive Components All tuned circuits and many ilters are made up of inductive and capacitive elements, including discrete components such as coils and capacitors and the stray and distributed inductance and capacitance that appear in all elec- tronic circuits. Both coils and capacitors offer an opposition to alternating- current low known as reactance, which is expressed in ohms (abbreviated Ω). Like resistance, reactance is an opposition that directly affects the amount of current in a circuit. In addition, reactive effects produce a phase shift between the currents and voltages in a circuit. Capacitance causes the current to lead the applied voltage, whereas inductance causes the current to lag the applied voltage. Coils and capacitors used together form tuned, or resonant, circuits. Capacitors. A capacitor used in an ac circuit continually charges and dis- Chip capacitors. charges. A capacitor tends to oppose voltage changes across it. This translates to an opposition to alternating current known as capacitive reactance XC. The reactance of a capacitor is inversely proportional to the value of capacitance C and operating frequency f. It is given by the familiar expression 1 XC 5 2πfC GOOD TO KNOW The reactance of a 100-pF capacitor at 2 MHz is Stray and distributed capaci- 1 tances and inductances can XC 5 6 5 796.2 V 6.28(2 3 10 ) (100 3 10212 ) greatly alter the operation and The formula can also be used to calculate either frequency or capacitance depending performance of a circuit. on the application. These formulas are 1 1 f5 and C5 2πXCC 2πf XC The wire leads of a capacitor have resistance and inductance, and the dielectric has Reactance leakage that appears as a resistance value in parallel with the capacitor. These character- Capacitor istics, which are illustrated in Fig. 2-8, are sometimes referred to as residuals or parasit- ics. The series resistance and inductance are very small, and the leakage resistance is Capacitive reactance Residual Figure 2-8 What a capacitor looks like at high frequencies. Lead resistance Lead inductance R L C L R Rleakage Electronic Fundamentals for Communications 41  very high, so these factors can be ignored at low frequencies. At radio frequencies, however, these residuals become noticeable, and the capacitor functions as a complex RLC circuit. Most of these effects can be greatly minimized by keeping the capacitor leads very short. This problem is mostly eliminated by using the newer chip capacitors, which have no leads as such. Capacitance is generally added to a circuit by a capacitor of a speciic value, but capacitance can occur between any two conductors separated by an insulator. For exam- ple, there is capacitance between the parallel wires in a cable, between a wire and a metal chassis, and between parallel adjacent copper patterns on a printed-circuit board. Stray (or distributed) capacitance These are known as stray, or distributed, capacitances. Stray capacitances are typically small, but they cannot be ignored, especially at the high frequencies used in communica- tion. Stray and distributed capacitances can signiicantly affect the performance of a circuit. Inductor (coil or choke) Inductors. An inductor, also called a coil or choke, is simply a winding of multiple turns of wire. When current is passed through a coil, a magnetic ield is produced around the coil. If the applied voltage and current are varying, the magnetic ield alternately expands and collapses. This causes a voltage to be self-induced into the coil winding, which has the effect of opposing current changes in the coil. This effect is known as Inductance inductance. The basic unit of inductance is the henry (H). Inductance is directly affected by the physical characteristics of the coil, including the number of turns of wire in the induc- tor, the spacing of the turns, the length of the coil, the diameter of the coil, and the type of magnetic core material. Practical inductance values are in the millihenry (mH 5 1023 H), microhenry (µH 5 10 26 H), and nanohenry (nH 5 10 29 H) regions. Fig. 2-9 shows several different types of inductor coils. Fig. 2-9(a) is an inductor made of a heavy, self-supporting wire coil. In Fig. 2-9(b) the inductor is formed of a copper spiral that is etched right onto the board itself. In Fig. 2-9(c) the coil is wound on an insulating form containing a powdered iron or ferrite core in the center, to increase its inductance. Fig. 2-9(d) shows another common type of inductor, one using turns of wire on a toroidal or doughnut-shaped form. Fig. 2-9(e) shows an inductor made by placing a small ferrite bead over a wire; the bead effectively increases the wire’s small inductance. Fig. 2-9( f ) shows a chip inductor. It is typically no more than 1⁄8 to 1⁄4 in long. A coil is contained within the body, and