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19 BASICFILTERS 19-1 19-2 19-3 19-4 19-5 Low-PassFilters High-Pass Filters Band-Pass Filters Band-StopFilters Technology Theoryinto Practice Workbench(EWB)and Electronics PSpiceTutorialsat http://www.prenhall.com/floyd r INTRODUCTION The concept of filters was introduced in Chapters 16, ll , and 18 to illustrate applications of RC RZ, and RIC circuits. This chapter is essentially an extension of the earlier material and provides additional coverage of the important topic of filters. Passivefilters are discussedin this chapter.Passive filters use various combinations of resistors, capacitors,and inductors. In a later course, you will study active filters which use passive components combined with amplifiers. You have already seenhow basic RC. RL. and RLC circuits can be used as filters. Now, you will learn that passivefilters can be placed in four general categoriesaccording to their response characteristics: low-pass,high-pass,band-pass,and band-stop.Within each category,there are several common types that will be examined. In the TECH TIP assignmentin Section 19-5, you will plot the frequency responsesof filters based on oscilloscope measurementsand identify the types of fllters. r TECHnology Theorv Into Practice r CHAPTER OBfECTTVES O Analyze the operation of RC and RZ low-pass filters tr Analyze the operation of RC and Rt high-pass filters O Analyze the operation of band-passfilters tr Analyze the operation of band-stopfilters 762 I BASICFILTERS FITTERS 19-1 r LOW.PASS A law-passfilter allows signals with lower frequencies to pass from input to output while rej ecting higher frequencies. After completing this section, you should be able to I Analyze the operation of RC and RL low-pass filters . Express the voltage and power ratios of a filter in decibels . Determine the critical frequency of a low-pass filter . Explain the difference between actual and ideal low-pass responsecurves . Define roll-off . Generatea Bode plot for a low-pass filter . Discuss phase shift in a low-pass filter Figure 19-1 shows a block diagram and a generalresponsecurve for a low-pass filter. The range of low frequenciespassedby a low-pass filter within a specified limit is called the passband of the fllter. The point consideredto be the upper end of the passbandis at the critical frequency,/f, as illustrated in Figure 19-1(b). The critical frequency is the frequency at which the fllter's output voltage is 10.77oof the maximum. The filter's critical frequency is also called the cutoff frequency, breakfrequency, or -3 dB frequency because the output voltage is down 3 dB from its maximum at this frequency.The term dB (decibel) is a commonly used one that you should understandbecausethe decibel unit is used in filter measurements. vor, Vn 0.70'1u" (a) (b) F I G U R E1 9 - 1 Low-passfi.lter block diagram and generalresponsecurve. Decibels The basis for the decibel unit stemsfrom the logarithmic responseof the human ear to the intensity of sound. The decibel is a logarithmic measurementof the ratio of one power to another or one voltage to another,which can be used to expressthe input-to-output relationship of a filter. The following equation expressesa power ratio in decibels: dB= 1o V i n / "r(,?) (19-1) From the properties of logarithms, the following decibel formula for a voltage ratio is derived. This formula is correct only when both voltases are measured in the same imoedance. dB=2o "'(+) LOW-PASS FILTERS. EXAMPTE 19-1 763 At a ceftain frequency, the output voltage of a fllter is 5 V and the input is 10 V. Expressthe voltageratio in decibels. so l u ti o n 'u 'o tltv^' Z \= zo loef ' \ *l' v/ r ol= / Related Problem z0togr o.s)- 6.02 = dB Express the ratio VourlVin= 0.85 in decibels. RC Low-Pass Filter A basic RC low-pass filter is shown in Figure 19-2. Notice that the output voltage is taken acrossthe caoacitor. F I C U R E1 9 - 2 I 0.005pF When the input is dc (0 Hz), the output voltage equals the input voltage becauseX6 is infinitely large. As the input frequency is increased,X6.decreasesand, as a result, Vo,, gradually decreasesuntil a frequency is reached where X6 = R. This is the critical frequency,f,, of the filter. Xc= R I _ 2nf"C D " 1 r,= ,n*- (1e-3) At any frequency,the output voltage magnitude is ,, v o u-t -/-{a:)r,,,, \ /nr*x?, by application of the voltage-divider formula. Since X6 = R atf,, the output voltage at the critical frequency can be expressedas = v"'=( v# R''' l#)'t =(Fhy' =(+)vn=0r07v' These calculations show that the output is l0.7%aof the input when Xc= R. The frequency at which this occurs is, by deflnition, the critical frequency. The ratio of output voltage to input voltage at the critical frequency can be expressedin decibelsas