MODULE TITLE : OPEATIONAL AMPLIFIES TOPIC TITLE : FILTES LESSON : FILTES OA - 4 - Teesside University 0
INTODUCTION An electrical filter is a device which is designed t pass sme frequencies and reject thers. Filters are fund in abundance in analgue systems, in applicatins ranging frm simple pwer supplies t the cmplexities f the clur televisin receiver. Traditinally, filters have been cnstructed frm passive cmpnents using the resnant prperties f inductrs and capacitrs t give sharply tuned circuits. Hwever, inductrs are large, heavy and cstly cmpnents and, mrever, they cannt be manufactured t a clse tlerance. In this lessn, we see hw an active device, the p-amp, can be used t cnstruct 'inductr-less' filters. Active filters find many applicatins in signal prcessing, particularly at lwer frequencies. At high frequencies (abve, say, MHz) the limited gain-bandwidth prduct f p-amps precludes their use in active filters; then inductrs have t be used, but physically an inductr at very high frequencies might just be a few turns f wire. YOU AIMS After cmpleting this lessn, yu shuld be able t: describe the ideal amplitude-frequency respnse f the fur basic types f filter cmpare the respnses f the fur basic Sallen & Key type filters with thse f LC circuits Teesside University 0
describe the effect f gain upn a Sallen & Key filter state what is meant by the rder f a filter shw hw first and secnd rder filters can be cascaded t prduce higher rder filters describe the principle features f Butterwrth and Chebychev filters shw hw a filter's phase distrtin can alter the shape f a signal. STUDY ADICE The analysis f the behaviur f filters is highly mathematical and cnsequently this is a lng and fairly arduus lessn. T prevent the text frm becming verburdened sme f the mre cmplex ideas have been develped in Appendices t the lessn. This material will nt be assessed. Other areas are develped in the Self-Assessment Questins and, t this end, yu are encuraged t tackle Self-Assessment Questins t 5 as yu encunter references t them in the lessn, rather than leaving them t the very end. Teesside University 0
3 IDEAL FILTES The ideal filter has a 'brick wall' respnse that passes the required frequencies withut attenuatin and cmpletely suppresses all ther frequencies. FIGUE gives the ideal respnses and blck diagram symbls f the fur basic types f filter : lw-pass, high-pass, band-pass and band-stp. The ideal respnse is suggested by the brick hatching; n current can pass thrugh a brick wall! The transitin between the filter pass and stpband ccurs at the cut-ff frequency f c. The ideal lw and high-pass filters have a single cut-ff frequency, whereas the band-pass and band-stp filters have lwer and upper cut-ff frequencies (f c and f c respectively). Teesside University 0
4 (a) Lw-pass filter i Passband Stp-band f c f i i Stp-band Pass-band (b) High-pass filter f c f i i Pass-band (c) Band-pass filter f c f c f i (d) Band-stp filter i Stp-band i f c f c f FIG. Teesside University 0
5 The practical filter falls smewhat shrt f the ideal. FIGUE shws the type f respnse we might expect frm a real band-pass filter; the sketch als shws mre f the terminlgy used t describe filters. /db 3dB ipple band Stp-band B 3dB Pass-band Stp-band Skirt f filter f FIG. Nte that the passband may well have a rippled respnse. The skirt f the respnse shuld rll ff as fast as pssible, but it is fund that the mre rapid the rll-ff the greater the passband ripple. C FILTES Simple filters can be built frm C elements. FIGUE 3 shws hw the fur basic filters can be cnstructed frm resistrs and capacitrs. The filters are crude because their respnses are a lng way frm the ideal; remember that a single C element will nly give a 6 db rll-ff per ctave. The band-pass filter can be frmed frm cascaded lw and high-pass filters. The lw-pass filter lets thrugh all frequencies belw the upper cut-ff Teesside University 0
6 frequency f c and the high-pass filter lets thrugh all frequencies abve the lwer cut-ff frequency f c. The band-stp filter can be frmed frm parallel lw and high-pass filters. The lw-pass filter rejects all frequencies abve the lwer cut-ff frequency f c and the high-pass filter rejects all frequencies belw the upper cut-ff frequency f c. i C FIG. 3(a) Lw-pass C i FIG. 3(b) High-pass Teesside University 0
7 C i C FIG. 3(c) Band-pass C 3 i C 4 FIG. 3(d) Band-stp The C circuit f FIGUE 3(d) perhaps warrants a little further explanatin. The lw-pass filter is frmed by the cmpnents and C and the high-pass Teesside University 0
