Decibels, Filters, and Bode Plots

Transcription

1 23 db Decibels, Filters, and Bode Plots 23. LOGAITHMS The use o logarithms in industry is so extensive that a clear understanding o their purpose and use is an absolute necessity. At irst exposure, logarithms oten appear vague and mysterious due to the mathematical operations required to ind the logarithm and antilogarithm using the longhand table approach that is typically taught in mathematics courses. However, almost all o today s scientiic calculators have the common and natural log unctions, eliminating the complexity o applying logarithms and allowing us to concentrate on the positive characteristics o the unction. Basic elationships Let us irst examine the relationship between the variables o the logarithmic unction. The mathematical expression N (b) x states that the number N is equal to the base b taken to the power x. A ew examples: () 2 27 (3) (e) 4 where e I the question were to ind the power x to satisy the equation 2 () x the value o x could be determined using logarithms in the ollowing manner: x log revealing that

2 8 DECIBELS, FILTES, AND BODE PLOTS db Note that the logarithm was taken to the base the number to be taken to the power o x. There is no limitation to the numerical value o the base except that tables and calculators are designed to handle either a base o (common logarithm, LOG ) or base e (natural logarithm, IN ). In review, thereore, I N (b) x, then x log b N. (23.) The base to be employed is a unction o the area o application. I a conversion rom one base to the other is required, the ollowing equation can be applied: log e x 2.3 log x (23.2) The content o this chapter is such that we will concentrate solely on the common logarithm. However, a number o the conclusions are also applicable to natural logarithms. Some Areas o Application The ollowing is a short list o the most common applications o the logarithmic unction:. This chapter will demonstrate that the use o logarithms permits plotting the response o a system or a range o values that may otherwise be impossible or unwieldy with a linear scale. 2. Levels o power, voltage, and the like, can be compared without dealing with very large or very small numbers that oten cloud the true impact o the dierence in magnitudes. 3. A number o systems respond to outside stimuli in a nonlinear logarithmic manner. The result is a mathematical model that permits a direct calculation o the response o the system to a particular input signal. 4. The response o a cascaded or compound system can be rapidly determined using logarithms i the gain o each stage is known on a logarithmic basis. This characteristic will be demonstrated in an example to ollow. Graphs Graph paper is available in the semilog and log-log varieties. Semilog paper has only one log scale, with the other a linear scale. Both scales o log-log paper are log scales. A section o semilog paper appears in Fig Note the linear (even-spaced-interval) vertical scaling and the repeating intervals o the log scale at multiples o. The spacing o the log scale is determined by taking the common log (base ) o the number. The scaling starts with, since log. The distance between and 2 is determined by log 2.3, or approximately 3% o the ull distance o a log interval, as shown on the graph. The distance between and 3 is determined by log 3.477, or about 48% o the ull width. For uture reerence, keep in mind that almost 5% o the width o one log interval isrepresented by a 3 rather than by the 5 o a linear scale. In addition, note that the number 5 is about 7% o the ull width, and 8 is about 9%. emembering the percentage o ull width o the

3 db LOGAITHMS Linear scale 3 2 % o ull width log 3 = % log 2 =.3 3% log 4 =.62 6% 7% 95% 78% 9% 85% FIG. 23. Semilog graph paper. Log scale lines 2, 3, 5, and 8 will be particularly useul when the various lines o a log plot are let unnumbered. Since log log log 2 log 3 the spacing between and, and, and, and so on, will be the same as shown in Figs. 23. and khz khz khz MHz khz 3 khz 5 khz 8 khz 3 khz 5 khz 8 khz 3 khz 5 khz 8 khz (log scale) FIG Frequency log scale.

4 2 DECIBELS, FILTES, AND BODE PLOTS db x d d 2 FIG Finding a value on a log plot. x Note in Figs. 23. and 23.2 that the log scale becomes compressed at the high end o each interval. With increasing requency levels assigned to each interval, a single graph can provide a requency plot extending rom Hz to MHz, as shown in Fig. 23.2, with particular reerence to the 3%, 5%, 7%, and 9% levels o each interval. On many log plots, the tick marks or most o the intermediate levels are let o because o space constraints. The ollowing equation can be used to determine the logarithmic level at a particular point between known levels using a ruler or simply estimating the distances. The parameters are deined by Fig Value x d /d 2 (23.3) The derivation o Eq. (23.3) is simply an extension o the details regarding distance appearing on Fig EXAMPLE 23. Determine the value o the point appearing on the logarithmic plot o Fig using the measurements made by a ruler (linear). 7/ 6 " Solution: 2 3/ 4 " 3 d 7/ d2 3/4.75 FIG Example 23.. Using a calculator: d /d Applying Eq. (23.3): Value x d /d POPETIES OF LOGAITHMS There are a ew characteristics o logarithms that should be emphasized:. The common or natural logarithm o the number is. log (23.4) just as x requires that x. 2. The log o any number less than is a negative number. log 2 log.5.3 log log. 3. The log o the product o two numbers is the sum o the logs o the numbers. log ab log a log b (23.5) 4. The log o the quotient o two numbers is the log o the numerator minus the log o the denominator.

5 db DECIBELS 2 log a b log a log b (23.6) 5. The log o a number taken to a power is equal to the product o the power and the log o the number. log a n n log a (23.7) Calculator Functions On most calculators the log o a number is ound by simply entering the number and pressing the LOG or IN key. For example, log 8 8 LOG with a display o.93. For the reverse process, where N, or the antilogarithm, is desired, the unction x is employed. On most calculators x appears as a second unction above the LOG key. For the case o.6 log N the ollowing keys are employed: 2ND F x with a display o Checking: log EXAMPLE 23.2 Evaluate each o the ollowing logarithmic expressions: a. log.4 b. log 25, c. log (.8)(24) 4 d. log 4 e. log () 4 Solutions: a b c. log (.8)(24) log.8 log d. log log 4 log 4 4 (4) 4 8 e. log 4 4 log 4() DECIBELS Power Gain Two levels o power can be compared using a unit o measure called the bel, which is deined by the ollowing equation:

6 22 DECIBELS, FILTES, AND BODE PLOTS db P2 B log P (bels) (23.8) However, to provide a unit o measure o less magnitude, a decibel is deined, where bel decibels (db) (23.9) The result is the ollowing important equation, which compares power levels P 2 and P in decibels: P2 db log P (decibels, db) (23.) I the power levels are equal (P 2 P ), there is no change in power level, and db. I there is an increase in power level (P 2 > P ), the resulting decibel level is positive. I there is a decrease in power level (P 2 < P ), the resulting decibel level will be negative. For the special case o P 2 2P, the gain in decibels is db log P 2 P log 2 3 db Thereore, or a speaker system, a 3-dB increase in output would require that the power level be doubled. In the audio industry, it is a generally accepted rule that an increase in sound level is accomplished with 3-dB increments in the output level. In other words, a -db increase is barely detectable, and a 2-dB increase just discernible. A 3-dB increase normally results in a readily detectable increase in sound level. An additional increase in the sound level is normally accomplished by simply increasing the output level another 3 db. I an 8-W system were in use, a 3-dB increase would require a 6-W output, whereas an additional increase o 3 db (a total o 6 db) would require a 32-W system, as demonstrated by the calculations below: P 2 6 db log log log 2 3 db P 8 P 2 32 db log log log 4 6 db P 8 For P 2 P, P 2 db log log () db P resulting in the unique situation where the power gain has the same magnitude as the decibel level. For some applications, a reerence level is established to permit a comparison o decibel levels rom one situation to another. For communication systems a commonly applied reerence level is P re mw (across a 6- load) Equation (23.) is then typically written as

7 db DECIBELS 23 P db m log m (23.) W 6 Note the subscript m to denote that the decibel level is determined with a reerence level o mw. In particular, or P 4 mw, db m log whereas or P 4 W, db m log 4 mw mw 4 mw mw log 4 (.6) 6 db m log 4 (3.6) 36 db m Even though the power level has increased by a actor o 4 mw/ 4 mw, the db m increase is limited to 2 db m. In time, the signiicance o db m levels o 6 db m and 36 db m will generate an immediate appreciation regarding the power levels involved. An increase o 2 db m will also be associated with a signiicant gain in power levels. Voltage Gain Decibels are also used to provide a comparison between voltage levels. Substituting the basic power equations P 2 V 2 2/ 2 and P V 2 / into Eq. (23.) will result in P 2 V 2 2/ 2 db log log V 2 P / log V 2 2/V 2 2 / log 2 log V 2 V V 2 V and db 2 log log 2 2 For the situation where 2, a condition normally assumed when comparing voltage levels on a decibel basis, the second term o the preceding equation will drop out (log ), and V2 db v 2 log V (db) (23.2) Note the subscript v to deine the decibel level obtained. EXAMPLE 23.3 Find the voltage gain in db o a system where the applied signal is 2 mv and the output voltage is.2 V. Solution: db v 2 log. 2V 2 log 2 log db Vi 2 mv or a voltage gain A v / o 6. EXAMPLE 23.4 I a system has a voltage gain o 36 db, ind the applied voltage i the output voltage is 6.8 V.

8 24 DECIBELS, FILTES, AND BODE PLOTS db Solution: db v 2 log Vi 36 2 log Vi.8 log Vi From the antilogarithm: o V V 6.8 V and 7.77 mv i TABLE 23. / db 2 log ( / ) db 2 6 db 2 db 2 26 db 4 db, 6 db, db Table 23. compares the magnitude o speciic gains to the resulting decibel level. In particular, note that when voltage levels are compared, a doubling o the level results in a change o 6 db rather than 3 db as obtained or power levels. In addition, note that an increase in gain rom to, results in a change in decibels that can easily be plotted on a single graph. Also note that doubling the gain (rom to 2 and to 2) results in a 6-dB increase in the decibel level, while a change o to (rom to, to, and so on) always results in a 2-dB decrease in the decibel level. The Human Auditory esponse One o the most requent applications o the decibel scale is in the communication and entertainment industries. The human ear does not respond in a linear ashion to changes in source power level; that is, a doubling o the audio power level rom /2 W to W does not result in a doubling o the loudness level or the human ear. In addition, a change rom 5 W to W will be received by the ear as the same change in sound intensity as experienced rom /2 W to W. In other words, the ratio between levels is the same in each case ( W/.5 W W/5 W 2), resulting in the same decibel or logarithmic change deined by Eq. (23.7). The ear, thereore, responds in a logarithmic ashion to changes in audio power levels. To establish a basis or comparison between audio levels, a reerence level o.2 microbar (mbar) was chosen, where mbar is equal to the sound pressure o dyne per square centimeter, or about millionth o the normal atmospheric pressure at sea level. The.2-mbar level is the threshold level o hearing. Using this reerence level, the sound pressure level in decibels is deined by the ollowing equation: P db s 2 log. 2 mbar (23.3) where P is the sound pressure in microbars. The decibel levels o Fig are deined by Eq. (23.3). Meters designed to measure audio levels are calibrated to the levels deined by Eq. (23.3) and shown in Fig

9 db DECIBELS 25 Output Power. db s Average value 6 in watts Jet engine Community siren Pneumatic riveter Threshold o pain µ bar o pressure Loud rock band, punch press Chain saw Very loud music Loud music, heavy truck, subway train Orchestra, highway traic Average conversation Quiet music Average residence, computer system Background music Quiet oice, computer hard drive Whispering Faint sounds, paper rustling Threshold o hearing Dynamic range 2 db s FIG Typical sound levels and their decibel levels. A common question regarding audio levels is how much the power level o an acoustical source must be increased to double the sound level received by the human ear. The question is not as simple as it irst seems due to considerations such as the requency content o the sound, the acoustical conditions o the surrounding area, the physical characteristics o the surrounding medium, and o course the unique characteristics o the human ear. However, a general conclusion can be ormulated that has practical value i we note the power levels o an acoustical source appearing to the let o Fig Each power level is associated with a particular decibel level, and a change o db in the scale corresponds with an increase or a decrease in power by a actor o. For instance, a change rom 9 db to db is associated with a change in wattage rom 3 W to 3 W. Through experimentation it has been ound that on an average basis the loudness level will double or every -db change in audio level a conclusion somewhat veriied by the examples to the right o Fig Using the act that a -db change corresponds with a tenold increase in power level supports the ollowing conclusion (on an approximate basis): Through experimentation it has been ound that on an average basis, the loudness level will double or every -db change in audio level. To double the sound level received by the human ear, the power rating o the acoustical source (in watts) must be increased by a actor o.