follows: Vo'' = 0'707V;' I v in = 0.101 / v \ = -3 dB 20 log{-j!l = 20 logt0.707r \ vi' / 764 I BASICFILTERS EXAMPLE 19-2 for thelow-passRC filter in Figure19-2. Determinethecriticalfrequency 1 | = = 318kHz " f'= 2nRC trloo orroJosrrF) Solution The output voltage is 3 dB below V;, atthis frequency (Vourhasa maximum value of Vr,). Related Problem A certain low-pass RC filter has R = 1.0 kO and C = 0.022 pF. Determine its critical frequencv. "Roll-Off" of the ResponseCurve The blue line in Figure 19-3 shows an actual responsecurve for a low-pass filter. The maximum output is defined to be 0 dB as a reference. Zero decibels correspondsto Vo,, = Vin, because20 ).og(Vo,,/V;,)= 20 log 1 = 0 dB. The output drops from 0 dB to -3 dB at the critical frequency and then continues to decreaseat a flxed rate. This pattern of decreaseis called the roll-off of the frequency response.The red line shows an ideal output responsethat is consideredto be "flat" out tojf. The output then decreasesat the fixed rate. ro't \ F I C U R E1 9 - 3 Actual and ideal responsecurvesfor a low-passfihen As you have seen, the output voltage of a low-pass filter decreasesby 3 dB when the frequency is increasedto the critical value f,. As the frequency continues to increase abovef,, the output voltage continuesto decrease.In fact, for each tenfold increasein frequency abovef,, there is a20 dB reduction in the output, as shown in the following steps. Let's take a frequency that is ten times the critical frequency (f = ljf").Since R = X, at f,, then R = l)Xc at l)f" becauseof the inverse relationship of X6 andf. The attenuation of the RC circuit is the ratio Vo6/V;, and is developed as follows: Vout - Xc v,,- ffi**r- - XC - XC ^,FoxjiE, XC =-L=-+=-1 f rooxT-T,-,Vxiloo.D y xct}t v 101 The dB attenuationis = -20 dB 20 bs(Y:-\= "\ V,"J 2olos(o.l) 10 =g.1 LOW-PASS FILTERS. 765 A tenfold changein frequency is called a decade. So, for the RC circuit, the output voltage is reduced by 20 dB for each decadeincrease in frequency.A similar result can be derived for the high-pass circuit. The roll-off is a constant -20 dB/decadefor a basic RC or RL filter. Figure 19-4 shows the ideal frequency responseplot on a semilog scale, where each interval on the horizontal axis representsa tenfold increasein frequency.This responsecurve is called a Bode plot. *10 -20 -30 tt -9 -40 tael FtcuRE't9-4 RCfilter (Bodeplot). Frequency roll-offfor a low-pass 19-3 EXAMPTE Make a Bode plot for the filter in Figure 19-5 for three decadesof frequency. Use semilog graph paper. F I G U R E1 9 - 5 I Solution 0.005prF The critical frequency for this low-pass filter is | = 31.8kHz t,' -- 2 n R C 2n(r.0 pF) ka)(0.005 The idealized Bode plot is shown with the red line on the semilog graph in Figure 19-6. The approximate actual response curve is shown with the blue line. Notice flrst that the horizontal scale is logarithmic and the vertical scale is linear. The frequency is on the logarithmic scale, and the filter output in decibels is on the vertical. 766 r BASICFILTERS O A U O O\{@€, qO\{-\O, S 5 u O\{@\O 0dB -3 dB N -10 dB 'out \ vrin \ -20 dB \ \ \ \ -30 dB \ \ -40 dB 10kHz f" 100kHz t000 kHz FICURE19-6 Bode plot for Figure 19-5. The red.line representsthe ideal responsecurtte and the blue line repfesentsthe actual response. The output is flat belowf, (31.8 kHz). As the frequency is increased abovef", output drops at a -20 dB/decade rate. Thus, for the ideal curve, every time the quency is increasedby ten, the output is reducedby 20 dB. A slight variation from occurs in actual practice. The output is actually at -3 dB rather than 0 dB at the cal frequency. Related Problem What happens to the critical frequency and roll-off rate if C i reducedto 0.001rrF in Fisure 19-5? LOW-PASS FILTERS. 767 Rl Low-PassFilter A basic Rl low-passfilter is shown in Figure 19-7. Notice that the output voltage is taken acrossthe resistor. F I C U R E1 9 _ 7 RL low-passfilter. Vi, When the input is dc (0 Hz), the output voltage ideally equals the input voltage because X7 is a short (if R1y is neglected). As the input frequency is increased, X1 increasesand, as a result, Vou,grad:uallydecreasesuntil the critical frequency is reached. At this point, X. = R and the frequency is 2nf,L = R " , - r'- R 2nL | (1e-4) }IT,\L/R) Just as in the RC low-pass filter, Vour= 0.70JVi, and, thus, the output voltage is -3 dB below the input voltage at the critical frequency. 19_4 EXAMPLE Make a Bode plot for the filter in Figure 19-8 for three decadesof frequency. Use semilog graph paper. F I G U R E1 9 - 8 Solution The critical frequency for this low-pass filter is t'= | I 2n(L/R) 2n(4.7 rtll2.2 =?4.