analog filter

PDA SpeakerResponseWith&WithoutCompensation f-Frequency-Hz C o m p e n s a ti o n - d B -60 -55 -50 -45 -40 -35 -30 -25 -20 -15 -10 -5 0 5 399 502 632 796 1002 1262 1589 2000 2518 3170 3991 5024 6325 7962 10024 S p e a k e r R e s p o n s e - d B -20 -15 -10 -5 0 5 10 15 20 25 30 35 40 45 InitialSpeaker Response Compensated Response BiquadFilter Compensation Application Report SLOA152–December 2010 Analog Active Audio Filters Stephen Crump .......................................................................................................... Audio Products ABSTRACT Analog active audio filters can be used to compensate frequency response problems in a variety of systems. Their responses are examined here to simplify the filter design task for design engineers. Contents 1 Introduction .................................................................................................................. 1 2 First Order Filters ........................................................................................................... 2 3 Second Order Filters ....................................................................................................... 3 4 Biquadratic Filters ........................................................................................................... 4 5 Analog Filter Implementations: Second-Order Filters. ................................................................. 5 6 Analog Filter Implementations: Biquadratic Filters. ..................................................................... 7 1 Introduction Active audio filters may be used to compensate problems in frequency response of audio systems and loudspeakers. This paper deals with analog filters. These filters can produce a response that is approximately the inverse of a system response or a loudspeaker acoustic response so that when the two are summed the result is nearly flat. They may also be used to produce the difference of a target response minus the system or loudspeaker response, so that when the two are summed the result is nearly the target. In either case the final response is more accurate or more pleasing. Cell phone and PDA speaker responses like the one shown below often have annoying peaks that reduce intelligibility. This is compensated with the response of a biquad filter, also shown. The sum, the compensated response, is much more pleasing and far more intelligible than the original. 1SLOA152–December 2010 Analog Active Audio Filters Submit Documentation Feedback © 2010, Texas Instruments Incorporated Biquad Valley Filter + - VI VO 43kW 22nF22nF 10kW 5.6kW 1.2kW1.5kW First Order Filters www.ti.com SPACER The schematic for the biquad filter is at right. This and other filters are discussed in detail in this paper. This paper begins with the relationship between filter singularities and their responses. Then it considers a number of possible filter implementations. (It’s typically easier to adjust a set of filter singularities to achieve the best result than to adjust the numerous component values in a circuit. Once the singularities are decided the filter usually can be implemented relatively easily.) It provides equations for responses and the parameters in them and discusses optimizing component choices. Audio filters may be first, second or higher order. First and second order analog filters are generally well understood and their audio uses are somewhat limited, so they are examined briefly. The paper also examines biquadratic filters, or biquads, in more depth because they are more powerful tools for response compensation or EQ. Bridged-T filters will be added in a later version. 