10 26 DECIBELS, FILTES, AND BODE PLOTS db In other words, doubling the sound level available rom a -W acoustical source would require moving up to a -W source. Instrumentation A number o modern VOMs and DMMs have a db scale designed to provide an indication o power ratios reerenced to a standard level o mw at 6. Since the reading is accurate only i the load has a characteristic impedance o 6, the -mw, 6 reerence level is normally printed somewhere on the ace o the meter, as shown in Fig The db scale is usually calibrated to the lowest ac scale o the meter. In other DB mw, DB 2.5 3VAC 3 FIG Deining the relationship between a db scale reerenced to mw, 6 and a 3-V-rms voltage scale. words, when making the db measurement, choose the lowest ac voltage scale, but read the db scale. I a higher voltage scale is chosen, a correction actor must be employed that is sometimes printed on the ace o the meter but always available in the meter manual. I the impedance is other than 6 or not purely resistive, other correction actors must be used that are normally included in the meter manual. Using the basic power equation P V 2 / will reveal that mw across a 6- load is the same as applying.775 V rms across a 6- load; that is, V P ( mw)(6).775 V. The result is that an analog display will have db [deining the reerence point o mw, db log P 2 /P log ( mw/ mw(re)] db] and.775 V rms on the same pointer projection, as shown in Fig A voltage o 2.5 V across a 6- load would result in a db level o db 2 log V 2 /V 2 log 2.5 V/ db, resulting in 2.5 V and.7 db appearing along the same pointer projection. A voltage o less than.775 V, such as.5 V, will result in a db level o db 2 log V 2 /V 2 log.5 V/.775 V 3.8 db, as is also shown on the scale o Fig Although a reading o db will reveal that the power level is times the reerence, don t assume that a reading o 5 db means that the output level is5mw.the: ratio is a special one in logarithmic circles. For the 5-dB level, the power level must be ound using the antilogarithm (3.26), which reveals that the power level associated with 5 db is about 3. times the reerence or 3. mw. A conversion table is usually provided in the manual or such conversions FILTES Any combination o passive (, L, and C) and/or active (transistors or operational ampliiers) elements designed to select or reject a band o

11 db FILTES 27 requencies is called a ilter. In communication systems, ilters are employed to pass those requencies containing the desired inormation and to reject the remaining requencies. In stereo systems, ilters can be used to isolate particular bands o requencies or increased or decreased emphasis by the output acoustical system (ampliier, speaker, etc.). Filters are employed to ilter out any unwanted requencies, commonly called noise, due to the nonlinear characteristics o some electronic devices or signals picked up rom the surrounding medium. In general, there are two classiications o ilters:. Passive ilters are those ilters composed o series or parallel combinations o, L, and C elements. 2. Active ilters are ilters that employ active devices such as transistors and operational ampliiers in combination with, L, and C elements. Low-pass ilter: V max.77v max Pass-band c (a) Stop-band High-pass ilter: V max.77v max c Stop-band (b) Pass-band Pass-band ilter: V max.77v max Stop-band o 2 Stop-band Pass-band (c) Stop-band ilter: V max.77v max o 2 Pass-band Stop-band Pass-band (d) FIG Deining the our broad categories o ilters.

12 28 DECIBELS, FILTES, AND BODE PLOTS db Since this text is limited to passive devices, the analysis o this chapter will be limited to passive ilters. In addition, only the most undamental orms will be examined in the next ew sections. The subject o ilters is a very broad one that continues to receive extensive research support rom industry and the government as new communication systems are developed to meet the demands o increased volume and speed. There are courses and texts devoted solely to the analysis and design o ilter systems that can become quite complex and sophisticated. In general, however, all ilters belong to the our broad categories o low-pass, high-pass, pass-band, and stop-band, as depicted in Fig For each orm there are critical requencies that deine the regions o pass-bands and stop-bands (oten called reject bands). Any requency in the pass-band will pass through to the next stage with at least 7.7% o the maximum output voltage. ecall the use o the.77 level to deine the bandwidth o a series or parallel resonant circuit (both with the general shape o the pass-band ilter). For some stop-band ilters, the stop-band is deined by conditions other than the.77 level. In act, or many stop-band ilters, the condition that /V max (corresponding with 6 db in the discussion to ollow) is used to deine the stop-band region, with the passband continuing to be deined by the.77-v level. The resulting requencies between the two regions are then called the transition requencies and establish the transition region. At least one example o each ilter o Fig will be discussed in some detail in the sections to ollow. Take particular note o the relative simplicity o some o the designs. C FIG Low-pass ilter. = FIG C low-pass ilter at low requencies. = V FIG C low-pass ilter at high requencies C LOW-PASS FILTE The -C ilter, incredibly simple in design, can be used as a low-pass or a high-pass ilter. I the output is taken o the capacitor, as shown in Fig. 23.8, it will respond as a low-pass ilter. I the positions o the resistor and capacitor are interchanged and the output is taken o the resistor, the response will be that o a high-pass ilter. A glance at Fig. 23.7(a) reveals that the circuit should behave in a manner that will result in a high-level output or low requencies and a declining level or requencies above the critical value. Let us irst examine the network at the requency extremes o Hz and very high requencies to test the response o the circuit. At Hz, X C 2pC and the open-circuit equivalent can be substituted or the capacitor, as shown in Fig. 23.9, resulting in. At very high requencies, the reactance is X C 2pC and the short-circuit equivalent can be substituted or the capacitor, as shown in Fig. 23., resulting in V. A plot o the magnitude o versus requency will result in the curve o Fig Our next goal is now clearly deined: Find the requency at which the transition takes place rom a pass-band to a stop-band.

13 db -C LOW-PASS FILTE 29 =.77 Pass-band c Stop-band (log scale) FIG. 23. versus requency or a low-pass -C ilter. For ilters, a normalized plot is employed more oten than the plot o versus requency o Fig Normalization is a process whereby a quantity such as voltage, current, or impedance is divided by a quantity o the same unit o measure to establish a dimensionless level o a speciic value or range. A normalized plot in the ilter domain can be obtained by dividing the plotted quantity such as o Fig. 23. with the applied voltage or the requency range o interest. Since the maximum value o or the low-pass ilter o Fig is, each level o in Fig. 23. is divided by the level o. The result is the plot o A v / o Fig Note that the maximum value is and the cuto requency is deined at the.77 level. A v =.77 Pass-band c Stop-band (log scale) FIG Normalized plot o Fig At any intermediate requency, the output voltage o Fig can be determined using the voltage divider rule: or and X X A v C 9 C 9 Vi j XC 2 X C 2 / tan (X C /) X C X C A v 9 tan Vi 2 X 2 C X C 9 j XC

14 3 DECIBELS, FILTES, AND BODE PLOTS db The magnitude o the ratio / is thereore determined by V X A v o C (23.4) Vi 2 X C 2 X C 2 and the phase angle is determined by v 9 tan X C tan X C (23.5) For the special requency at which X C, the magnitude becomes A v.77 Vi 2 X C 2 which deines the critical or cuto requency o Fig The requency at which X C is determined by 2p c C and c (23.6) 2p C The impact o Eq. (23.6) extends beyond its relative simplicity. For any low-pass ilter, the application o any requency less than c will result in an output voltage that is at least 7.7% o the maximum. For any requency above c, the output is less than 7.7% o the applied signal. Solving or and substituting gives X C v v 2 X C 2 X C and v 2 X C 2 The angle v is, thereore, the angle by which leads. Since v tan /X C is always negative (except at Hz), it is clear that will always lag, leading to the label lagging network or the network o Fig At high requencies, X C is very small and /X C is quite large, resulting in v tan /X C approaching 9. At low requencies, X C is quite large and /X C is very small, resulting in v approaching. At X C, or c, tan /X C tan 45. A plot o v versus requency results in the phase plot o Fig The plot is o leading, but since the phase angle is always negative, the phase plot o Fig ( lagging ) is more appropriate. Note that a change in sign requires that the vertical axis be changed to the angle by which lags. In particular, note that the phase angle between and is less than 45 in the pass-band and approaches X C 2 X C 2

15 db -C LOW-PASS FILTE 3 v ( leads ) Pass-band c Stop-band (log scale) 45 9 FIG Angle by which leads. v ( lags ) 9 45 Pass-band c Stop-band (log scale) FIG Angle by which lags. at lower requencies. In summary, or the low-pass -C ilter o Fig. 23.8: c 2p C For < c, >.77 whereas or > c, <.77 At c, lags by 45 The low-pass ilter response o Fig. 23.7(a) can also be obtained using the -L combination o Fig with c (23.7) 2p L In general, however, the -C combination is more popular due to the smaller size o capacitive elements and the nonlinearities associated with inductive elements. The details o the analysis o the low-pass -L will be let as an exercise or the reader. L FIG Low-pass -L ilter. k EXAMPLE 23.5 a. Sketch the output voltage versus requency or the low-pass -C ilter o Fig b. Determine the voltage at khz and MHz, and compare the results to the results obtained rom the curve o part (a). c. Sketch the normalized gain A v /. = 2 V C FIG Example pf

16 32 DECIBELS, FILTES, AND BODE PLOTS db Solutions: a. Eq. (23.6): c 38.3 khz 2pC 2p( k)(5 pf) At c,.77(2 V) 4.4 V. See Fig = 2 V.77 (volts) 9.8 V 4.4 V V 6.V 38.3 khz khz khz c MHz MHz (log scale) Pass-band Stop-band FIG Frequency response or the low-pass -C network o Fig b. Eq. (23.4): X C 2 At khz: X C 3.8 k 2pC 2p( khz)(5 pf) 2 V and 9.8 V 3. k 8 k 2 At MHz: X C.32 k 2pC 2p( MHz)(5 pf) 2 V and 6. V. k 32 2 k Both levels are veriied by Fig c. Dividing every level o Fig by 2 V will result in the normalized plot o Fig

17 db -C HIGH-PASS FILTE A v = khz khz khz c MHz MHz (log scale) FIG Normalized plot o Fig C HIGH-PASS FILTE As noted early in Section 23.5, a high-pass -C ilter can be constructed by simply reversing the positions o the capacitor and resistor, as shown in Fig At very high requencies the reactance o the capacitor is very small, and the short-circuit equivalent can be substituted, as shown in Fig The result is that. C = V FIG High-pass ilter. FIG C high-pass ilter at very high requencies. At Hz, the reactance o the capacitor is quite high, and the open-circuit equivalent can be substituted, as shown in Fig In this case, V. A plot o the magnitude versus requency is provided in Fig , with the normalized plot in Fig = V = =.77 FIG C high-pass ilter at Hz. Stop-band c Pass-band (log scale) FIG versus requency or a high-pass -C ilter. At any intermediate requency, the output voltage can be determined using the voltage divider rule: jxc