-skHz kf2) The idealized Bode plot is shown with the red line on the semilog graph in Figure 19-9. The approximate actual responsecurve is shown with the blue line. Notice first that the horizontal scale is logarithmic and the vertical scale is linear. The frequency is on the logarithmic scale, and the filter output in decibels is on the vertical. 768 I BASICFILTERS 0dB -3 dB -10 dB vor, u, -20 dB -30 dB -40 dB 10kHz 1000kHz 10,000 kHz F I C U R E1 9 - 9 Bodeplot for Figure 19-8. The red line is the i.deal response cufle and the blue line is the actual responsecurve. The output is flat below/f (74.5 kIIz). As the frequency is increasedabove/f, the output drops at a -20 dB/decade rate. Thus, for the ideal curve, every time the frequency is increasedby ten, the output is reducedby 20 dB. A slight variation from this occurs in actual practice. The output is actually at -3 dB rather than 0 dB at the critical frequency. Related Problem What happens to the critical frequency and roll-off rate if I is reducedto I mH in Figure 19-8? PhaseShift in a Low-PassFilter The RC low-pass filter acts as a lag network. Recall from Chapter 16 that the phaseshift from input to output is expressedas *,un '(*) e = _-eo" \ R / At the critical frequency, Xc = R and, therefore, 6 = -45'. As the input frequencyis reduced, @ decreasesand approaches0o when the frequency approacheszero. Figure 19-10, illustratesthis phasecharacteristic. FILTERS. HICH-PASS 769 F I G U R E1 9 - 1 0 Phasecharacteristic of a low-passfilter. The Rt low-passfilter alsoactsas a lag network.Recallfrom Chapter17 that the as phaseshift is expressed ,lX,\ 0 = _tan |r_J As in the RC filter, the phaseshift from input to outputis -45" at the critical frequency belowf,. for frequencies anddecreases sEcTtoN19-1 REVIEW 1. In a certain low-pass filter, f, = 2.5 WIz What is its passband? 2. In a certainlow-passfilter.R = 100 Q andX6= 2 Cl at a frequency./. Determine Y ou,at fl when V;, = 5/0" V rms. 3. Vout= 400mV, and Vn= 1.2V. Expressthe ratio VouylVi,in dB' FILTERS 19_2I HIGH.PASS A high-passfilter allows signals with higher frequencies to pass ftom input to output while rej ecting lower fre quencies. After completing this section, you should be able to I Analyze the operation of RC and RL high-pass filters . Determine the critical frequency of a high-pass filter . Explain the difference between actual and ideal responsecurves . Generatea Bode plot for a high-pass filter . Discuss phase shift in a high-pass filter Figure 19-1 I shows a block diagram and a general responsecurve for a high-pass filter. The frequency consideredto be the lower end of the passbandis called the critical frequency.Just as in the low-pass filter, it is the frequency at which the output isl0.7%o of the maximum, as indicated in the f,gure. 770 I BASICFILTERS Vo't Vin 0.701un G;;itn;t frequencies +Passesthese frequencies F I G U R E1 9 - 1 1 High-passfiher blnck diagram and responsecune. RC High-Pass Filter A basic RC high-passfilter is shown in Figure I9-I2. Notice that the output voltageis taken acrossthe resistor. F I G U R E1 9 - 1 2 RC hish-pass " filter. L r/ vi,, "-lF-1-- V ou, I =R -L : When the input frequency is at its critical value, Xc = R and the output voltageis O.707vi,,just as in the caseof the low-pass filter. As the input frequency increasesabove f,, Xs decreasesand, as a result, the output voltage increasesand approachesa value equal to V;,. The expressionfor the critical frequency of the high-pass filter is the sameas for the low-pass filter. , t"- | 2nRC Belowf,, the output voltage decreases(rolls ofl at a rate of -20 dB/decade.Figure 19-13 showsan actualand an ideal responsecurve for a high-passfilter. Ideal 0.07fc 0-3 o.tf, dB *20 dB -40 dB v^,,. -f (dB) F I G U R E1 9 - 1 3 Actual and i.dealresponsecumesfor a high-passfilter 19_5 EXAMPLE Make a Bode plot for the fllter in Figure 19-14 for three decadesof frequency. Use semilog graph paper. F I G U R E1 9 - 1 4 vtr4 0.047pF Solution The critical frequency for this high-pass filter is t "' - l 2nRC 2n(3300)(0.047pF) = 10.3kHz = 10kHz The idealized Bode plot is shown with the red line on the semilog graph in Figure 19-15. The approximate actual responsecurve is shown with the blue line. Notice first that the horizontal scale is logarithmic and the vertical scale is linear. The frequency is on the logarithmic scale, and the filter output in decibels is on the vertical. a A U O,{@€ a ! @€. a 5 u 6 {@\O Vou, u, 100Hz 1000Hz F I G U R E1 9 - 1 5 Bode plot for Figure 19-14. The reil line is the i.deal response cume and the blue line is the actual response curae. The output is flat beyond/| (approximately 10 kHz). As the frequency is reduced below /|, the output drops at a -20 dB/decade rate. Thus, for the ideal curve, every time the frequency is reduced by ten, the output is reduced by 20 dB. A slight variation from this occurs in actual practice. The output is actually at -3 dB rather than 0 dB at the critical frequency. Related Problem If the frequency for the high-pass filter is decreasedto 10 Hz, what is the output to input ratio in decibels? 