2 First Order Filters First-order filters implement responses with single poles as their denominators. They have limited response bands, either low-pass or high-pass, which are described below. • Low pass: H(s) = Ho wo / ( s + wo ). For small s, this is Ho, flat low-frequency response; for large s, it is Ho wo / s, a first-order rolloff. • High pass: H(s) = Ho s / ( s + wo ). For large s, this is Ho, flat high-frequency response; for small s, it is Ho s / wo, a first-order rollup. 2 Analog Active Audio Filters SLOA152–December 2010 Submit Documentation Feedback © 2010, Texas Instruments Incorporated 2nd-ORDERFILTERRESPONSES,H =1, =0.2/0.5/1/2O a -50 -40 -30 -20 -10 0 10 20 f-Frequency-Hz 100 158.5 251.2 398.1 631 1000 1584.9 2511.9 3981.1 6309.6 0 R e s p o n s e - d B www.ti.com Second Order Filters 3 Second Order Filters Second-order filters implement responses with quadratic terms as their denominators. They have limited response bands, either low-pass, band-pass or high-pass, which are described below. (1) • Low pass: H(s) = Ho wo 2 / (s2 + a wo s + wo 2). For small s, this is Ho, flat low-frequency response; for large s, it is Ho wo 2 / s2, a second-order rolloff. Response magnitude is G(w) = sqrt(Ho 2 wo 4 / (w4 + w2 wo 2 (a2 – 2) + wo 4) ). Phase is Φ(w) = p/2 – arctan( a wo w / (wo 2 – w2)). SPACER • Band pass: H(s) = Ho a wo s / ( s2 + a wo s + wo 2 ). For small s, this is Ho a s / wo, an increasing first-order response or rollup; for large s, it is Ho a wo / s, a decreasing first-order response or rolloff. At s = jwo, it is Ho, band-center response. Response magnitude is G(w) = sqrt(Ho 2 a2 w2 wo 2 / (w4 + w2 wo 2 (a2 – 2) + wo 4) ). Phase is Φ(w) – arctan( a wo w / (wo 2 – w2)). SPACER • High pass: H(s) = Ho s2 / (s2 + a wo s + wo 2). For large s, this is Ho, flat high-frequency response; for small s, it is Ho s2 / wo 2, a second-order rollup. Response magnitude is G(w) = sqrt(Ho 2 w4 / (w4 + w2 wo 2 (a2 – 2) + wo 4) ). Phase is Φ(w) = p – arctan( a wow / (wo 2 – w2)). SPACER Ho scales response magnitude, while wo sets the characteristic frequency, the frequency at which the filter operates. The variable a sets the sharpness of the peak the filter produces, which varies inversely with a (sharper peak with smaller a). (1) Reference: Operational Amplifiers, Design and Applications, Graeme, Tobey and Huelsman, Burr-Brown, McGraw-Hill Book Company, 1971, ISBN 07-064917-0, pages 284-286. 3SLOA152–December 2010 Analog Active Audio Filters Submit Documentation Feedback © 2010, Texas Instruments Incorporated BIQUAD RESPONSES vs. FILTER FORM -20 -15 -10 -5 0 5 10 15 20 Single Peak Single Valley Low Frequency Boost High Frequency Boost f - Frequency - Hz 100 158.5 251.2 398.1 631 1000 1584.9 2511.9 3981.1 6309.6 0 R e s p o n s e - d B Biquadratic Filters www.ti.com 4 Biquadratic Filters Biquadratic or biquad filters implement responses with quadratic terms for both their numerators and their denominators. They have wide responses, with a single peak or valley or with low or high frequency boost with or without peaking. Their responses have the following form. H(s) = Ho ( s2 + aZ wZ s + wZ 2 ) / ( s2 + aP wP s + wP 2 ). Filter response is influenced by relationships between wZ and wP and between aZ and aP. • Single peak or valley. The filter produces this response when wZ and wP are equal. If aZ is greater than aP the filter produces a peak. If aP is greater than aZ the filter produces a valley. SPACER • Low frequency boost. The filter produces this response when wZ is greater than wP. If aZ is less than about 1, the response includes a valley above the boost frequency. If aP is less