18 34 DECIBELS, FILTES, AND BODE PLOTS db A v =.77 Stop-band c Pass-band (log scale) FIG Normalized plot o Fig or Vi jxc V and o tan (X C /) Vi 2 X C 2 The magnitude o the ratio / is thereore determined by and the phase angle v by 2 X 2 C tan (X C /) V A v o (23.8) Vi 2 X C 2 X C 2 v tan X C (23.9) For the requency at which X C, the magnitude becomes.77 Vi 2 X C 2 as shown in Fig The requency at which X C is determined by X C 2pc C and c (23.2) 2p C For the high-pass -C ilter, the application o any requency greater than c will result in an output voltage that is at least 7.7% o the magnitude o the input signal. For any requency below c, the output is less than 7.7% o the applied signal. For the phase angle, high requencies result in small values o X C, and the ratio X C / will approach zero with tan (X C /) approaching, as shown in Fig At low requencies, the ratio X C / becomes

19 db -C HIGH-PASS FILTE 35 v ( leads ) 9 45 Stop-band c Pass-band (log scale) FIG Phase-angle response or the high-pass -C ilter. quite large, and tan (X C /) approaches 9. For the case X C, tan (X C /) tan 45. Assigning a phase angle o to such that, the phase angle associated with is v, resulting in v and revealing that v is the angle by which leads. Since the angle v is the angle by which leads throughout the requency range o Fig , the high-pass -C ilter is reerred to as a leading network. In summary, or the high-pass -C ilter: c 2p C For < c, <.77 whereas or > c, >.77 At c, leads by 45 L The high-pass ilter response o Fig can also be obtained using the same elements o Fig but interchanging their positions, as shown in Fig FIG High-pass -L ilter. EXAMPLE 23.6 Given 2 k and C 2 pf: a. Sketch the normalized plot i the ilter is used as both a high-pass and a low-pass ilter. b. Sketch the phase plot or both ilters o part (a). c. Determine the magnitude and phase o A v / at c or the high-pass ilter. Solutions: a. c 2pC (2p)(2 k)(2 pf) Hz The normalized plots appear in Fig b. The phase plots appear in Fig c. c ( Hz) Hz 2 2 X C 2pC 4 k (2p)( Hz)(2 pf)

20 36 DECIBELS, FILTES, AND BODE PLOTS db A v = A v =.77 Low-pass.77 High-pass c = Hz (log scale) c = Hz (log scale) FIG Normalized plots or a low-pass and a high-pass ilter using the same elements. θ ( lags ) θ ( leads ) 9 9 High-pass 45 Low-pass 45 c = Hz (log scale) c = Hz (log scale) FIG Phase plots or a low-pass and a high-pass ilter using the same elements. A v Vi (2)2 X C 2 5 X C.4472 v tan tan 4 k tan k Vi 4 2 k k 2 and A v PASS-BAND FILTES A number o methods are used to establish the pass-band characteristic o Fig. 23.7(c). One method employs both a low-pass and a high-pass ilter in cascade, as shown in Fig The components are chosen to establish a cuto requency or the high-pass ilter that is lower than the critical requency o the low-pass ilter, as shown in Fig A requency may pass through the low- High-pass Low-pass ilter ilter FIG Pass-band ilter.

21 db PASS-BAND FILTES 37 Low-pass V max.77v max BW High-pass c o c 2 (High-pass) (Low-pass) FIG Pass-band characteristics. pass ilter but have little eect on due to the reject characteristics o the high-pass ilter. A requency 2 may pass through the high-pass ilter unmolested but be prohibited rom reaching the high-pass ilter by the low-pass characteristics. A requency o near the center o the passband will pass through both ilters with very little degeneration. The network o Example 23.7 will generate the characteristics o Fig However, or a circuit such as the one shown in Fig. 23.3, there is a loading between stages at each requency that will aect the level o. Through proper design, the level o may be very near the level o in the pass-band, but it will never equal it exactly. In addition, as the critical requencies o each ilter get closer and closer together to increase the quality actor o the response curve, the peak values within the pass-band will continue to drop. For cases where max max the bandwidth is deined at.77 o the resulting max. C.5 nf High-pass ilter 2 4 k k C 2 4 pf Low-pass ilter EXAMPLE 23.7 For the pass-band ilter o Fig. 23.3: a. Determine the critical requencies or the low- and high-pass ilters. b. Using only the critical requencies, sketch the response characteristics. c. Determine the actual value o at the high-pass critical requency calculated in part (a), and compare it to the level that will deine the upper requency or the pass-band. FIG Pass-band ilter. Solutions: a. High-pass ilter: c 6. khz 2p C 2p( k)(.5 nf) Low-pass ilter: c khz 2p2 C 2 2p(4 k)(4 pf) b. In the mid-region o the pass-band at about 5 khz, an analysis o the network will reveal that.9 as shown in Fig The bandwidth is thereore deined at a level o.77(.9 ).636 as also shown in Fig c. At khz, X C 2pC 7

22 38 DECIBELS, FILTES, AND BODE PLOTS db Pass-band c 6 khz c 995 khz Actual c Actual c FIG Pass-band characteristics or the ilter o Fig and X C2 2 4 k 2pC2 resulting in the network o Fig X C k = k V' X C2 4 k FIG Network o Fig at khz. The parallel combination ( 2 jx C2 ) is essentially.976 k because the 2 X C 2 combination is so large compared to the parallel resistor. Then.976 k ( ) V kj.7 k with (4 k 9 )( ) 4 kj4 k so that.73 at khz Since the bandwidth is deined at.636 the upper cuto requency will be higher than khz as shown in Fig The pass-band response can also be obtained using the series and parallel resonant circuits discussed in Chapter 2. In each case, however, will not be equal to in the pass-band, but a requency range in which will be equal to or greater than.77v max can be deined. For the series resonant circuit o Fig , X L X C at resonance, and max (23.2) s l

23 db PASS-BAND FILTES 39 Pass-band ilter V V omaxi l L C.77max s 2 BW FIG Series resonant pass-band ilter. and s (23.22) 2p LC X L with Q s (23.23) l s and BW (23.24) Q s For the parallel resonant circuit o Fig , Z Tp value at resonance, and is a maximum ZT max V p i ZT p p (23.25) Pass-band ilter Z Tp l L C V omax.77max = V C p 2 BW FIG Parallel resonant pass-band ilter. with Z Tp Q 2 l l Q l (23.26) and p 2pLC (23.27) Q l

24 4 DECIBELS, FILTES, AND BODE PLOTS db For the parallel resonant circuit L Q p X l (23.28) p and BW (23.29) Q p As a irst approximation that is acceptable or most practical applications, it can be assumed that the resonant requency bisects the bandwidth. l 2 = 2 mv L mh C. mf 33 FIG Series resonant pass-band ilter or Example EXAMPLE 23.8 a. Determine the requency response or the voltage or the series circuit o Fig b. Plot the normalized response A v /. c. Plot a normalized response deined by A v A v /A vmax. Solutions: a. s 5,329.2 Hz 2pLC 2p( mh)(. mf) X L 2p(5,329.2 Hz)( mh) Q s 9.4 l 33 2 s 5,329.2 Hz BW 5.57 khz Qs 9.4 At resonance: 33 ( ) max (2 mv) l mv At the cuto requencies: (.77)(.943 ) (2 mv) 3.34 mv Note Fig mv 3.34 mv BW = 5.57 khz s 5.3 khz FIG Pass-band response or the network. (log scale) b. Dividing all levels o Fig by 2 mv will result in the normalized plot o Fig (a).

25 db STOP-BAND FILTES 4 A v = = 2 mv A A v = v = A vmax A v BW{.77 BW{ s (log scale) s (log scale) (a) (b) FIG Normalized plots or the pass-band ilter o Fig c. Dividing all levels o Fig (a) by A vmax.943 will result in the normalized plot o Fig (b) STOP-BAND FILTES Stop-band ilters can also be constructed using a low-pass and a highpass ilter. However, rather than the cascaded coniguration used or the pass-band ilter, a parallel arrangement is required, as shown in Fig A low-requency can pass through the low-pass ilter, and a higher-requency 2 can use the parallel path, as shown in Figs and However, a requency such as o in the reject-band is higher than the low-pass critical requency and lower than the high-pass critical requency, and is thereore prevented rom contributing to the levels o above.77v max. (low) o Low-pass ilter (low) o High-pass ilter 2 (high) 2 (high) FIG Stop-band ilter. max.77max BW c o c 2 (log scale) (Low-pass) (High-pass) FIG Stop-band characteristics.

26 42 DECIBELS, FILTES, AND BODE PLOTS db Since the characteristics o a stop-band ilter are the inverse o the pattern obtained or the pass-band ilters, we can employ the act that at any requency the sum o the magnitudes o the two waveorms to the right o the equals sign in Fig will equal the applied voltage. = Pass-band Stop-band o o FIG Demonstrating how an applied signal o ixed magnitude can be broken down into a pass-band and stop-band response curve. For the pass-band ilters o Figs and 23.34, thereore, i we take the output o the other series elements as shown in Figs and 23.42, a stop-band characteristic will be obtained, as required by Kirchho s voltage law. For the series resonant circuit o Fig. 23.4, Equations (23.22) through (23.24) still apply, but now, at resonance, Stop-band ilter l L max =.77 BW C min s 2 FIG Stop-band ilter using a series resonant circuit. Stop-band ilter l L C max =.77 BW min p 2 FIG Stop-band ilter using a parallel resonant network.

27 db DOUBLE-TUNED FILTE 43 l min l (23.3) For the parallel resonant circuit o Fig , Equations (23.26) through (23.29) are still applicable, but now, at resonance, Vi min Z T p (23.3) The maximum value o or the series resonant circuit is at the low end due to the open-circuit equivalent or the capacitor and at the high end due to the high impedance o the inductive element. For the parallel resonant circuit, at Hz, the coil can be replaced by a short-circuit equivalent, and the capacitor can be replaced by its open circuit and /( l ). At the high-requency end, the capacitor approaches a short-circuit equivalent, and increases toward DOUBLE-TUNED FILTE Some network conigurations display both a pass-band and a stop-band characteristic, such as shown in Fig Such networks are called double-tuned ilters. For the network o Fig (a), the parallel resonant circuit will establish a stop-band or the range o requencies not permitted to establish a signiicant V L. The greater part o the applied voltage will appear across the parallel resonant circuit or this requency range due to its very high impedance compared with L. For the pass-band, the parallel resonant circuit is designed to be capacitive (inductive i L s is replaced by C s ). The inductance L s is chosen to cancel the eects o the resulting net capacitive reactance at the resonant pass-band requency o the tank circuit, thereby acting as a series resonant circuit. The applied voltage will then appear across L at this requency. Double-tuned ilter Double-tuned ilter C L s C L s L p L p L V L L V L (a) (b) FIG Double-tuned networks. For the network o Fig (b), the series resonant circuit will still determine the pass-band, acting as a very low impedance across the parallel inductor at resonance. At the desired stop-band resonant requency,

