A Differential 3 Erik Margan rd -order Bessel Filter Experimental Particle Physics department, Jožef Stefan Institute, Ljuljana, Slovenia The circuit in Fig.1 has een presented in Linear Audio as part of a differential line drive system, with only the final result of the analysis of the system transfer function. Here we show the complete analysis and the procedure of finding the suitale component values from the system poles, and verify the design y calculating the frequency domain and time domain (unit step response) performance. p 1 A 1 a R 3 R 2 R 1 C 1 o 2 a 750 750 750 R 4 390pF a C 4 4n7 C 3 1n8 3k0 R 8 R A 4 R 3 + R 2 R 7 R 6 R 5 3k0 C 2 2 750 750 750 390pF A 2 Fig.1: Differential third-order Bessel filter with a 50 khz cut off, and a gain of two, inverting. To make the circuit analysis easier, we cut the differential filter in half, taking only the upper arm, and grounding the lower side of capacitors G and G%. In the analysis we shall follow the node laels as are indicated in Fig.3. IMPORTANT: In a differential filter the capacitors G and G% are driven y the resistors in oth arms, so in order to have the same system time constants those capacitors should e of doule value in the single-ended circuit, as indicated in Fig.3. -1-
Real or Ideal Opamps? First we need to show that we are not making a ig mistake if, instead of a real opamp, having a limited gain and andwidth, we use an idealized model in the circuit analysis. A good audio grade opamp has a very high input resistance (>1M H), and a very low input capacitance (~1pF), so at audio frequencies (and even a decade eyond) its loading on the filter components can e neglected. To show this more clearly let us compare the amplifier s open loop performance to the highest circuit time constant set y V and G in the feedack loop. But efore we set off, let us make a short digression and consider the general polynomial form of the system transfer function as required y the circuit theory. This will hopefully clarify a couple of points, which sadly often remain oscure even in some of the est textooks. The general Cauchy 8 th -order polynomial form, y which we descrie the performance of a circuit with 8 reactive components (capacitors or inductors) in the frequency domain is: J 8 a œ a a â a 8 Here,, á, 8 are the circuit poles set y the various system time constants. In general, a circuit may also have some zeros, ad etc., in the numerator, ut here we are dealing with pole-only circuits. When the expression (1) is multiplied, the last polynomial coefficient, which is a product of all the poles, â 8, will determine the system gain, which will oviously e different for systems with different numer of poles. This is undesirale, ecause we want to treat the real (DC) gain of the circuit separately from its frequency dependence. We therefore normalize the characteristic circuit polynomial y dividing it with the same polynomial evaluated at œ (DC). So the frequency dependence is: J 8 a a a â a 8 a a âa 8 œ œ J 8 a a a â a 8 a a âa 8 In this way it is ovious that for œ, as well as for very low frequencies, the frequency response will e equal to 1 (it will deviate from unity as approaches the lowest of the poles and progressively onward). Consequently the frequency independent gain can e attriuted to a separate factor multiplying the expression (2). The amplifier s transfer function E acan e modeled y assuming its DC gain to e E, and its single dominant pole œ. Every pole causes an additional 90 phase shift at high frequencies, so in order to remain stale at closed feedack loop, all other amplifier poles must e at very high frequencies, where the amplifier s gain is less than unity. This prevents the feedack from ecoming positive with (1) (2) -2-
enough gain to cause oscillations. On the other hand, any pole defined y external circuitry must e suitaly damped, so that the resistive (energy dissipating) part of the poles dominate, and the poles are positioned on the left hand side of the complex plane (they have their real part negative, whilst their imaginary part can e either zero or forming complex conjugate pairs). Poles too close to the imaginary axis (very low real part value) or poles in the right hand side of the complex plane (positive real part) are sure signs of troule. With the frequency variale œ4, and the non-inverting input grounded, @ p œ, we have the open-loop gain determined y: @ o œ a@ p @ n E a œ a@ p @ n E œ@ ne (3) From (1) we can express the open loop gain as a function of the complex frequency œ5 4 in the Laplace space: @ o E aœ œe @ and along the imaginary frequency axis: E4 a œe 4 n (4) (5) 10 5 10 4 A 0 A 0 2 f 0 i n p A( s) o 10 A( f ) 10 3 2 A( f ) ϕ ( f ) [ ] 180 10 1 ϕ ( ) f f T f 0 2 A 0 1 225 10 0 A 1 270 1 10 315 0 1 2 3 4 5 6 7 8 10 10 10 10 10 10 10 10 10 f [Hz] Fig.2: A typical open loop gain magnitude E0 a and phase : a0 of an inverting amplifier. At the dominant pole frequency 0 the magnitude is lower y Î È, and falls off y a factor of 10 for each decade increase of the frequency. The dashed lines show the influence of a possile second pole, which for a stale unity gain closed loop must e higher than the transition frequency 0TÞ Note that for practical reasons we usually show the graphs as functions of the full cycle frequency (in Hz), instead of the angular frequency (in radîs). -3-