8 filter and C. The resistr 4 prevents C frm shrt-circuiting the utput at high frequencies. The resistr 3 is included simply t give equal utput either side f the stpband; withut it, wuld have a much greater amplitude abve the stpband than belw it. The cmpund filters f FIGUE 3(c) and (d) will nly wrk prperly if the cut-ff frequencies f c and f c are spaced well apart, s that f c is at least 8f c. The resulting minimum bandwidth f 7f c may well be t wide fr many applicatins. FIGUE 4 gives typical frequency respnses fr the C band-pass and bandstp filters. It can be seen that the filters are far frm ideal. Nte in particular the 0 db attenuatin f the band-stp filter in the passbands. Estimate the 3 db bandwidth f the C band-pass and band-stp filters. Teesside University 0
9 0 4 6 8 0 4 6 8 0 Frequency/Hz Frequency/Hz 0 00 k 0 k 00 k 0 00 k 0 k 00 k 8 0 4 i /db 6 8 0 Band-stp filter Band-pass filter (a) (b) FIG. 4 /db i
0 FIGUE 5 is a reprductin f FIGUE 4 shwing the 3 db bandwidths. In the case f the band-pass filter the 3 db pints are measured 3 db dwn frm the peak respnse f the filter. This gives f c as 60 Hz and f c as khz, the bandwidth, B, f the filter is 840 Hz. It is nt quite s bvius where t determine the bandwidth f the band-stp filter. D we measure it frm 3 db abve the bttm f the trugh r 3 db dwn frm the pass-bands? It seems t make mre sense t say that the filter shuld reject all frequencies 3 db belw the pass-bands than t pass all frequencies 3 db abve the stpband. This is hw the crner frequencies have been estimated in FIGUE 5, t give a bandwidth f abut 400 Hz. Teesside University 0
0 4 6 8 0 4 6 8 0 Frequency/Hz Frequency/Hz 0 00 k 0 k 00 k 0 00 k 0 k 00 k f c f c f c f c 8 0 3dB B 4 i /db 3dB B 6 8 0 Band-stp filter Band-pass filter (a) (b) FIG. 5 /db i
LC FILTES esistance, inductance and capacitance can be used t frm LC circuits with very high Q-factrs. The steep-sided respnse f such sharply tuned circuits gives very gd rejectin f unwanted frequencies and very narrw bandwidths can be achieved. C L FIG. 6 Series LC High-pass Filter FIGUE 6 shws a series LC circuit, the utput being taken acrss the inductr. The transfer functin f this arrangement is : ()... j Q where is the natural, undamped, resnant frequency f the circuit and Q the quality factr f the circuit, i.e. Teesside University 0
3 L and Q LC C (Self-Assessment Questin asks yu t derive equatin ()). In FIGUE 7, the amplitudes f the transfer functins (expressed in decibels) have been pltted against the nrmalized angular frequency fr varius values f Q. The respnse curves are thse f a high-pass filter where a high Q gives a very peaked respnse and where a lw Q (less than ) gives a mre gentle respnse with n versht f the x-axis. Nte als that, fr a Q f r greater, the minimum rll-ff is abut 40 db per decade. The respnse f a simple C circuit (r L fr that matter), with its 0 db per decade rll-ff, has als been pltted fr cmparisn. Teesside University 0
4 0 Q 0 0 /db 0 0 0. 0 00 C Q Q 0. 0 30 40 FIG. 7 Series LC Lw-pass Filter The utput f the series LC circuit culd, f curse, be taken frm acrss ne f the ther cmpnents t give a different respnse. Teesside University 0
5 Make a rugh sketch f the transfer functin if the utput is taken acrss the capacitr fr: (a) a Q factr f. (b) a Q factr f 0. Teesside University 0
6 FIGUE 8 shws the respnse is that f a lw-pass filter. Q 0 /db 0 Q 0. 0 FIG. 8 Series LC Band-pass Filter A band-pass filter is btained if the utput is taken frm acrss the resistr. FIGUE 9 shws the amplitude/frequency respnse fr Q-factrs f and 0. Teesside University 0
7 0 0..0 0 00 /db 0 0 Q 30 Q 0 40 FIG. 9 Estimate the 3 db bandwidths f bth f the band-pass filter respnse curves f FIGUE 9 if f is khz. Teesside University 0
8 FIGUE 0 shws the relevant parts f the respnse curves. B 0 0..0 0 3dB /db 0 Q 0 B Q 0 FIG. 0 Fr Q, the half pwer pints ccur at f 600 Hz and f 600 Hz. (Nte that, despite appearances given by the lgarithmic scale, the half-pwer pints are nt symmetrically placed abut the resnant frequency.) Thus the bandwidth is abut khz. Estimatin f the bandwidth fr Q 0 is trickier due t the sharply peaked respnse. ery rughly, the half pwer pints ccur at f 900 Hz and f 060 Hz, giving a bandwidth f abut 60 Hz. We discver, later, a much mre cnvenient methd f estimating bandwidth when the Q-factr is knwn. Teesside University 0
9 The transfer functin f the series LC circuit when used as a band-pass filter can be shwn t be (see Self-Assessment Questin ): jq... ( ) where again L and Q LC C The 3dB-bandwidth f the band-pass filter The vltage gain f the band-pass filter falls by 3 db when the mdulus f the denminatr f equatin () is equal t ; that is when: Q Slutin f this equatin (Self-Assessment Questin 3) gives nrmalized halfpwer angular frequencies f : and 4 Q Q 4Q Q The nrmalized bandwidth and B B, that is t say. Q is given by the difference between Teesside University 0
