576 Active Filter Circuits Thus, we can design a fourth-order low-pass filter with any arbitrary cutoff frequency by starting with a fourth-order cascade consisting of prototype low-pass filters and then scaling the components by kt = w c /0.435 to place the cutoff frequency at any value of w c desired. Note that we can build a higher order low-pass filter with a nonunity gain by adding an inverting amplifier circuit to the cascade. Example 15.7 illustrates the design of a fourth-order low-pass filter with nonunity gain. Example 15.7 Designing a Fourth-Order Low-Pass Op Amp Filter Design a fourth-order low-pass filter with a cutoff frequency of 500 Hz and a passband gain of 10. Use 1 fxF capacitors. Sketch the Bode magnitude plot for this filter. Finally, add an inverting amplifier stage with a gain of Rf/Ri = 10. As usual, we can arbitrarily select one of the two resistor values. Because we are already using 138.46 11 resistors, let /¾ = 138.46 ft; then, Solution We begin our design with a cascade of four proto- type low-pass filters. We have already used Eq. 15.23 to calculate the cutoff frequency for the resulting fourth-order low-pass filter as 0.435 rad/s. A frequency scale factor of kf = 7222.39 will scale the component values to give a 500 Hz cutoff fre- quency. A magnitude scale factor of k, n = 138.46 permits the use of 1 (JLF capacitors. The scaled com- ponent values are thus R = 138.46 0; C = l/xF. R f = 10/¾ 1384.6 tt. The circuit for this cascaded the fourth-order low-pass filter is shown in Fig. 15.18. It has the transfer function H(s) = -10 7222.39 s + 7222.39 The Bode magnitude plot for this transfer func- tion is sketched in Fig. 15.19. Figure 15.18 A The cascade circuit for the fourth-order low-pass filter designed in Example 15.7. 15.4 Higher Order Op Amp Filters 577 30 20 PQ — 10 -20 -30 \ \ \ 10 50 100 500 1000 /(Hz) 500010,000 Figure 15.19 A The Bode magnitude plot for the fourth-order low-pass fitter designed in Example 15.7. By cascading identical low-pass filters, we can increase the asymptotic slope in the transition and control the location of the cutoff frequency, but our approach has a serious shortcoming: The gain of the filter is not con- stant between zero and the cutoff frequency oo c . Remember that in an ideal low-pass filter, the passband magnitude is 1 for all frequencies below the cutoff frequency. But in Fig. 15.16, we see that the magnitude is less than 1 (0 dB) for frequencies much less than the cutoff frequency. This nonideal passband behavior is best understood by looking at the magnitude of the transfer function for a unity-gain low-pass nth-order cas- cade. Because H(s) the magnitude is given by 1//0)1 = (* + "«,)"' co cn Va? + (x)l r \/(co/a> cn ) 2 + 1 (15.24) As we can see from Eq. 15.24, when (o <5C (o a „ the denominator is approximately 1, and the magnitude of the transfer function is also nearly 1. But as a) —* (o cn , the denominator becomes larger than 1, so the magnitude becomes smaller than 1. Because the cascade of low-pass filters results in this nonideal behavior in the passband, other approaches are taken in the design of higher order filters. One such approach is examined next. Butterworth Filters A unity-gain Butterworth low-pass filter has a transfer function whose magnitude is given by \H(ja>)\ = , =====, (15.25) Vl + (co/co c ) 2 " where n is an integer that denotes the order of the filter. 1 When studying Eq. 15.25, note the following: 1. The cutoff frequency is o) c rad/s for all values of n. 2. If n is large enough, the denominator is always close to unity when a) < co c . 3. In the expression for | H(ja>)\, the exponent of co/co c is always even. This last observation is important, because an even exponent is required for a physically realizable circuit (see Problem 15.26). Given an equation for the magnitude of the transfer function, how do we find H(s)? The derivation for H(s) is greatly simplified by using a proto- type filter.Therefore, we set o) c equal to 1 rad/s in Eq. 15.25. As before, we will use scaling to transform the prototype filter to a filter that meets the given filtering specifications. To find H(s), first note that if N is a complex quantity, then \N\ 2 = NN*, where N* is the conjugate of N. It follows that \H(jto)\ 2 = H{jo))H{-ia>). (15.26) But because s = jco, we can write \H(jo))\ 2 = H(S)H(S). (15.27) Now observe that s 2 = — a?. Thus, \H(j<o)\ 2 = 1 + 1 + 1 + a> 2 " 1 (<o 2 y 1 is 2 )" 1 1 + (-l)V' or H(s)H(-s) = , , , -^. (15.28) 1 + (-l)'V The procedure for finding H(s) for a given value of n is as follows: 1. Find the roots of the polynomial 1 + (-l)\v 2 * = 0. 