The magnitude (asolute value) of the transfer function is the square root of the product of the gain y its own complex conjugate: @ o º º œ È Ea4 Ea4 œ E Œ Œ @ Ë 4 4 n œe Ë œe ËŒ (6) Fig.2 shows graphically the expression (7). At DC a œ the gain magnitude equals E. When the input frequency ecomes œ the open loop gain falls to EÎ È. Aove the gain magnitude decreases y a factor of 10 for every 10-fold increase in frequency (or db Î0, or ' db Î0). And when œ ÈE, the open loop gain decreases to unity (and decreasing further eyond that frequency). This frequency is the transition frequency, often laeled T. Of course, the gain magnitude is not the whole story, the change of the phase angle of the output signal referred to the input is also important. From equation (1) the phase angle is expressed as the arctangent of the ratio of the imaginary to the real part of the gain function: eee a f :a œ arctan Œ (7) dee a f To separate the real and the imaginary part we need to rationalize the denominator, so we multiply oth the numerator and the denominator y the complex conjugate of the denominator: 4 E4 a a œe œe a 4 a4 a4 (8) Now the phase along the imaginary axis is: Î E a Ñ : a4 œ arctanð Óœ arctan Œ E Ï Ò The phase plot indicates the initial phase inversion ( 180 ) at low frequencies, since the signal is eing fed to the amplifier s inverting input. However, note that the positive direction of the phase rotation is defined as counterclockwise, so the negative frequency sign, in (9), indicates a clockwise rotation. With the frequency increasing towards the dominat pole the phase angle rotates y a further 45 to 225, and another 45 eyond, reaching 270 two decades higher. (9) -4-
Good audio grade opamps usually have 0 œ Î 1 Hz, and E, thus 0T œ TÎ 1 MHz. Oserve that V and G in our circuit set the shortest time constant, equivalent to 0 œ Î 1 V G 647 khz, which is almost 20 lower than T, so there will e still some 24 db of feedack at that frequency. Thus we can e sure that neither the amplifier s input impedance, nor its gain and andwidth limitations will influence much the VG time constant, and at audio frequencies the situation will e etter still. & Circuit Analysis p 3 2 1 A 1 a R 3 R 2 R 1 C 1 o 2 a 750 750 750 R 4 390pF C 4 4n7 C 4a 4n7 C 3a 1n8 3k0 C 3 1n8 A R 4 R 3 + R 2 Fig.3: The single-ended filter used for analysis. Note the douled G and G. % We want to find the closed loop response of the circuit in Fig.3. With a real opamp the current sum equation at node @ would e a@ @ ÎV œ a@ @ o G, and we would have to use equation (3) to express @ y @ o. But ecause at the top of the audio and the open loop gain is still aout 500, we can use an ideal amplifier model, and set @ @ œ, which simplifies the @ node equation to: From (10) we express @ : p Next, the current sum at the node @ @ V œ@g o (10) @ œ @ o G V (11) is: @ @ @ @ @ o œ @ G V V V % We can separate the voltages: @ œ @ Œ V V V G V @ o V V V (13) % % (12) -5-
By sustituting @ from (11) we otain: V V V @ œ @ og V Œ G V @ o (14) V V V % % By separating the voltage variales, and grouping the various powers of we get: V V V @ œ@ o GV Œ GGVV (15) V V V % % Finally we set the current sum equation at node @ : @ a @ @ @ œ @ G% (16) V V This we rearrange as: @ a œ @ Œ V V G% V @ og V V V (17) We now sustitute @ from (15) and group the various powers of : @ a œ@ oˆ T U VW (18) where the polynomial coefficients are: TœGGGVVV % V V V UœGGVV Œ GGVV % Œ V V V % V V V V V VœGV Œ GV % V V V V V % % % V Wœ V Œ (22) V% V The system transfer function is: @ o œ (23) @ T U V W a In order to otain a canonical polynomial expression, we need to make the coefficient at the highest power of equal to 1, so we must divide all the coefficients y T. By sustituting O œ UÎT, O œ VÎT, and O œ WÎT we otain: where we have: @ @ o a GGGVVV % œ O O O V V V O œ Œ Œ GV V GV V V % % (19) (20) (21) (23) (24) -6-