772 T BASICFILTERS RLHigh-Pass Filter A basic Rl high-pass filter is shown in Figure 19-16. Notice that the output is taken acrossthe inductor. FIGURE 19-16 RL high-passfilter. When the input frequency is at its critical value, X, = R, and the output voltageis 0.707V1. As the frequency increases abovef,, X; increasesand, as a result, the output voltage increasesuntil it equals V,,. The expressionfor the critical frequency of the highpass fllter is the same as for the low-pass fllter. - l t'- 2n1ua1 PhaseShiftin a High-Pass Filter Both the RC and the RI high-pass filters act as lead networks. Recall from Chapters16 and l7 that the phase shift from input to output for the RC lead network is -t(X.\ o=tan (n/ and for the RL lead network is d = 90"-,""- /4\ \ n / At the critical frequency,Xr= R and, therefore,6 = 45o.As the frequencyis increased, f decreasestoward 0o, as shown in Figure 19-17. F I G U R E1 9 _ 1 7 Phasecharacteristic of a high-passfilten BAND-PASS FILTERS. EXAMPLE 19-6 773 (a) In Figure 19-18, find the value of C so that X6 is ten times less than R at an input frequencyof l0 kHz. (b) If a 5 V sine wave with a dc level of 10 V is applied, what are the output voltage magnitude and the phase shift? F I C U R E1 9 - 1 8 c ,*"___)l_1_r*, l*'" t_ Solution (a) The valueof C is determinedasfollows: f2)= 68 O Xc = 0.1R= 0.1(680 C= | 2nlXs 2n(10kHzX68A) =0.234uF The neareststandardvalueof C is 0.22 uF. (b) The magnitudeof the sinewaveoutputis determinedasfollows: \-R 680f) I \., / V o ,=, ,l l\',,,=l--+15V=4.98V t Y R z +X t t \ V t o 8 o Q ) ' + i 6 8Q ) ' r The phaseshift is -,1X.\ , ' / 6 8O \ @=ran o \ /=,un \680o/=t., = At 7 19 kHz, which is a decadeabovethe critical frequency,the sinusoidaloutput is almostequalto theinputin magnitude,andthephaseshift is very small.The l0 V dc level hasbeenfilteredout anddoesnot appearat the output. RelatedProblem Repeatparts(a) and (b) of the exampleif R is changedto 220 Q. stcTtoN19-2 REVIEW 1. The input voltage of a high-pass filter is 1 V. What is Vo,, at the critical frequency? 2. In a certain high-pass filter, V;, = 1020" V R = 1.0 kO, and Xy = 15 kfl. Determine V,,,. FILTERS 19_3r BAND.PASS A band-passfiIter allows a certein band offreqaencies to pass snd attenuates or rejects allfrequencies below und ubove the passband. After completing this section, you should be able to I Analyze the operation ofband-pass filters . Show how a band-passfilter is implemented with low-pass and high-pass fllters . Define bandwidth 774 T BASICFILTERS . Explain the series-resonantband-passfilter . Explain the parallel-resonantband-passfilter . Calculate the bandwidth and output voltage of a band-passfilter Figure 19-19 showsa typical band-passresponsecurve. vor, v 0;70'7v^N (- 3 dB) F I G U R E1 9 - 1 9 Typicalband-passresponsecurve, Low-Pass/H igh-PassFilter A combination of a low-pass and a high-pass filter can be used to form a band-pass fil. ter, as illustrated in Figure 19-20. The loading effect of the secondfilter on the first must be taken into account. F I G U R E1 9 - 2 0 Low-passand high-passfilters usedto form a band-pass filter. If the critical frequency,fat1, of the low-pass filter is higher than the critical frequency,,fanl, of the high-pass filter, the responsesoverlap. Thus, all frequencies except those betweenf,q1 andf,11,are eliminated, as shown in Figure 19-21. F I G U R E1 9 - 2 1 Overlappingresponsecumes of a high-pass/low-pass filter. BAND.PASS FILTERS. 775 The bandwidth of a band-pass filter is the range of frequencies for which the current, and therefore the output voltage, is equal to or greater than 70.7Vo of its value at the resonant frequency. As vou know. bandwidth is often abbreviatedBW EXAMPLE19-7 filter withf, = 2 kllz anda low-passfilter withf, = 2.5 kHz areusedto A high-pass filter.Assumingno loadingeffect,what is the bandwidthof the constructa band-pass passband? = 2.5 kllz - 2 kHz = 500 Hz BW = f<o - [email protected]) RelatedProhlem If fdn = 9 kHz andthe bandwidthis I .5 kHz, what is f,q,y? Solution Filter SeriesResonantBand-Pass A type of series resonant band-passfilter is shown in Figure 19-22. As you learned in Chapter 18, a series resonant circuit has minimum impedance and maximum current at the resonantfrequency,/|. Thus, most of the input voltage is dropped across the resistor at the resonant frequency. Therefore, the output acrossR has a band-passcharacteristic with a maximum output at the frequency of resonance.The resonant frequency is called the center frequency,/6. The bandwidth is determinedby the quality