28 44 DECIBELS, FILTES, AND BODE PLOTS db the series resonant circuit is capacitive. The inductance L p is chosen to establish parallel resonance at the resonant stop-band requency. The high impedance o the parallel resonant circuit will result in a very low load voltage V L. For rejected requencies below the pass-band, the networks should appear as shown in Fig For the reverse situation, L s in Fig (a) and L p in Fig (b) are replaced by capacitors. EXAMPLE 23.9 For the network o Fig (b), determine L s and L p or a capacitance C o 5 pf i a requency o 2 khz is to be rejected and a requency o 6 khz accepted. Solution: For series resonance, we have s 2pLC and L s 4.7 mh 4p 2 2 sc 4p 2 (6 khz) 2 (5 pf) At 2 khz, X Ls ql 2p s L s (2p)(2 khz)(4.7 mh) 76.8 and X C 59.5 qc (2p)(2 khz)(5 pf) For the series elements, American (Madison, WI; Summit, NJ; Cambridge, MA) (958) V.P. at Bell Laboratories Proessor o Systems Engineering, Harvard University j (X Ls X C ) j ( ) j 44.7 j X C At parallel resonance (Q l assumed), X Lp X C X Lp 44.7 and L p q.3 mh (2p)(2 khz) The requency response or the preceding network appears as one o the examples o PSpice in the last section o the chapter. Courtesy o AT&T Archives In his early years at Bell Laboratories, Hendrik Bode was involved with electric ilter and equalizer design. He then transerred to the Mathematics esearch Group, where he specialized in research pertaining to electrical networks theory and its application to long-distance communication acilities. In 946 he was awarded the Presidential Certiicate o Merit or his work in electronic ire control devices. In addition to the publication o the book Network Analysis and Feedback Ampliier Design in 945, which is considered a classic in its ield, he has been granted 25 patents in electrical engineering and systems design. Upon retirement, Bode was elected Gordon McKay Proessor o Systems Engineering at Harvard University. He was a ellow o the IEEE and American Academy o Arts and Sciences. FIG Hendrik Wade Bode 23. BODE PLOTS There is a technique or sketching the requency response o such actors as ilters, ampliiers, and systems on a decibel scale that can save a great deal o time and eort and provide an excellent way to compare decibel levels at dierent requencies. The curves obtained or the magnitude and/or phase angle versus requency are called Bode plots (Fig ). Through the use o straight-line segments called idealized Bode plots, the requency response o a system can be ound eiciently and accurately. To ensure that the derivation o the method is correctly and clearly understood, the irst network to be analyzed will be examined in some detail. The second network will be treated in a shorthand manner, and inally a method or quickly determining the response will be introduced.

29 db BODE PLOTS 45 High-Pass -C Filter Let us start by reexamining the high-pass ilter o Fig The highpass ilter was chosen as our starting point because the requencies o primary interest are at the low end o the requency spectrum. The voltage gain o the system is given by C A v Vi jxc j X C j 2pC j 2p C FIG High-pass ilter. I we substitute c (23.32) 2p C which we recognize as the cuto requency o earlier sections, we obtain A v (23.33) j ( c / ) We will ind in the analysis to ollow that the ability to reormat the gain to one having the general characteristics o Eq. (23.33) is critical to the application o the Bode technique. Dierent conigurations will result in variations o the ormat o Eq. (23.33), but the desired similarities will become obvious as we progress through the material. In magnitude and phase orm: A v A v v tan ( c / ) (23.34) Vi (c /) 2 providing an equation or the magnitude and phase o the high-pass ilter in terms o the requency levels. Using Eq. (23.2), A vdb 2 log A v and, substituting the magnitude component o Eq. (23.34), A vdb 2 log 2 log 2 log c / 2 c / 2 and A vdb 2 log 2 c ecognizing that log x log x /2 2 log x, we have A vdb (2) log 2 2 c log c 2

30 46 DECIBELS, FILTES, AND BODE PLOTS db For requencies where K c or ( c / ) 2 k, c 2 2 and A vdb log 2 c c but log x 2 2 log x resulting in A vdb 2 log c However, logarithms are such that log b log and substituting b c /, we have b A vdb 2 log (23.35) K c c First note the similarities between Eq. (23.35) and the basic equation or gain in decibels: G db 2 log /. The comments regarding changes in decibel levels due to changes in / can thereore be applied here also, except now a change in requency by a 2 ratio will result in a 6-dB change in gain. A change in requency by a ratio will result in a 2-dB change in gain. Two requencies separated by a 2: ratio are said to be an octave apart. For Bode plots, a change in requency by one octave will result in a 6-dB change in gain. Two requencies separated by a : ratio are said to be a decade apart. For Bode plots, a change in requency by one decade will result in a 2-dB change in gain. One may wonder about all the mathematical development to obtain an equation that initially appears conusing and o limited value. As speciied, Equation (23.35) is accurate only or requency levels much less than c. First, realize that the mathematical development o Eq. (23.35) will not have to be repeated or each coniguration encountered. Second, the equation itsel is seldom applied but simply used in a manner to be described to deine a straight line on a log plot that permits a sketch o the requency response o a system with a minimum o eort and a high degree o accuracy. To plot Eq. (23.35), consider the ollowing levels o increasing requency: For c /, / c. and 2 log. 2 db For c /4, / c.25 and 2 log.25 2 db For c /2, / c.5 and 2 log.5 6 db For c, / c and 2 log db

31 db BODE PLOTS 47 Note rom the above equations that as the requency o interest approaches c, the db gain becomes less negative and approaches the inal normalized value o db. The positive sign in ront o Eq. (23.35) can thereore be interpreted as an indication that the db gain will have a positive slope with an increase in requency. A plot o these points on a log scale will result in the straight-line segment o Fig to the let o c. A v(db) (linear scale) Idealized Bode plot 2 c c log c 2 c 2 c 3 c 5 c c 2 log = db c db Actual requency response 6 db/octave or 2 db/decade 7 db (log scale) FIG Idealized Bode plot or the low-requency region. For the uture, note that the resulting plot is a straight line intersecting the -db line at c. It increases to the right at a rate o 6 db per octave or 2 db per decade. In other words, once c is determined, ind c /2, and a plot point exists at 6 db (or ind c /, and a plot point exists at 2 db). Bode plots are straight-line segments because the db change per decade or octave is constant. The actual response will approach an asymptote (straight-line segment) deined by A vdb db since at high requencies k c and c / with A vdb 2 log 2 log (c /) 2 2 log db The two asymptotes deined above will intersect at c, as shown in Fig , orming an envelope or the actual requency response. At c, the cuto requency, A vdb 2 log 2 log 2 log (c /) db At 2 c, c A vdb 2 log log log.25 db as shown in Fig c

32 48 DECIBELS, FILTES, AND BODE PLOTS db At c /2, A vdb 2 log c 2 log 5 c / 2 2 log (2) db separating the idealized Bode plot rom the actual response by 7 db 6 db db, as shown in Fig eviewing the above, at c, the actual response curve is 3 db down rom the idealized Bode plot, whereas at 2 c and c /2, the actual response curve is dbdown rom the asymptotic response. The phase response can also be sketched using straight-line asymptotes by considering a ew critical points in the requency spectrum. Equation (23.34) speciies the phase response (the angle by which leads ) by v tan c (23.36) For requencies well below c ( K c ), v tan ( c / ) approaches 9 and or requencies well above c ( k c ), v tan ( c / ) will approach, as discovered in earlier sections o the chapter. At c, v tan ( c / ) tan 45. Deining K c or c / (and less) and k c or c (and more), we can deine an asymptote at v 9 or K c /, an asymptote at v or k c, and an asymptote rom c / to c that passes through v 45 at c. The asymptotes deined above all appear in Fig Again, the Bode plot or Eq.(23.36) is a straight line because the change in phase angle will be 45 or every tenold change in requency. Substituting c / into Eq. (23.36), v tan c tan c / 9 θ ( leads ) θ = 9 Dierence = Actual response 45 Dierence = 5.7 c c c c c (log scale) θ = FIG Phase response or a high-pass -C ilter.

33 db BODE PLOTS 49 or a dierence o rom the idealized response. Substituting c, c v tan tan 5.7 c In summary, thereore, at c, v 45, whereas at c / and c, the dierence between the actual phase response and the asymptotic plot is 5.7. EXAMPLE 23. a. Sketch A vdb versus requency or the high-pass -C ilter o Fig b. Determine the decibel level at khz. c. Sketch the phase response versus requency on a log scale. Solutions: a. c Hz 2pC (2p)( k)(. mf) The requency c is identiied on the log scale as shown in Fig A straight line is then drawn rom c with a slope that will intersect 2 db at c / 59.5 Hz or 6 dbat c / Hz. A second asymptote is drawn rom c to higher requencies at db. The actual response curve can then be drawn through the 3-dB level at c approaching the two asymptotes o Fig Note the -db dierence between the actual response and the idealized Bode plot at 2 c and.5 c. C. mf FIG Example 23.. k db c = 59.5 Hz c 2 = Hz c = Hz Hz 3 2 Hz 3 Hz khz 2 khz 2 c 5 khz db khz (log scale) 6 9 db 3 db at = c 2 5 Actual response curve 8 2 db 2 24 FIG Frequency response or the high-pass ilter o Fig Note that in the solution to part (a), there is no need to employ Eq. (23.35) or to perorm any extensive mathematical manipulations.

34 5 DECIBELS, FILTES, AND BODE PLOTS db b. Eq. (23.33): A vdb 2 log 2 log H z 2 c 2 2 log 2 log db (.592)2 as veriied by Fig c. See Fig Note that v 45 at c Hz, and the dierence between the straight-line segment and the actual response is 5.7 at c / 59.2 Hz and c 5,923.6 Hz. θ ( leads ) 9 Dierence = Dierence = 5.7 (log scale) Hz Hz c = 59.5 Hz khz c = Hz FIG Phase plot or the high-pass -C ilter. khz c = 5,95.5 Hz khz Low-Pass -C Filter C FIG Low-pass ilter. For the low-pass ilter o Fig. 23.5, jx C A v Vi jxc j X C j j j XC 2p C 2p C and A v (23.37) j ( /c ) with c (23.38) 2p C as deined earlier. Note that now the sign o the imaginary component in the denominator is positive and c appears in the denominator o the requency ratio rather than in the numerator, as in the case o c or the high-pass ilter.

35 db BODE PLOTS 5 In terms o magnitude and phase, A v A v v tan ( / c ) Vi (/ ) 2 c (23.39) An analysis similar to that perormed or the high-pass ilter will result in A vdb 2 log k c (23.4) Note in particular that the equation is exact only or requencies much greater than c, but a plot o Eq. (23.4) does provide an asymptote that perorms the same unction as the asymptote derived or the high-pass ilter. In addition, note that it is exactly the same as Eq. (23.35), except or the minus sign, which suggests that the resulting Bode plot will have a negative slope [recall the positive slope or Eq. (23.35)] or increasing requencies beyond c. A plot o Eq. (23.4) appears in Fig or c khz. Note the 6-dB drop at 2 c and the 2-dB drop at c. c db. khz 2 c c khz 2 c 2 khz (log scale) c khz 3 6 db dierence 6 db db dierence (log scale) Actual requency response 2 db FIG Bode plot or the high-requency region o a low-pass -C ilter. At k c,the phase angle v tan ( / c ) approaches 9, whereas at K c, v tan ( / c ) approaches. At c, v tan 45, establishing the plot o Fig Note again the 45 change in phase angle or each tenold increase in requency. v ( leads ) c / v = c / 45 Dierence = 5.7 c 45 c c (log scale) 9 Dierence = 5.7 v = 9 FIG Phase plot or a low-pass -C ilter.