0 The term means that the respnse curve is nt symmetrical 4Q abut the centre frequency. This effect is illustrated in the curves (A) and (B) f FIGUE. Hwever, when the Q-factr exceeds 4, the ffset is less than % f ; fr high Q circuits, we can assume the bandwidth t be equally distributed abut as shwn in curve (C). (Nw refer t Self-Assessment Questin 4). Increasing Q A C B FIG. Teesside University 0
Calculate the nrmalized half-pwer frequencies, f and f, f a band-pass filter having a Q-factr f : (i) (ii) 0. Teesside University 0
f f and f 4Q Q f 4Q Q (i) When Q : f f 6 03 0 5 0 78... 4 f f 6 03 0 5 8... 4 (ii) When Q 0 : f f f f 400 400 00 0 05 0 95... 0. 00 0. 05. 05 0 We have seen that the LC cmbinatin can give sme very useful frequency respnses and wuld be suitable fr use as a filter. At resnance, the circuit can be sharply tuned (that is f a high Q factr) t give gd rejectin f unwanted frequencies. Unless wrking at very high frequencies thugh, inductrs are large, heavy, cstly and bth the subject f and the surce f magnetic interference. Mrever, they are nt readily cmpatible with micr-electrnic circuits. Teesside University 0
3 ACTIE FILTES The type f filters we have lked at s far in this lessn have all been passive, that is t say, they have nly cnsisted f passive cmpnents, namely resistrs, capacitrs and inductrs. We will nw shw hw the use f an active device, in cnjunctin with nly resistrs and capacitrs, can give the respnse f an LC circuit. The active device used is the ubiquitus p-amp. First cnsider the integratr circuit f FIGUE. C v v FIG. Summing the currents at the inverting input, using the 'virtual earth' cncept: where Xc X 0 c jc Teesside University 0
4 This gives a transfer functin f: jc j c Sketch a rugh gain/frequency respnse (Bde plt) f the integratr. At what frequency is its utput a maximum? FIGUE 3 shws the frequency respnse f the integratr. The gain rlls ff at 0 db per decade, crssing the 0 db axis at a frequency f f. Maximum utput ccurs fr a frequency f 0 Hz (d.c.) when, in thery, the utput vltage is infinite but, in practice, is limited by the supply rail. Teesside University 0
5 /db ll-ff at 0 db per decade 0 f f 0 0 f 0 0f 0 FIG. 3 The integratr can be cnverted int a lw-pass filter by placing a resistr in parallel with the feedback capacitr, as shwn in FIGUE 4. C FIG. 4 Teesside University 0
6 The transfer functin f the circuit is given by: where jc This yields the nw familiar frm: G ( ) j Q c... 3 where G (the d.c. gain) and c C. The filter represented by equatin (3) is said t have a first-rder respnse, as the highest pwer f is. A first-rder filter will give a rll-ff f 0 db per decade. Higher rder filters will give a mre ideal 'brick wall' respnse with steeper rll-ffs. A secnd-rder transfer functin, giving a rll-ff f 40 db per decade, will be f the frm: j Q Again, is the natural, undamped, resnant frequency f the filter. The equatin shuld be cmpared with the transfer functin f the high-pass LC filter given earlier, which als is secnd rder. Teesside University 0
7 THE SALLEN & KEY LOW PASS FILTE FIGUE 5(a) shws the circuit f a Sallen & Key* lw-pass filter. Nte that there are tw feedback paths; in this particular example 00% feedback has been applied t the inverting input, giving unity lw-frequency gain. Feedback t the nn-inverting input is via the capacitr C, the amunt f feedback applied increasing with frequency and the feedback rati H being a functin f. C C FIG. 5(a) What type f psitive feedback des the circuit apply (i.e. current-derived, series-fed, etc.)? * Sallen & L Key published their classical paper "A Practical Methd f Designing C Active Filters" in 955. Their riginal designs wuld have used therminic valves rather than p-amps as the active devices! Teesside University 0
8 The circuit applies vltage-derived, shunt-fed feedback. 4 3 FIG. 5 (b) It is shwn in Appendix that the vltage transfer functin f a circuit having this general frm is: 3 3 4... ( A) The impedances,, 3 and 4 are as shwn in FIGUE 5(b). Fr cnvenience, we shall assume that all the resistrs and all the capacitrs are f equal value. Thus, fr this particular filter circuit,,, 3 and 4 jc jc Teesside University 0
9 Substituting fr,, 3 and 4 int (A), jc jc ( jc) Therefre, C j C... 4 ( ) This is a secnd-rder transfer functin f the fllwing frm: j Q... 5 ( ) The transfer functin f the Sallen & Key filter is f exactly the same frm as that f the LC lw-pass circuit and we have thus realized a secnd-rder filter withut using an inductr. Cmparing the cefficients f equatins (4) and (5) reveals that the filter has a natural frequency f: f πc and a Q factr f: Q FIGUE 6 shws the nrmalized frequency respnse f ur secnd-rder lw-pass filter. The respnse has a decided drp that may nt be acceptable fr sme applicatins. A sharper respnse can be achieved by raising the Q- factr. We see hw this can be dne in the next sectin. The natural frequency, f, f a secnd rder filter is nt t be cnfused with its 'half-pwer' frequency. See, fr example, Self-Assessment Questin 3. Teesside University 0
