Decibels, Filters, and Bode Plots

Transcription

1 Decibels, Filters, and Bode Plots 2 Objectives Develop conidence in the use o logarithms and decibels in the description o power and voltage levels. Become amiliar with the requency response o high- and low-pass ilters. Learn to calculate the cuto requency and describe the phase response. Be able to calculate the cuto requencies and sketch the requency response o a pass-band, stop-band, or double-tuned ilter. Develop skills in interpreting and establishing the Bode response o any ilter. Become aware o the characteristics and operation o a crossover network. 2. 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: e2 4 where e To ind the power x to satisy the equation 2 () x you can determine the value o x using logarithms in the ollowing manner: x log revealing that db

2 98 DEIBELS, 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, IN ). In review, thereore, LOG ) or base e (natural logarithm, I N b2 x, then x log b N. (2.) The base to be used 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 (2.2) In this chapter, we concentrate solely on the common logarithm. However, a number o the conclusions are also applicable to natural logarithms. Some Areas o Application The ollowing are some o the most common applications o the logarithmic unction:. The response o a system can be plotted 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 is demonstrated in an example to ollow. Graphs Graph paper is available in semilog and log-log varieties. Semilog paper has only one log scale, with the other a linear scale. Both scales o loglog paper are log scales. A section o semilog paper appears in Fig. 2.. 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 is represented 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 lines 2, 3, 5, and 8 is particularly useul when the various lines o a log plot are let unnumbered.

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% Log scale Since log log log 2 log 3 FIG. 2. Semilog graph paper. the spacing between and, and, and, and so on, is the same as shown in Figs. 2. and 2.2. Note in Figs. 2. and 2.2 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. 2.2, with particular reerence to the 3%, 5%, 7%, and 9% levels o each interval 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. 2.2 Frequency log scale.

4 92 DEIBELS, FILTES, AND BODE PLOTS db x d d 2 x 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 (2.3) FIG. 2.3 Finding a value on a log plot. 2 7/ 6 " 3/ 4 " FIG. 2.4 Example The derivation o Eq. (2.3) is simply an extension o the details regarding distance appearing in Fig. 2.. EXAMPLE 2. Determine the value o the point appearing on the logarithmic plot in Fig. 2.4 using the measurements made by a ruler (linear). Solution: Using a calculator: d 7> d 2 3>4.75 d >d Applying Eq. (2.3): 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 (2.4) 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 (2.5) 4. The log o the quotient o two numbers is the log o the numerator minus the log o the denominator. log a b log a log b (2.6)

5 db DEIBELS 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 (2.7) alculator Functions Using the TI-89 calculator, the common logarithm o a number is determined by irst selecting the ATALOG key and then scrolling to ind the common logarithm unction. The time involved in scrolling through the options can be reduced by irst selecting the key with the irst letter o the desired unction in this case, L as shown below to ind the common logarithm o the number 8. ATALOG L 4 LOG ( ENTE 8 ) ENTE.93 For the reverse process, to ind N, or the antilogarithm, use the unction. In this case, the unction is ound ater the letter Z in the catalog. It is easier i you irst select A to start at the top o the listing and then go backwards to the power o ten unction by using the upward moving scroll option. The antilogarithm o the number.6 is ound as ollows: ATALOG A ^ ( ENTE. 6 ) ENTE 3.98 EXAMPLE 2.2 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 4 log e. log 4 4 log 4() DEIBELS Power Gain Two levels o power can be compared using a unit o measure called the bel, which is deined by the ollowing equation: B log P 2 P (bels) (2.8)

6 922 DEIBELS, FILTES, AND BODE PLOTS db However, to provide a unit o measure o less magnitude, a decibel is deined, where bel decibels db2 (2.9) The result is the ollowing important equation, which compares power levels P 2 and P in decibels: db log P 2 P (decibels, db) (2.) 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 Thereore, or a speaker system, a 3 db increase in output requires 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: For P 2 P, db log P 2 P log 2 3 db db log P 2 P log 6 8 log 2 3 db db log P 2 P log 32 8 log 4 6 db db log P 2 P log 2 db 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) Eq. (2.) is then typically written as db m log P mw ` 6 (2.)

7 db DEIBELS 923 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 log db m 4 mw mw log 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 is also 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 and P V 2 2 V 2 2 > 2 > into Eq. (2.) results in db log P 2 log V 2 2 > 2 P V 2 > log V >V log 2 > a V 2 2 b V log a 2 b and db 2 log V 2 V log 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 drops out (log ), and db y 2 log V 2 V (db) (2.2) Note the subscript y to deine the decibel level obtained. EXAMPLE 2.3 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 y 2 log 2 log.2 V 2 mv 2 log db or a voltage gain A y / o 6.

8 924 DEIBELS, FILTES, AND BODE PLOTS db EXAMPLE 2.4 I a system has a voltage gain o 36 db, ind the applied voltage i the output voltage is 6.8 V. Solution: db y 2 log 36 2 log From the antilogarithm:.8 log and V 7.77 mv TABLE 2. / db 2 log ( / ) db 2 6 db 2 db 2 26 db 4 db, 6 db, db Table 2. 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. 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 is 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. (2.7). 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: db s 2 log P.2 mbar (2.3) where P is the sound pressure in microbars. The decibel levels in Fig. 2.5 are deined by Eq. (2.3). Meters designed to measure audio levels are calibrated to the levels deined by Eq. (2.3) and shown in Fig. 2.5.

9 db DEIBELS 925 Output Power. db s Average value 6 in watts Jet engine ommunity siren Pneumatic riveter Threshold o pain μbar o pressure Loud rock band, punch press hain 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. 2.5 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 in 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 doubles or every db change in audio level a conclusion somewhat veriied by the examples to the right in Fig 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. In other words, doubling the sound level available rom a W acoustical source requires 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

10 926 DEIBELS, FILTES, AND BODE PLOTS db DB mw, DB 2.5 3VA 3 FIG. 2.6 Deining the relationship between a db scale reerenced to mw, 6 Ω and a 3 V rms voltage scale. 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 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 used, which is sometimes printed on the ace o the meter but is 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 / reveals that mw across a 6 load is the same as applying.775 V rms across a 6 load; that is, V 2P 2 mw 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 results 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, results in a db level o db 2 log V 2 /V 2 log.5 V/.775 V 3.8 db, also shown on the scale in Fig Although a reading o db reveals that the power level is times the reerence, don t assume that a reading o 5 db means that the output level is 5 mw. 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. 2.4 FILTES Any combination o passive (, L, and ) and/or active (transistors or operational ampliiers) elements designed to select or reject a band o requencies is called a ilter. In communication systems, ilters are used to pass those requencies containing the desired inormation and to reject the remaining requencies. In stereo systems, ilters can isolate particular bands o requencies or increased or decreased emphasis by the output acoustical system (ampliier, speaker, and so on). Filters are used 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 elements.

11 db FILTES Active ilters are ilters that employ active devices such as transistors and operational ampliiers in combination with, L, and elements. Since this text is limited to passive devices, the analysis o this chapter is limited to passive ilters. In addition, only the most undamental orms are 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 stopband, 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 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. 2.7 Deining the our broad categories o ilters.