the unit is soldered to the circuit board with the end connections. These devices look exactly like chip resistors and capacitors. In a dc circuit, an inductor will have little or no effect. Only the ohmic resistance of the wire affects current low. However, when the current changes, such as during the time the power is turned off or on, the coil will oppose these changes in current. When an inductor is used in an ac circuit, this opposition becomes continuous and Inductive reactance constant and is known as inductive reactance. Inductive reactance XL is expressed in ohms and is calculated by using the expression XL 5 2πfL For example, the inductive reactance of a 40-µH coil at 18 MHz is XL 5 6.28(18 3 106 )(40 3 1026 ) 5 4522 V In addition to the resistance of the wire in an inductor, there is stray capacitance between the turns of the coil. See Fig. 2-10(a). The overall effect is as if a small capac- itor were connected in parallel with the coil, as shown in Fig. 2-10(b). This is the equivalent circuit of an inductor at high frequencies. At low frequencies, capacitance may be 42 Chapter 2 Figure 2-9 Types of inductors. (a) Heavy self-supporting wire coil. (b) Inductor made as copper pattern. (c) Insulating form. (d) Toroidal inductor. (e) Ferrite bead inductor. (f ) Chip inductor. Printed circuit (PC) PC board board (a) (b) Powdered Toroidal iron or ferrite core core Insulating Core moves in or form out to vary inductance Turns (c) Toroidal of wire inductor (d ) Ferrite Solder Component bead connection lead or wire Body (e) (f ) ignored, but at radio frequencies, it is suficiently large to affect circuit operation. The coil then functions not as a pure inductor, but as a complex RLC circuit with a self- resonating frequency. Any wire or conductor exhibits a characteristic inductance. The longer the wire, the greater the inductance. Although the inductance of a straight wire is only a fraction of Figure 2-10 Equivalent circuit of an inductor at high frequencies. (a) Stray capacitance between turns. (b) Equivalent circuit of an inductor at high frequencies. Winding (coil) resistance R L ⫽ inductor C ⫽ stray capacitance Stray capacitance between turns (a) (b) Electronic Fundamentals for Communications 43  a microhenry, at very high frequencies the reactance can be signiicant. For this reason, it is important to keep all lead lengths short in interconnecting components in RF circuits. This is especially true of capacitor and transistor leads, since stray or distributed induc- tance can signiicantly affect the performance and characteristics of a circuit. Quality factor Q Another important characteristic of an inductor is its quality factor Q, the ratio of inductive power to resistive power: I2XL XL Q5 2 5 IR R This is the ratio of the power returned to the circuit to the power actually dissipated by the coil resistance. For example, the Q of a 3-µH inductor with a total resistance of 45V at 90 MHz is calculated as follows: 2πfL 6.28(90 3 106 )(3 3 1026 ) 1695.6 Q5 5 5 5 37.68 R 45 45 Resistor Resistors. At low frequencies, a standard low-wattage color-coded resistor offers nearly pure resistance, but at high frequencies its leads have considerable inductance, and stray capacitance between the leads causes the resistor to act as a complex RLC circuit, as shown in Fig. 2-11. To minimize the inductive and capacitive effects, the leads are kept very short in radio applications. The tiny resistor chips used in surface-mount construction of the electronic circuits preferred for radio equipment have practically no leads except for the metallic end pieces soldered to the printed-circuit board. They have virtually no lead inductance and little stray capacitance. Many resistors are made from a carbon-composition material in powdered form sealed inside a tiny housing to which leads are attached. The type and amount of carbon material determine the value of these resistors. They contribute noise to the circuit in which they are used. The noise is caused by thermal effects and the granular nature of the resistance material. The noise contributed by such resistors in an ampliier used to amplify very low level radio signals may be so high as to obliterate the desired signal. To overcome this problem, ilm resistors were developed. They are made by depos- iting a carbon or metal ilm in spiral form on a ceramic form. The size of the spiral and the kind of metal ilm determine the resistance value. Carbon ilm resistors are quieter than carbon-composition resistors, and metal ilm resistors are quieter than carbon ilm resistors. Metal ilm resistors should be used in ampliier circuits that must deal with very low level RF signals. Most surface-mount resistors are of the metallic ilm type. Skin Efect. The resistance of any wire conductor, whether it is a resistor