30 0 0..0 0 /db 0 Q 0 FIG. 6 Filter with gain The abve analysis has been fr 00% feedback. FIGUE 7 shws hw the negative feedback path f the filter circuit can be mdified t give the circuit a measure f gain. Teesside University 0
3 C C FIG. 7 If, in equatin A derived in Appendix, 3 and 4 jc, then the transfer functin f the lw-pass filter can be shwn t be: G 3 j ( G)... ( 6) where G By cmparing the cefficients f the abve equatin (6) with the 'standard' secnd rder equatin (5), find an expressin fr Q in terms f the gain G. Teesside University 0
3 Cmparisn f the tw equatins shws that Q 3. G Thus we set the value f G, fr desired Q, by suitable values f and. In practice, thugh, G has t be limited t less than abut.66 (Q less than abut 3) fr this circuit; larger values f G make the circuit's Q-factr susceptible t variatin due t cmpnent tlerances. The curves f FIGUE 8 shw the effect f G upn the nrmalized frequency respnse (nte that the gain has als been nrmalized by dividing by G). Higher gain leads t a mre rapid initial rll-ff but with a peaked respnse. S, ne f the tasks f filter design is t achieve an acceptable balance between the cnflicting interests f rapid rll-ff and an even respnse in the passband. 0 G.9 0 G.8 /db G 0 0 G 0. 0 G 0 FIG. 8 Teesside University 0
33 Calculate the Q-factr fr each f the curves f FIGUE 8. Q 3 G Fr when G, Q 3 G 05. Fr when G, Q 3 0. Fr when G 8., Q 3. 8 5 and finally, Fr when G 9., Q 3. 9 0 A Design Example A lw-pass filter is t have an undamped, natural resnant frequency, f, f 3.4 khz. Design a suitable equal C/equal, Sallen and Key filter having a Q-factr f. The capacitrs are t have a value f 0 nf. Teesside University 0
34 Slutin The angular resnant frequency,, is given by: C Thus C πf C Fr C 0 nf and f 3.4 khz, π 3400 0 09 468 Ω The required gain is given by : G 3 Q Q frm 3 G and fr Q : G 3 5. The gain is set by the rati f the resistrs in the negative feedback lp. G Thus 5. 5. Teesside University 0
35 T minimize the effects f ffset currents, the resistances n the inverting and nn-inverting inputs f the p-amp shuld be f the same value; that is t say: 468 468 936 Ω Making the substitutin.5 : 5. 5. 936 yielding 5 603 Ω and 3 405 Ω In practice, f curse we wuld have t cmprmise ur design by selecting resistrs t preferred values. Thus, and wuld have t be 4.7, 6 and 4 kω respectively, when selected frm the E4 range. Frtuitusly, thugh, the crrect rati fr and is maintained. SALLEN AND KEY HIGH-PASS FILTE A Sallen & Key high-pass filter can be cnstructed by the simple expedient f interchanging the resistrs and capacitrs in the psitive feedback lp. FIGUE 9 shws an equal C/equal, unity gain circuit. Teesside University 0
36 C C FIG. 9 Fllwing frm equatin A in Appendix I, the transfer functin f the circuit is: j where the angular resnant frequency, as usual. C The Q-factr is the same as fr the unity gain lw-pass filter, i.e. Q 0.5. FIGUE 0 gives the nrmalized frequency respnse f the filter. Teesside University 0
37 0 0..0 0 /db 0 0 30 Q 40 FIG. 0 As fr the case f the lw-pass filter, the rll-ff can be imprved by intrducing gain int the circuit t raise the Q-factr. The transfer functin, with gain G, becmes: G 3 j ( G) FIGUE shws the effect f gain upn the nrmalized frequency respnse. Teesside University 0
38 0 Q 0 0 Q 5 0 Q 0. 0 00 /db 0 Q 0.5 G 0 30 40 FIG. SALLEN & KEY BAND-PASS FILTE A wide band-pass filter can be simply cnstructed frm cascaded lw-pass and high-pass p-amp filters. This methd, hwever, is nt suited t narrw bandpass filters as very clse tlerance cmpnents wuld have t be used. FIGUE gives the circuit f a narrw band-pass filter using a single pamp stage. Nte that fr this particular circuit the feedback is t the inverting input. Teesside University 0
39 C C FIG. In Appendix it is shwn that the filter has the transfer functin: jc jc... ( A7) emember that in this derivatin we impsed the design cnstraint that the capacitrs will be f equal value, i.e.: C C C. A little algebraic manipulatin can get this equatin int the 'standard frm' f equatin (). (eferring back yu will see that equatin () is the respnse f the LC band-pass filter). Equatin (A7) can be put int this frm by dividing thrugh by the factr : Teesside University 0
40 jc jc j C jc S that: C j jc This equatin can be given the same treatment as that fr the equatin fr series resnance in the LC band-pass filter. (Nw cmplete Self-Assessment Questin 5.) esnance ccurs, at an angular frequency, when is real, that is when: C C Thus C Teesside University 0