than about 1, the response includes a peak below the boost frequency. SPACER • High frequency boost. The filter produces this response when wP is greater than wZ. If aZ is less than about 1, the response includes a valley below the boost frequency. If aP is less than about 1, the response includes a peak above the boost frequency. SPACER Response magnitude is G(w) = sqrt( (w4 + w2 wz 2 (az 2 – 2) + wz 4) / (w4 + w2 wp 2 (ap 2 – 2) + wp 4) ). Phase is Φ(w) = arctan( aZ wZ w / (wZ 2 – w 2) ) – arctan( aP wP w / (wP 2 – w2) ). Traces in the graph that follows illustrate something of the range of responses a biquad filter can generate. The responses are arranged as follows for clarity. • Single peak and valley responses are presented in the order of decreasing peaks, or decreasing aZ with respect to aP. For all these responses Ho is 1 and wZ and wP are 1 kHz. SPACER • Low and high frequency boost responses are presented in the order of decreasing peaks and valleys, or increasing aZ and aP. Also, aZ and aP are made equal, wZ and wP are placed symmetrically around 1kHz, and Ho is set to 0.5 for low frequency boost and 2 for high frequency boost, to make the responses symmetrical around zero dB and 1kHz. SPACER 4 Analog Active Audio Filters SLOA152–December 2010 Submit Documentation Feedback © 2010, Texas Instruments Incorporated BIQUADRESPONSESvs.H , , , ,o z z p pa v a v -20 -15 -10 -5 0 5 10 15 20 f-Frequency-Hz 100 158.5 251.2 398.1 631 1000 1584.9 2511.9 3981.1 6309.6 10000 R e s p o n s e - d B a az1 p0.18 a az3 p2 a az2 p3 a az0.18 p1 a az0.28 p0.28 a az0.75 p0.75 a az0.4 p0.4 a az1.5 p1.5 a az0.28 p0.28 a az0.75 p0.75 a az0.4 p0.4 a az1.5 p1.5 - + VI VO R1 R2 C1 C2 Inverting R3 + - VI VO R1 R2 C2 C2 Rs Rf Non-Inverting K=1+Rf/Rs V +I V -O R1 R2 C1 C2 Differential R3 + - V -I R4 R5 R6 C3 + - V +O R4=R1;R5=R2;R6=R3;C3=C2 R3 R1 R3 R1 f s R K = 1+ R æ ö ç ÷ è ø 1 (R1 R2 C1 C2)´ ´ ´ 1 (R2 R3 C1 C2)´ ´ ´ ( ) 1 R2 R3 (C1 2) C2´ ´ ´ ´ R1 C1 R2 C2 R1 C2 (1-K) R2 C2 R1 C1 R2 C1 ´ ´ ´ ´ + + ´ ´ ´ ( ) ( ) R2 R3 C2 R2 R3 + + C1 2 R3 R2 R1 é ù ´ æ ö ê ú ´ ç ÷ ê ú ´ è ø ê ú ë û ( ) ( ) R2 R3 C2 R2 R3 + + C1 R3 R2 R1 é ù ´ ê ú ´ ê ú ê ú ë û www.ti.com Analog Filter Implementations: Second-Order Filters. Note that aZ and aP do not have to be equal! Varying aZ and aP can create responses that range among and beyond the extremes in the graph. 5 Analog Filter Implementations: Second-Order Filters. Second-order filters may be non-inverting or inverting. The schematics below show single-ended forms, both non-inverting and inverting, and an inverting, differential form, with equations for their Ho, a and wo. Low-Pass Filters Non-Inverting Inverting Differential Inverting Ho wo a 5SLOA152–December 2010 Analog Active Audio Filters Submit Documentation Feedback © 2010, Texas Instruments Incorporated - + VI VO R1 C1 C2 Inverting R3 R 2 + - VI VO R1 C2 C1 Rs Rf R2 Non-Inverting K=1+Rf/Rs R 3 Differential - + + - R1 R 2 R3 R4 R5 C1 C2 C3 C4 V -O V +O V +I V -I R4=R1;R5=R3;C3=C1;C4=C2 f s R K = 1+ R æ ö ç ÷ è ø R3 / R1 C2 1+ C1 æ ö ç ÷ è ø R3 / R1 C2 1+ C1 æ ö ç ÷ è ø 1 R1 R2 C1 C2´ ´ ´ 1 R3 C1 C2 R1 R2 R1+R2 ´ ´ ´ ´ æ ö ç ÷ è ø ( ) 1 R3 C1 C2 R1 R2/2 R1+R2/2 æ ö ´ ´ ´ ´ ç ÷ ç ÷ è ø R1 C1 R2 C2 R1 C2 (1-K) R2 C2 R1 C1 R2 C1 ´ ´ ´ ´ + + ´ ´ ´ ( ) ( ) C1 C2 R1×R2 C2 C1 R1+R2 R3 æ ö é ù ç ÷ + ´ ê ú ç ÷ ê ú ë û è ø ( ) ( ) C1 C2 R1×R2/2 C2 C1 R1+R2 R3 æ ö é ù ç ÷ + ´ ê ú ç ÷ ê ú ë û è ø - + Inverting R 1 R2 C1 C2 C3 VO VI + - Rs Rf Non-Inverting K=1+Rf/Rs R1 R 2 VO C1 C2 VI Differential - + + -R 1 R2 R3 C1 C2 C3 C4 C5 C6 V -O V +O V -I V +I C4 = C1; C5 = C2; C6 = C3; R3 = R2 C1 C3 f s R K = 1+ R æ ö ç ÷ è ø R3 / R1 C2 1+ C1 æ ö ç ÷ è ø 1 (R1 R2 C1 C2)´ ´ ´ 1 (R1 R2 C2 C3)´ ´ ´ 1 R1 R2 C2 C3 2 æ ö ´ ´ ´ ç ÷ è ø ( ) ( ) ( ) R1 C1 R2 C2 R1 C2 (1-K) R2 C2 R1 C1 R2 C1 æ ö æ ö æ ö ´ ´ ´ ´ + + ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ è ø è ø è ø ( ) R1 / 2C1 C2 C3 + + C3 C2 R2C2 C3 æ öé ù ´ ç ÷ ê ú ç ÷ ´ ê ú ë û è ø C1 C2 C3 R1 + + × C3 C2 R2C2xC3 é ù ê ú ë û Analog Filter Implementations: Second-Order Filters. www.ti.com Band-Pass Filters Non-Inverting Inverting Differential Inverting Ho wo a High-Pass Filters Non-Inverting Inverting Differential Inverting Ho wo a 6 Analog Active Audio Filters SLOA152–December 2010 Submit Documentation Feedback © 2010, Texas Instruments Incorporated - + R1 R2 R3 R4 C1 C2 VO VI R5 K=-R5/R4 Inverting - + + - R1 R4 C1 C2 V - O V + O V - I V + I C1 C2 R1 R5 2 * R 3 R2 R4 R2 R5 K=R5/R4 Differential + - R1 R2 R3 R 4C1 C2 Rs Rf VO VI K=1+Rf/Rs Non-Inverting R5 (NEGATIVE) R4 - R5 R4 f s R 1+ R R3 (R2 R3)+ 1 (R1 DR2 C1 C2) 1 R1DR2 R4 C1C2 R1+R4 æ ö ç ÷ è ø ( ) (R1 + R4) DR2 (C1 +C2)+R1R4C2 (1-KD) R1 DR2 R4 C1 C2 × (R1 + R4) ( )C1 + C2 DR2 (R1 C1 C2) www.ti.com Analog Filter Implementations: Biquadratic Filters. 6 Analog Filter Implementations: Biquadratic Filters. Biquadratic filters also may be non-inverting, inverting, or differential inverting. The schematics below show all these forms with equations for their Ho, a and wo. The equations begin with a factor K, the gain of the inner opamp circuit, for each of the filter forms. The remaining quantities, D, Ho, wZ, aZ, wP and aP, are common to all the filter forms. D is a multiplier used to simplify the following equations. Biquad Filters – High Frequency Boost Non-Inverting Inverting Differential Inverting K D Multplier for R2 and K Ho K Response at high frequency wz wP az ap Schematics for low frequency boost filters follow. The input circuit in each filter includes a series chain of capacitors to ground or virtual ground. This load could destabilize either the opamp in the filter or an opamp driving the input, so a small value resistor, maybe 100Ω, is added in series with the final cap in the chain. Also, feedback elements in the inverting and differential filters are capacitors. These provide no path for DC bias, so a large value resistor, maybe 1 to 10 M, is added in parallel with the feedback cap. Of course, these resistances will have a small effect on filter responses, but they should not degrade them significantly. 7SLOA152–December 2010 Analog Active Audio Filters Submit Documentation Feedback © 2010, Texas Instruments Incorporated - + Inverting S R1 R2 C1 C2 C3 C4 VO VI L R3 K = -C4/C3 L = Large-Value Resistor 3 Places Differential - + + - R1 R2 2 * R 3 C1 C2 C3 C4 V - O V + O V - I L C4 L S S C1 C2 C3 R2 R1 V + I K=C4/C3 + - Non-Inverting R1 R2 R3 C1 C2 C3 Rs Rf VO VI S=Small-Value Resistor4places S K=1+Rf/Rs C4 (NEGATIVE) C3 - C4 C3 f s R 1+ R R3 (R2 R3)+ ( ) K C1C2 C1C2+C1C3+C1C3 ´ 1 (R1 DR2 C1 C2) ( )( ) 1 R1DR2 C1C2 + C1C3 + C1C3´ ( ) ( ) DR2 (C1 + C2)+R1C3 + R1C2 (1-KD) R1 DR2 C1 C2 + C1C3+C2C3 ´ ´ ´ ( ) ( ) C1 + C2 DR2 (R1 C1 C2) Analog Filter Implementations: Biquadratic Filters. www.ti.com Biquad Filters – Low Frequency Boost Non-Inverting Inverting Differential Inverting K D Multplier for R2 and K Ho Response at low frequency wz wP az ap 8 Analog Active Audio Filters SLOA152–December 2010 Submit Documentation Feedback © 2010, Texas Instruments Incorporated + - Non-Inverting ThereareNOInvertingor Differentialversions becauseeachofthese loadstheoutputofthe inputnetworkand producessomelowor highfrequencyshelf. R1 R2 R3 C1 C2 Rs Rf VO VI K=1+Rf/Rs f s R 1+ R ( ) R3 R2 + R3 1 (R1 DR2 C1C2) 1 (R1 DR2 C1 C2) ( ) DR2 (C1 + C2)+ R1 