or capacitor lead or the wire in an inductor, is primarily determined by the ohmic resistance of the Skin effect wire itself. However, other factors inluence it. The most signiicant one is skin effect, the tendency of electrons lowing in a conductor to low near and on the outer surface Figure 2-11 Equivalent circuit of a resistor at high (radio) frequencies. Resistor Lead inductance Stray capacitance 44 Chapter 2 Figure 2-12 Skin efect increases wire and inductor resistance at high frequencies. Wire Electrons flow on or near surface No current flow in center of wire of the conductor frequencies in the VHF, UHF, and microwave regions (Fig. 2-12). This has the effect of greatly decreasing the total cross-sectional area of the conductor, thus increasing its resistance and signiicantly affecting the performance of the circuit in which the conductor is used. For example, skin effect lowers the Q of an inductor at the higher frequencies, causing unexpected and undesirable effects. Thus many high-frequency coils, particularly those in high-powered transmitters, are made with cop- per tubing. Since current does not low in the center of the conductor, but only on the surface, tubing provides the most eficient conductor. Very thin conductors, such as a copper pattern on a printed-circuit board, are also used. Often these conductors are sil- ver- or gold-plated to further reduce their resistance. Tuned Circuits and Resonance A tuned circuit is made up of inductance and capacitance and resonates at a speciic frequency, the resonant frequency. In general, the terms tuned circuit and resonant circuit Tuned (resonant) circuit are used interchangeably. Because tuned circuits are frequency-selective, they respond best at their resonant frequency and at a narrow range of frequencies around the resonant frequency. Series Resonant Circuits. A series resonant circuit is made up of inductance, Series resonant circuit capacitance, and resistance, as shown in Fig. 2-13. Such circuits are often referred to as LCR circuits or RLC circuits. The inductive and capacitive reactances depend upon the LCR circuit frequency of the applied voltage. Resonance occurs when the inductive and capacitive RLC circuit reactances are equal. A plot of reactance versus frequency is shown in Fig. 2-14, where fr is the resonant frequency. Figure 2-13 Series RLC circuit. Figure 2-14 Variation of reactance with frequency. VL VC XC XL Reactance XL XC XL ⫽ XC Vs R VR fr Frequency Electronic Fundamentals for Communications 45  The total impedance of the circuit is given by the expression Z 5 2R2 1 (XL 2 XC ) 2 When XL equals XC, they cancel each other, leaving only the resistance of the circuit to oppose the current. At resonance, the total circuit impedance is simply the value of all series resistances in the circuit. This includes the resistance of the coil and the resistance of the component leads, as well as any physical resistor in the circuit. The resonant frequency can be expressed in terms of inductance and capacitance. A formula for resonant frequency can be easily derived. First, express XL and XC as an equivalence: XL 5 XC. Since 1 XL 5 2πfr L and XC 5 2πfr C we have 1 2πfr L 5 2πfr C Solving for f r gives 2π 1LC 1 fr 5 In this formula, the frequency is in hertz, the inductance is in henrys, and the capacitance is in farads. Example 2-14 What is the resonant frequency of a 2.7-pF capacitor and a 33-nH inductor? 2π 1LC 1 1 fr 5 5 6.28233 3 10 3 2.7 3 10212 29 5 5.33 3 108 Hz or 533 MHz It is often necessary to calculate capacitance or inductance, given one of those val- ues and the resonant frequency. The basic resonant frequency formula can be rearranged to solve for either inductance and capacitance as follows: 1 1 L5 2 2 and C5 2 2 4π f C 4π f L For example, the capacitance that will resonate at a frequency of 18 MHz with a 12-µH inductor is determined as follows: 1 1 C5 5 4π 2fr2L 39.478(18 3 106 ) 2 (12 3 1026 ) 1 5 5 6.5 3 10212 F or 6.5 pF 39.478(3.24 3 1014 )(12 3 1026 ) 46 Chapter 2 Example 2-15 What value of inductance will resonate with a 12-pF capacitor at 49 MHz? 1 1 L5 2 2 5 4π fr C 39.478(49 3 106 ) 2 (12 3 10212 ) 5 8.79 3 1027 H or 879 nH As indicated earlier, the basic deinition of resonance in a series tuned circuit is the point at which XL equals XC. With this condition, only the resistance of the circuit impedes the current. The total circuit impedance at resonance is Z 5 R. For this reason, resonance in a series tuned circuit can also be deined as the point at which the circuit impedance is lowest and the circuit current is highest. Since the circuit is resistive at resonance, the current is in phase with the applied voltage. Above the resonant fre- quency, the inductive reactance is higher than the capacitive reactance, and the induc- tor voltage drop is