factor, Q, of the circuit and the resonantfrequency,as was discussedin Chapter 18. Recall rhat Q = ytlp. r, *---lu[f! '*l'i v*, l--1__=4 c** Series resonant + R t l_ : n " t t t , f fo F I G U R E1 9 - 2 2 Seriesresonantband-pass filter. A higher value of Q results in a smaller bandwidth. A lower value of Q causesa larger bandwidth. A formula for the bandwidth of a resonantcircuit in terms of B is stated in the following equation: u*=t (1e-5) 776 T BASICFILTERS EXAMPLE 19-B Determine the output voltage magnitude at the center frequency ffi) and the bandwidth for the filter in Figure 19-23. F I G U R E1 9 - 2 3 Solution At /0, the impedance of the resonant circuit is equal to the winding reslstance,Ry,. By the voltage-divider formula, * -=\i rloo r o?)'o a)'u " = 9.09 v.,,-=\(n--!-\u, v R*)''' The center frequency is to- znyTc - 2nY (l mHX0.0022pF) = 107kHz At/6, the inductive reactanceis Xr= 2nfL = 2n(107 kHz)(l mH) = 672 I and the total resistance is R , o , =R + R w = 1 0 0f 2 + 1 0 O = 1 1 0Q Therefore.the circuit Q is x, =e ' t t 0^ = ; : , =f672f) fi The bandwidthis U*= 107kHz = l 7 . 5 k H z f" e=- RelatedProhlem If a I mH coil with a winding resistanceof 18 f2 replacesthe existingcoil in Figure 19*23,how is the bandwidthaffected? ParallelResonant Band-Pass Filter A type of band-passfilter using a parallel resonant circuit is shown in Figlre 19-24. Recall that a parallel resonantcircuit has maximum impedance at resonance.The circuit in Figure 19-24 actsas a voltage divider. At resonance,the impedanceof the tank is much greater than the resistance.Thus, most of the input voltage is acrossthe tank, producing a maximum output voltage at the resonant (center) frequency. FIGURE19-24 Parallel resonant band-passfilter. vi, BAND-PASS FILTERS. 777 For frequenciesabove or below resonance,the tank impedancedrops off, and more of the input voltage is acrossR. As a result, the output voltage acrossthe tank drops off, creating a band-passcharacteristic. EXAMPLE 19-9 What is the center lrequency of the filter in Figure 19-25?AssumeRw = 0 Q. F I G U R E1 9 - 2 5 The center frequency of the filter is its resonantfrequency. Solution = 5.03MHz ft= --!2nYLC 2nY(r0 pH)(100 pF) Related Prohlem EXAMPLE 19-10 Determine/6 in Figure 19-25 if C is changed to 1000 pF. Determine the center frequency and bandwidth for the band-passfilter in Figure 19-26 if the inductor has a windins resistanceof 15 Q. F I G U R E1 9 - 2 6 Solution Recall from Chapter 18 (Eq. 18-15) that the resonant(center) frequency of a nonideal tank circuit is " ft-6a*c/D @ - - - - r ! l . f f - 2tYLC 2n\/(50 mHX0.01p.F The Q of thecoil at resonance is , , _ I L _ 2 n f o L _ 2n(7.12kHz)(50mH) = 149 Rw Rw 15f) The bandwidth of the filter is 7.12 kITz u * = 6fn= - - * - = 4 7 . 8 H 2 Note that since Q > 10, the simpler formula (Eq. 18-6) could have been used to calculate/6. Related Problem Knowing the value of Q, recalatlate/. using the simpler formula. 778 r BASICFILTERS sEcTroN19-3 REVIEW 1. For a band-passfilter, f4ny = 29.8 kIIz andf"o = 30.2 kHz. What is the bandwidth? 2. A parallelresonantband-passfilter has the following values:Rw = l5 Q. L = 50 pH, and C = 470 pF. Determinethe approximatecenterfrequency. 19_4 I BAND.STOPFITTERS A band-stopfilter is essentially the opposite of a band-passfilter in terms of the responses.A band-stopfilter ullows all frequencies to pass except those lying within a certain stopband, After completing this section, you should be able to r Analyze the operation of band-stop filters . Show how a band-stop filter is implemented with low-pass and high-pass filters . Explain the series-resonantband-stop filter . Explain the parallel-resonantband-stop filter . Calculate the bandwidth and output voltage of a band-stop filter Figure 19-27 showsa generalband-stopresponsecurve. F I G U R E1 9 - 2 7 Generalband-stopresponsecurve. Vout V*o" 0.707vnax (-3 dB) igh-PassFilter Low-Pass/H A band-stop filter can be formed from a low-pass and a high-pass filter, as shown in Figure 19-28. F I G U R E1 9 - 2 8 Low-passand high-passfilters usedto form a band-stopfilter. If the low-pass critical frequency,,f,ai, is set lower than the high-pass critical frequency,f,17r,a band-stop characteristicis formed as illustrated in Figure 19-29. BAND-STOP FILTERSI 779 F I G U R E1 9 - 2 9 Band-stopresponsecurve. SeriesResonantBand-StopFilter A series resonant circuit used in a band-stop configuration is shown in Figure 19-30. Basically, it works as follows: At the resonantfrequency,the impedanceis minimum, and therefore the output voltage is minimum. Most of the input voltage is dropped acrossR. At frequencies above and below resonance,the impedance increases,causing more voltage acrossthe output. F I G U R E1 9 - 3 0 Seriesresonant band-stopfilter. EXAMPLE 19-11 vi, Find the output voltage magnitude at fs and the bandwidth in Figure 19-31. FIGURE,I9-31 R vi, 100mV 5 6 O ,p. w 2Q L 100 mH C 0.01rrF- + Solution Since X1 = Xc &t resonance,the output voltage is 2a v*, = (-3*-\r. = ( \too " " -,, " = 3.45mv \ n + n r / " - - \ s 8a 1 To determine the bandwidth, calculate the center frequency and Q of the coil. fo=--]-=+=5.03kH2 rv 2*Ee znr.