V V V V O œ Œ GGVV V V V V GGVV % % % % V V O œ Œ GGGVVV V V % % (25) (26) In the normalized canonical polynomial form, the numerator must e equal to the coefficient O. But y doing so we increase the numerator y av V ÎV%, so we must multiply the whole transfer function y an inverse of this factor: @ o V% O œ (27) @ V V O O O a It is immediately ovious that the system DC gain must e: E œ V% V V (28) System Poles Now that we have the transfer function expressed y the characteristic polynomial, and its coefficients determined, we must relate those coefficients to the rd system poles. By oserving the relations in a general 3 -order form: J a œe a a a a a a (29) œe a a a a a a a a (30) we can equate the coefficients O, O, O with the appropriate cominations of system poles (note that O œ ): O œ (31) O œ O œ a a a (32) (33) In order to express the poles ßß y the polynomial coefficients Oßß we have to solve a system of three equations with three unknowns. An easy way to accomplish the task would e y realizing that in (32) we have a sum and a product of two poles, say and, and those are readily availale from the other two equations. Specifically we reorder (32) as: O œ a (34) -7-
Then from (33) we have: O œ and from (31) we have: œo With (35) and (36) we return to (34) to otain: O O œ ao rd which, after a little reordering can e written as a general 3 -order equation: (35) (36) (37) (38) O O O œ Various forms of cuic equations have een solved already in the XVI century y the Venetian mathematician Noccolò Fontana Tartaglia, and pulished in 1545 y Gerolamo Cardano, using purely algeraic expressions. Later François Viète independently discovered the trigonometric solutions. Here we shall use the general algeraic form. A general solution can have either three real roots, or three coincident real roots, or one real and two complex conjugate roots. In order to shorten the long expressions let us replace the polynomial coefficients as follows: + œo, œo - œo Since there is a rather long common term, we shall replace it y the symol H: (39) H œ É '+, )- )+ È, +, &%+,- )- + - (40) The real polynomial root is then: < H ', + + œ Œ ' H * and the two complex-conjugate roots are: < ß H, + + È H ', + œ Œ 4 Œ H * ' H * (41) (42) It should e pointed out that the expression under the square root in H must e non-negative in order to e applicale to solutions with at least one real root, as required y any realizale electronic circuit. So once the real solution is known, with œ <, we can go ack to equations (35) and (36) and solve a quadratic equation, -8-
using its well known general solution (known already to Baylonian mathematicians). From (36) we have: œo And from (35) we have: œ O Thus y equating (43) to (44): O œo we have the quadratic equation for : with the general solution: (43) (44) (45) O ao œ (46) O O % O a Êa Š œ (47) ß We otain here either two real solutions (if the expression under the square root is positive), or two coincident real solutions (if the expression under the square root is zero), or a complex conjugate pair (if the expression under the square root is negative). It is not necessary to solve for from either (43) or (44), ecause if we assign to then œ. Actually, we may spare ourselves all that troule, since œ < and œ <. It is now possile to express the poles ßß y the actual VG time constants from the coefficients O ßß in equations (24 26). However, this is a tedious work, and does not offer much insight in the working of our circuit, and anyway, those rd expressions will e different for different types of 3 -order filters, so we may as well e satisfied y numerical values for the poles. ßß However, we still have to find a way of calculating the circuit components, as rd well as find such values which will conform to the desired Bessel 3 -order system. Apparently the expressions given y the polynomial coefficients O ßß are more simple to work with. But first let us see the relations for the Bessel poles and the polynomial rd coefficients of a 3 -order system. The general form for Bessel coefficients of order 8 can e calculated as: - œ a83x 3xa83 x 3 83 º 3œâ8 (48) -9-
rd For a 3 -order system ( 8œ, and 3œÑ,,, the coefficients are: - œ& (49) - œ& (50) - œ' (51) - œ (52) Our characteristic polynomial is now - - -, and we want to find its roots. We can again use (40 42), y making the following sustitutions: + œ-, œ- - œ- (53) rd then y solving equations (40 42) we arrive at the Bessel poles of a 3 -order system: < œ Þ [radîs] (54) < œ Þ))* 4Þ(&%% [radîs] (55) œ Þ))* 4Þ(&%% [radîs] (56) < Note that the Bessel poles were derived on the assumption that the envelope delay is normalized to unity, and not the andwidth as in the Butterworth case The andwidth (the angular frequency at which the magnitude response falls to ÎÈ ) will thus e somewhat higher than 1 radîs, so we shall have to account for this, in addition to the actual andwidth, when specifying the values of resistors and the capacitors. Now for Butterworth systems the andwidth is simply equal to the 8 th root of the asolute value of the product of all the 8 poles. Because all Butterworth poles lie on a circle, the andwidth is also equal to the asolute value of any single pole. Unfortunately