greater than the capacitor voltage drop. Therefore, the circuit is inductive, and the current will lag the applied voltage. Below resonance, the capacitive reactance is higher than the inductive reactance; the net reactance is capacitive, thereby producing a leading current in the circuit. The capacitor voltage drop is higher than the inductor voltage drop. The response of a series resonant circuit is illustrated in Fig. 2-15, which is a plot of the frequency and phase shift of the current in the circuit with respect to frequency. At very low frequencies, the capacitive reactance is much greater than the induc- tive reactance; therefore, the current in the circuit is very low because of the high impedance. In addition, because the circuit is predominantly capacitive, the current leads the voltage by nearly 90°. As the frequency increases, XC goes down and XL goes up. The amount of leading phase shift decreases. As the values of reactances approach one another, the current begins to rise. When XL equals XC, their effects cancel and the impedance in the circuit is just that of the resistance. This produces a current peak, where the current is in phase with the voltage (0°). As the frequency Figure 2-15 Frequency and phase response curves of a series resonant circuit. ⫹90⬚ (lead) ␪ Phase shift angle (␪) 0⬚ Circuit current I I ⫺90⬚ (lag) Below resonance (capacitive) fr Above resonance (inductive) Electronic Fundamentals for Communications 47  Figure 2-16 Bandwidth of a series resonant circuit. Ipeak ⫽ 2 mA 0.707Ipeak ⫽ 1.414 mA f1 fr f2 BW = f2 ⫺ f1 continues to rise, XL becomes greater than XC. The impedance of the circuit increases and the current decreases. With the circuit predominantly inductive, the current lags the applied voltage. If the output voltage were being taken from across the resistor in Fig. 2-13, the response curve and phase angle of the voltage would correspond to those in Fig. 2-15. As Fig. 2-15 shows, the current is highest in a region centered on the resonant frequency. The narrow frequency range over which the current is highest is Bandwidth called the bandwidth. This area is illustrated in Fig. 2-16. The upper and lower boundaries of the bandwidth are deined by two cutoff frequen- cies designated f1 and f2. These cutoff frequencies occur where the current amplitude is 70.7 percent of the peak current. In the igure, the peak circuit current is 2 mA, and the current at both the lower ( f1 ) and upper ( f2 ) cutoff frequency is 0.707 of 2 mA, or 1.414 mA. Half-power points Current levels at which the response is down 70.7 percent are called the half-power points because the power at the cutoff frequencies is one-half the power peak of the curve. P 5 I 2R 5 (0.707 Ipeak ) 2R 5 0.5 Ipeak2R The bandwidth BW of the tuned circuit is deined as the difference between the upper and lower cutoff frequencies: BW 5 f2 2 f1 For example, assuming a resonant frequency of 75 kHz and upper and lower cutoff frequencies of 76.5 and 73.5 kHz, respectively, the bandwidth is BW 5 76.5 2 73.5 5 3 kHz. The bandwidth of a resonant circuit is determined by the Q of the circuit. Recall that the Q of an inductor is the ratio of the inductive reactance to the circuit resistance. This holds true for a series resonant circuit, where Q is the ratio of the inductive reactance to the total circuit resistance, which includes the resistance of the inductor plus any additional series resistance: XL Q5 RT Recall that bandwidth is then computed as fr BW 5 Q If the Q of a circuit resonant at 18 MHz is 50, then the bandwidth is BW 5 18/50 5 0.36 MHz 5 360 kHz. 48 Chapter 2 Example 2-16 What is the bandwidth of a resonant circuit with a frequency of 28 MHz and a Q of 70? fr 28 3 106 BW 5 5 5 400,000 Hz 5 400 kHz Q 70 The formula can be rearranged to compute Q, given the frequency and the bandwidth: fr Q5 BW Thus the Q of the circuit whose bandwidth was computed previously is Q 5 75 kHz /3kHz 5 25. Since the bandwidth is approximately centered on the resonant frequency, f1 is the same distance from fr as f2 is from fr. This fact allows you to calculate the resonant frequency by knowing only the cutoff frequencies: fr 5 2f1 3 f2 For example, if f1 5 175 kHz and f2 5 178 kHz, the resonant frequency is fr 5 2175 3 103 3 178 3 103 5 176.5 kHz For a linear frequency scale, you can calculate the center or resonant frequency by using an average of the cutoff frequencies. f1 1 f2 fr 5 2 If the circuit Q is very high (.100), then the response curve is approximately sym- metric around the resonant frequency. The cutoff frequencies will then be roughly equi- distant from the resonant frequency by the amount of BW/2. Thus the cutoff frequencies can be calculated if the bandwidth and the resonant frequency are known: BW BW f1 5 fr 2 and f2 5 fr 1 2 2 For instance, if the resonant