C2 (1-KD) R1 DR2 C1 C2 ´ ´ ( ) ( ) C1 + C2 DR2 R1C1C2 ´ www.ti.com Analog Filter Implementations: Biquadratic Filters. Biquad Filters – Single Peak or Valley Non-Inverting K D Multiplier for R2 and K Ho K Response at very low and high frequencies wz wP (= wz) az ap The most complicated quantities are aZ and aP, so we will look at these in some detail. As we will see, it is more difficult to achieve low values of aZ and aP than high, so we will concentrate on reducing these quantities. We will consider how to produce values as small as about 0.5, a value that provides significant peaking. aZ is the same for all 3 filter configurations. The ratio (C1+C2) / √(C1C2) in aZ ranges from about 3.5 to 2 to about 3.5 again as (C1/C2) ranges from 0.1 to 1 to 10, so aZ is reduced by making C1 and C2 different in value. So it is typically best to make C1 and C2 similar in value. aZ can be controlled by varying the ratio √(DR2/R1). If R1 = 20×DR2 and C1 = C2, aZ = 2 / √(20) = 0.45, probably close to the lowest value needed. For aP we face similar constraints with the first term or two in the numerators, but we have the advantage of the last term, which is negative in non-inverting and differential filters for any KD product greater than 1. So we can use this term to reduce aP if we need to do so. (Beware, however: if KD is made large enough, the last term will cancel the rest of the numerator, aP will equal zero and the filter will oscillate at wP!) Note that we do NOT have this advantage in inverting biquads! In those, since K is negative, the sum (1- KD) is always greater than 1. As a result, it is likely to be very difficult to achieve low values of aP. For this reason inverting biquads are not likely to be generally useful in EQ work. Single peak or valley filters are a special case. In these, aP = aZ + R1C2×(1-KD) / √(R1DR2C1C2). If C1 and C2 are similar in value, aP ~= aZ + (1-KD) √(R1/DR2) ~= aZ + 2×(1-KD)/aZ. If KD is less than 1, (1-KD) is positive, aP is greater than aZ and the filter creates a valley. If KD equals 1, aP equals aZ and there is no peak or valley. If KD is greater than 1, the second term is negative, aP is smaller than aZ and the filter creates a peak. Note that, for single peak or valley filters, reducing aZ increases the magnitude of the second term in aP. So in valley filters, with KD < 1, reducing aZ tends to increase aP. This tends to make the resulting valley broad and deep. Make (1-KD) smaller to narrow or reduce the valley. In peak filters, with KD > 1, reducing aZ tends to reduce aP by making the last term in aP more negative. This tends to make the resulting peak narrow and high. Make (1-KD) less negative to broaden or reduce the peak. 9SLOA152–December 2010 Analog Active Audio Filters Submit Documentation Feedback © 2010, Texas Instruments Incorporated IMPORTANT NOTICE Texas Instruments Incorporated and its subsidiaries (TI) reserve the right to make corrections, modifications, enhancements, improvements, and other changes to its products and services at any time and to discontinue any product or service without notice. Customers should obtain the latest relevant information before placing orders and should verify that such information is current and complete. All products are sold subject to TI’s terms and conditions of sale supplied at the time of order acknowledgment. TI warrants performance of its hardware products to the specifications applicable at the time of sale in accordance with TI’s standard warranty. Testing and other quality control techniques are used to the extent TI deems necessary to support this warranty. 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