4 At resnance: This is the mid-band gain f the filter. The terms C and in the denminatr represent the nrmalized C vltages acrss the 'L' and 'C' f the circuit (where the '' represents the nrmalized vltage acrss the ''). The nrmalized vltage acrss 'L' r 'C' f the filter at the resnant frequency will equal the Q-factr f the circuit reactive (remember Q at resnance and in this instance resistive ). resistive S the Q-factr f the filter can be fund by putting int either f the abve tw terms. Thus: C Q C C C Q If we make the further restrictin that, find the values f mid-band gain, f and Q in terms f the circuit parameters C and. Teesside University 0
4 Obvusly, mid-band gain Q C C f πc BAND-STOP FILTE A wide bandwidth band-stp filter can be built frm lw and high-pass filters placed in parallel as shwn in FIGUE 3 (ppsite), the third p-amp acting as a summing amplifier. Narrw band-stp filters cannt, thugh, be realized using this methd. FIGUE 4 (ppsite), shws the circuit f a narrw band-stp filter using just a single p-amp stage. Teesside University 0
43 FIG. 3 3 C C C 3 FIG. 4 Teesside University 0
44 The input circuit cnsists f a 'twin-tee' netwrk, the 'tees' being frmed by / /C 3 and C /C / 3. The transfer functin f the circuit is given withut prf; fr the cnditin 3, and C C C 3 C then: j where C The similarity f this equatin with that fr the band-pass filter will nt escape yu. Estimate the utput f the filter when: (i) << (ii) (iii) >>. Teesside University 0
45 (i) When <<, then << and the transfer functin can be apprximated t: j j and as <<, then >>, s that: The filter is perating in the lwer passband and lets all frequencies thrugh unattenuated. (ii) When, then 0 and the imaginary term in the denminatr tends t inifinity. Thus the stpband. 0, the filter gives n utput and is perating in (iii) When >>, then >> and the transfer functin can be apprximated t: j and as >>, then >>, s that : The filter is perating in the upper passband and lets all frequencies thrugh unattenuated. Teesside University 0
46 FIGUE 5 gives the respnse f the band-stp filter. The ideal respnse is als shwn and explains why this type f filter is ften referred t as a ntch band-stp filter. 0 0..0 0 /db 0 Ideal respnse 0 30 40 FIG. 5 HIGHE ODE FILTES Till nw, the filters we have analyzed have all been f first r secnd rder. Higher rder filters can be used t give repnses clser t the ideal 'brick' wall, as they will have much sharper rll-ffs. Teesside University 0
47 The rder f passive LC filters can be determined by the number f energystrage elements they pssess. The energy strage elements in an LC filter are, f curse, the inductrs and capacitrs. Thus the filters f FIGUE 6 are f first, secnd and third rder. FIG. 6(a) FIG. 6(b) FIG. 6(c) Teesside University 0
48 In active filters, nly ne type f energy strage element is used, the capacitr. Thus an n-th rder active filter will have n capacitrs acting as energy strage elements. Higher rder active filters can be built by cascading lwer rder filters. Fr example, the Sallen & Key filter is f secnd rder, s cascading tw such filters, having the same cut-ff frequency, can give a furth rder filter. The idea is depicted in FIGUE 7. 3rd rder nd rder st rder 4th rder nd rder nd rder 5th rder nd rder nd rder st rder FIG. 7 Filters can be categrized by the mathematical frm f their transfer functin. We have seen that a first rder transfer functin has the frm: j Teesside University 0
49 and a secnd rder functin the frm: j Q Similarly, a third rder functin has the frm: 3 a j b j c j The denminatrs in each f the abve examples are plynmials f n-th rder filter will have an n-th rder plynmial in its transfer functin: j. An a n n j a n n j a n n j... a... 7 j ( ) Tw f the mst widely used types f filter have transfer functins invlving Butterwrth* and the Chebychev* plynmials. * Butterwrth, a pineer in electrnics, published his paper 'On the Thery f Filter Amplifiers' in Wireless Engineer in 930. Chebychev was a ussian mathematician, 8-894. Teesside University 0
50 The Butterwrth espnse The Butterwrth filter has a magnitude respnse f the frm: n... ( 8) where n is the rder f the filter. In rder t realize this respnse, the value f the cefficients a n, a n, a n... in equatin (7) must be thse f a Butterwrth plynmial. As Butterwrth filters are widely used, the values f the Butterwrth plynmial cefficients have been evaluated and given in tabular frm fr use by circuit designers. The cefficients up t the rder f 4 are given in the table belw. 0 a a a 3 a 4.000.44.000 3.000.000.000 4.63 3.44.63.000 TABLE A : Butterwrth Cefficents (t 3 decimal places) Dubtless yu are still a little mystified abut the the cnnectin between the cefficients and equatin (7). An example wuld be timely! Teesside University 0