frequency is 49 MHz (49,000 kHz) and the bandwidth is 10 kHz, then the cutoff frequencies will be 10k f1 5 49,000 kHz 2 5 49,000 kHz 2 5 kHz 5 48,995 kHz 2 f2 5 49,000 kHz 1 5 kHz 5 49,005 kHz Keep in mind that although this procedure is an approximation, it is useful in many applications. The bandwidth of a resonant circuit deines its selectivity, i.e., how the circuit Selectivity responds to varying frequencies. If the response is to produce a high current only over a narrow range of frequencies, a narrow bandwidth, the circuit is said to be highly selec- tive. If the current is high over a broader range of frequencies, i.e., the bandwidth is wider, the circuit is less selective. In general, circuits with high selectivity and narrow bandwidths are more desirable. However, the actual selectivity and bandwidth of a circuit must be optimized for each application. Electronic Fundamentals for Communications 49  The relationship between circuit resistance Q and bandwidth is extremely impor- tant. The bandwidth of a circuit is inversely proportional to Q. The higher Q is, the smaller the bandwidth. Low Qs produce wide bandwidths or less selectivity. In turn, Q is a function of the circuit resistance. A low resistance produces a high Q, a nar- row bandwidth, and a highly selective circuit. A high circuit resistance produces a low Q, wide bandwidth, and poor selectivity. In most communication circuits, circuit Qs are at least 10 and typically higher. In most cases, Q is controlled directly by the resistance of the inductor. Fig. 2-17 shows the effect of different values of Q on bandwidth. Example 2-17 The upper and lower cutoff frequencies of a resonant circuit are found to be 8.07 and 7.93 MHz. Calculate (a) the bandwidth, (b) the approximate resonant frequency, and (c) Q. b. fr 5 1f1 f2 5 1(8.07 3 106 ) (7.93 3 106 ) 5 8 MHz a. BW 5 f2 2 f1 5 8.07 MHz 2 7.93 MHz 5 0.14 MHz 5 140 kHz fr 8 3 106 c. Q 5 5 5 57.14 BW 140 3 103 Example 2-18 What are the approximate 3-dB down frequencies of a resonant circuit with a Q of 200 at 16 MHz? fr 16 3 106 BW 5 5 5 80,000 Hz 5 80 kHz Q 200 BW 80,000 f1 5 fr 2 5 16,000,000 2 5 15.96 MHz 2 2 BW 80,000 f2 5 fr 1 5 16,000,000 1 5 16.04 MHz 2 2 Resonance produces an interesting but useful phenomenon in a series RLC circuit. Consider the circuit in Fig. 2-18(a). At resonance, assume XL 5 XC 5 500 V. The total circuit resistance is 10 V. The Q of the circuit is then XL 500 Q5 5 5 50 R 10 If the applied or source voltage Vs is 2 V, the circuit current at resonance will be Vs 2 I5 5 5 0.2 A R 10 50 Chapter 2 Figure 2-17 The efect of Q on bandwidth and selectivity in a resonant circuit. BW 3 BW 2 High Q1, narrow Gain, dB BW 1 bandwidth ⫺3 dB points Medium Q2, medium bandwidth Low Q3, wide bandwidth fr Figure 2-18 Resonant step-up voltage in a series resonant circuit. VL ⫽ 100 V I ⫽ 0.2 A VC ⫽ 100 V XC ⫽ 500 ⍀ XL ⫽ 500 ⍀ VL ⫽ 100 V Equal and 180° Vs⫽ 2 V I out of phase VR ⫽ Vs ⫽ 2 V R ⫽ 10 ⍀ VR ⫽ 2 V VC ⫽ 100 V (a ) (b) When the reactances, the resistances, and the current are known, the voltage drops across each component can be computed: VL 5 IXL 5 0.2(500) 5 100 V VC 5 IXC 5 0.2(500) 5 100 V VR 5 IR 5 0.2(10) 5 2 V As you can see, the voltage drops across the inductor and capacitor are signiicantly higher than the applied voltage. This is known as the resonant step-up voltage. Although Resonant step-up voltage the sum of the voltage drops around the series circuit is still equal to the source voltage, at resonance the voltage across the inductor leads the current by 90° and the voltage across the capacitor lags the current by 90° [see Fig. 2-18(b)]. Therefore, the inductive and reac- tive voltages are equal but 180° out of phase. As a result, when added, they cancel each other, leaving a total reactive voltage of 0. This means that the entire applied voltage appears across the circuit resistance. The resonant step-up voltage across the coil or capacitor can be easily computed by multiplying the input or source voltage by Q: VL 5 VC 5 QVs In the example in Fig. 2-18, VL 5 50(2) 5 100 V. Electronic Fundamentals for Communications 51  This interesting and useful phenomenon means that small applied voltages can essen- tially be stepped up to a higher voltage — a form of simple ampliication without active circuits that is widely applied in communication circuits. Example 2-19 A series resonant circuit has a Q of 150 at 3.5 MHz. The applied voltage is 3 µV. What is the voltage across the capacitor? VC 5 QVs 5 150(3 3 1026 ) 5 450 3 1026 5 450 µV Parallel resonant circuit Parallel Resonant Circuits. A parallel resonant circuit is formed when the inductor and capacitor are connected in parallel with the applied voltage, as shown in Fig. 2-19(a). In general, resonance in a parallel tuned circuit can also be deined as the