36 52 DECIBELS, FILTES, AND BODE PLOTS db Even though the preceding analysis has been limited solely to the -C combination, the results obtained will have an impact on networks that are a great deal more complicated. One good example is the highand low-requency response o a standard transistor coniguration. Some capacitive elements in a practical transistor network will aect the low-requency response, and others will aect the high-requency response. In the absence o the capacitive elements, the requency response o a transistor would ideally stay level at the midband value. However, the coupling capacitors at low requencies and the bypass and parasitic capacitors at high requencies will deine a bandwidth or numerous transistor conigurations. In the low-requency region, speciic capacitors and resistors will orm an -C combination that will deine a low cuto requency. There are then other elements and capacitors orming a second -C combination that will deine a high cuto requency. Once the cuto requencies are known, the 3-dB points are set, and the bandwidth o the system can be determined. 23. SKETCHING THE BODE ESPONSE In the previous section we ound that normalized unctions o the orm appearing in Fig had the Bode envelope and the db response db Low-pass: j c 3 db c 6 db/octave (or increasing ) (a) db High-pass: c db j c 3 db 6 db/octave (or increasing ) (b) FIG db response o (a) low-pass ilter and (b) high-pass ilter. k nf 2 4 k FIG High-pass ilter with attenuated output. C indicated in the same igure. In this section we introduce additional unctions and their responses that can be used in conjunction with those o Fig to determine the db response o more sophisticated systems in a systematic, time-saving, and accurate manner. As an avenue toward introducing an additional unction that appears quite requently, let us examine the high-pass ilter o Fig which has a high-requency output less than the ull applied voltage. Beore developing a mathematical expression or A v /, let us irst make a rough sketch o the expected response.

37 db SKETCHING THE BODE ESPONSE 53 At Hz, the capacitor will assume its open-circuit equivalence, and V. At very high requencies, the capacitor can assume its short-circuit equivalence, and 2 4 k.8 2 k4 k The resistance to be employed in the equation or cuto requency can be determined by simply determining the Thévenin resistance seen by the capacitor. Setting Vand solving or Th (or the capacitor C) will result in the network o Fig , where it is quite clear that Th 2 k4 k5 k Th = V 2 FIG Determining Th or the equation or cuto requency. Thereore, c 3.83 khz 2pTh C 2p(5 k)( nf) A sketch o versus requency is provided in Fig (a). A normalized plot using as the normalizing quantity will result in the response o Fig (b). I the maximum value o A v is used in the normalization process, the response o Fig (c) will be obtained..8 2 = A v = 2 2 A v = A v A vmax = A v c c c (a) (b) (c) FIG Finding the normalized plot or the gain o the high-pass ilter o Fig with attenuated output. For all the plots obtained in the previous section, was the maximum value, and the ratio / had a maximum value o. For many situations, this will not be the case, and we must be aware o which ratio is being plotted versus requency. The db response curves or the plots o Figs (b) and 23.57(c) can both be obtained quite directly using the oundation established by the conclusions depicted in Fig , but wemust be aware o what to expect and how they will dier. In Fig (b) we are comparing the output level to the input voltage. In

38 54 DECIBELS, FILTES, AND BODE PLOTS db Fig (c) we are plotting A v versus the maximum value o A v. On most data sheets and or the majority o the investigative techniques commonly employed, the normalized plot o Fig (c) is used because it establishes db as an asymptote or the db plot. To ensure that the impact o using either Fig (b) or Fig (c) in a requency plot is understood, the analysis o the ilter o Fig will include the resulting db plot or both normalized curves. For the network o Fig : jx C 2 jx C Dividing the top and bottom o the equation by 2 results in 2 2 XC j 2 X C but j j j 2 q( 2 )C 2p( 2 )C c j with c and Th 2 2pTh C so that I we divide both sides by, we obtain Vi 2 2 j( c / ) 2 2 j( c / ) A v (23.4) rom which the magnitude plot o Fig (b) can be obtained. I we divide both sides by A vmax 2 /( 2 ), we have A Av v A j ( c / ) v m ax (23.42) rom which the magnitude plot o Fig (c) can be obtained. Based on the past section, a db plot o the magnitude o A v A v /A vmax is now quite direct using Fig (b). The plot appears in Fig For the gain A v /, we can apply Eq. (23.5): 2 log ab 2 log a 2 log b where 2 log 2 2 j( c / ) 2 log 2 2 log (c /) 2 2 The second term will result in the same plot o Fig , but the irst term must be added to the second to obtain the total db response. Since 2 /( 2 ) must always be less than, we can rewrite the irst term as 2 log log 2 log 2 2 log 2 2 2

39 db SKETCHING THE BODE ESPONSE 55 A v db = A v A vmax db c = 3.83 khz 3 db FIG db plot or A v or the high-pass ilter o Fig and 2 log 2 log 2 (23.43) 2 2 providing the drop in db rom the -db level or the plot. Adding one log plot to the other at each requency, as permitted by Eq. (23.5), will result in the plot o Fig log 2 2 db 2 log 2 db A vdb = db c = 3.83 khz = c = 3.83 khz.94 db 2 log 2 =.94 db 2.94 db 3 db = 4.94 db FIG Obtaining a db plot o A vdb db. Vi For the network o Fig , the gain A v / can also be ound in the ollowing manner: 2 2 j X C 2 j 2 j 2 /X C A v Vi 2 jx C j ( 2 ) X C j ( 2 )/X C j q 2 C j 2p 2 C j q( 2 )C j 2p ( 2 )C

40 56 DECIBELS, FILTES, AND BODE PLOTS db and A v j ( / ) (23.44) V j ( / c ) with and c 2p2 C 2p( 2 )C The bottom o Eq. (23.44) is a match o the denominator o the lowpass unction o Fig (a). The numerator, however, is a new unction that will deine a unique Bode asymptote that will prove useul or a variety o network conigurations. Applying Eq. (23.5): 2 log 2 log Vi ( c / ) 2 2 log ( / ) 2 log ( c / ) 2 Let us now consider speciic requencies or the irst term. At : 2 log 2 log db i At 2 : At : 2 2 log 2 log 2 6 db 2 log 2 log.5 6 db 2 log db 6 db/octave A db plot o 2 log ( / ) is provided in Fig Note that the asymptote passes through the -db line at and has a positive slope o 6 db/octave (or 2 db/decade) or requencies above and below or increasing values o. I we examine the original unction A v, we ind that the phase angle associated with j / / 9 is ixed at 9, resulting in a phase angle or A v o 9 tan ( / c ) tan ( c / ). Now that we have a plot o the db response or the magnitude o the unction /, we can plot the db response o the magnitude o A v using a procedure outlined by Fig Solving or and c : FIG db plot o / khz 2p2 C 2p(4 k)( nf) with c 3.83 khz 2p( 2 )C 2p(5 k)( nf) For this development the straight-line asymptotes or each term resulting rom the application o Eq. (23.5) will be drawn on the same requency axis to permit an examination o the impact o one line section on the other. For clarity, the requency spectrum o Fig has been divided into two regions. In region we have a -db asymptote and one increasing at 6 db/octave or increasing requencies. The sum o the two as deined by Eq. (23.5) is simply the 6-dB/octave asymptote shown in the igure.

41 db LOW-PASS FILTE WITH LIMITED ATTENUATION 57 A v db 2 c.94 db (log scale) 3 db Actual response FIG Plot o A v db or the network o Fig In region 2 one asymptote is increasing at 6 db, and the other is decreasing at 6 db/octave or increasing requencies. The net eect is that one cancels the other or the region greater than c, leaving a horizontal asymptote beginning at c.acareul sketch o the asymptotes on a log scale will reveal that the horizontal asymptote is at.94 db, as obtained earlier or the same unction. The horizontal level can also be determined by simply plugging c into the Bode plot deined by / ;that is, c 3.83 khz 2 log 2 log 2 log khz 2 log db The actual response can then be drawn using the asymptotes and the known dierences at c (3 db) and at.5 c or 2 c ( db). In summary, thereore, the same db response or A v / can be obtained by isolating the maximum value or deining the gain in a dierent orm. The latter approach permitted the introduction o a new unction or our catalog o idealized Bode plots that will prove useul in the uture LOW-PASS FILTE WITH LIMITED ATTENUATION Our analysis will now continue with the low-pass ilter o Fig , which has limited attentuation at the high-requency end. That is, the output will not drop to zero as the requency becomes relatively high. The ilter is similar in construction to Fig , but note that now includes the capacitive element. At Hz, the capacitor can assume its open-circuit equivalence, and. At high requencies the capacitor can be approximated by a short-circuit equivalence, and 2 C FIG Low-pass ilter with limited attenuation.

42 58 DECIBELS, FILTES, AND BODE PLOTS db 2 A plot o versus requency is provided in Fig (a). A sketch o A v / will appear as shown in Fig (b). 2 A v = c (log scale) c (log scale) (a) (b) FIG Low-pass ilter with limited attenuation. An equation or in terms o can be derived by irst applying the voltage divider rule: ( 2 jx C ) 2 jx C 2 jx C 2 /X C j and A v Vi 2 jx C ( 2 )/X C j ( j)( 2 X C j) ( j)(( 2 )/X C j) j( 2 /X C ) j(( 2 )/X C ) so that A v j ( / ) (23.45) Vi j ( / c) with and c 2p2 C 2p( 2 )C The denominator o Eq. (23.45) is simply the denominator o the low-pass unction o Fig (a). The numerator, however, is new and must be investigated. Applying Eq. (23.5): A vdb 2 log 2 log (/ ) 2 2 log Vi (/ c ) 2 For k, ( / ) 2 k, and the irst term becomes 2 log (/ ) 2 2 log (( / ) 2 ) /2 2 log ( / ) k which deines the idealized Bode asymptote or the numerator o Eq. (23.45). At, 2 log db, and at 2, 2 log 2 6 db. For requencies much less than, ( / ) 2 K, and the irst term o the Eq. (23.5) expansion becomes 2 log 2 log db, which establishes the low-requency asymptote. j 2p 2 C j 2p( 2 )C

43 db LOW-PASS FILTE WITH LIMITED ATTENUATION 59 The ull idealized Bode response or the numerator o Eq. (23.45) is provided in Fig log 2 Actual response 3 db 6 db/octave db FIG Idealized and actual Bode response or the magnitude o ( j (/ )). We are now in a position to determine A v db by plotting the asymptote or each unction o Eq. (23.45) on the same requency axis, as shown in Fig Note that c must be less than since the denominator o includes only 2, whereas the denominator o c includes both 2 and. A v = db 2 3 db c (log scale) Actual response 2 log 2 2 FIG A vdb versus requency or the low-pass ilter with limited attenuation o Fig Since 2 /( 2 ) will always be less than, we can use an earlier development to obtain an equation or the drop in db below the -db axis at high requencies. That is, 2 log 2 /( 2 ) 2 log /(( 2 )/ 2 ) 2 log 2 log (( 2 )/ 2 )