12 928 DEIBELS, FILTES, AND BODE PLOTS db 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 pass-band 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 in Fig. 2.7 is discussed in some detail in the sections to ollow. Take particular note o the relative simplicity o some o the designs. FIG. 2.8 Low-pass ilter. = FIG low-pass ilter at low requencies. = V LOW-PASS FILTE The - 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. 2.8, it responds 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 is that o a high-pass ilter. A glance at Fig. 2.7(a) reveals that the circuit should behave in a manner that results 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 2p and the open-circuit equivalent can be substituted or the capacitor, as shown in Fig. 2.9, resulting in. At very high requencies, the reactance is X 2p and the short-circuit equivalent can be substituted or the capacitor, as shown in Fig. 2., resulting in V. A plot o the magnitude o versus requency results in the curve in Fig. 2.. Our next goal is now clearly deined: Find the requency at which the transition takes place rom a pass-band to a stop-band. FIG low-pass ilter at high requencies. =.77 Pass-band c Stop-band (log scale) FIG. 2. versus requency or a low-pass - ilter.

13 db - LOW-PASS FILTE 929 For ilters, a normalized plot is used more oten than the plot o versus requency in Fig. 2.. 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 in Fig. 2. with the applied voltage or the requency range o interest. Since the maximum value o or the low-pass ilter in Fig. 2.8 is, each level o in Fig. 2. is divided by the level o. The result is the plot o A y / in 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. 2.2 Normalized plot o Fig. 2.. At any intermediate requency, the output voltage in Fig. 2.8 can be determined using the voltage divider rule: or and A y X 9 j X X 9 j X X X 2 ltan X >2 A y X 2 2 X 2 9 tan a X b The magnitude o the ratio / is thereore determined by A y X 2 2 X 2 B a X b 2 (2.4) and the phase angle is determined by u 9 tan X tan X (2.5) For the special requency at which X, the magnitude becomes

14 93 DEIBELS, FILTES, AND BODE PLOTS db A y B a 2 b X which deines the critical or cuto requency in Fig The requency at which X is determined by 2p c and c (2.6) 2p The impact o Eq. (2.6) extends beyond its relative simplicity. For any low-pass ilter, the application o any requency less than c results 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 2 2 X u d V X i c X u d 2 X and 2 2 X u 2 The angle u is, thereore, the angle by which leads. Since u tan /X is always negative (except at Hz), it is clear that will always lag, leading to the label lagging network or the network in Fig At high requencies, X is very small and /X is quite large, resulting in u tan /X approaching 9. At low requencies, X is quite large and /X is very small, resulting in u approaching. At X, or c, tan /X tan 45. A plot o u versus requency results in the phase plot in Fig The plot is o leading, but since the phase angle is always negative, the phase plot in Fig. 2.4 ( 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 at lower requencies. v ( leads ) Pass-band c Stop-band (log scale) 45 9 FIG. 2.3 Angle by which leads.

15 db - LOW-PASS FILTE 93 v ( lags ) 9 45 Pass-band c Stop-band (log scale) FIG. 2.4 Angle by which lags. In summary, or the low-pass - ilter in Fig. 2.8: c 2p For 6 c, 7.77 whereas or 7 c, 6.77 At c, lags by 45 The low-pass ilter response in Fig. 2.7(a) can also be obtained using the -L combination in Fig. 2.5 with L c 2pL (2.7) In general, however, the - 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 can be an exercise or independent study. FIG. 2.5 Low-pass -L ilter. EXAMPLE 2.5 a. Sketch the output voltage versus requency or the low-pass - ilter in Fig b. Determine the voltage at khz and MHz, and compare the results to the results obtained rom the curve in part (a). c. Sketch the normalized gain A y /. Solutions: = 2 V k 5 pf a. Eq. (2.6): c 2p 38.3 khz 2p k25 pf2 FIG. 2.6 Example 2.5. At c,.77(2 V) 4.4 V. See Fig b. Eq. (2.4): At khz: B a 2 b X X 2p 3.8 k 2p khz25 pf2

16 932 DEIBELS, FILTES, AND BODE PLOTS db = 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. 2.7 Frequency response or the low-pass - network in Fig and At MHz: and 2 V 9.8 V B a k k b X 2p.32 k 2p MHz25 pf2 2 V 6. V B a k 2.32 k b Both levels are veriied by Fig c. Dividing every level in Fig. 2.7 by 2 V results in the normalized plot in Fig A v = khz khz khz c MHz MHz (log scale) FIG. 2.8 Normalized plot o Fig HIGH-PASS FILTE FIG. 2.9 High-pass ilter. As noted in Section 2.5, a high-pass - 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. At Hz, the reactance o the capacitor is quite high, and the opencircuit equivalent can be substituted, as shown in Fig In this case, V.

17 db - HIGH-PASS FILTE 933 = = V FIG high-pass ilter at very high requencies. FIG high-pass ilter at Hz. A plot o the magnitude versus requency is provided in Fig. 2.22, with the normalized plot in Fig = =.77 Stop-band c Pass-band (log scale) FIG versus requency or a high-pass - ilter. A v =.77 Stop-band c Pass-band (log scale) FIG Normalized plot o Fig At any intermediate requency, the output voltage can be determined using the voltage divider rule: or and j X j X 2 2 X 2 tan X >2 2 2 X 2 tan X >2 The magnitude o the ratio / is thereore determined by A y 2 2 X 2 B a X b 2 (2.8)

18 934 DEIBELS, FILTES, AND BODE PLOTS db and the phase angle u by u tan X (2.9) For the requency at which X, the magnitude becomes B a X b as shown in Fig The requency at which X is determined by X 2p c and c (2.2) 2p For the high-pass - ilter, the application o any requency greater than c results 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, and the ratio X / approaches zero with tan (X /) approaching, as shown in Fig At low requencies, the ratio X / becomes quite large, and tan (X /) approaches 9. For the case X, tan (X /) tan 45. Assigning a phase angle o to such that, the phase angle associated with is u, resulting in u and revealing that u is the angle by which leads. Since the angle u is the angle by which leads throughout the requency range in Fig. 2.24, the high-pass - ilter is reerred to as a leading network. 9 v ( leads ) 45 Stop-band c Pass-band (log scale) FIG Phase-angle response or the high-pass - ilter. In summary, or the high-pass - ilter: c 2p For 6 c, 6.77 whereas or 7 c, 7.77 At c, leads by 45

19 db - HIGH-PASS FILTE 935 The high-pass ilter response in Fig can also be obtained using the same elements in Fig. 2.5 but interchanging their positions, as shown in Fig L EXAMPLE 2.6 Given 2 k and 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 in part (a). c. Determine the magnitude and phase o A y / at 2 c or the high-pass ilter. FIG High-pass -L ilter. Solutions: a. c 2p 2p22 k22 pf Hz The normalized plots appear in Fig b. The phase plots appear in Fig 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. c. 2 c Hz Hz 2 X 2p 2p Hz22 pf2 4 k