point at which the inductive and capacitive reactances are equal. The resonant frequency is therefore calculated by the resonant frequency formula given earlier. If we assume lossless compo- nents in the circuit (no resistance), then the current in the inductor equals the current in the capacitor: IL 5 IC Although the currents are equal, they are 180° out of phase, as the phasor diagram in Fig. 2-19(b) shows. The current in the inductor lags the applied voltage by 90°, and the current in the capacitor leads the applied voltage by 90°, for a total of 180°. Now, by applying Kirchhoff’s current law to the circuit, the sum of the individual branch currents equals the total current drawn from the source. With the inductive and capacitive currents equal and out of phase, their sum is 0. Thus, at resonance, a parallel tuned circuit appears to have ininite resistance, draws no current from the source and thus has ininite impedance, and acts as an open circuit. However, there is a high circulating current between the inductor and capacitor. Energy is being stored and transferred between the inductor and capacitor. Because such a circuit acts as a kind of storage vessel for electric energy, it is Tank circuit often referred to as a tank circuit and the circulating current is referred to as the tank current. Tank current In a practical resonant circuit where the components do have losses (resistance), the circuit still behaves as described above. Typically, we can assume that the capacitor has practically zero losses and the inductor contains a resistance, as illustrated in Fig. 2-20(a). At resonance, where XL 5 XC, the impedance of the inductive branch of the circuit is higher than the impedance of the capacitive branch because of the coil resistance. The capacitive current is slightly higher than the inductive current. Even if the reactances are Figure 2-19 Parallel resonant circuit currents. (a) Parallel resonant circuit. (b) Current relationships in parallel resonant circuit. ⫹90° IT ⫽冪苴苴苴苴苳 (IL) 2 ⫹ (IC ) 2 IC Line current IC IL Vs Vs IL ⫺90° (a) (b) 52 Chapter 2 Figure 2-20 A practical parallel resonant circuit. (a) Practical parallel resonant circuit with coil resistance RW. (b) Phase relationships. IT IC Line current (Vector sum of IC and IL L IT leads Vs because IC ⬎ IL) Vs Vs IC C RW IL IL (lags Vs by less than 90° because of RW) (a) (b) equal, the branch currents will be unequal and therefore there will be some net current low in the supply line. The source current will lead the supply voltage, as shown in Fig. 2-20(b). Nevertheless, the inductive and capacitive currents in most cases will cancel because they are approximately equal and of opposite phase, and consequently the line or source current will be signiicantly lower than the individual branch currents. The result is a very high resistive impedance, approximately equal to Vs Z5 IT The circuit in Fig. 2-20(a) is not easy to analyze. One way to simplify the mathematics involved is to convert the circuit to an equivalent circuit in which the coil resistance is translated to a parallel resistance that gives the same overall results, as shown in Fig. 2-21. The equivalent inductance Leq and resistance Req are calculated with the formulas L(Q2 1 1) Leq 5 and Req 5 RW (Q2 1 1) Q2 and Q is determined by the formula XL Q5 RW where RW is the coil winding resistance. If Q is high, usually more than 10, Leq is approximately equal to the actual induc- tance value L. The total impedance of the circuit at resonance is equal to the equivalent parallel resistance: Z 5 Req Figure 2-21 An equivalent circuit makes parallel resonant circuits easier to analyze. Lossless resonant circuit L R eq C C RW L eq Actual circuit Equivalent circuit R eq = R W (Q 2 + 1) L eq = L (Q 2 + 1) 2 Z= R Q eq Electronic Fundamentals for Communications 53  Example 2-20 What is the impedance of a parallel LC circuit with a resonant frequency of 52 MHz and a Q of 12? L 5 0.15 µH. XL Q5 RW XL 5 2πf L 5 6.28(52 3 106 )(0.15 3 1026 ) 5 49 V XL 49 RW 5 5 5 4.1 V Q 12 Z 5 Req 5 RW (Q2 1 1) 5 4.1(122 1 1) 5 4.1(145) 5 592 V If the Q of the parallel resonant circuit is greater than 10, the following simpliied formula can be used to calculate the resistive impedance at resonance: L Z5 CRW The value of RW is the winding resistance of the coil. Example 2-21 Calculate the impedance of the circuit given in Example 2-20 by using the formula Z 5 L/CR. fr 5 52 MHz RW 5 4.1 V L 5 0.15 µH 1 1 C5 5 4π 2fr2L 39.478(52 3 106 ) 2 (0.15 3 1026 ) 5 6.245 3 10211 L 0.15 3 1026 Z5 5 5 586 V CRW (62.35 3 10212 )(4.1) This is close to the previously computed value of 592 V. The formula Z 5 L/CRW is an approximation. GOOD TO KNOW A frequency and phase response curve of a parallel resonant circuit is shown in Fig. 2-22. Below the resonant frequency, XL is