44 6 DECIBELS, FILTES, AND BODE PLOTS db 2 and 2 log 2 log 2 (23.46) 2 as shown in Fig In region o Fig , both asymptotes are at db, resulting in a net Bode asymptote at db or the region. At c, one asymptote maintains its -db level, whereas the other is dropping by 6 db/octave. The sum o the two is the 6-dB drop per octave shown or the region. In region 3 the 6-dB/octave asymptote is balanced by the 6-dB/octave asymptote, establishing a level asymptote at the negative db level attained by the c asymptote at. The db level o the horizontal asymptote in region 3 can be determined using Eq. (23.46) or by simply substituting into the asymptotic expression deined by c. The ull idealized Bode envelope is now deined, permitting a sketch o the actual response by simply shiting 3 db in the right direction at each corner requency, as shown in Fig The phase angle associated with A v can be determined directly rom Eq. (23.45). That is, 2 v tan / tan / c (23.47) A ull plot o v versus requency can be obtained by simply substituting various key requencies into Eq. (23.47) and plotting the result on a log scale. The irst term o Eq. (23.47) deines the phase angle established by the numerator o Eq. (23.45). The asymptotic plot established by the numerator is provided in Fig Note the phase angle o 45 at and the straight-line asymptote between / and. 9 v j v = 9 45 Actual phase angle v = FIG Phase angle or ( j ( / )). Now that we have an asymptotic plot or the phase angle o the numerator, we can plot the ull phase response by sketching the asymptotes or both unctions o Eq. (23.45) on the same graph, as shown in Fig The asymptotes o Fig clearly indicate that the phase angle will be in the low-requency range and (9 9 ) in the high-requency range. In region 2 the phase plot drops below due to the impact o the c asymptote. In region 4 the phase angle increases since the asymptote due to c remains ixed at 9, whereas that due to is increasing. In the midrange the plot due to is balancing the

45 db HIGH-PASS FILTE WITH LIMITED ATTENUATION 6 v (A v ) c / / c c 45 9 FIG Phase angle or the low-pass ilter o Fig continued negative drop due to the c asymptote, resulting in the leveling response indicated. Due to the equal and opposite slopes o the asymptotes in the midregion, the angles o and c will be the same, but note that they are less than 45. The maximum negative angle will occur between and c. The remaining points on the curve o Fig can be determined by simply substituting speciic requencies into Eq. (23.45). However, it is also useul to know that the most dramatic (the quickest) changes in the phase angle occur when the db plot o the magnitude also goes through its greatest changes (such as at and c ) HIGH-PASS FILTE WITH LIMITED ATTENUATION The ilter o Fig is designed to limit the low-requency attenuation in much the same manner as described or the low-pass ilter o the previous section. At Hz the capacitor can assume its open-circuit equivalence, and [ 2 /( 2 )]. At high requencies the capacitor can be approximated by a short-circuit equivalence, and. The resistance to be employed when determining c can be ound by inding the Thévenin resistance or the capacitor C, as shown in Fig A careul examination o the resulting coniguration will reveal that Th 2 and c /2p( 2 )C. A plot o versus requency is provided in Fig. 23.7(a), and a sketch o A v / appears in Fig. 23.7(b). An equation or A v / can be derived by irst applying the voltage divider rule: and A v Vi 2 2 jx C ( j XC) j X 2 2 jx C 2 2 ( jx C ) 2 ( jx C ) j X C 2 2 j 2 X C 2 j 2 X C j X C C C 2 FIG High-pass ilter with limited attenuation. Vi = V Th FIG Determining or the c calculation or the ilter o Fig

46 62 DECIBELS, FILTES, AND BODE PLOTS db A v = c 2 2 c (a) (b) FIG High-pass ilter with limited attenuation. j 2X C 2 2 j 2 X C j ( 2 ) 2 j( 2 )X C X C 2 j X C j X C j 2p C XC j O IXCOI j j 2 2 2p( 2 )C 2 so that A v j ( / ) (23.48) V j ( / ) with and c 2p C 2p( 2 )C i The denominator o Eq. (23.48) is simply the denominator o the high-pass unction o Fig (b). The numerator, however, is new and must be investigated. Applying Eq. (23.5): V A vdb 2 log o 2 log ( /) 2 2 log Vi (c / ) 2 For K, ( / ) 2 k, and the irst term becomes 2 log ( / ) 2 2 log ( / ) K which deines the idealized Bode asymptote or the numerator o Eq. (23.48). At, 2 log db At.5, 2 log 2 6 db At., 2 log 2 db For requencies greater than, / K and 2 log db, which establishes the high-requency asymptote. The ull idealized Bode plot or the numerator o Eq. (23.48) is provided in Fig c

47 We are now in a position to determine A vdb by plotting the asymptotes or each unction o Eq. (23.48) on the same requency axis, as shown in Fig Note that c must be more than since 2 must be less than. When determining the linearized Bode response, let us irst examine region 2, where one unction is db and the other is dropping at 6 db/octave or decreasing requencies. The result is a decreasing asymptote rom c to.atthe intersection o the resultant o region 2 with,weenter region, where the asymptotes have opposite slopes and cancel the eect o each other. The resulting level at is determined by 2 log ( 2 )/ 2,asound in earlier sections. The drop can also be determined by simply substituting into the asymptotic equation deined or c.inregion 3 both are at db, resultdb HIGH-PASS FILTE WITH LIMITED ATTENUATION 63 2 log 2 6 db/octave Actual response 3 db db (log scale) FIG Idealized and actual Bode response or the magnitude o ( j ( /)). A vdb 2 3 c (log scale) 2 log 2 2 Actual response FIG A vdb versus requency or the high-pass ilter with limited attenuation o Fig

48 64 DECIBELS, FILTES, AND BODE PLOTS db ing in a -db asymptote or the region. The resulting asymptotic and actual responses both appear in Fig The phase angle associated with A v can be determined directly rom Eq. (23.48); that is, v tan tan c (23.49) A ull plot o v versus requency can be obtained by simply substituting various key requencies into Eq. (23.49) and plotting the result on a log scale. The irst term o Eq. (23.49) deines the phase angle established by the numerator o Eq. (23.48). The asymptotic plot resulting rom the numerator is provided in Fig Note the leading phase angle o 45 at and the straight-line asymptote rom / to. v j 45 / v = Actual response (log scale) 9 v = 9 FIG Phase angle or ( j ( /)). Now that we have an asymptotic plot or the phase angle o the numerator, we can plot the ull phase response by sketching the asymptotes or both unctions o Eq. (23.48) on the same graph, as shown in Fig The asymptotes o Fig clearly indicate that the phase angle will be 9 in the low-requency range and (9 9 ) in the high-requency range. In region 2 the phase angle is increasing above because one angle is ixed at 9 and the other is becoming less negative. In region 4 one is and the other is decreasing, resulting in a decreasing v or this region. In region 3 the positive angle is always greater than the negative angle, resulting in a positive angle or the entire region. Since the slopes o the asymptotes in region 3 are equal but opposite, the angles at c and are the same. Figure reveals that the angle at c and will be less than 45. The maximum angle will occur between c and, as shown in the igure. Note again that the greatest change in v occurs at the corner requencies, matching the regions o greatest change in the db plot.

49 db HIGH-PASS FILTE WITH LIMITED ATTENUATION 65 v (A v ) / c / c c (log scale) 45 9 FIG Phase response or the high-pass ilter o Fig EXAMPLE 23. For the ilter o Fig : a. Sketch the curve o A vdb versus requency using a log scale. b. Sketch the curve o v versus requency using a log scale. Solutions: a. For the break requencies: 37.2 Hz 2p C 2p(9. k)(.47 mf) c Hz 2 2p(.9 k)(.47 mf) 2p C 2 9. k C.47 mf 2 k FIG Example 23.. The maximum low-level attentuation is 2 9. k k 2 log 2 log 2 k 2 log. 2.9 db The resulting plot appears in Fig b. For the break requencies: At 37.2 Hz, v tan tan c tan tan Hz 37.2 Hz At c Hz, v tan 37.2 Hz tan Hz

50 66 DECIBELS, FILTES, AND BODE PLOTS db A vdb 2 db 3 db Hz 37.2 Hz Hz c Hz 3 db Hz db (log scale) 7 db 2 db FIG A vdb versus requency or the ilter o Fig At a requency midway between c and on a log scale, or example, 2 Hz: v tan 37.2 Hz 2 Hz tan The resulting phase plot appears in Fig Hz 2 Hz v Hz 37.2 Hz 2 Hz Hz Hz c Hz (log scale) FIG v (the phase angle associated with A v ) versus requency or the ilter o Fig OTHE POPETIES AND A SUMMAY TABLE Bode plots are not limited to ilters but can be applied to any system or which a db-versus-requency plot is desired. Although the previous sections did not cover all the unctions that lend themselves to the idealized linear asymptotes, many o those most commonly encountered have been introduced. We now examine some o the special situations that can develop that will urther demonstrate the adaptability and useulness o the linear Bode approach to requency analysis.

51 db OTHE POPETIES AND A SUMMAY TABLE 67 In all the situations described in this chapter, there was only one term in the numerator or denominator. For situations where there is more than one term, there will be an interaction between unctions that must be examined and understood. In many cases the use o Eq. (23.5) will prove useul. For example, i A v should have the ormat 2( j 2 / )( j/ ) (a)(b)(c) A v (23.5) ( j / )( j/ 2 ) (d)(e) we can expand the unction in the ollowing manner: (a)(b)(c) A vdb 2 log (d)(e) 2 log a 2 log b 2 log c 2 log d 2 log e revealing that the net or resultant db level is equal to the algebraic sum o the contributions rom all the terms o the original unction. We will, thereore, be able to add algebraically the linearized Bode plots o all the terms in each requency interval to determine the idealized Bode plot or the ull unction. I two terms happen to have the same ormat and corner requency, as in the unction A v ( j / )( j / ) the unction can be rewritten as A v so that A vdb 2 log ( j / ) 2 ( ( /) 2 ) 2 2 log ( ( / ) 2 ) or K, ( / ) 2 k, and A vdb 2 log ( / ) 2 4 log / versus the 2 log ( / ) obtained or a single term in the denominator. The resulting db asymptote will drop, thereore, at a rate o 2 db/ octave (4 db/decade) or decreasing requencies rather than 6 db/octave. The corner requency is the same, and the high-requency asymptote is still at db. The idealized Bode plot or the above unction is provided in Fig Note the steeper slope o the asymptote and the act that the actual curve will now pass 6 dbbelow the corner requency rather than 3 db, as or a single term. Keep in mind that i the corner requencies o the two terms in the numerator or denominator are close but not exactly equal, the total db drop is the algebraic sum o the contributing terms o the expansion. For instance, consider the linearized Bode plot o Fig with corner requencies and 2. In region 3 both asymptotes are db, resulting in an asymptote at dbor requencies greater than 2.For region 2, one asymptote is at db, whereas the other drops at 6 db/octave or decreasing requencies. The net result or this region is an asymptote dropping at 6 db, as shown in the same igure. At,weind two asymptotes dropping o at 6 dbor decreasing requencies. The result is an asymptote dropping o at 2 db/octave or this region.

52 68 DECIBELS, FILTES, AND BODE PLOTS db A vdb 2 db 6 db/octave 6 db 6 db Actual response 2 db/octave 2 db 2 db FIG Plotting the linearized Bode plot o. ( j ( /)) 2 A vdb 2 db 3 db or 4 2 (2 octaves below) 6 db/octave 3 db or 4 2 (2 octaves below) 2 db/octave Actual response FIG Plot o A vdb or with < 2. ( j( /))( j( 2 /)) I and 2 are at least two octaves apart, the eect o one on the plotting o the actual response or the other can just about be ignored. In other words, or this example, i < 4 2, then the actual response will be down 3 db at 2 and. The above discussion can be expanded or any number o terms at the same requency or in the same region. For three equal terms in the denominator, the asymptote will drop at 8 db/octave, and so on. In time the procedure will be somewhat sel-evident and relatively straightorward to apply. In many cases the hardest part o inding a solution is to put the original unction in the desired orm.