2. Assign the left-half plane roots to H(s) and the right-half plane roots to H(-s). 3. Combine terms in the denominator of H(s) to form first- and second-order factors. Example 15.8 illustrates this process. 1 This filter was developed by the British engineer S. Butterworth and reported in Wireless Engineering 7 (1930): 536-541. 15.4 Higher Order Op Amp Filters 579 Example 15.8 Calculating Butterworth Transfer Functions Find the Butterworth transfer functions for n = 2 and n = 3. Solution For n = 2, we find the roots of the polynomial 1 + (-l)V = 0. Rearranging terms, we find s 4 =-1 = 1/180°. Therefore, the four roots are *i = 1/4T = 1/V5 + j/V2, s 2 = 1/135° = -1/V2 + j/V2, s 3 = 1/225° = -1/V2 + -//V2, s 4 = 1/315° = 1/V2 + -//V2. Roots s 2 and .y 3 are in the left-half plane. Thus, H{$) = (s + 1/V2 - j/V2)(s + 1/V5 + ;/V5) _ 1 " (5 2 + V5s + 1)' For n = 3, we find the roots of the polynomial l + (-i)V = o. Rearranging terms, s 6 = 1/ry = 1/360°. Therefore, the six roots are s t = l/o: = l, .v 2 = 1/60° = 1/2 + /V5/2, 5 3 = 1/120° = -1/2+ /V3/2, s 4 = 1/180° = -1 +/0, .v 5 = 1/240° = -1/2 + -/V3/2, .v A = 1/300° = 1/2 + -;'V3/2. Roots S3, \ 4 , and s$ are in the left-half plane. Thus, H ^ ~ (s + 1)(5 + 1/2 - j\/3/2)(s + 1/2 + /V3/2) 1 " Cv+l)(.v 2 + .v+ 1)' We note in passing that the roots of the Butterworth polynomial are always equally spaced around the unit circle in the s plane. To assist in the design of Butterworth filters, Table 15.1 lists the Butterworth polynomials up to n = 8. TABLE 15.1 Normalized (so that co c = 1 rad/s) Butterworth Polynomials up to the Eighth Order /i wth-Order Butterworth Polynomial 1 (s + 1) 2 (r + V2.v + 1) 3 (s + 1)(5 2 + s + 1) 4 (.v 2 + 0.7655 + l)(r + 1.848s + 1) 5 (J + l)(r + 0.618s + 1)(5 2 + 1.6185 + 1) 6 (5 2 + 0.5185 + 1)(5 2 + V2 + 1)(5 2 + 1.9325 + 1) 7 (5 + 1)(5 2 + 0.4455 + 1)(5 2 + 1.2475 + 1)(5 2 + 1.8025 + 1) 8 (5 2 + 0.3905 + 1)(5 2 + 1.111s + l)(s 2 + 1.6663s + l)(s 2 + 1.962s + 1) Butterworth Filter Circuits Now that we know how to specify the transfer function for a Butterworth filter circuit (either by calculating the poles of the transfer function directly or by using Table 15.1), we turn to the problem of designing a circuit with 580 Active Filter Circuits Figure 15.21 A A circuit that provides the second-order transfer function for the Butterworth filter cascade. such a transfer function. Notice the form of the Butterworth polynomials in Table 15.1. They are the product of first- and second-order factors; therefore, we can construct a circuit whose transfer function has a Butterworth polynomial in its denominator by cascading op amp circuits, each of which provides one of the needed factors. A block diagram of such a cascade is shown in Fig. 15.20, using a fifth-order Butterworth poly- nomial as an example. All odd-order Butterworth polynomials include the factor (s + 1), so all odd-order Butterworth filter circuits must have a subcircuit that provides the transfer function H(s) = 1/(5 + 1). This is the transfer function of the prototype low-pass op amp filter from Fig. 15.1. So what remains is to find a circuit that provides a transfer function of the form H(s) = l/(s 2 +bis + 1). Such a circuit is shown in Fig. 15.21. The analysis of this circuit begins by writing the s-domain nodal equations at the noninverting terminal of the op amp and at the node labeled V a : y-y V - V + (V a - V 0 )sC x + ^^ = 0, R V 0 sC 2 + R Vn - V* R = 0. (15.29) (15.30) Simplifying Eqs. 15.29 and 15.30 yields (2 + RC h s)V a -(1 + RC lS )V 0 = V, , (15.31) -V a + (1 + RC 2 s)V a = 0. (15.32) Using Cramer's rule with Eqs. 15.31 and 15.32, we solve for V a : V n = 2+RC^s V, -1 0 2+RC h s -(1+RC t s) -1 l+RC 2 s V; i? 2 C,C 2 5 2 + 2RC 2 s + 1 (15.33) Then, rearrange Eq. 15.33 to write the transfer function for the circuit in Fig. 15.21: *<'> £ 1 R 2 CiC 2 s L + s + RC, R 2 C X C 2 (15.34) 1 s 2 +0.6185+ 1 1 5 2 + 1.6185+ 1 Figure 15.20 A A cascade of first- and second-order circuits with the indicated transfer functions yielding a fifth-order low-pass Butterworth filter with o) r = 1 rad/s. 15.4 