for Bessel systems there is no simple formula to relate the pole values to the system andwidth. Bessel poles lie on a family of ellipses ecoming larger with increasing system order, ut all with the near focus at the complex plane origin and the other focus on the positive part of the real axis. To otain the andwidth we must input the values of < ßß into equation (29), calculate the frequency response magnitude for a range of frequencies, then find the frequency at which the magnitude is down y ÎÈ, and then divide all the polynomial roots y that frequency. For a greater precision it might e necessary to reiterate this process once or twice. rd Using that procedure the andwidth of the 3 -order Bessel system normalized to a unit envelope delay is 3N Þ(&&' radî s. By dividing the roots < ßß y N a andwidth of radîs will result. For audio work we need a 50 khz andwidth, or H œ 1 & radîs, so we must multiply the roots y a factor. Thus our poles will e: HÎ N ßß œ < ßß H N (43-57) -10-
F(s) rd s 3 4 10 5 3 2 1 s 1 s 0 0 jω [rad/s] 1 2 s 2 3 4 8 7 6 5 4 3 2 1 0 σ [rad/s] 10 5 Fig.4: The layout of the system poles ßß. The pole at the complex plane origin, œ, is owed to the Laplace transform of the input unit step signal. The system poles are on the left hand side (negative real part) of the complex plane. The poles and form a complex conjugate pair, the imaginary parts are controlled dominantly y the ratio GÎG. The real pole is controlled dominantly y G. % 15 10 5 0 10 10 5 8 6 σ 4 2 0 10 5 s σ + jω Fig.5: A waterfall plot of the magnitude (asolute value) of the transfer function, lj a l, aove the complex plane. The approximately Gaussian shape of the surface along the imaginary axis is the frequency response lj a4 l in linear (negative and positive) frequency scale. Aove each pole the magnitude goes to infinity. The shape of each curve shows what would the frequency response look like if the poles were moved closer to the imaginary axis. 0 jω 5 10 5 10-11-
Calculating the Resistors and Capacitors Now we have everything necessary to calculate the values of resistors and capacitors. However, there are 4 resistors and 3 capacitors in the filter, ut we have a system of only 3 equations with 3 unknowns in (31 33). Fortunately, the circuit time constants are defined y appropriate VG products, which allows us to reduce the numer of unknowns. We have already estalished that the system gain E œ V% ÎaV V. In order to have a gain of 2 (to compensate the loss owed to the 600 H line impedance matching), we can make V œ V œ V, and V% œ %V. The remaining resistor V can e of any suitale value, so to minimize the system variations we set it also to V œ V. This determines the resistor ratios, and y making V œ H we can disregard it in the equations and determine the capacitance ratios. Once we do that, we can increase V to any suitale value and decrease the capacitors in proportion. Note that y using the poles from (57) in relations for the polynomial coefficients O ßß (31 33) we would have to solve the 3-equation system once to express the poles y the coefficients, and then again to find the VG component values from the poles. It is thus more economic here to express the components directly y the coefficients, so we have to solve the system of equations only once. From (24), (25) and (26) we have the normalized component values: V V V ' œ Œ Œ GV V GV V V % % (58) V V V V & œ Œ GGVV V V V V GGVV % % % % (59) V V & œ Œ GGGVVV V V % % (60) By replacing V œ V œ V œ V and V% œ %V (as required for a gain of 2): V V V ' œ Œ Œ GV V GV V %V % V V V V & œ Œ GGVV V V %V %V GG%VV % V V & œ Œ GGGVVV %V V % (61) (62) (63) By making Vœ we have: -12-
' œ a Œ G G % % & œ Œ G G % % %G G % & œ a GGG % Consequently: ' œ % * G %G % (64) (65) (66) (67) & œ ( G G %G G % (68) & œ GGG % From (69) we can express, say, G : and insert it into (68); G œ & œ G G % ( G G% % G G G % (69) (70) (71) From (71) we express G We insert G from (72) into (67): G œ ' œ ( &G a G % % * G% ( % &G a G % % (72) (73) Now our only variale is ') &' G% G% G% & & rd G %. By reordering we arrive at the 3 -order polynomial: œ We can solve (74) y the same relations as for the roots < ßß (40 42). (74) -13-