20 936 DEIBELS, FILTES, AND BODE PLOTS db A y B a X b k a B 2 k b u tan X tan 4 k 2 k tan and A y PASS-BAND FILTES A number o methods are used to establish the pass-band characteristic in Fig. 2.7(c). One method uses both a low-pass and a high-pass ilter in cascade, as shown in Fig High-pass Low-pass ilter ilter FIG Pass-band ilter. 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 lowpass 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 pass-band passes through both ilters with very little degeneration. Low-pass V max.77v max BW High-pass c o c 2 (High-pass) (Low-pass) FIG Pass-band characteristics. The network in Example 2.7 generates the characteristics o Fig However, or a circuit such as the one shown in Fig. 2.3, there is a loading between stages at each requency that aects the level o. Through proper design, the level o may be very near the level o

21 db PASS-BAND FILTES 937 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 passband continue to drop. For cases where max max the bandwidth is deined at.77 o the resulting max. EXAMPLE 2.7 For the pass-band ilter in Fig. 2.3: 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 deines the upper requency or the pass-band..5 nf 2 4 k k 2 4 pf Solutions: a. High-pass ilter: Low-pass ilter: c 6. khz 2p 2p k2.5 nf2 c khz 2p 2 2 2p4 k24 pf2 b. In the mid-region o the pass-band at about 5 khz, an analysis o the network reveals that.9 as shown in Fig The bandwidth is thereore deined at a level o.77(.9 ).636 as also shown in Fig High-pass ilter Low-pass ilter FIG. 2.3 Pass-band ilter Pass-band c 6 khz c 995 khz Actual c Actual c FIG. 2.3 Pass-band characteristics or the ilter in Fig c. At khz, and X 2p 7 X 2 2p k resulting in the network in Fig The parallel combination 2 j X 2 2 is essentially.976 k because the 2 X 2 combination is so large compared to the parallel resistor. X k = k V' X 2 4 k FIG Network o Fig. 2.3 at khz.

22 938 DEIBELS, FILTES, AND BODE PLOTS db with Then V.976 k k j.7 k k k j4 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 hapter 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 in Fig. 2.33, X L X at resonance, and max l s (2.2) and s (2.22) 2p2L with Q s (2.23) l X L and BW s (2.24) Q s Pass-band ilter V V omaxi l L.77max s 2 BW FIG Series resonant pass-band ilter.

23 db PASS-BAND FILTES 939 Pass-band ilter Z Tp l L V omax.77max = V p 2 BW FIG Parallel resonant pass-band ilter. For the parallel resonant circuit in Fig. 2.34, at resonance, and Z Tp is a maximum value max Z T p Z Tp p (2.25) with Z Tp Q l 2 l Q l (2.26) and p 2p2L Q l (2.27) For the parallel resonant circuit Q p X L l (2.28) and BW p (2.29) As a irst approximation that is acceptable or most practical applications, it can be assumed that the resonant requency bisects the bandwidth. Q p EXAMPLE 2.8 a. Determine the requency response or the voltage or the series circuit in Fig b. Plot the normalized response A y /. c. Plot a normalized response deined by A y A y >A ymax. Solutions: a. s Q s 2p2L 5,329.2 Hz 2p2 mh2. mf2 X L 2p5,329.2 Hz2 mh2 9.4 l 33 2 BW s 5,329.2 Hz 5.57 khz Q s 9.4 l 2 = 2 mv L mh. mf 33 FIG Series resonant pass-band ilter or Example 2.8.

24 94 DEIBELS, FILTES, AND BODE PLOTS db At resonance: max 8.86 mv At the cuto requencies: mv2 Note Fig l mv mv 8.86 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 in Fig by 2 mv results in the normalized plot in Fig. 2.37(a). c. Dividing all levels in Fig. 2.37(a) by A ymax.943 results in the normalized plot in Fig. 2.37(b). 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 in Fig STOP-BAND FILTES Stop-band ilters can also be constructed using a low-pass and a high-pass ilter. However, rather than the cascaded coniguration used or the passband ilter, a parallel arrangement is required, as shown in Fig A low-requency can pass through the low-pass ilter, and a higherrequency 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 lowpass critical requency and lower than the high-pass critical requency, and is thereore prevented rom contributing to the levels o above.77v max. The characteristics o a stop-band ilter are the inverse o the pattern obtained or the pass-band ilters; thereore, at any requency, the sum o

25 db STOP-BAND FILTES 94 (low) o Low-pass ilter (low) o 2 (high) High-pass ilter 2 (high) FIG Stop-band ilter. max.77max BW c o c 2 (log scale) (Low-pass) (High-pass) FIG Stop-band characteristics. the magnitudes o the two waveorms to the right o the equals sign in Fig. 2.4 equals the applied voltage. For the pass-band ilters in Figs and 2.34, thereore, i we take the output o the other series elements as shown in Figs. 2.4 and 2.42, a stop-band characteristic is obtained, as required by Kirchho s voltage law. = Pass-band Stop-band o o FIG. 2.4 Demonstrating how an applied signal o ixed magnitude can be broken down into a pass-band and stop-band response curve. Stop-band ilter l L max =.77 BW min s 2 FIG. 2.4 Stop-band ilter using a series resonant circuit.

26 942 DEIBELS, FILTES, AND BODE PLOTS db Stop-band ilter l L max =.77 BW min p 2 FIG Stop-band ilter using a parallel resonant network. For the series resonant circuit in Fig. 2.4, Eqs. (2.22) through (2.24) still apply, but now, at resonance, min l l (2.3) For the parallel resonant circuit in Fig. 2.42, Eqs. (2.26) through (2.29) are still applicable, but now, at resonance, min Z Tp (2.3) 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 inductor 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. 2.9 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 ilter Double-tuned ilter L s L s L p L p L V L L V L (a) (b) FIG Double-tuned networks.

27 db BODE PLOTS 943 double-tuned ilters. For the network in Fig. 2.43(a), the parallel resonant circuit establishes a stop-band or the range o requencies not permitted to establish a signiicant V L. The greater part o the applied voltage appears 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 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 then appears across L at this requency. For the network in Fig. 2.43(b), the series resonant circuit still determines the pass-band, acting as a very low impedance across the parallel inductor at resonance. At the desired stop-band resonant requency, 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 results 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. 2.43(a) and L p in Fig. 2.43(b) are replaced by capacitors. EXAMPLE 2.9 For the network in Fig. 2.43(b), determine L s and L p or a capacitance 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 2p2L and L s 4p 2 2 s 4.7 MH 4p 2 6 khz2 2 5 pf2 At 2 khz, X Ls vl 2p s L s 2p22 khz24.7 mh and X v p22 khz25 pf2 For the series elements, j X Ls X 2 j j 44.7 j X At parallel resonance (Q l assumed), X Lp X X Lp and L p v mh 2p22 khz2 The requency response or the preceding network appears as one o the examples o PSpice in the last section o the chapter. 2. 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.