(toO..rHxo.otrrp') kHzXl0OmH) n _ 4, _ 2nfL _ 2n(5.03 \t=-=R R 58fi 3.16kO = - 5 + . 5 58f) n w =4 = t' o= ' ,9"= 92.3:H2 o Related Prohlem width. s4.s Assume Rw= 10 f) in Figure 19-31. Determine Vou,ardthe band- 7BO r BASICFILTERS ParallelResonantBand-StopFilter A parallel resonantcircuit used in a band-stop configuration is shown in Figure 19-32.At the resonant frequency, the tank impedance is maximum, and so most of the input volt age appearsacrossit. Very little voltage is acrossR at resonance.As the tank impedance decreasesabove and below resonance.the output voltage increases. F I G U R E1 9 - 3 2 Parallel resonantband-stopfilten EXAMPLE19_12 Find the center frequency of the filter in Figure 19-33. Sketch the output response curve showing the minimum and maximum voltages. F I C U R E1 9 _ 3 3 Solution C The center frequency is , ^- = @ - @ = 5 . 7 9 M H 2 r0 znr,(spplrso pg z*Frc At the center (resonant)frequency, Xr= 2nfoL= 2n(5;79MHz)(5 pH) = 182 fl - 182-Q=ll.g o . =x, R* 8Cl Z,= Rv,{Qz+ 1) = 8 Q(22.82+ l) = 4.17 pg (purely resistive) Next, use the voltage-divider formula to find the minimum output voltage magnitude. v o , , \ .-i/, R \ , ,' ' - / 5 6 9 . 3 ) l o v = r . r 8 v " \ R+ z , ) " \ + . l l k o / ' " At zero frequency, the impedance of the tank is R1ybecauseXg = * and Xt = 0 Q. Therefore, the maximum output voltage below resonanceis v o u r '\ "m--' l, R-\,, -- / t 9 9 * ) l o V = 9 . 8 6 v * \ R R * ) ' ' ' \ s o so / ' " ' As the frequency increasesmuch higher than ftt, X6 approaches0 f), and Vou, approachesVi"(70 V). Figure 19-34 showsthe responsecurve. TECHNOLOGY THEORYINTO PRACTICE. FIGURE19-34 781 vo*(y) 10 9.86 / (MHz) Related Problem What is the minimum output voltage if R = 1.0 kfl in 19-33? sEcTtoN19-4 REVIEW 1. How does a band-stop filter differ from a band-passfilter? 2. Name threebasic ways to constructa band-stopfilter. 19-5 r TECHnologyTheory Into Practice In this TECH TIP,you will plot thefrequencyresponses of two typesof fi.ltersbased on u seriesof oscilloscope me&surements and i.ilentifythe typeof filter in euchcase. The filters are contained in sealedmodules as shown in Figure 19-35. You are concerned only with determining the filter resDonsecharacteristics and not the types of internal comDonents. F I G U R E1 9 - 3 5 Fiher modules. FilterMeasurement and Analysis I Refer to Figure 19-36. Based on the seriesof four oscilloscope measurements,create a Bode plot for the filter under test, specify applicable frequencies, and identify the type of filter. r Refer to Figure 19-31 . Based on the seriesof six oscilloscope measurements,create a Bode plot for the filter under test, specify applicable frequencies,and identify the type of filter. BASICFILTERS 2 V peak-to-peaksignal from function generator F I G U R E1 9 - 3 6 T E C H N O L O CT YH E O R YI N T O P R A C T I C E. Filter module 2 IN GND OUT 2 V peak-to-peaksignal fiom function generator F l c u R E1 9 - 3 7 SECTION19-5 REVIEW l. Explain how the waveforms in Figure 19*36 indicate the type of filter. 2. Explain how the wavefonns in Figure 19-37 indicate the type of filter. 783 784 I BASICFILTERS r SUMMARY I In an RC low-pass filter, the output voltage is taken acrossthe capacitor and the output lagsthe input. r In an RI low-pass filter, the output voltage is taken across the resistor and the output lags the input. r In an RC high-pass f,lteq the output is taken acrossthe resistor and the output leads the input. r In an Rl high-pass filter, the output is taken acrossthe inductor and the output leads the input. r The roll-off rate of a basic RC or RZ filter is -20 dB per decade. r A band-passfilter passesfrequenciesbetween the lower and upper critical frequenciesand rejects all others. r A band-stop filter rejects frequenciesbetween its lower and upper critical frequencies and passes all others. r The bandwidth of a resonant filter is determined by the quality factor (Q) of the circuit andthe resonant frequency. I Critical frequencies are also called -3 dB frequencies. r The output voltage is 70.'