However, here we need only the real solution, ecause once we determine G % we already have real relations for finding G and G. Our coefficients in (74) are now: ') + œ, œ - œ & &' & We insert these into (41) and the resulting < will e our normalized G%. By putting the numers into a mathematical computer program, such as Matla, MathCAD, Mathematica, or other suitale software, we otain: G % (75) œ Þ)& F (76) It is useful to make a program for equations (40 42) since the general rd solutions can e used for any other 3 -order system calculation. With the value of G from (76) we return to (72) and calculate G : G % ( œ œ Þ&') F (77) G ˆ G % % and with the value of G we return to (70) and otain G : G œ œ Þ& F (78) G G % Rememer, these values have een calculated y assuming a Bessel system with the envelope delay of 1 s with the upper cut off frequency N Þ(& radîs and with a normalized value V œ H. Because œ ÎVG, and ecause we require an actual cut off frequency H œ 1 & radîs (or 0H œ 50 khz), we now have to multiply these capacitor values y NÎ H, and divide them y the actual value of V to otain the required values for a system upper cut off frequency equal to. We can choose any standard value of V (making sure that our amplifiers are capale of driving such an impedance), and then check if the resulting capacitor values are within their tolerance limits to their standard values. Without a special order the capacitors availale on the market are usually of a 10% tolerance, with the E12 set of values, whilst the resistors are easily otainale with 1% values and the E48 set. Therefore there is a good chance of finding easily several suitale V values. One such suitale value is V œ (& H. This makes V% œ H, and the capacitors: G % œ *Þ&%) nf G œ Þ((' nf G œ G œ ) pf (79) As already emphasized, in the differential filter G% and G are eing driven y the resistors in oth filter arms, so they effectively see a doule resistance. Thus the differential filter uses G and G of half the value of the single-ended example. % We can approximate the values to the following nearest standard values: H -14-
G % œ%þ(nf G œþ) nf G œ G œ * pf (80) Note that for our purpose it is not particularly important to have the upper cut off frequency exactly 50 khz, it is much more important to have the envelope delay sustantially flat up to the filter cut off frequency. With a flat envelope delay all the relevant frequencies will pass through the system with equal delay, and that preserves the shape of the time-domain transient (step and impulse) response as close as possile to the ideal for the chosen andwidth. Design Verification Because the circuit is very simple we can verify the design y uilding it on a proto-oard, and measure its performance y inserting a low frequency square wave to the input and oserve the waveform on an oscilloscope, checking the rise time (the 10% to 90% transition should e of the order of 10 12 µs) and the amount of overshoot (this should e ideally 0.4%, not exceeding 1%). We can also uild the filter model in one of the many circuit simulation programs and run the AC and transient simulations. We can also verify the design y calculating the magnitude, the phase angle and the envelope delay as functions of frequency and plot the results. We can also check the design y calculating and plotting the transient (step) response. The procedure example is shown here riefly. Frequency Domain Performance As already shown in equation (6), the magnitude (asolute value) of the transfer function is calculated y taking the square root of the product of the transfer function with its own complex conjugate. We take the transfer function from (27) and express it as a function of purely imaginary frequency 4 : Ja4 œe O a4 a4 O 4 O O The gain E is the same as in (28), and the coefficients Oßß are the same as in equations (24 26). Taking the appropriate power of the imaginary unit yields: Ja4 œe O 4 O 4 O O We need to separate the real and imaginary parts in the denominator: (81) (82) -15-
Ja4 œe O a O O 4a O (83) We rationalize the denominator y multiplying oth the numerator and the denominator y the complex conjugate of the denominator: Ja4 œ E O ca O O 4a O d ca O O 4a O dca O O 4a O d By multiplying the terms in the denominator we otain: Ja4 œeo So the magnitude is: Qa œ E ÈJa4 Ja4 a O O 4a O a O O a O (84) (85) (86) œeo œeo Éa O O a O a O O a O È ' % ao O ao O O O ' % ao O ao O O O To otain the phase as the function of frequency use (85), and take the arctangent of the imaginary to real ratio: eeja4 f : a œ arctan deja4 f (87) (88) EO : a œ arctan EO a O a O O a O a O O a O O a O (89) O œ arctan O O The envelope delay is calculated as the phase derivative in frequency. Because the phase angle is measured in radians and the angular frequency in radians per second, the envelope delay is measured in seconds. œ. : a 7 ea. (90) (91) -16-
M( f ) [db] rd By using the phase expression of (90) the envelope delay is: œ. O 7 ea Œarctan. O O (92) œ œ œ a O a O O a O a O O O O Š a O O % % O aoo OOO O O O Š a O O % a O O % O ao O O O O a O O a O a O O œ % % O ao O O O O ' % ao O ao O O O (93) 10 5 0 5 10 15 p M( f ) ϕ ( f ) τ e ( f ) ϕ ( f ) [ ] τ e ( f ) [µs] 180 0 240 5 300 10 360 20 a R 3 R 2 R 1 1 A 1 C 1 o 420 25 30 35 750 750 750 C 4 4n7 C 4a 4n7 C 3a 1n8 R 4 390pF 3k0 R A 4 C 3 R 3 + R 2 1n8 480 40 10 3 10 4 10 5 10 6 f [Hz] Fig.6: Frequency domain performance of the (single ended) 3rd-order Bessel filter. Note the gain of 2 (+6 db), the ) db Î0 roll off, the phase inversion ( 180 ) at low frequencies, and the envelope delay of ~6 µs eing essentially flat up to the cut off frequency (~50 khz). The graphs were plotted using actual component values rounded to the standard values. The difference from the ideal Bessel filter is mostly notale from the % envelope delay, which increases very slightly etween & and Hz. -17-