53 db OTHE POPETIES AND A SUMMAY TABLE 69 EXAMPLE 23.2 A transistor ampliier has the ollowing gain: A v j 5 Hz j 2 Hz j khz j 2 khz a. Sketch the normalized response A v A v /A vmax, and determine the bandwidth o the ampliier. b. Sketch the phase response, and determine a requency where the phase angle is close to. Solutions: A a. A v A v v Avmax j 5 Hz j 2 Hz j khz j 2 khz (a)(b)(c)(d) a b c d and A vdb 2 log a 2 log b 2 log c 2 log d clearly substantiating the act that the total number o decibels is equal to the algebraic sum o the contributing terms. A careul examination o the original unction will reveal that the irst two terms in the denominator are high-pass ilter unctions, whereas the last two are low-pass unctions. Figure 23.8 demonstrates how the combination o the two types o unctions deines a bandwidth or the ampliier. The high-requency ilter unctions have deined the low cuto requency, and the low-requency ilter unctions have deined the high cuto requency. A v = A v = A v = A vmax.77a vmax High-pass A vmax.77a vmax Low-pass 2 A vmax.77a vmax BW 2 BW = 2 FIG Finding the overall gain versus requency or Example Plotting all the idealized Bode plots on the same axis will result in the plot o Fig Note or requencies less than 5 Hz that the resulting asymptote drops o at 2 db/octave. In addition, since 5 Hz and 2 Hz are separated by two octaves, the actual response will be down by only about 3 db at the corner requencies o 5 Hz and 2 Hz. For the high-requency region, the corner requencies are not separated by two octaves, and the dierence between the idealized plot

54 7 DECIBELS, FILTES, AND BODE PLOTS db A vdb Hz 3 db 6 db Hz Hz 2 Hz khz khz 2 khz khz BW 2 db 2 db FIG A vdb versus requency or Example and the actual Bode response must be examined more careully. Since khz is one octave below 2 khz, we can use the act that the dierence between the idealized response and the actual response or a single corner requency is db. I we add an additional -db drop due to the 2-kHz corner requency to the 3-dB drop at khz, we can conclude that the drop at khz will be 4 db, as shown on the plot. To check the conclusion, let us write the ull expression or the db level at khz and ind the actual level or comparison purposes. A vdb 2 log 5 H k H z 2 z 2 log 2 k H H z 2 z 2 log k k H H z 2 z 2 log 2 k k H H z z 2. db.7 db 3. db.969 db 3.98 db 4 db as beore An examination o the above calculations clearly reveals that the last two terms predominate in the high-requency region and essentially eliminate the need to consider the irst two terms in that region. For the low-requency region an examination o the irst two terms is suicient. Proceeding in a similar ashion, we ind a 4-dB dierence at 2 khz, resulting in the actual response appearing in Fig Since the bandwidth is deined at the 3-dB level, a judgment must be made as to where the actual response crosses the 3-dB level in the high-requency region. A rough sketch suggests that it is near 8.5 khz. Plugging this requency into the high-requency terms results in A vdb 2 log 8.5 khz 2 2 log 8.5 khz 2 khz 2 khz 2.48 db.645 db 2.8 db

55 db OTHE POPETIES AND A SUMMAY TABLE 7 which is relatively close to the 3-dB level, and BW high low 8.5 khz 2 Hz 8.3 khz In the midrange o the bandwidth, A vdb will approach db. At khz: A vdb 2 log 5 Hz 2 2 log 2 Hz 2 khz khz 2 log khz 2 2 log khz 2 khz 2 khz.8 db.73 db.432 db.8 db.235 db 5 db which is certainly close to the -db level, as shown on the plot. b. The phase response can be determined by simply substituting a number o key requencies into the ollowing equation, derived directly rom the original unction A v : v tan 5 Hz tan 2 Hz tan tan khz 2 khz However, let us make ull use o the asymptotes deined by each term o A v and sketch the response by inding the resulting phase angle at critical points on the requency axis. The resulting asymptotes and phase plot are provided in Fig Note that at 5 Hz, the sum o the two angles determined by the straight-line asymptotes is (actual 2 ). At khz, i we subtract 5.7 or one corner requency, we obtain a net angle o (actual 5.6 ). 8 Hz Hz Hz 2 Hz khz khz 2 khz khz FIG Phase response or Example At khz the asymptotes leave us with v (actual 7.56 ). The net phase plot appears to be close to at about 3 Hz. As a check on our assumptions and the use o the asymptotic approach, let us plug in 3 Hz into the equation or v:

56 72 DECIBELS, FILTES, AND BODE PLOTS db v tan 5 Hz tan 2 Hz tan 3 Hz 3 Hz tan 3 Hz 3 Hz khz 2 khz as predicted In total, the phase plot appears to shit rom a positive angle o 8 ( leading )toanegative angle o 8 as the requency spectrum extends rom very low requencies to high requencies. In the midregion the phase plot is close to ( in phase with ), much like the response to a common-base transistor ampliier. In an eort to consolidate some o the material introduced in this chapter and provide a reerence or uture investigations, Table 23.2 was developed; it includes the linearized db and phase plots or the unctions appearing in the irst column. These are by no means all the unctions encountered, but they do provide a oundation to which additional unctions can be added. eviewing the development o the ilters o Sections 23.2 and 23.3, it is probably evident that establishing the unction A v in the proper orm is the most diicult part o the analysis. However, with practice and an awareness o the desired ormat, methods will surace that will signiicantly reduce the eort involved COSSOVE NETWOKS The topic o crossover networks is included primarily to present an excellent demonstration o ilter operation without a high level o complexity. Crossover networks are employed in audio systems to ensure that the proper requencies are channeled to the appropriate speaker. Although less expensive audio systems have to rely on one speaker to cover the ull audio range rom about 2 Hz to 2 khz, better systems will employ at least three speakers to cover the low range (2 Hz to about 5 Hz), the midrange (5 Hz to about 5 khz), and the high range (5 khz and up). The term crossover comes rom the act that the system is designed to have a crossover o requency spectrums or adjacent speakers at the 3-dB level, as shown in Fig Depending on the design, each ilter can drop o at 6 db, 2 db, or 8 db, with complexity increasing with the desired db drop-o rate. The three-way crossover network o Fig is quite simple in design, with a lowpass -L ilter or the wooer, an -L-C pass-band ilter or the midrange, and a high-pass -C ilter or the tweeter. The basic equations or the components are provided below. Note the similarity between the equations, with the only dierence or each type o element being the cuto requency. L low 2p L mid 2p 2 (23.5) C mid 2p C high 2p 2 (23.52)

57 db COSSOVE NETWOKS 73 TABLE 23.2 Idealized Bode plots or various unctions. Function db Plot Phase Plot A v j A vdb v ( leads ) 6 db/octave 9 45 db / A v j A vdb db 6 db/octave 9 45 v ( leads ) A v j A vdb 6 db/octave 9 45 v ( leads ) 9 A v j c A vdb 6 db/octave c db 9 45 v ( leads ) 9 45 A v j c A vdb db c 6 db/octave v ( leads ) 2 /

58 74 DECIBELS, FILTES, AND BODE PLOTS db db 4 Hz 3 db db 3 db 4 Hz L low = 3.3 mh 8 5 khz 3 db C mid = 47 F L mid = 27 H 8 db 5 khz 3 db C high = 3.9 F 8 FIG Three-way, 6-dB-per-octave, crossover network. For the crossover network o Fig with three 8- speakers, the resulting values are 8 L low 3.83 mh 3.3 mh 2p 2p(4 Hz) (commercial value) 8 L mid mh 27 mh 2p2 2p(5 khz) (commercial value) C mid mf 47 mf 2p 2p(4 Hz)(8 ) (commercial value) C high mf 3.9 mf 2p2 2p(5 khz)(8 ) (commercial value) as appearing on Fig For each ilter, a rough sketch o the requency response is included to show the crossover at the speciic requencies o interest. Because all three speakers are in parallel, the source voltage and impedance or each are the same. The total loading on the source is obviously a unction o the requency applied, but the total delivered is determined solely by the speakers since they are essentially resistive in nature. To test the system, let us apply a 4-V signal at a requency o khz (a predominant requency o the typical human auditory response curve) and see which speaker will have the highest power level. At khz, X Llow 2pL low 2p( khz)(3.3 mh) 2.74 (Z )( ) (8 )(4 V ) ZT 8 j V68.9

59 db APPLICATIONS 75 X Lmid 2pL mid 2p( khz)(27 mh).696 X Cmid pCmid 2p( khz)(47 mf) (Z )( ) ZT (8 )(4 V ) 8 j.696 j V.93 X Chigh 4.8 2pChigh 2p( khz)(3.9 mf) (Z )( ) (8 )(4 V ) ZT 8 j V 78.9 Using the basic power equation P V 2 /, thepower to the wooer is V 2 (.44 V) 2 P low.259 W 8 to the midrange speaker, V 2 (3.94 V) 2 P mid.94 W 8 and to the tweeter, V 2 (.77 V) 2 P high.74 W 8 resulting in a power ratio o 7.5 between the midrange and the wooer and 26 between the midrange and the tweeter. Obviously, the response o the midrange speaker will totally overshadow the other two. 5-ohm coax attenuator IN db 3 db db 2 db 5 db 2 db FIG Passive coax attenuator. OUT 23.6 APPLICATIONS Attenuators Attenuators are, by deinition, any device or system that can reduce the power or voltage level o a signal while introducing little or no distortion. There are two general types: passive and active. The passive type uses only resistors, while the active type uses electronic devices such as transistors and integrated circuits. Since electronics is a subject or the courses to ollow, our attention here will be only on the resistive type. Attenuators are commonly used in audio equipment (such as the graphic and parametric equalizers introduced in the previous chapter), antenna systems, AM or FM systems where attenuation may be required beore the signals are mixed, and any other application where a reduction in signal strength is required. The unit o Fig has coaxial input and output terminals and switches to set the level o db reduction. It has a lat response rom dc to about 6 GHz, which essentially means that its introduction into the network will not aect the requency response or this band o requencies. The design is rather simple with resistors connected in either a tee (T) or a wye (Y) coniguration as shown in Figs and 23.86, respectively, or a 5- coaxial system. In each case the resistors are 2 Attentuation 2 db 2 db 3 db 5 db db 2 db 2.9 Ω 5.7 Ω 8.5 Ω 4. Ω 26. Ω 4. Ω FIG Tee (T) coniguration Ω 25.2 Ω 4.9 Ω 82.2 Ω 35. Ω. Ω

60 76 DECIBELS, FILTES, AND BODE PLOTS db 2 Attentuation 2 db 2 db 3 db 5 db db 2 db 87. Ω 436. Ω 292. Ω 78.6 Ω 96.2 Ω 6. Ω FIG Wye (Y) coniguration. 5.8 Ω.6 Ω 7.6 Ω 3.4 Ω 7.2 Ω Ω chosen to ensure that the input impedance and output impedance match the line. That is, the input and output impedances o each coniguration will be 5. For a number o db attenuations, the resistor values or the T and Y are provided in Figs and Note in each design that two o the resistors are the same, while the third is a much smaller or larger value. For the -db attenuation, the resistor values were inserted or the T coniguration in Fig (a). Terminating the coniguration with a 5- load, we ind through the ollowing calculations that the input impedance is, in act, 5 : i 2( L ) (2.9 5 ) Looking back rom the load as shown in Fig (b) with the source set to zero volts, we ind through the ollowing calculations that the output impedance is also 5 : o 2( s ) (2.9 5 ) s 2.9 i = 5 Ω L V s = V o = 5 Ω -db attenuator (a) (b) FIG db attenuator: (a) loaded; (b) inding o. In Fig , a 5- load has been applied, and the output voltage is determined as ollows: 2( L ) 47.4 rom above and V V 2 s 47.4 V s V s with LV 2 5 (.942V V L s ) L.89V s s V s L 5 V L FIG Determining the voltage levels or the -db attenuator o Fig (a).