Higher Order Op Amp Filters 581 Finally, set R = 1 fi in Eq. 15.34; then H(s) = C,C H-2 (15.35) r + —s + C\ C\C 2 Note that Eq. 15.35 has the form required for the second-order circuit in the Butterworth cascade. In other words, to get a transfer function of the form H(s) = 1 s 2 + b h <> + 1 we use the circuit in Fig. 15.21 and choose capacitor values so that Z?i = —- and 1 = C, c t c 2 (15.36) We have thus outlined the procedure for designing an /ith-order Butterworth low-pass filter circuit with a cutoff frequency of w t . = 1 rad/s and a gain of 1 in the passband. We can use frequency scaling to calculate revised capacitor values that yield any other cutoff frequency, and we can use magnitude scaling to provide more realistic or practical component values in our design. We can cascade an inverting amplifier circuit to pro- vide a gain other than 1 in the passband. Example 15.9 illustrates this design process. Example 15.9 Designing a Fourth-Order Low-Pass Butterworth Filter Design a fourth-order Butterworth low-pass filter with a cutoff frequency of 500 Hz and a passband gain of 10. Use as many 1 k£l resistors as possible. Compare the Bode magnitude plot for this Butterworth filter with that of the identical cascade filter in Example 15.7. Solution From Table 15.1, we find that the fourth-order Butterworth polynomial is (r + 0.7655 + 1)(5 2 + 1.848^ + 1). We will thus need a cascade of two second-order fil- ters to yield the fourth-order transfer function plus an inverting amplifier circuit for the passband gain of 10. The circuit is shown in Fig. 15.22. Let the first stage of the cascade implement the transfer function for the polynomial (s 2 + 0.765s + 1). From Eq. 15.36, Cj = 2.61 F, C 2 = 0.38 F. Let the second stage of the cascade implement the transfer function for the polynomial (s 2 + 1.8485 + 1). From Eq. 15.36, C 3 = 1.08 F, C 4 = 0.924 F. The preceding values for C^, C 2 , C 3 , and C4 yield a fourth-order Butterworth filter with a cutoff frequency of 1 rad/s. A frequency scale factor of kf = 3141.6 will move the cutoff frequency to 500 Hz. A magnitude scale factor of k m = 1000 will permit the use of 1 kft resistors in place of 1 ft resistors. The resulting scaled component values are R = 1 kft, C, = 831 nF, C 2 = 121 nF, C 3 = 344 nF, C, = 294 nF. 582 Active Filter Circuits Finally, we need to specify the resistor values in the inverting amplifier stage to yield a passband gain of 10. Let 7?, = 1 kil; then R MR, = 10 kO. Figure 15.23 compares the magnitude responses of the fourth-order identical cascade fil- ter from Example 15.7 and the Butterworth filter we just designed. Note that both filters provide a passband gain of 10 (20 dB) and a cutoff frequency of 500 Hz, but the Butterworth filter is closer to an ideal low-pass filter due to its flatter passband and steeper rolloff at the cutoff frequency. Thus, the Butterworth design is preferred over the identical cascade design. Figure 15.22 • A fourth-order Butterworth filter with non-unity gain. 30 20 10 P2 Bu tei ! worth \\ \ \ ;de :as< \ \ ntical :ade \ \ -10 -20 -30 10 50 100 500 1000 500010,000 /(Hz) Figure 15.23 • A comparison of the magnitude responses for a fourth-order low-pass filter using the identical cascade and Butterworth designs. The Order of a Butterworth Filter It should be apparent at this point that the higher the order of the Butterworth filter, the closer the magnitude characteristic comes to that of an ideal low-pass filter. In other words, as n increases, the magnitude stays close to unity in the passband, the transition band narrows, and the magni- tude stays close to zero in the stopband. At the same time, as the order increases, the number of circuit components increases. It follows then that 15.4 Higher Order Op Amp Filters 583 a fundamental problem in the design of a filter is to determine the small- est value of n that will meet the filtering specifications. In the design of a low-pass filter, the filtering specifications are usually given in terms of the abruptness of the transition region, as shown in Fig. 15.24. Once A p , co p , A x , and &> s are specified, the order of the Butterworth filter can be determined. For the Butterworth filter, 20 log,,, 1 Vl + co p " = -101og 10 (l +4"), A s = 20 logio" (15.37) Vl + cuj" = -101og I0 (l +(*>;"). (15.38) It follows from the definition of the logarithm