28 944 DEIBELS, FILTES, AND BODE PLOTS db The curves obtained or the magnitude and/or phase angle versus requency are called Bode plots (Fig. 2.44). 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 is examined in some detail. The second network is treated in a shorthand manner, and inally a method or quickly determining the response is introduced. FIG Hendrik Wade Bode. ourtesy o AT&T Archives. American (Madison, Wl; Summit, NJ; ambridge, MA) (958) V.P. at Bell Laboratories Proessor o Systems Engineering, Harvard University 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 longdistance communication acilities. In 948 he was awarded the Presidential ertiicate 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. High-Pass - Filter Let us start by reexamining the high-pass ilter in 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 A y j X j X j a 2p b I we substitute c (2.32) 2p which we recognize as the cuto requency o earlier sections, we obtain A y j c > 2 j 2p (2.33) We will ind in the analysis to ollow that the ability to reormat the gain to one having the general characteristics o Eq. (2.33) is critical to the application o the Bode technique. Dierent conigurations result in variations o the ormat o Eq. (2.33), but the desired similarities become obvious as we progress through the material. In magnitude and phase orm: FIG High-pass ilter. A y A y u 2 c > 2 2 tan c > 2 (2.34) providing an equation or the magnitude and phase o the high-pass ilter in terms o the requency levels. Using Eq. (2.2), A ydb 2 log A y and, substituting the magnitude component o Eq. (2.34), Ay db 2 log 2 log 2 log c / 2 c / 2 and A ydb 2 log B a c b 2

29 db BODE PLOTS 945 ecognizing that log 2x log x >2 2 log x, we have A ydb 2 22log c a c b 2 d log c a c b 2 d For requencies where << c or ( c / ) 2 >>, a c b 2 a c b 2 and but resulting in A ydb log a 2 c b log x 2 2 log x A ydb 2 log c However, logarithms are such that and substituting b c /, we have log b log b A ydb 2 log c << c (2.35) First note the similarities between Eq. (2.35) 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 results in a 6 db change in gain. A change in requency by a : ratio results 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, Eq. (2.35) is accurate only or requency levels much less than c. First, realize that the mathematical development o Eq. (2.35) does not have to be repeated or each coniguration encountered. Second, the equation itsel is seldom applied but simply used 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. (2.35), 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

30 946 DEIBELS, FILTES, AND BODE PLOTS db 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. (2.35) 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 results in the straight-line segment in 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 6dB(orind 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 approaches an asymptote (straight-line segment) deined by A ydb db since at high requencies >> c and c / with A ydb 2 log 2 c >2 2 log 2 2 The two asymptotes deined above intersect at c, as shown in Fig. 2.46, orming an envelope or the actual requency response. At c, the cuto requency, A ydb 2 log 2 c >2 2 log log 22 3 db At 2 c, A ydb 2 log a 2 c b 2 log a 2 B 2 c B 2 b 2 log 2.25 db as shown in Fig log db

31 db BODE PLOTS 947 At c /2, A ydb 2 log B a c c >2 b 2 2 log log 25 7 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 db down 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. Eq. (2.34) speciies the phase response (the angle by which leads )by u tan c (2.36) For requencies well below c ( << c ), u tan ( c / ) approaches 9 and or requencies well above c ( >> c ), u tan ( c / ) will approach, as discovered in earlier sections o the chapter. At c, u tan ( c / ) tan 45. Deining << c or c / (and less) and >> c or c (and more), we can deine an asymptote at u 9 or << c /, an asymptote at u or >> c, and an asymptote rom c / to c that passes through u 45 at c. The asymptotes deined above all appear in Fig Again, the Bode plot or Eq. (2.36) is a straight line because the change in phase angle will be 45 or every tenold change in requency. Substituting c / into Eq. (2.36), c u tan a c > b tan or a dierence o rom the idealized response. 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 - ilter.

32 948 DEIBELS, FILTES, AND BODE PLOTS db Substituting c, u tan a c b tan 5.7 c In summary, thereore, at c, u 45, whereas at c / and c, the dierence between the actual phase response and the asymptotic plot is mf FIG Example 2.. k EXAMPLE 2. a. Sketch A ydb versus requency or the high-pass - ilter in Fig b. Determine the decibel level at khz. c. Sketch the phase response versus requency on a log scale. Solutions: a. c 2p Hz 2p2 k2. mf2 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 db at 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. 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 in Fig Note that in the solution to part (a), there is no need to use Eq. (2.35) or to perorm any extensive mathematical manipulations.

33 db BODE PLOTS 949 b. Eq. (2.33): A ydb 2 log B a 2 log 2 2 c b Hz a b B 2 log log db as veriied by Fig c. See Fig Note that u 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. 2.5 Phase plot or the high-pass - ilter. khz c = 5,95.5 Hz khz Low-Pass - Filter For the low-pass ilter in Fig. 2.5, A y j X j X j X j j j X 2p 2p FIG. 2.5 Low-pass ilter. and A y (2.37) j > c 2 with c (2.38) 2p 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.

34 95 DEIBELS, FILTES, AND BODE PLOTS db In terms o magnitude and phase, A y A y u 2 > c 2 2 tan > c 2 (2.39) An analysis similar to that perormed or the high-pass ilter results in A ydb 2 log c >> c (2.4) Note in particular that the equation is exact only or requencies much greater than c, but a plot o Eq. (2.4) 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. (2.35), except or the minus sign, which suggests that the resulting Bode plot will have a negative slope [recall the positive slope or Eq. (2.35)] or increasing requencies beyond c. A plot o Eq. (2.4) appears in Fig or c khz. Note the 6 db drop at 2 c and the 2 db drop at 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 - ilter. At >> c, the phase angle u tan (/ c ) approaches 9, whereas at << c, u tan (/c) approaches. At c, u tan 45, establishing the plot in 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 - ilter.

35 db SKETHING THE BODE ESPONSE 95 Even though the preceding analysis has been limited solely to the - combination, the results obtained will have an impact on networks that are a great deal more complicated. One good example is the high- and low-requency response o a standard transistor coniguration. Some capacitive elements in a practical transistor network aect the lowrequency response, and others aect the high-requency response. In the absence o the capacitive elements, the requency response o a transistor ideally stays level at the midband value. However, the coupling capacitors at low requencies and the bypass and parasitic capacitors at high requencies deine a bandwidth or numerous transistor conigurations. In the low-requency region, speciic capacitors and resistors orm an - combination that deines a low cuto requency. There are then other elements and capacitors orming a second - combination that 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. 2. SKETHING 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 indicated in the same igure. In this section, we introduce additional unctions and their responses that can be used in conjunction with those in Fig to determine the db response o more sophisticated systems in a systematic, time-saving, and accurate manner. db Low-pass: j c 3 db c 6 db/octave (or increasing ) (a) db High-pass: j c c db 3 db 6 db/octave (or increasing ) (b) FIG db response o (a) low-pass ilter and (b) high-pass ilter. As an avenue toward introducing an additional unction that appears quite requently, let us examine the high-pass ilter in Fig which has a high-requency output less than the ull applied voltage. Beore developing a mathematical expression or A y /, let us irst make a rough sketch o the expected response. At Hz, the capacitor assumes its open-circuit equivalence, and V. At very high requencies, the capacitor can assume its shortcircuit equivalence, and k nf 2 4 k 2 4 k V i 2 k 4 k.8 FIG High-pass ilter with attenuated output.