less than XC; thus the inductive current The bandwidth of a circuit is in- is greater than the capacitive current, and the circuit appears inductive. The line current versely proportional to the circuit Q. lags the applied voltage. Above the resonant frequency, XC is less than XL; thus the The higher the Q, the smaller capacitive current is more than the inductive current, and the circuit appears capacitive. the bandwidth. Low Q values Therefore, the line current leads the applied voltage. The phase angle of the impedance will be leading below resonance and lagging above resonance. produce wide bandwidths or At the resonant frequency, the impedance of the circuit peaks. This means that the less selectivity. line current at that time is at its minimum. At resonance, the circuit appears to have a very high resistance, and the small line current is in phase with the applied voltage. 54 Chapter 2 Figure 2-22 Response of a parallel resonant circuit. +90° (leading) in degrees (red curve only) Phase shift of line current Impedance (blue curve only) Z 0° -90° (lagging) Below fr Above resonance resonance (inductive) (capacitive) Note that the Q of a parallel circuit, which was previously expressed as Q 5 XL /RW, can also be computed with the expression RP Q5 XL where RP is the equivalent parallel resistance, Req in parallel with any other parallel resistance, and XL is the inductive reactance of the equivalent inductance Leq. You can set the bandwidth of a parallel tuned circuit by controlling Q. The Q can be determined by connecting an external resistor across the circuit. This has the effect of lowering RP and increasing the bandwidth.d Example 2-22 What value of parallel resistor is needed to set the bandwidth of a parallel tuned circuit to 1 MHz? Assume XL 5 300 V, RW 5 10 V, and fr 5 10 MHz. XL 300 Q5 5 5 30 RW 10 RP 5 RW (Q2 1 1) 5 10(302 1 1) 5 10(901) 5 9010 V (equivalent resistance of the parallel circuit at resonance) fr BW 5 Q fr 10 MHz Q5 5 5 10 (Q needed for 1-MHz bandwidth) BW 1 MHz RPnew 5 QXL 5 10(300) 5 3000 V (this is the total resistance of the circuit RPnew made up of the original RP and an externally connected resistor Rext ) RPRext RPnew 5 RP 1 Rext RPnewRP 9010(3000) Rext 5 5 5 4497.5V RP 2 RPnew 9010 2 3000 Electronic Fundamentals for Communications 55  2-3 Filters Filter A ilter is a frequency-selective circuit. Filters are designed to pass some frequencies and reject others. The series and parallel resonant circuits reviewed in Section 2-2 are exam- ples of ilters. There are numerous ways to implement ilter circuits. Simple ilters created by using Passive filter resistors and capacitors or inductors and capacitors are called passive ilters because they use passive components that do not amplify. In communication work, many ilters are of the passive LC variety, although many other types are used. Some special types of ilters are active ilters that use RC networks with feedback in op amp circuits, switched capacitor ilters, crystal and ceramic ilters, surface acoustic wave (SAW) il- ters, and digital ilters implemented with digital signal processing (DSP) techniques. The ive basic kinds of ilter circuits are as follows: Low-pass ilter. Passes frequencies below a critical frequency called the cutoff frequency and greatly attenuates those above the cutoff frequency. High-pass ilter. Passes frequencies above the cutoff but rejects those below it. Bandpass ilter. Passes frequencies over a narrow range between lower and up- per cutoff frequencies. Band-reject ilter. Rejects or stops frequencies over a narrow range but allows frequencies above and below to pass. All-pass ilter. Passes all frequencies equally well over its design range but has a fixed or predictable phase shift characteristic. RC filter RC Filters A low-pass ilter allows the lower-frequency components of the applied voltage to develop output voltage across the load resistance, whereas the higher-frequency compo- nents are attenuated, or reduced, in the output. A high-pass ilter does the opposite, allowing the higher-frequency components of the applied voltage to develop voltage across the output load resistance. The case of an RC coupling circuit is an example of a high-pass ilter because the ac component of the input voltage is developed across R and the dc voltage is blocked by the series capacitor. Furthermore, with higher frequencies in the ac component, more ac voltage is coupled. Any low-pass or high-pass ilter can be thought of as a frequency-dependent voltage divider because the amount of output voltage is a function of frequency. RC ilters use combinations of resistors and capacitors to achieve the desired response. Most RC ilters are of the low-pass or high-pass type. Some band-reject or notch ilters are also made with RC circuits. Bandpass ilters can be