Example Cnsider a third rder Butterwrth filter. Its plynmial will be: where, frm TABLE A, a, a and a 3. Thus the transfer functin becmes: Thus 3 4 4 4 4 6 4 4 3 j j j 3 j 3 3 a j a j a j 5 Teesside University 0
5 6 Using the general frm f the equatin. n where n 3 Nw, the plynmial 3 j j j can be factrized t j s that the transfer functin becmes: j j j j j j j The 3rd rder filter can nw be built by cascading a Sallen & Key secnd rder lw-pass filter with a lw-pass first rder filter, as shwn in FIGUE 8. Teesside University 0
53 C 3 C C 3 FIG. 8 In FIGUE 8, the lw-pass filter is frmed by 3 /C 3, the simple C element being driven by the p-amp vltage fllwer. Determine, frm the equatin n the half-pwer frequency f a Butterwrth filter. Teesside University 0
54 At half pwer: n That is: n n Thus at half pwer. It is ne f the prperties f a Butterwrth filter that the half-pwer frequency is always, irrespective f the rder f the filter. Anther mst imprtant prperty f the Butterwrth filter is that it gives the maximally flat passband respnse. By this we mean that the respnse rlls ff as steeply as pssible, cmmensurate with n versht in the pass-band. FIGUE 9 gives the nrmalized respnse f a Butterwrth lw-pass filter fr n t 4. Nte that fr n the respnse rlls ff at 0 db per decade; this respnse crrespnds t the Sallen and Key filter having a Q-factr f. Teesside University 0
55 0 0..0 0 /db 0 n 0 30 n 4 n 40 Butterwrth respnse n 3 FIG. 9 The Chebychev espnse The mathematics f the Chebychev respnse are, unfrtunately, even mre cmplex than that fr the Butterwrth, and we will give them but the mst cursry glance here. The magnitude respnse f a Chebychev filter is f the frm: ( ε C n ) Teesside University 0
56 where ε is the ripple factr f the filter and C n is a Chebychev plynmial f rder n. In ur applicatin the plynmial is a functin f and the first fur in the series are: C C C C 3 4 3 4 3 8 4 8 FIGUE 30 shws Chebychev frequency respnses fr values f n frm t 4 and with ε. The respnse f a first rder Chebychev filter with ε is the same as that f a first rder Butterwrth filter. This means that a specified rll-ff can be achieved in fewer stages by using a Chebychev filter. There is, f curse, a penalty thugh : the filter's respnse becmes uneven, r rippled, in the passband. The amunt f passband ripple is cntrlled by the value f ε, the larger the value f ε the greater the ripple. Fr example, setting ε t gives a ripple f 3 db, this can be reduced t db by setting ε t 0.5. In ding s, thugh, the skirt f the respnse is lifted by abut 6 db. A design cmprmise has t be reached between the rate f rll-ff and permissible ripple in the passband. Teesside University 0
57 0 0..0 0 0..0 /db 0 0 n 0 n n n 3 3 n 4 30 40 n 4 n n 3 FIG. 30 FILTES AND PHASE DISTOTION Phase distrtin ccurs when the phase relatinships present in a signal are upset. As a simple illustratin f this pint, FIGUE 3(a) shws a cmpsite signal frmed frm a fundamental frequency f and its third harmnic f 3. The signal is nw passed thrugh an electrnic device that has unity gain but a phase respnse f that shwn in FIGUE 3(c). The utput f the device will be that f FIGUE 3(b), the signal has been distrted. It can be seen that phase distrtin alters the shape f the signal. Teesside University 0
58 Electrnic f device f (a) (b) 0 0. 0 f f Phase shift 90 80 (c) FIG. 3 Unfrtunately, bth Butterwrth and Chebychev filters will have a phase repnse similar t ur electrnic device and will, therefre, cause phase distrtin. In sme applicatins, such distrtin will be f little cnsequence; fr example, speech will tlerate large amunts f phase distrtin as the ear is insensitive t the phase relatinships present in the cmplex signal impressed upn it. In ther applicatins, thugh, such as in the prcessing f clur vide signals r in sme data transmissin systems, phase distrtin must be kept t a minimum and the use f Butterwrth r Chebychev filters may have t be avided. Teesside University 0
59 Fr such applicatins there is ne type f filter, the Bessel* filter, that has been designed t minimize phase distrtin in the passband. The disadvantage f the Bessel filter is its inferir magnitude respnse. FIGUE 3 cmpares the magnitude respnses f 5th rder Bessel, Butterwrth and Chebychev filters. Chebychev /db Bessel Butterwrth f FIG. 3 * Bessel, 784-846, was a German mathematician and astrnmer, wh, besides his many cntributins t mathematics, was able t predict the existence f unknwn stars frm the erratic mtin f knwn stars. These predictins were nt verified until after his death. Teesside University 0