Time Domain Performance (Transient Response) It is eyond the scope of this text to provide the full mathematical ackground on the forward and inverse Laplace transform, the complex numer theory, the contour integration of Cauchy or Laurent types of complex functions, and the theory of residues, which are necessary for understanding the relationship etween the frequency domain and time domain performance of electronic circuits. Here we shall only give a cookery ook recipe for otaining the (linear) time domain performance via residue calculation. More information can e found in standard mathematical textooks, and in a compressed form suitale for electronics engineers in Ref.6. A residue ( residuum in Latin, meaning a remainder) of the transfer function is found y eliminating one of its poles and calculating the value of the transfer function at that pole from the remaining poles. By doing so for all the poles we otain all the residues and the sum of all the residues is the time domain response of the system. It is worth mentioning that the residue pairs of complex conjugated pole pairs are also complex conjugated, thus summing them up results in the doule value of the real part only (imaginary components, eing of opposite sign, cancel y the addition). Thus only 8Î residues need to e calculated for even order systems, and a8 Î for odd order systems. Here we offer a full calculation as an example. The calculation of residues is easy if we express the transfer function y (29) with the values of the poles from (57). Then the general form for a 5 th residue is: 8 3œ V 5 œ lim a p 5 5 8 3œ a 3 3 > e (94) a Note: The expression (94) is valid only for functions containing simple non-repeating poles. For functions containing multiple (coincident) poles, the procedure is different. If a function 7 contains 7 coincident poles, say 3 œá œ 37 œ+, there is a term a+ in the characteristic polynomial of the transfer function, and to otain the residue at the pole œ+ the limiting process should e performed on the a7 th derivative: a7. 7 V 5 œ lim + K p+ a 7 x a a. a7 Note that circuits containing coincident poles (equal VG components) are never optimal in nd any sense, except maye in the case of a 2 -order system, if critical damping is asolutely necessary. In practice it is easy to disregard optimization and use equal VG components throughout the system, ut in such a way we sacrifice too much, since in order to achieve the same andwidth in a system with 8 stages those VG components must e selected for the cut off frequency higher y È8, with a consequent increase in system noise caused y the larger andwidth in the first stages. It is therefore always preferale to use staggered (complex conjugated) poles wherever possile. This is especially true for multi-pole Bessel systems, since (owed to the flat envelope delay) their theoretical overshoot of the step response never exceeds 0.5%, a design goal also easily achievale in practice. -18-
So for systems with simple staggered poles we can always use equation (94) for the evaluation of the residues. One more important fact to note is that the sum of residues of the transfer function J a will give us the system s impulse response. The system s time domain response to an input signal is equal to the convolution integral of that signal with the system s impulse response. In the frequency domain the convolution integral is transformed into a simple multiplication of the transformed signal y the transformed impulse response. Because the Laplace transform of a unit impulse a> (the Dirac s delta) is equal to H aœ1, and the multiplication y 1 does not change the transfer function, it is ovious that the inverse Laplace transform of the transfer function J a must return the system s impulse response 0> a. However for the step response the situation is different. The Laplace transform of the Heaviside s unit step 2 a > œ l> is equal to L a œ Î, thus to otain the step response we need the residues of the composite function K aœ J a. Because of this there will e an additional residue owed to the pole at œ. The meaning of this pole and its residue is easily comprehended y realizing that frequency is inverse of time, thus œ means >œ_. In other words, the pole at the complex plane origin determines the final value of the system s output to which it settles to when all transient phenomena have