/Voof its maximum at the critical frequencies. r GLOSSARY These terms are also in the end-of-book glossary. Attenuation A reduction of the output signal compared to the input signal, resulting in a ratio with a value of less than I for the output voltage to the input voltage of a circuit. Band-pass filter A filter that passesa range of frequencieslying between two critical frequencies and rejects frequencies above and below that range. Band-stop filter A filter that rejects a range of frequencieslying between two critical frequencies and passesfrequencies above and below that range. Bode plot The graph of a filter's frequency responseshowing the change in the output voltageto input voltage ratio expressedin dB as a function of frequency for a constant input voltage. Center frequency ffi) The resonant frequency of a band-passor band-stop filter. Criticaf frequency The frequency at which a filter's output voltage is 70.7Voof the maximum. Decade A tenfold change in frequency or other parameter. Decibel A logarithmic measurement of the ratio of one power to another or one voltage t0 another,which can be used to expressthe input-to-output relationship of a filter. Passband The range of frequenciespassedby a filter. Roll-off r FORMULAS The rate of decreaseof a filter's frequencv resDonse. (1e-1) dB= 1oron " \/fu\ P-l Power ratio in decibels (re-2) dB=2or"* - fp) Voltage ratio in decibels \V* I (1e-3) , t'- 7 2nRC Critical frequency (re-4) t "' - I 2n(LlR) Critical frequency (1e-s) BW=b o Bandwidth PROBLEMS. r SEIF.TEST 785 1 . The maximum output voltage of a certain low-pass filter is 10 V. The output voltage at the critical frequency is (a) 10 V (b) 0 V (c) 7.07Y (d) 1.414v 2. A sinusoidal voltage with a peak-to-peak value of 15 V is applied to an RC low-pass filter. If the reactanceat the input frequency is zero, the output voltage is (a) 15 V peak-to-peak (b) zero (c) 10.6 V peak+o-peak (d) 7.5 V peak-to-peak 3. The same signal in Question 2 is applied to an RC high-pass filter. If the reactanceis zero at the input frequency, the output voltage is (a) 15 V peak-to-peak (b) zero (c) 10.6V peak-to-peak (d) 7.5 V peak-to-peak 4. At the critical frequency, the output of a filter is down from its maximum by (b) -3 dB (a) 0 dB (c) -20 dB (d) -6 dB If the output of a low-pass RC fllter is 12 dB below its maximum at f = 1 kJIz, then at f = l0 kHz, the output is below its maximum by (a) 3 dB (b) l0 dB (c) 20 dB (d) 32 dB 6. In a passive filter, the ratio Vou,lVi, is called (a) roll-off 1 (b) gain (c) attenuation (d) critical reduction For each decadeincreasein frequency above the critical frequency, the output of a low-pass filter decreasesby (a) 20 dB (b) 3 dB (c) 10 dB (d) 0 dB 8. At the critical frequency, the phase shift through a high-pass filter is (a) 90" (b) 0' (c) 45" (d) dependenton the reactance 9. In a seriesresonantband-passfilter, a higher value of Q results in (a) a higher resonant frequency (b) a smaller bandwidth (c) a higher impedance (d) a larger bandwidth 10. At seriesresonance, (a)Xr=Yt (b) Xc>XL G) Xg<XL 1 1 . In a certain parallel resonant band-passfilter the resonant frequency is 10 kHz. If the bandwidth is 2 kJ1z, the lower critical frequency is (a) 5 kHz t2. (b) 12 kIIz (c) 9 kHz (d) not determinable In a band-passfilter, the output voltage at the resonant frequency is (a) minimum (b) maximum (c) 10.7Voof maximum (d) 70.'7Voof minimum 13. ln a band-stop filter, the output voltage at the critical frequenciesis (a) minimum (b) maximum (c) 10.7Voof maximum (d) 70.77aof minimum 14, At a suffrciently high value of Q, the resonant frequency for a parallel resonant filter is ideally (a) much greater than the resonant frequency of a series resonant filter (tr) much less than the resonant frequency of a series resonant filter (c) equal to the resonant frequency of a seriesresonant filter r PROBLEMS More dfficult problems are indicated by an asterisk (*). SECTION19-1 Low-Pass Filters 1. In a certain low-pass filter, Xc = 500 O and R = 2.2 kQ. What is the output voltage (Vo,,) when the input is l0 V rms? 2. A certain low-pass filter has a critical frequency of 3 kHz. Determine which of the following frequenciesare passedand which are rejected: (a) 100 Hz (b) I kHz (c) 2WIz (d) 3 kHz (e) 5 kHz 786 T BASICFILTERS Determine the output voltage (V,,,) of each filter in Figure 19-38 at the specified frequency when V;" = 10 V. u" (a)f = 60Hz I Vou, (.b)f = 400H2 (c)/= u, (d) f = zkJlz lkHz F I G U R E1 9 - 3 8 4. What is/f for each f,lter in Figure 19-38? Determine the output voltage atlf in each case whenVr"=5Y. For the filter in Figure 19-39, calculate the value of C required for each of the following critical frequencies: (a) 60 Hz (b) 500 Hz (c) I kHz (d) 5 kHz F I G U R E1 9 - 3 9 14,*----ML-----?_----------o%. 22oa I T . l_ 6. Determinethe critical frequency for each switch position on the switched filter network of Figure 19-40. F I G U R E1 9 - 4 0 R 10ko 7. Sketch a Bode plot for each part of Problem 5. 8. For each following case,expressthe voltage ratio in dB: (a)V*=lY,Vo,,=l\ ( b ' )v i " = 5 Y , V o u , = J \ / ( c ) V * = l 0 v , V o n = ' 7 . 