made by combining low-pass and high-pass RC sections, but this is rarely done. Low-pass filter (high cut filter) Low-Pass Filter. A low-pass ilter is a circuit that introduces no attenuation at frequen- cies below the cutoff frequency but completely eliminates all signals with frequencies above the cutoff. Low-pass ilters are sometimes referred to as high cut ilters. The ideal response curve for a low-pass ilter is shown in Fig. 2-23. This response curve cannot be realized in practice. In practical circuits, instead of a sharp transition at the cutoff frequency, there is a more gradual transition between little or no attenuation and maximum attenuation. The simplest form of low-pass ilter is the RC circuit shown in Fig. 2-24(a). The circuit forms a simple voltage divider with one frequency-sensitive component, in this case the capacitor. At very low frequencies, the capacitor has very high reactance com- pared to the resistance and therefore the attenuation is minimum. As the frequency increases, the capacitive reactance decreases. When the reactance becomes smaller than the resistance, the attenuation increases rapidly. The frequency response of the basic 56 Chapter 2 Figure 2-23 Ideal response curve of a low-pass filter. Cutoff frequency Output Signals in the passband pass unattenuated Signals above fco are eliminated Frequency fco Figure 2-24 RC low-pass filter. (a) Circuit. (b) Low-pass filter. Vout(max) (0 dB) 0.707 Vout(max) 6 dB/octave or (⫺3 dB) 20 dB/decade rate Vout 6 dB Vin ⫺9 dB R 20 dB C ⫺23 dB 1 XC ⫽ R fco ⫽ fco 6 kHz Hz 2␲RC 600 1200 (a) (b) circuit is illustrated in Fig. 2-24(b). The cutoff frequency of this ilter is that point where R and XC are equal. The cutoff frequency, also known as the critical frequency, is deter- mined by the expression XC 5 R 1 5R 2πfc 1 fco 5 2πRC For example, if R 5 4.7 kV and C 5 560 pF, the cutoff frequency is 1 fco 5 5 60,469 Hz or 60.5 kHz 2π(4700) (560 3 10212 ) Example 2-23 What is the cutoff frequency of a single-section RC low-pass ilter with R 5 8.2 kV and C 5 0.0033 µF? 1 1 fco 5 5 3 2πRC 2π(8.2 3 10 )(0.0033 3 1026 ) fco 5 5881.56 Hz or 5.88 kHz Electronic Fundamentals for Communications 57 Figure 2-25 Two stages of RC filter improve the response but increase signal loss. (a) Circuit. (b) Response curve. 12 dB/octave or Buffer amplifier to isolate 40 dB/decade roll-off rate Vout(max) RC sections 3 dB R R Vout 12 dB Vin 40 dB C C fco 6 kHz Hz 600 1200 (a) (b) At the cutoff frequency, the output amplitude is 70.7 percent of the input amplitude at lower frequencies. This is the so-called 3-dB down point. In other words, this ilter has a voltage gain of 23 dB at the cutoff frequency. At frequencies above the cutoff frequency, Octave the amplitude decreases at a linear rate of 6 dB per octave or 20 dB per decade. An octave Decade is deined as a doubling or halving of frequency, and a decade represents a one-tenth or times-10 relationship. Assume that a ilter has a cutoff of 600 Hz. If the frequency doubles to 1200 Hz, the attenuation will increase by 6 dB, or from 3 dB at cutoff to 9 dB at 1200 Hz. If the frequency increased by a factor of 10 from 600 Hz to 6 kHz, the attenuation would increase by a factor of 20 dB from 3 dB at cutoff to 23 dB at 6 kHz. If a faster rate of attenuation is required, two RC sections set to the same cutoff frequency can be used. Such a circuit is shown in Fig. 2-25(a). With this circuit, the rate of attenuation is 12 dB per octave or 40 dB per decade. Two identical RC circuits are used, but an isolation or buffer ampliier such as an emitter-follower (gain < 1) is used between them to prevent the second section from loading the irst. Cascading two RC sections without the isolation will give an attenuation rate less than the theoretically ideal 12-dB octave because of the loading effects. If the cutoff frequency of each RC section is the same, the overall cutoff frequency for the complete ilter is somewhat less. This is caused by added attenuation of the second section. With a steeper attenuation curve, the circuit is said to be more selective. The disad- vantage of cascading such sections is that higher attenuation makes the output signal considerably smaller. This signal attenuation in the passband of the ilter is called Insertion loss insertion loss. A low-pass ilter can also be implemented with an inductor and a resistor, as shown in Fig. 2-26. The response curve for this RL ilter is the same as that shown in Fig. 2-24(b). The cutoff frequency is determined by using the formula R fco 5 2πL Figure 2-26 A low-pass filter implemented with an inductor. L XL  R R fco  R 2L 58 Chapter 2 Figure 2-27 Frequency response curve of a high-pass filter. (a) Ideal. (b) Practical. 0 dB 3 dB Output 6 dB/octave or Passband 20 dB/decade 1 fco⫽ 2␲RC fco fco Frequency Frequency