vanished, after some long time (several times longer than the largest system s time constant). For low pass systems this will e the system s DC value, equal to 1 for unit gain systems, and equal to E for systems with a DC gain of E. For and pass and high pass systems this value will usually e zero (or a small DC offset). In spite of this eing a general rule, it is a good practice to check the result Here we now calculate the step response. We shall use the following numerical values of the poles, taken from (54 56) and multiplied y the ratio of the desired to the normalized system cut off frequency, as in (57): % H 1 & & œ < œ Þ œ %Þ&&& radîs (95) Þ(&&' œ < œ < N H 1 & œ aþ))* 4Þ(&%% N Þ(&&' & œ aþ*( 4Þ*% radîs (96) H 1 & œ aþ))* 4Þ(&%% N Þ(&&' & œ aþ*( 4Þ*% radîs (97) Since we want the step response we must multiply our transfer function J a, equation (29) y the Laplace transform of the input unit step signal, Î, so we shall have an additional pole at the complex plane origin: œ radîs (98) and our new function will e: % % -19-
a a a K aœ J a œ E a a a Our first residue for p will e: V a > œ lima E p a a a e > (99) (100) œe œe a a a e> (101) (102) ecause the product a œ a œ œ efore the limiting process, and after the limiting we have the product of the poles in oth the numerator and the denominator, which equals 1, and also e œ. The second residue for p will e: a a a V1a> œ lim a E p a a a e > (103) œ lim E p a a e> œ E a a e> (104) (105) œ E a a e> (106) ecause a cancels efore the limiting, and cancels after the limiting. Likewise for p : a a a V a > œ lim a E p a a a e > (107) œ lim E p a a e> œ E a a e> (108) (109) œ E a a e> (110) and finally for p : -20-
a a a V a > œ lim a E p a a a e œ lim E p a a e> œ E a a e> > (111) (112) (113) œ E a a e> (114) The time domain response is the sum of all residues: 8 ga> œ V a > œ V a > V a > V a > V a > 3œ 3 (115) which in our case means: > ga > œ E e > a a e > a a e a a (116) We may now insert the appropriate numers into the computer and calculate the step response for the required time range and resolution. According to a well known relation, which has een derived for a simple VG low pass system, the step rise time from 10% to 90% of the final amplitude is related to the cut off frequency as: > Þ& < 0 H (117) Since Bessel systems follow closely the frequency response of the 1st-order system down to db, the same approximation is valid for Bessel systems of any order. Thus, for a 50 khz cut off we expect a rise time of aout ( µs, and since the system must settle close to the final value within some & larger time, it will e enough to select a time range from to & µs. A resolution of Þ µs should e suitale, giving us a total of & time samples (longer time range andîor smaller time increments can e chosen at will, though this means more time samples and thus longer calculation). However, efore leaving the suject, there is another important insight to realize, which is difficult to see from the complex exponential expression in (116). In order to see this let us write the poles in terms of their real and imaginary part. The pole is real, and ß form a complex conjugate pair: œ 5 œ 5 4 œ 5 4 (118) -21-
Here the real and imaginary components represent the numerical values as in (95 97). The time domain response can then e written as: a5 4 a5 4 ga> œ E a5 5 4 a5 5 4 e5 > 5 a5 4 4 a5 4 5 a5 4 5 4 ea 5 > 5 a5 4 4 e a5 > a5 4 5 a5 4 5 4 As we sum and multiply the various terms we otain: ga> œ E 5 a5 5 e5 > 4 5 a5 4 e a5 5 5 4 4 > 4 (119) 5 a5 4 e a5 > (120) 5 5 4 4 To reorder this properly we must rationalize the denominators of the last two terms, so we multiply those numerators and denominators y the denominator s complex conjugate: ga> œ E 5 a5 5 e5 > 4 > 5 a5 4 a5 5 4 e a5 a5 5 4 a5 5 4 4 5 a5 4 a5 5 4 e a5 > (121) a5 5 4 a5 5 4 4 The denominators are now real (except for the exponential terms, which will e dealt with later): 5 ga> œ E a5 5 e5 > 4 5 a5 4 a5 5 4 e a5 > a5 5 4 4 5 a5 4 a5 5 4 e a5 > a5 5 4 Now we perform the multiplications in the numerators: ga> œ E œ 5 a5 5 e5 > 4 (122) -22-
4 > 5 c5 5 5 4 a 5 5 d e a5 a5 5 4 4 5 c5 5 5 4 a 5 5 d e a5 > (123) a5 5 4 and we separate the real and imaginary parts, ut we leave the imaginary unit in the denominators of the exponential parts, as we shall need them later: ga> œ E 5 a5 5 e5 > a 4 > a 4 > 5 5 a5 5 5 e 5 5 a 5 5 e 4 5 5 4 a a5 5 4 a5 4 > a5 4 > 5 a5 5 5 e 5 a 5 5 e 4 5 5 4 a 5 5 4 a (124) We group together the two real parts and the two imaginary parts: ga> œ E 5 a5 5 e5 > a5 4 > a5 4 > 5 a5 5 5 e e a5 5 4 4 4 4 5 a 5 5 e e 4 a5 5 4 4 a5 > a5 > We can now cancel the imaginary unit in the last term: ga> œ E œ 5 a5 5 e5 > (125) a5 4 > a5 4 > 5 a5 5 5 e e a5 5 4 4 4 5 a 5 5 e e a5 5 Now we separate the real and imaginary exponents: ga> œ E œ 5 a5 5 e5 > a5 > a5 4 > 5 > 4 > 5 > 4 > 5 a5 5 5 e e e e Œ a5 5 4 4 (126) -23-