60 SELF-ASSESSMENT QUESTIONS. Derive the transfer functin f the circuit f FIGUE 6, i.e. shw that: where... j Q L and Q LC C. (). Derive the transfer functin f the circuit f FIGUE 6, when the utput is taken frm acrss the resistr, i.e. shw that: jq Q ( )... where L and Q LC C. 3. Slve the quadratic equatin t find the half-pwer frequencies f the band-pass LC filter: Q Teesside University 0
6 4. Frm the results f Self-Assessment Questin 3, shw that f ff, where f and f are the lwer and upper half pwer frequencies. 5. The transfer functin (equatin ()) f the series LC band-pass filter can be written as: L j C Sketch a rugh graph t shw hw the imaginary terms L, C and L C vary with frequency. 6. FIGUE 33 shws the circuit f a unity gain Sallen & Key high pass filter using equal C and equal. I f I C X I x C x FIG. 33 Teesside University 0
6 (i) Write dwn the relatinship between, and. (ii) Express x in terms f, and the capacitr's reactance X c (nte that the resistr and capacitr frm a ptential divider n the nninverting input). (iii) Write dwn the equatin representing Kirchhff's current law fr the pint 'X' n the circuit. (iv) Nte that I x X c Express the currents I f and I x in a similar fashin. (v) Hence, rewrite the equatin derived in (iii), abve, in terms f the vltages and cmpnent values. Thus, (and this is the really hard part!) derive the transfer functin f the filter. 7. A Sallen & Key band-pass filter f the type shwn in FIGUE is required t have a passband f 00 Hz centred n a frequency f khz. In rder t match the signal surce t the input f the filter, the resistr is required t have a value f kω. Calculate suitable values fr the remaining cmpnents if the capacitrs are t be f equal value. Teesside University 0
63 ANSWES TO SELF-ASSESSMENT QUESTIONS. The impedance f the series circuit is: j L jc and treating the circuit as a ptential divider, the vltage acrss the inductr is: j L j L jc j L j L jc Dividing the right-hand side thrugh by jl: j L j LC But and Q LC L S that: j Q Teesside University 0
64. Fllwing the same prcedure as in Self-Assessment Questin, we btain: j L jc and dividing the right-hand side thrugh by : j L jc But Q L C s that: jq 3. We have: Q and squaring bth sides: Q Teesside University 0
65 Q and nw taking the square rt f each side: ± Q ± Q 0 Multiplying thrugh by gives a quadratic in the frm ax bx c 0 where x : ± 0 Q We can nw apply 'the frmula' t slve fr : ± ± Q Q 4 Nte that 4 will always be greater than and as we are nly Q Q interested in psitive frequencies we can ignre the case when 4 is negative. Thus Q Teesside University 0
The term represents the (nrmalized) centre frequency and the term gives the tw half pwer pints with respect t the central frequency. Thus: 4. We have, frm Self-Assessment Questin 3: s that f f f f Q Q Q 4 4 4 4 Q Q Q f f Q Q f f Q 4 4 and Q 4 4 and Q Q Q Q Q 4 Q ± ± 4 4 Q Q Q Q 66 Teesside University 0
67 f f f f and therefre f f f In mathematical parlance, the resnant frequency is equal t the gemetric mean f the half-pwer frequencies. If the Q f the filter is high, the resnant frequency is apprximated by the arithmetic mean f f f the half-pwer frequencies, that is f. 5. FIGUE 34 shws the required curves, drawn perhaps a little mre accurately than yu were called upn t d! Teesside University 0
68 X L L C C 0 FIG. 34 The curves are f interest because they reveal why the LC filter's respnse is asymmetric. Nte als that resnance ccurs when: L C 0 that is when L C r when LC Teesside University 0
6. (i). (ii) (iii) I I f I x 0 (iv) (v) Substituting fr x : s that: X X X X c c c c X X c c X X X X c c c c X X X X c c c c 0 X c 0 X X 0 x c x x c I X I I X,, x c f x x x c X X X x c x c c s 69 Teesside University 0
70 X X c c X c X c X c X c c c X X We can nw make the substitutin X c j C : j C jc This can be put int the 'standard' frm j where C 7. The Q-factr f the circuit can be determined frm the relatinship: B f where B is the bandwidth. Q 000 Thus Q 00 0. Teesside University 0
7 Frm the text we have: Q s that Q 0 4 00 400 3 ( ) kω We can nw use the relatinship f πc tfindc. C π f π 0 0 400 0 3 3 3 C 8 nf Teesside University 0
7 SUMMAY First rder transfer functins are f the frm: G j c Secnd rder transfer functins have the frm: G j Q Higher rder transfer functins can be btained by multiplying first and secnd rder functins tgether. The transfer functin f a bandpass filter is f the frm: jq and the nrmalized half-pwer frequencies are: and 4 Q Q 4Q Q The nrmalized bandwidth B. Q Teesside University 0