36 952 DEIBELS, FILTES, AND BODE PLOTS db = V Th FIG Determining Th or the equation or cuto requency. 2 The resistance to be used in the equation or cuto requency can be determined by determining the Thévenin resistance seen by the capacitor. Setting V and solving or Th (or the capacitor ) results in the network in Fig. 2.56, where it is quite clear that Th 2 k4 k5 k Thereore, c 2p Th 3.83 khz 2p5 k2 nf2 A sketch o versus requency is provided in Fig. 2.57(a). A normalized plot using as the normalizing quantity results in the response in Fig. 2.57(b). I the maximum value o A y is used in the normalization process, the response in Fig. 2.57(c) is obtained. 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 in Figs. 2.57(b) and 2.57(c) can both be obtained quite directly using the oundation established by the conclusions depicted in Fig. 2.54, but we must be aware o what to expect and how they will dier. In Fig. 2.57(b), we are comparing the output level to the input voltage. In Fig. 2.57(c) we are plotting A y versus the maximum value o A y. On most data sheets and or the majority o the investigative techniques commonly used, the normalized plot in Fig. 2.57(c) is used because it establishes db as an asymptote or the db plot. To ensure that the impact o using either Fig. 2.57(b) or Fig. 2.57(c) in a requency plot is understood, the analysis o the ilter in Fig includes the resulting db plot or both normalized curves..8 2 =.8V i 2.8 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 in Fig with attenuated output. For the network in Fig. 2.55: 2 2 j X 2 c d V 2 j X i Dividing the top and bottom o the equation by 2 results in 2 2 c X d j 2

37 db SKETHING THE BODE ESPONSE 953 but j X j 2 v 2 2 j 2p 2 2 j c with c 2p Th and Th 2 so that 2 c 2 j c > 2 d I we divide both sides by, we obtain A y 2 c 2 j c > 2 d (2.4) rom which the magnitude plot in Fig. 2.57(b) can be obtained. I we divide both sides by A ymax 2 > 2 2, we have A y A y A ymax j c > 2 (2.42) rom which the magnitude plot in Fig. 2.57(c) can be obtained. Based on the past section, a db plot o the magnitude o A y A y >A ymax is now quite direct using Fig. 2.54(b). The plot appears in Fig A v db = A v A vmax db c = 3.83 khz 3 db FIG db plot or or the high-pass ilter in Fig A y For the gain A y /, we can apply Eq. (2.5): 2 log ab 2 log a 2 log b where 2 2 log e c 2 j c >2 d 2 log log 2 c > 2 2

38 954 DEIBELS, FILTES, AND BODE PLOTS db The second term results in the same plot in Fig. 2.58, 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 2 2 log 2 2 log 2 2 log and 2 log 2 log 2 (2.43) 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. (2.5), results in the plot in 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 Vi Obtaining a db plot o A vdb db. For the network in Fig. 2.55, the gain A y / can also be ound in the ollowing manner: A y 2 2 j X 2 2 j X j 2 >X j 2 2>X j 2p 2 j 2p 2 2 j 2 j 2 2 X j v 2 j v 2 2 and A y j > 2 (2.44) j > c 2 with 2p 2 and c 2p 2 2 The bottom o Eq. (2.44) is a match o the denominator o the lowpass unction in Fig. 2.54(a). The numerator, however, is a new unction

39 db SKETHING THE BODE ESPONSE 955 that deines a unique Bode asymptote that will prove useul or a variety o network conigurations. Applying Eq. (2.5): 2 log 2 log V c dc i 2 > c 2 d 2 2 log > 2 2 log 2 > c 2 2 Let us now consider speciic requencies or the irst term. At : 2 log 2 log db At 2 : At 2 : 2 log 2 log 2 6 db 2 log 2 log.5 6 db 2 log 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 y, we ind that the phase angle associated with j / / 9 is ixed at 9, resulting in a phase angle or A y 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 y using a procedure outlined by Fig db FIG. 2.6 db plot o /. 6 db/octave A v db 2 c.94 db (log scale) 3 db Actual response FIG. 2.6 Plot o A y db or the network in Fig

40 956 DEIBELS, FILTES, AND BODE PLOTS db Solving or and c : with 2p khz 2p4 k2 nf2 c 2p khz 2p5 k2 nf2 For this development, the straight-line asymptotes or each term resulting rom the application o Eq. (2.5) are 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 in Fig. 2.6 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. (2.5) is simply the 6 db/octave asymptote shown in the igure. 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. A careul sketch o the asymptotes on a log scale reveals 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, 2 log 2 log c 3.83 khz 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 y / 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. 2.2 LOW-PASS FILTE WITH LIMITED ATTENUATION 2 FIG Low-pass ilter with limited attenuation. Our analysis now continues with the low-pass ilter in Fig. 2.62, which has limited attenuation 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. 2.55, 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 2 A plot o versus requency is provided in Fig. 2.63(a). A sketch o A y / appears as shown in Fig. 2.63(b). An equation or in terms o can be derived by irst applying the voltage divider rule: 2 j X 2 2 j X

41 db LOW-PASS FILTE WITH LIMITED ATTENUATION 957 A v = c (log scale) c (log scale) (a) (b) FIG Low-pass ilter with limited attenuation. and A y 2 j X 2 >X j 2 j X 2 2>X j j2 2 X j2 j2 2 2>X j2 j 2 >X 2 j 2 2>X 2 j 2p 2 j 2p 2 2 so that A y j > 2 (2.45) j > c 2 with 2p 2 and c 2p 2 2 The denominator o Eq. (2.45) is simply the denominator o the lowpass unction in Fig. 2.54(a). The numerator, however, is new and must be investigated. Applying Eq. (2.5): A ydb 2 log 2 log 2 > log For >>,(/ ) 2 >>, and the irst term becomes 2 log 2 > log > >2 2 log > 2 W which deines the idealized Bode asymptote or the numerator o Eq. (2.45). At, 2 log db, and at 2, 2 log 2 6 db. For requencies much less than,(/ ) 2 <<, and the irst term o the Eq. (2.5) expansion becomes 2 log 2 2 log db, which establishes the low-requency asymptote. The ull idealized Bode response or the numerator o Eq. (2.45) is provided in Fig We are now in a position to determine A y db by plotting the asymptote or each unction o Eq. (2.45) 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. 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 > c 2 2

42 958 DEIBELS, FILTES, AND BODE PLOTS db 2 2 log Actual response 3 db 6 db/octave db FIG Idealized and actual Bode response or the magnitude o ( j (/ )). A v = db 2 3 db c Actual response 2 log 2 2 (log scale) FIG versus requency or the low-pass ilter with limited attenuation o Fig A ydb 2 log 2 /( 2 ) 2 log /(( 2 )/ 2 ) 2 log 2 log (( 2 )/ 2 ) 2 and 2 log 2 log 2 (2.46) 2 as shown in Fig In region in Fig. 2.65, 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 2