5 a 5 5 e e e e a5 5 Œ and we extract the common real exponential part: ga> œ E œ 5 a5 5 e5 > 5 > 4 > 5 > 4 > 4 > 4 > 5 a5 5 5 5 e e > e Œ a5 5 4 4 (127) 4 4 5 a 5 5 e > e > 5 (128) a5 5 > e Œ The imaginary exponential parts can e rearranged to form the Euler s representation of trigonometric functions, a sine and a cosine: ga> œ E œ 5 a5 5 e5 > 4 > 4 > 5 a5 5 5 5 e e > e a5 5 4 5 a 5 5 5 e e 5 5 > e a and so we have a purely real time function: ga> œ E œ 5 a5 5 e5 > 4 > 4 > 5 a5 5 5 e 5 > sina a5 5 5 a 5 5 a5 5 e 5 > cosa > > (129) (130) What we have here are some real scaling coefficients in front of real exponential functions with negative exponents (decaying with time), the last two multiplying (damping) a sine and a cosine function. The resulting step response is plotted in Fig.7, and it clearly shows that the design goal of otaining a maximally steep rise time with minimal overshoot, characteristic of Bessel systems, has een met. Of educational importance is Fig.8, where each row of equation (130) has een drawn separately in the red-green-lue order, with the gray curve the resulting sum (ut without the multiplication y E ). Note the purely exponential first row (red), and the heavily damped sine and cosine of the second (green) and third (lue) row. It is precisely the correct damping which ensures the desired response. -24-
0.5 0 p 1 A 1 a R 3 R 2 R 1 C 1 o o [V] 0.5 1.0 750 750 750 C 4 4n7 C 4a 4n7 C 3a 1n8 R 4 390pF 3k0 A R 4 C 3 R 3 + R 2 1n8 1.5 o 2.0 2.5 0 5 10 15 t [µs] 20 25 30 35 Fig.7: Time domain response to a unit step (going from V to V at > œ ). Note the gain of, the &% level delay of ~ ' µs (as indicated y the envelope delay), the rise time (from 10% to 90%, or from Þ V to Þ) V) of ~ ( µs, and the overshoot at ~ & µs of % aove the final value. 1.0 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1.0 0 5 10 15 20 25 30 35 t [µs] Fig.8: Plot of the row y row components of equation (130) in the red-green-lue order. The gray plot is their sum (as in Fig.7, ut without E ). Note the purely exponential red line (first row), and the heavily damped sine and cosine (green and lue) components (second and third row). -25-
This completes our filter analysis. I hope that the analysis has een presented with enough detail and without gaps, so that it should e easy to follow, and the general design principles shown are clear and easily applicale to other similar circuits. Those readers who will encounter some difficulties in the required theoretical or mathematical ackground should consult the relevant literature, some good examples are indicated in the references, ut of course the list of availale literature is much much roader. -26-
References: [1] Zverev, A.I. : Handook of Filter Synthesis, John Wiley and Sons, New York, 1967. <http://www.amazon.com/handook-filter-synthesis-anatol- Zverev/dp/0471749427/refsr_1_1?sooks&ieUTF8&qid1356282748&sr1-1&keywordsHandook+of+filter+synthesis> [2] Kraniauskas, P.: Transforms in Signals and Systems, Addison-Wesley, Wokingham, 1992. <http://www.amazon.com/transforms-signals-systems-applications- Mathematics/dp/0201196948/refsr_1_1?sooks&ieUTF8&qid1356282878&sr1-1&keywordsTransforms+in+Signals+and+Systems> [3] Churchil, R.W., Brown, J.W. : Complex Variales and Applications, Fourth Edition, International Student Edition, McGraw-Hill, Auckland, 1984. <http://www.amazon.com/complex-variales-applications-international- Churchill/dp/7111133048/refsr_1_6?sooks&ieUTF8&qid1356282998&sr1-6&keywordscomplex+variales+and+applications+churchill> [4] O Flynn, M., Moriarthy, E.: Linear Systems, Time Domain and Transform Analysis, John Wiley, New York, 1987. <http://www.amazon.com/linear-systems-domain-transform- Analysis/dp/006044925X/refsr_1_1?sooks&ieUTF8&qid1356283231&sr1-1&keywordsLinear+Systems%2C+Time+Domain+and+Transform+Analysis> [5] Scott, D.E. : An Introduction to System Analysis, A System Approach, McGraw-Hill, New York, 1987. <http://www.amazon.com/introduction-circuit-analysis-mcgraw-hill- Engineering/dp/0070561273/refsr_1_1?sooks&ieUTF8&qid1356283312&sr1-1&keywordsAn+Introduction+to+System+Analysis%2C+A+System+Approach+y+Scott> [6] Starič, P., Margan, E.: Wideand Amplifiers, Springer Verlag, 2005. < http://www.springer.com/engineering/circuits+%26+systems/ook/978-0-387-28340-1> <http://www.amazon.com/wideand-amplifiers-peter- Staric/dp/0387283404/refsr_1_1?sooks&ieUTF8&qid1356283458&sr1-1&keywordswideand+amplifiers> Some Useful Related Wikipedia Pages: < http://en.wikipedia.org/wiki/network_analysis_%28electrical_circuits%29> < http://en.wikipedia.org/wiki/symolic_circuit_analysis> < http://en.wikipedia.org/wiki/active_filter> < http://en.wikipedia.org/wiki/bessel_filter> < http://en.wikipedia.org/wiki/step_response> -27-