61 db APPLICATIONS 77 Calculating the drop in db will result in the ollowing: A db 2 log V L 2 log.8 9V s Vs Vs 2 log.89. db substantiating the act that there is a -db attenuation. As mentioned earlier, there are other methods or attenuation that are more sophisticated in design and beyond the scope o the coverage o this text. However, the above designs are quite eective and relatively inexpensive, and they perorm the task at hand quite well. Noise Filters Noise is a problem that can occur in any electronic system. In general, it is the presence o any unwanted signal that can aect the overall operation o a system. It can come rom a power source (6-Hz hum), rom eedback networks, rom mechanical systems connected to electrical systems, rom stray capacitive and inductive eects, or possibly rom a local signal source that is not properly shielded the list is endless. The manner in which the noise is eliminated or handled is normally analyzed by someone with a broad practical background and with a sense or the origin or the unwanted noise and how to remove it in the simplest and most direct way. In most cases the problem will not be part o the original design but a second eort in the testing phase to remove unexpected problems. Although sophisticated methods can be applied when the problem can be serious in nature, most situations are handled simply by the proper placement o an element or two o a value sensitive to the problem. In Fig two capacitors have been strategically placed in the tape recording and playback sections o a tape recorder to remove the C n pf C Short circuit to high-requency noise Compensation control o high requencies ecording head 2 CC Applied signal P Coupling capacitor 3 P Playback network Playback head 3 P P Filter to reduce stray pickup s pf C C C s Ampliier and notch ilter ecord phase Coupling capacitor (a) Playback phase (b) FIG Noise reduction in a tape recorder.

62 78 DECIBELS, FILTES, AND BODE PLOTS db undesirable high-requency noise (rushing sound) that can result rom unexpected, randomly placed particles on a magnetic tape, noise coming down the line, or noise introduced rom the local environment. During the record mode, with the switches in the positions shown (), the -pf capacitor at the top o the schematic will act as a short circuit to the high-requency noise. The capacitor C is included to compensate or the act that recording on a tape is not a linear process versus requency. In other words, certain requencies are recorded at higher amplitudes than others. In Fig a sketch o recording level versus requency has been provided, clearly indicating that the human audio range o about 4 Hz to 2 khz is very poor or the tape recording process, starting to rise only ater 2 khz. Thus, tape recorders must include a ixed biasing requency which when added to the actual audio signal will bring the requency range to be ampliied to the region o high-amplitude recording. On some tapes the actual bias requency is provided, while on others the phrase normal bias is used. Even ater you pass the bias requency, there is a requency range that ollows that drops o considerably. Compensation or this drop-o is provided by the parallel combination o the resistor and the capacitor C mentioned above. At requencies near the bias requency, the capacitor is designed to act essentially like an open circuit (high reactance), and the head current and voltage are limited by the resistors and 2. At requencies in the region where the tape gain drops o with requency, the capacitor begins to take on a lower reactance level and reduce the net impedance across the parallel branch o and C. The result is an increase in head current and voltage due to the lower net impedance in the line, resulting in a leveling in the tape gain ollowing the bias requency. Eventually, the capacitor will begin to take on the characteristics o a short circuit, eectively shorting out the resistance, and the head current and voltage will be a maximum. During playback this bias requency is eliminated by a notch ilter so that the original sound is not distorted by the highrequency signal. ecording level High requency drop o 2 khz 3 khz Bias requency FIG Noise reduction in a tape recorder. During playback (P), the upper circuit o Fig is set to ground by the upper switch, and the lower network comes into play. Again note the second -pf capacitor connected to the base o the transistor to short to ground any undesirable high-requency noise. The resistor is there to absorb any power associated with the noise signal when the capacitor takes on its short-circuit equivalence. Keep in mind that the capacitor was chosen to act as a short-circuit equivalent or a particular requency range and not or the audio range where it is essentially an open circuit.

63 db APPLICATIONS 79 Alternators in a car are notorious or developing high-requency noise down the line to the radio, as shown in Fig. 23.9(a). This problem is usually alleviated by placing a high-requency ilter in the line as shown. The inductor o H will oer a high impedance or the range o noise requencies, while the capacitor ( mf to 47, mf) will act as a short-circuit equivalent to any noise that happens to get through. For the speaker system in Fig. 23.9(b), the push-pull power arrangement o transistors in the output section can oten develop a short period o time between pulses where the strong signal voltage is zero volts. During this short period the coil o the speaker rears its inductive eects, sees an unexpected path to ground like a switch opening, and quickly cuts o the speaker current. Through the amiliar relationship v L L(di L /dt), an unexpected voltage will develop across the coil and set a high-requency oscillation on the line that will ind its way back to the ampliier and cause urther distortion. This eect can be subdued by placing an -C path to ground that will oer a low-resistance path rom the speaker to ground or a range o requencies typically generated by this signal distortion. Since the capacitor will assume a short-circuit equivalence or the range o noise disturbance, the resistor was added to limit the current and absorb the energy associated with the signal noise. V Push-pull response t High-requency noise L C C V s V Car alternator H C adio ( µf to 47, µf) Push-pull ampliier C b (a) (b) Short-circuit path or unwanted high-requency oscillation FIG Noise generation: (a) due to a car alternator; (b) rom a push-pull ampliier. In regulators, such as the 5-V regulator o Fig (a), when a spike in current comes down the line or any number o reasons, there will be a voltage drop along the line, and the input voltage to the regulator will drop. The regulator, perorming its primary unction, will sense this drop in input voltage and will increase its ampliication level through a eedback loop to maintain a constant output. However, the spike is o such short duration that the output voltage will have a spike o its own because the input voltage has quickly returned to its normal level, and with the increased ampliication the output will jump to a higher level. Then the regulator senses its error and quickly cuts its gain. The sensitivity to changes in the input level has caused the output level to go through a number o quick oscillations that can be a real

64 8 DECIBELS, FILTES, AND BODE PLOTS db i 5 V High-requency noise i 5 V i Input t 5 V egulator t Feedback Output i t Input 5 V egulator Output High-requency µf noise stabilizer bypass to ground (Open-circuit or 5 V dc level) (a) Filter (b) FIG egulator: (a) eect o spike in current on the input side; (b) noise reduction. problem or the equipment to which the dc voltage is applied: A highrequency noise signal has been developed. One way to subdue this reaction and, in act, slow the system response down so that very short interval spikes have less impact is to add a capacitor across the output as shown in Fig (b). Since the regulator is providing a ixed dc level, a large capacitor o mf can be used to short-circuit a wide range o high-requency disturbances. However, you don t want to make the capacitor too large or you ll get too much damping, and large overshoots and undershoots can develop. To maximize the input o the added capacitor, you must place it physically closer to the regulator to ensure that noise is not picked up between the regulator and capacitor and to avoid developing any delay time between output signal and capacitive reaction. In general, as you examine the schematic o working systems and see elements that don t appear to be part o any standard design procedure, you can assume that they are either protective devices or due to noise on the line that is aecting the operation o the system. Noting their type, value, and location will oten reveal their purpose and modus operandi COMPUTE ANALYSIS PSpice High-Pass Filter The computer analysis will begin with an investigation o the high-pass ilter o Fig The cuto requency is determined by c /2pC.592 khz, with the voltage across the resistor approaching V at high requencies at a phase angle o. For this analysis, the ac voltage source VAC was used. Within the Property Editor, the quantities deined appear next to the source in Fig Otherwise, building the circuit is quite straightorward. Our interest lies in the eect o requency on the magnitude o the output voltage across the resistor and the resulting phase angle. Following the selection o AC Sweep under the Analysis type heading, the Start Frequency should be set at Hz so that we have some data points at the very low end and an End Frequency o khz so that we extend well into the high requencies. Here is the obvious advantage

65 db COMPUTE ANALYSIS 8 FIG High-pass -C ilter to be investigated using PSpice. o the logarithmic axis: a wide range o values on a plot o limited size. To ensure suicient data points, the Points/Decade was set at k. When V(:) is selected, the resulting SCHEMATIC appears as shown in Fig Note the nice transition rom one region to the other. At low requencies when the reactance o the capacitor ar outweighs that o the resistor, most o the applied voltage appears across the capacitor, and very little across the resistor. At much higher requencies the reactance o the capacitor drops o very quickly, and the voltage across the resistor picks up toward a maximum o V. Select Plot-Add Plot to Window-Trace-Add Trace-P(V(:))- OK, and the phase plot o Fig results showing a shit rom 9, when the network is highly capacitive in nature, to, when it becomes resistive. I we select the phase plot SEL>> and click on the Toggle cursor pad, a let click will place a cursor on the screen that can deine the requency at which the phase angle is 45. At 45.2, which is the closest we can come with the available data points, we ind that the corresponding requency is.585 khz which is a very close match with the.592 khz calculated above. The right-click cursor can be placed at khz to show that the phase angle has dropped to.9, which certainly deines the network as resistive at this requency. Double-Tuned Filter Our analysis will now turn to a airly complexlooking ilter or which an enormous amount o time would be required to generate a detailed plot o gain versus requency using a handheld calculator. It is the same ilter examined in Example 23.9, so we have a chance to test our theoretical solution. The schematic appears in Fig with VAC again chosen since the requency range o interest will be set by the Simulation Proile. Again, the attributes or the source

66 82 DECIBELS, FILTES, AND BODE PLOTS db FIG Magnitude and phase plot or the high-pass -C ilter o Fig FIG Using PSpice to analyze a double-tuned ilter.

67 db COMPUTE ANALYSIS 83 were set in the Property Editor box rather than by selecting the components rom the screen. Note the need or the two resistors in series with the inductors since inductors cannot be considered ideal elements. The small value o the resistive elements, however, will have no eect on the results obtained. In the Simulation Settings dialog box, AC Sweep was again selected with a Start Frequency o Hz and an End Frequency o MHz (be sure to enter this value as MEGHZ) to ensure that the ull-range eect is provided. We can then use the axis controls to close in on the desired plot. The Points/Decade remains at k, although with this range o requencies it may take a ew seconds to simulate. Once the SCHEMATIC appears, Trace-Add Trace-V(L:)-OK will result in the plot o Fig Quite obviously there is a reject-band around 2 khz and a pass-band around 6 khz. Isn t it interesting that up to khz we have another pass-band as the inductor L p provides an almost direct path o low impedance rom input to output. At requencies approaching MHz, there is a continous stop-band due to the open-circuit equivalence o the L p inductor. Using the cursor option, we can place the let-click cursor on the minimum point o the graph by using the Cursor Trough key pad (the second pad to the right o the Toggle cursor pad). The right click can be used to identiy the requency o the maximum point on the curve near 6 khz. The results appearing in the Probe Cursor box clearly support our theoretical calculations o 2 khz or the band-stop minimum (A 2.2 khz with a magnitude o essentially V) and 6 khz or the pass-band maximum (A khz with a magnitude o V). FIG Magnitude plot versus requency or the voltage across L o the network o Fig