43 db LOW-PASS FILTE WITH LIMITED ATTENUATION 959 region 3 can be determined using Eq. (2.46) or by substituting into the asymptotic expression deined by c. The ull idealized Bode envelope is now deined, permitting a sketch o the actual response by shiting 3 db in the right direction at each corner requency, as shown in Fig The phase angle associated with A y can be determined directly rom Eq. (2.45). That is, u tan > tan > c (2.47) A ull plot o u versus requency can be obtained by substituting various key requencies into Eq. (2.47) and plotting the result on a log scale. The irst term o Eq. (2.47) deines the phase angle established by the numerator o Eq. (2.45). 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. (2.45) on the same graph, as shown in Fig v (A v ) c / / c c 45 9 FIG Phase angle or the low-pass ilter in Fig

44 96 DEIBELS, FILTES, AND BODE PLOTS db 2 FIG High-pass ilter with limited attenuation. Vi = V Th FIG Determining or the c calculation or the ilter in Fig The asymptotes in Fig clearly indicate that the phase angle will be in the low-requency range and (9 9 ) in the highrequency 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 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. (2.45). 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 ). 2.3 HIGH-PASS FILTE WITH LIMITED ATTENUATION The ilter in 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 used when determining c can be ound by inding the Thévenin resistance or the capacitor, as shown in Fig A careul examination o the resulting coniguration reveals that Th 2 and c >2p 2 2. A plot o versus requency is provided in Fig. 2.7(a), and a sketch o A y / appears in Fig. 2.7(b). An equation or A y / can be derived by irst applying the voltage divider rule: 2 2 j X A v = V i 2 c 2 2 c (a) (b) FIG. 2.7 High-pass ilter with limited attenuation.

45 db HIGH-PASS FILTE WITH LIMITED ATTENUATION 96 and A y j X 2 j X 2 j X 2 j X 2 2 j 2 X 2 j X 2 j X 2 j 2 X j X 2 j 2 X 2 j 2 2X j 2X 2 j 2 2 X 2 j X j X j X X j j j p 2p 2 2 so that A y j > 2 (2.48) j c > 2 with 2p and c 2p 2 2 The denominator o Eq. (2.48) is simply the denominator o the highpass unction in Fig. 2.54(b). The numerator, however, is new and must be investigated. Applying Eq. (2.5): A ydb 2 log 2 log 2 > log For <<,( / ) 2 >>, and the irst term becomes 2 log 2 > log > 2 V which deines the idealized Bode asymptote or the numerator o Eq. (2.48). 2 c > 2 2 At, At.5, At., 2 log db 2 log 2 6 db 2 log 2 db For requencies greater than, / << and 2 log db, which establishes the high-requency asymptote. The ull idealized Bode plot or the numerator o Eq. (2.48) is provided in Fig We are now in a position to determine A ydb by plotting the asymptotes or each unction o Eq. (2.48) 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. At the intersection o the resultant o region 2 with, we enter region, where the asymptotes have opposite slopes and cancel the eect o each other. The resulting level at is determined by

46 962 DEIBELS, FILTES, AND BODE PLOTS db 2 log 2 6 db/octave Actual response 3 db db (log scale) FIG. 2.7 Idealized and actual Bode response or the magnitude o ( j ( /)). A vdb 2 3 c (log scale) 2 log 2 2 Actual response FIG versus requency or the high-pass ilter with limited attenuation in Fig A ydb 2 log ( 2 )/ 2, as ound in earlier sections. The drop can also be determined by substituting into the asymptotic equation deined or c. In region 3, both are at db, resulting in a db asymptote or the region. The resulting asymptotic and actual responses both appear in Fig The phase angle associated with A y can be determined directly rom Eq. (2.48); that is, A ull plot o u versus requency can be obtained by substituting various key requencies into Eq. (2.49) and plotting the result on a log scale. The irst term o Eq. (2.49) deines the phase angle established by the numerator o Eq. (2.48). The asymptotic plot resulting rom the numeru tan tan c (2.49)

47 db HIGH-PASS FILTE WITH LIMITED ATTENUATION 963 v j 45 / v = Actual response (log scale) 9 v = 9 FIG Phase angle or ( j( /)). ator is provided in Fig Note the leading phase angle o 45 at and the straight-line asymptote rom / to. 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. (2.48) on the same graph, as shown in Fig The asymptotes in Fig clearly indicate that the phase angle will be 9 in the low-requency range and (9 9 ) in the highrequency 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 u 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. Fig reveals that the angle at c and will be less than 45. The maximum angle occurs between c and, as shown in v (A v ) / c / c c (log scale) 45 9 FIG Phase response or the high-pass ilter in Fig

48 964 DEIBELS, FILTES, AND BODE PLOTS db the igure. Note again that the greatest change in u occurs at the corner requencies, matching the regions o greatest change in the db plot. EXAMPLE 2. For the ilter in Fig. 2.75: 9. k A ydb a. Sketch the curve o versus requency using a log scale. b. Sketch the curve o u versus requency using a log scale..47 mf 2 k FIG Example 2.. Solutions: a. For the break requencies: 2p 2p9. k2.47 mf 2 c 2p a 2 2 b 37.2 Hz Hz 2p.9 k2.47 mf 2 The maximum low-level attenuation is 2 log log 9. k k k 2 log. 2.9 db The resulting plot appears in Fig A vdb 2 db 3 db Hz 37.2 Hz Hz c Hz 3 db Hz db (log scale) 7 db 2 db FIG versus requency or the ilter in Fig A ydb b. For the break requencies: At 37.2 Hz, u tan tan c tan Hz tan 37.2 Hz

49 db ADDITIONAL POPETIES OF BODE PLOTS 965 At c Hz, u tan At a requency midway between c and on a log scale, or example, 2 Hz: 37.2 Hz Hz u tan tan 2 Hz 2 Hz Hz Hz tan The resulting phase plot appears in Fig v Hz 37.2 Hz 2 Hz Hz Hz c Hz (log scale) FIG u (the phase angle associated with A y ) versus requency or the ilter in Fig ADDITIONAL POPETIES OF BODE PLOTS 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 urther demonstrate the adaptability and useulness o the linear Bode approach to requency analysis. 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. (2.5) will prove useul. For example, i A y has the ormat A y 2 j 2> 2j > 2 j > 2 j > 2 2 a2b2c2 d2e2 (2.5) we can expand the unction in the ollowing manner: A ydb 2 log a2b2c2 d2e2 2 log a 2 log b 2 log c 2 log d 2 log e

50 966 DEIBELS, FILTES, AND BODE PLOTS db 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 can, thereore, 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 y the unction can be rewritten as j > 2 j > 2 A y j > 2 2 so that A ydb 2 log 2 > log > or <<,( l /) 2 >>, and A ydb 2 log > 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 now passes 6 db below the corner requency rather than 3 db, as or a single term. 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 > 22 2

51 db ADDITIONAL POPETIES OF BODE PLOTS 967 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 ydb or with < 2. j > 22 j 2 > 22 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 in Fig with corner requencies and 2. In region 3, both asymptotes are db, resulting in an asymptote at db or 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, we ind two asymptotes dropping o at 6 db or decreasing requencies. The result is an asymptote dropping o at 2 db/ octave or this region. I and 2 are at least two octaves apart, the eect o one on the plotting o the actual response or the other can almost be ignored. In other words, or this example, i 6 4 2, the actual response is 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. Eventually, the procedure will become 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. EXAMPLE 2.2 A transistor ampliier has the ollowing gain: A y a j 5 Hz ba j 2 Hz ba j khz ba j 2 khz b a. Sketch the normalized response A y A y >A ymax, and determine the bandwidth o the ampliier. b. Sketch the phase response, and determine a requency where the phase angle is close to.