0 7Y (.d)Vi"=25Y,Vo,,= 5y 9. The input voltage to a low-pass RC filter is 8 V rms. Find the output voltage at the following dB levels: (a) -1 dB (b) -3 dB (c) -6 dB (d) -20 dB 10. For a basic RC low-pass filter, find the output voltage in dB relative to a 0 dB input for the following frequencies (f" = | Wlz): (a) 10 kHz (b) 100 kHz (c) 1 MHz . 787 PROBLEMS SECTION19-2 High-PassFilters 11. In a high-pass fiItel X6 = 500 A and R = 2.2 kA. What is the output voltage (V,ur) when Vi,=lOVrms? 12. A high-pass filter has a critical frequency of 50 Hz. Determine which of the following frequencies are passedand which are rejected: (a) 1 Hz (b) 20 Hz (c) 50 Hz (d) 60 Hz (e) 30 kHz 13. Determine the output voltage of each filter in Figure 19-41 at the specified frequency when Vi,= l0Y' rin (a) f = 6ottz $) f = a00Hz (c)/= (d) f = zkHz lkHz FIGURE 19-41 14. What isf, for each filter in Figure 19-41? Determine the output voltage atf, in each case (V- = l0 V)' 15. Sketch the Bode plot for each filter in Figure 19-41. *16. Determinef, for each switch position in Figure 19-42. FIGURE19-42 Voutt R2 1.0ko '-'-ll Vout2 "2 o.otp v,, n) 3 0.015pF R R5 2.2kO l 4 VoutZ Voutl SECTION19-3 Band-Pass Filters 17. Determine the center frequencv for each filter in Fisure 19-43. FTGURE 19-43 r v'[email protected] lznrLI 0.01pF R 7sa (a) (b) 7BB T BASICFILTERS 18. Assuming that the coils in Figure 19-43 have a winding resistance of 10 Q, find the bandwidth for each filter. 19. What are the upper and lower critical frequenciesfor each filter in Figure 19-43? Assumethe responseis symmetrical about/6. 20. For each fllter in Figure 1944, find the center frequency of the passband.Neglect R1y. F I G U R E1 9 - 4 4 vi, 21. If the coils in Figure 1944 have a winding resistanceof 4 Q, what is the output voltageat resonancewhen V;, = 120Y? *,,) Determine the separationof center frequencies for all switch positions in Figure 19-45. Do any of the responsesoverlap? Assume Rw = 0 C) for each coil. F I G U R E1 9 - 4 5 Lr 50 pH R 4. 100prH L3 1000pF L2 0.001prF *23. Design a band-passfilter using a parallel resonant circuit to meet all of the following specifications: BI4z= 500 Hz; Q = 40; and [email protected]*\ = 20 mA, V61^*y= 2.5 Y. SECTION19-4 Band-Stop Filters 24. Determine the center frequencv for each filter in Fisure 19-46. PROBLEMS. 789 FrcuRE19-46 vi, r 0.0022 1lF (a) 0.041 1tF (b) 25. For each filter in Figure 1947 , find the center frequency of the stopband. 26. If the coils in Figtre 1941 have a winding resistanceof 8 Q, what is the output voltage resonancewhen V;n= 50 V? F I G U R E1 9 - 4 7 Vin *27. Determine the values of L1 and L2 in Figure 19-48 to pass a signal with a frequency of I2O0 kHz and stop (reject) a signal with a frequencv of 456 kHz. F I G U R E1 9 - 4 8 EWBTroubleshooting and Analysis Theseproblems require your EWB compact disk. 28. Open file PROI9-28.EWB and determine if there is a fault. If so, find the fault. 29, Open file PRO19-29.EWB and determine if there is a fault. If so, find the fault. 30. Open file PRO19-30.EWB and determine if there is a fault. If so, find the fault. 31. Open file PROI9-3l.EWB and determine if there is a fault. If so, find the fault. 32. Open file PRO19-32.EWB and determine if there is a fault. If so, find the fault. 33. Open file PRO19-33.EWB and determine if there is a fault. If so, find the fault. 34. Open file PRO19-34.EWB and determine the center frequency of the circuit. 35. Open file PRO19-35.EWB and determine the bandwidth of the circuit. 790 I BASICFILTERS r ANSWERS TO SECTION REVIEWS Section19-1 1. The passbandis 0 Hz to 2.5 kJ{z. 2. You,= 1002-88.9' mV rms 3, 20 Iog(V",,/V*) = -9.54 dB Section19-2 l. Vou,=0.707Y 2. You,= 9.9813.81"Y Section19-3 l. BW= 30.2kIlz - 29.8kJIz= 400Hz 2. f6= I.04MHz Section '19-4 l. A band-stop filter rejects, rather than passes,a certain band of frequencies. 2. High-pass/low-passcombination, series resonant circuit, and parallel resonant circuit Section19-5 l. The waveforms indicate that the output amplitude decreaseswith an increase in frequency asin a low-pass filter. 2. The waveforms indicate that the output amplitude is maximum at 10 kHz and drops off above and below as in a band-passfllter. r ANSWERS TO RELATED PROBLEMS FOR EXAMPLES t9-l -1.41 dB l9-2 7.23kHz l9-3 f, increases to 159kHz. Roll-off rateremains-20 dB/decade. 194 f" increases to 350 kHz. Roll-off rateremains-20 dB/decade. -60 19-s dB 19-6 C = 0.723y.F;Vou,= 4.98V; d = 5.7" l9-7 10.5kHz 19-8 B![/increases to 18.8kHz. l9-9 1.59MHz 19-10 7.12kHz (no significantdifference) l9-ll V"il = 15.2mV; BW = l05Hz t9-12 t94V r. (c) I ANSWERS TO SELF.TEST e. (b) 2. (b) 10. (a) 3. (a) 11. (c) 4. (b) 12. (b) s.(d) 13. (c) 6. (c) 1.4.(c) /. (al 8. (c)