73 Active filters can be used t replace passive filters, thereby replacing expensive inductrs. The p-amp in the active filter als acts as a buffer t prevent lading that wuld therwise adversely affect the filter's perfrmance. Sallen and Key circuits are widely used as secnd rder active filters. They can be cascaded t give higher rder filters. The Butterwrth filter gives a maximally flat respnse in the passband. The Chebychev filter gives a mre rapid rll-ff, but at the expense f ripple in the passband. Bessel filters can be used t minimize phase distrtin. Teesside University 0
74 APPENDIX DEIATION OF THE TANSFE FUNCTION OF THE SALLEN & KEY FILTE In this analysis, we adpt, perhaps, an uncnventinal apprach by viewing the circuit as that f a feedback lp frmed arund a nn-ideal amplifier as shwn in FIGUE A. The amplifier's input is shunted by and the feedback lp is applied by 4. The amplifier's vltage gain, G, is set by and s that: G We wish t find an expressin fr the circuit's clsed lp vltage transfer functin: G f The circuit applies vltage-derived, shunt-fed feedback t the amplifier via 4, the vltage causing a current I f t flw frm utput t input. Fr this type f feedback, in the general feedback equatin, G f G GH the feedback factr H is an admittance given by: H Y f I Teesside University 0
75 4 I f I 3 I Nn-ideal amplifier FIG. A The pen-lp gain G will be an impedance I, s that the lp-gain G H Y is dimensinless. The clsed-lp gain G f will als be an impedance I, (G f and G are smetimes referred t as transimpedances). Thus, fr the circuit, we can re-write the general feedback equatin as: G f G G H Y First we shall find the input vltage t the imperfect amplifier in terms f the utput vltage. The series netwrk acrss the input f the amplifier acts as a ptential divider s that: Teesside University 0
76 Nting that, as the amplifier has a gain G, then relatinship amplifier, then and als that the still hlds fr the ideal p-amp cntained within the G G and G Next, we find an expressin fr G. The pen-lp transimpedance f the imperfect amplifier, G, will be given by: G I We can nw get an expressin fr G in terms f the circuit parameters as: I and substituting fr : I I G G Teesside University 0
77 and therefre: G I G We can find an expressin fr G f in a similar fashin. transimpedance, G f, is given by: The clsed-lp G f I where I 3 and, again substituting fr : I G 3 Thus G f G 3 and dividing 'tp and bttm' f the right-hand side by : G f 3 G Teesside University 0
78 We shuld nte at this pint that the rati G f. is the clsed-lp vltage gain, Finally, the feedback factr H Y is als readily determined: H Y I f where If 4 G 4 Therefre: I f G 4 4 ( G ) G 4 G G Therefre: H Y I ( ) f G G 4 We are nw in a psitin t substitutue fr H Y, G and G f int the feedback equatin and s btain an expressin fr the vltage transfer functin. Befre we prceed thugh, a small, but imprtant, bservatin must be made. G The feedback equatin has been derived assuming that the feedback GH is negative fr psitive values f H. Fr the equatin t wrk with psitive Teesside University 0
79 feedback, either H must be made negative r the sign in the denminatr changed. We will change the sign, s: G f G G H Y We have established that G f 3 G, ( G ) Y G 4 G G and H Substituting int the feedback equatin: 3 G G G ( G ) G 4 G ( ) G 4 Crss-multiplying: 3 ( ) G 4 G G Dividing thrugh by G : G 3 ( ) G 4 G Teesside University 0
80 Frm which: 3 G ( G ) 4 G ( ) 3 3 G 3 G 4 3 3 G G 4 ( ) 3 4 and hence: 3 3 ( G ) 4 G 3 4... ( A) If 00% negative feedback is applied t the p-amp, s that G circuits transfer functin simplifies t:, then the 3 3 4... ( A) Teesside University 0
8 APPENDIX DEIATION OF THE TANSFE FUNCTION FO A SALLEN AND KEY BANDPASS FILTE In FIGUE A(a) the cmpnents f the bandpass filter f FIGUE have been represented as general impedances. 3 4 3 FIG. A(a) 3 T 3 FIG. A(b) Teesside University 0
8 As in Appendix, a 'system' apprach is als adpted in this analysis. The filter can be regarded as feedback amplifier (shwn shaded), having the transfer functin T, which has feedback applied by the cmpnents 3 and. This idea is represented in FIGUE A(b). The 'shaded' amplifier is a virtual-earth amplifier fr which we can immediately deduce: T 3... A3 4 ( ) We can nw g n t btain a transfer functin fr the cmplete circuit. Unfrtunately, thugh, the input t the 'shaded' amplifier is nt at virtual earth and s we have t start frm first principles. Summing the currents at the junctin f and 4 : 3 3 3 4 3 0 3 3 3 3 4 3 3... A4 3 4 ( ) Frm equatin A3: 3 4 Teesside University 0
83 and substituting fr 3 in (A4): 3 4 3 4 3 4 3 4 3 4 3 4 Finally, multiplying numeratr and denminatr by gives the general transfer functin f the circuit: 3 4 4 3... ( A5) In the circuit f FIGUE : 3 4 jc jc Teesside University 0
84 Substituting in (A5) fr these specific cmpnents gives: C jc C jc... ( A6) If the capacitrs are f equal value, say C, then (A6) becmes: jc jc... ( A7) Teesside University 0