52 968 DEIBELS, FILTES, AND BODE PLOTS db Solutions: a. A y A y A ymax A y a j 5 Hz ba j a2b2c2d2 a a ba b ba c ba d b and 2 Hz ba j khz ba j 2 khz b A ydb 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 reveals that the irst two terms in the denominator are high-pass ilter unctions, whereas the last two are low-pass unctions. Fig. 2.8 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. 2.8 Finding the overall gain versus requency or Example 2.2. Plotting all the idealized Bode plots on the same axis results in the plot in Fig Note or requencies less than 5 Hz that the result- A vdb Hz 3 db 6 db Hz Hz 2 Hz khz khz 2 khz khz BW 2 db 2 db FIG. 2.8 versus requency or Example 2.2. A ydb

53 db ADDITIONAL POPETIES OF BODE PLOTS 969 ing 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 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 dbdrop due to the 2 khz corner requency to the 3dBdropat 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. 2 5 Hz A ydb 2 log a B khz b 2 khz 2 log a B khz b. db.7 db 3. db.969 db 3.98 db 4 db as beore 2 2 Hz 2 log a B khz b 2 khz 2 log a B 2 khz b An examination o the above calculations 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, examining 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 3dBlevelin 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 khz A ydb 2 log a B khz b 8.5 khz 2 log a B 2 khz b 2.48 db.645 db 2.8 db 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, khz:.235 db 5 db A ydb approaches db. At Hz A ydb 2 log a B khz b 2 Hz 2 log a B khz b 2 log B a khz khz b 2 2 log B a khz 2 khz b 2.8 db.73 db.432 db.8 db which is certainly close to the db level, as shown on the plot. b. The phase response can be determined by substituting a number o key requencies into the ollowing equation, derived directly rom the original unction A y :

54 97 DEIBELS, FILTES, AND BODE PLOTS db 5 Hz 2 Hz u tan tan tan tan khz 2 khz However, let us make ull use o the asymptotes deined by each term o A y 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 2.2. At khz, the asymptotes leave us with u (actual 7.56 ). The net phase plot appears to be close to at about 3 Hz. To check on our assumptions and the use o the asymptotic approach, plug in 3 Hz into the equation or u: u tan 5 Hz 2 Hz tan 3 Hz 3 Hz as predicted tan 3 Hz khz tan 3 Hz 2 khz In total, the phase plot appears to shit rom a positive angle o 8 ( leading ) to a negative 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. Table 2.2 consolidates some o the material introduced in this chapter and provides a reerence or uture investigations. It includes the linearized db and phase plots or the unctions appearing in the irst column. There are many other unctions, but these provide a oundation to which others can be added. eviewing the development o the ilters in Sections 2.2 and 2.3 shows that establishing the unction A y in the proper orm is the most

55 db ADDITIONAL POPETIES OF BODE PLOTS 97 TABLE 2.2 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 A vdb 6 db/octave db 9 45 v ( leads ) 9 45 A v j A vdb db 6 db/octave v ( leads ) / 45 9

56 972 DEIBELS, FILTES, AND BODE PLOTS db diicult part o the analysis. However, with practice and an awareness o the desired ormat, you will discover methods that will signiicantly reduce the eort involved. 2.5 OSSOVE NETWOKS db 3 db db 4 Hz mid = 47 F db 4 Hz L low = 3.3 mh 5 khz high = 3.9 F 3 db 8 5 khz L mid = 27 H 8 3 db 8 3 db The topic o crossover networks is included primarily to present an excellent demonstration o ilter operation without a high level o complexity. rossover networks are used in audio systems to ensure that the proper requencies are channeled to the appropriate speaker. Although less expensive audio systems have only one speaker to cover the ull audio range rom about 2 Hz to 2 khz, better systems have 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 in Fig is quite simple in design, with a low-pass -L ilter or the wooer, an -L- pass-band ilter or the midrange, and a high-pass - 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. FIG Three-way, crossover network with 6 db per octave. L low 2p L mid 2p 2 (2.5) mid 2p high 2p 2 (2.52) For the crossover network in Fig with three 8 speakers, the resulting values are L low L mid mh S 3.3 mh 2p 2p4 Hz2 (commercial value) mh S 27 mh 2p 2 2p5 khz2 (commercial value) mid 2p mf S 47 mf 2p4 Hz28 2 (commercial value) high 2p mf S 3.9 mf 2p5 khz28 2 (commercial value) as shown in 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

57 db APPLIATIONS 973 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 has the highest power level. At khz, Using the basic power equation P V 2 /, the power to the wooer is to the midrange speaker, and to the tweeter, X Llow 2pL low 2p khz23.3 mh Z V 2 Z T 8 j V 68.9 X Lmid 2pL mid 2p khz227 mh2.696 X mid p mid 2p khz247 mf 2 Z V 2 Z T 8 j.696 j V.93 X high Z 2 2 Z T.77 V 78.9 P low V 2 P mid V 2 P high V 2 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 totally overshadows the other two. 2.6 APPLIATIONS Attenuators 4.8 2p high 2p khz23.9 mf 2.44 V V V V 2 8 j W.94 W.74 W 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, only the resistive type is covered here. Attenuators are commonly used in audio equipment (such as the graphic and parametric equalizers introduced in hapter 2), 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.

58 974 DEIBELS, FILTES, AND BODE PLOTS db 5 ohm coax attenuator IN db 3 db db 2 db 5 db 2 db OUT FIG Passive coax attenuator. The unit in 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 2.86, respectively, or a 5 coaxial system. In each case, the resistors are 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. 2 2 Attenuation 2 Attenuation 2 db db db db db db db db db 2 db db 2 db FIG Tee (T) coniguration. FIG Wye (Y) coniguration. For the db attenuation, the resistor values were inserted or the T coniguration in Fig. 2.87(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 s 2.9 i = L V s = V o = 5 db attenuator (a) (b) FIG db attenuator: (a) loaded; (b) inding o.

59 db APPLIATIONS 975 Looking back rom the load as shown in Fig. 2.87(b) with the source set to zero volts, we ind through the ollowing calculations that the output impedance is also 5 : o 2 s In Fig. 2.88, a 5 load has been applied, and the output voltage is determined as ollows: 2 L rom above and V 2 V s 47.4 V s V s with V L LV 2 L 5.942V s V s s V s L 5 V L FIG Determining the voltage levels or the db attenuator in Fig. 2.87(a). alculating the drop in db results in the ollowing: 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, relatively inexpensive, and perorm quite well. Noise Filters A ydb 2 log V L V s 2 log.89 V s 2 log.89. db Noise is a problem that can occur in any electronic system. In general, the presence o any unwanted signal 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. To solve a noise problem, an analyst needs a broad practical background, a sense or the origin or the unwanted noise, and the ability to remove it in the simplest, most direct way. Generally noise problems arise during the testing phase, not during the original design phase. Although sophisticated methods may be needed, most situations are resolved simply by rearranging an element or two o a value sensitive to the problem. V s

60 976 DEIBELS, FILTES, AND BODE PLOTS db n pf Short circuit to high-requency noise ompensation control o high requencies ecording head 2 Applied signal P oupling capacitor 3 P Playback network Playback head 3 P P Filter to reduce stray pickup s pf s Ampliier and notch ilter ecord phase oupling capacitor (a) Playback phase (b) FIG Noise reduction in a tape recorder. In Fig. 2.89, two capacitors have been strategically placed in the tape recording and playback sections o a tape recorder to remove the 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 acts as a short circuit to the highrequency noise. The capacitor 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. 2.9 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 brings the requency range to be ampliied to the region o high-amplitude recording. On some tapes, ecording level High requency drop-o 2 khz 3 khz Bias requency FIG. 2.9 Noise reduction in a tape recorder.

61 db APPLIATIONS 977 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. ompensation or this drop-o is provided by the parallel combination o the resistor and the capacitor 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. 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 begins 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 high-requency signal. During playback (P), the upper circuit in 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. Alternators in a car are notorious or developing high-requency noise down the line to the radio, as shown in Fig. 2.9(a). This problem is usually alleviated by placing a high-requency ilter in the line as shown. The inductor o H oers a high impedance or the range o noise requencies, while the capacitor ( mf to 47, mf) acts as a short-circuit equivalent to any noise that happens to get through. For the speaker system in Fig. 2.9(b), the push-pull power arrangement o transistors in the output section can oten develop a short period o time between pulses where V Push-pull response t High-requency noise L V s V ar alternator H adio ( μf to 47, μf) Push-pull ampliier b (a) (b) Short-circuit path or unwanted high-requency oscillation FIG. 2.9 Noise generation: (a) due to a car alternator; (b) rom a push-pull ampliier.

62 978 DEIBELS, FILTES, AND BODE PLOTS db 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 y L L(di L /dt), an unexpected voltage develops across the coil and sets a high-requency oscillation on the line that inds its way back to the ampliier and causes urther distortion. This eect can be subdued by placing an - path to ground that oers a low-resistance path rom the speaker to ground or a range o requencies typically generated by this signal distortion. Since the capacitor assumes 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. In regulators, such as the 5 V regulator in Fig. 2.92(a), when a spike in current comes down the line or any number o reasons, there is a voltage drop along the line, and the input voltage to the regulator drops. The regulator, perorming its primary unction, senses this drop in input voltage and increases its ampliication level through a eedback loop to maintain a constant output. However, the spike is o such short duration that the output voltage has a spike o its own because the input voltage has quickly returned to its normal level, and with the increased ampliication the output jumps 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 problem or the equipment to which the dc voltage is applied: A high-requency 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. 2.92(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 highrequency 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 oten reveals their purpose and modus operandi. 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.

63 db OMPUTE ANALYSIS OMPUTE ANALYSIS PSpice Double-Tuned Filter Our analysis now turns 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 2.9, so we have a chance to test our theoretical solution. The schematic appears in Fig with VA again chosen since the requency range o interest is set by the Simulation Proile. Again, the attributes or the source are 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, has no eect on the results obtained. FIG Using PSpice to analyze a double-tuned ilter. In the Simulation Settings dialog box, select A Sweep again 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. 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 SHEMATI appears, Trace-Add Trace-V(L:)-OK results in the plot in Fig Quite obviously, there is a reject-band around 2 khz and a pass-band around 6 khz. Note that up to khz, there is 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 stopband due to the open-circuit equivalence o the L p inductor. Using the cursor option, place the let-click cursor on the minimum point o the graph by using the ursor Trough keypad (the second pad to the right o the Toggle cursor pad). ight-click to identiy the requency o the maximum point on the curve near 6 khz. The results appearing in the Probe ursor box clearly support our theoretical calculations o 2 khz or the band-stop minimum (A 2.3 khz with a magnitude o essentially V) and 6 khz or the pass-band maximum (A khz with a magnitude o V).

64 98 DEIBELS, FILTES, AND BODE PLOTS db FIG Magnitude plot versus requency or the voltage across L o the network in Fig Let us now concentrate on the range rom khz to MHz where most o the iltering action is taking place. That was the advantage o choosing such a wide range o requencies when the Simulation Settings were set up. The data have been established or the broad range o requencies, and you can simply select a band o interest once the region o most activity is deined. I the requency range were too narrow in the original simulation, another simulation would have to be deined. Select Plot-Axis Settings-X Axis-User Deined-kHz to MEGHz-OK to obtain the plot at the bottom o Fig A db plot o the results can FIG db and magnitude plot or the voltage across L o the network in Fig

65 db OMPUTE ANALYSIS 98 also be displayed in the same igure by selecting Plot-Add Plot to Window-Trace-Add Trace-DB(V(L:))-OK, resulting in the plot at the top o the igure. Using the let-click cursor option and the ursor Trough key, you ind that the minimum is at db at a requency o 2 khz, which is an excellent characteristic or a band-stop ilter. Using the right-click cursor and setting it on 6 khz, you ind that the drop is 3. mdb or essentially db, which is excellent or the pass-band region. Multisim High-Pass Filter The Multisim computer analysis is an investigation o the high-pass ilter in Fig The cuto requency is determined by /2p.592 khz, with the voltage across the resistor approaching V at high requencies at a phase angle o. For this analysis, the omponent: A_POWE under POWE_ SOUES was chosen. The components o the source were set in the POWE_SOUES dialog box as shown in Fig Use the sequence Simulate-Analyses-A Analysis to obtain the A Analysis dialog box. Select the ollowing settings: Start requency: Hz, Stop requency: khz, Sweep type: Decade (logarithmic), Number o points per decade:, Vertical scale: Linear. Under the Output option, move $2 to Selected variables or analysis using the Add option and remove the $ option using the emove option. Select Simulate to obtain the response in Fig Add the grid option to each, and then select Show/Hide ursors to permit a determination o the magnitude and phase angle at the cuto requency. As shown in Fig. 2.96, the magnitude is.77 at.589 khz and the phase angle is 45.4 at.58 khz very close to the expected results. FIG High-pass - ilter to be investigated using PSpice.