that 10 0M '* = 1 + <4", (15.39) JQ-0.M, = i + w 2n (1540 j Now we solve for <a n p and to" and then form the ratio (toj(o p )". We gel o) s V Vl0"°- M « - 1 VuP M„ 1 (T, (T T (15.41) where the symbols a s and a p have been introduced for convenience. From Eq. 15.41 we can write n log u ,(w s /o;p) = log H) («- s /cr p ), or n = JoginWa>) logm(a> 5 /&)„) (15.42) We can simplify Eq. 15.42 if o) p is the cutoff frequency, because then A p equals -20 log ]( )V2, and <r p = 1. Hence login o\ One further simplification is possible. We are using a Butterworth fil- ter to achieve a steep transition region. Therefore, the filtering specifica- tion will make 10~ aM * » l.Thus C7 S « U)- (mA \ (15.44) log,o a s « -0.05/1,. (15.45) Therefore, a good approximation for the calculation of n is -0.05/4, Note that w 5 /w /} = f s /f p - so we can work with either radians per second or hertz to calculate n. \H(ja>)\ dB Stop band Figure 15.24 • Defining the transition region for a low-pass filter. 584 Active Filter Circuits The order of the filter must be an integer; hence, in using either Eq. 15.42 or Eq. 15.46, we must select the nearest integer value greater than the result given by the equation. The following examples illustrate the usefulness of Eqs. 15.42 and 15.46. Example 15.10 Determining the Order of a Butterworth Filter a) Determine the order of a Butterworth filter that has a cutoff frequency of 1000 Hz and a gain of no more than -50 dB at 6000 Hz. b) What is the actual gain in dB at 6000 Hz? Solution a) Because the cutoff frequency is given, we know a v = 1. We also note from the specification that 10 01( 50) is much greater than 1. Hence, we can use Eq. 15.46 with confidence: (-0.05)(-50) log 10 (6000/1000) = 3.21. Therefore, we need a fourth-order Butterworth filter. b) We can use Eq. 15.25 to calculate the actual gain at 6000 Hz. The gain in decibels will be K = 20 log ID 1 Vl + 6* -62.25 dB. Example 15.11 An Alternate Approach to Determining the Order of a Butterworth Filter a) Determine the order of a Butterworth filter whose magnitude is 10 dB less than the passband magnitude at 500 Hz and at least 60 dB less than the passband magnitude at 5000 Hz. b) Determine the cutoff frequency of the filter (in hertz). c) What is the actual gain of the filter (in decibels) at 5000 Hz? Solution a) Because the cutoff frequency is not given, we use Eq. 15.42 to determine the order of the filter: <r p = Vl0- a, <- 10 > - 1 = 3, = \/ 1(r o.i(-60) - i ps loot), ojtop = fs/fp = 5000/500 = 10, log 10 (1000/3) * " log 10 (10) " 2>52 ' Therefore we need a third-order Butterworth filter to meet the specifications. b) Knowing that the gain at 500 Hz is — 10 dB, we can determine the cutoff frequency. From Eq. 15.25 we can write -101og 10 [l + (co/co c f) = -10, where u> = 10007T rad/s. Therefore 1 + (w/o> c ) 6 = 10, and 0) = 2178.26 rad/s. It follows that f c = 346.68 Hz. c) The actual gain of the filter at 5000 Hz is K = -10 log 10 [l + (5000/346.68) 6 ] = -69.54 dB. 15.4 Higher Order Op Amp Filters 585 Butterworth High-Pass, Bandpass, and Bandreject Filters An nth-order Butterworth high-pass filter has a transfer function with the nth-order Butterworth polynomial in the denominator, just like the nth-order Butterworth low-pass filter. But in the high-pass filter, the numerator of the transfer function is s n , whereas in the low-pass filter, the numerator is 1. Again, we use a cascade approach in designing the Butterworth high-pass filter. The first-order factor is achieved by including a prototype high-pass filter (Fig. 15.4, with R { = R 2 = 1 XI, and C = 1 F) in the cascade. To produce the second-order factors in the Butterworth polynomial, we need a circuit with a transfer function of the form H(s) = s 2 s 2 + bis + I Such a circuit is shown in Fig. 15.25. This circuit has the transfer function V v 2 H(s) = -f = - — — . (15.47) Vi 2 2 s 2 + ^—:s •+ R 2 C R X R 2 C 2 Setting C = 1 F yields .v 2 H(s) = —. (15.48) s 2 + —s + R 2 R\R 2 Thus, we can realize any second-order factor in a Butterworth polynomial of the form (s 2 + b x s + 1) by including in the cascade the second-order circuit in Fig. 15.25 with resistor values that satisfy Eq. 15.49: h x = — and 1 = ——. (15.49) 1\ 2 *M^\> At this point, we pause to make a couple of observations relative to Figs. 15.21 and 15.25 and their prototype transfer functions l/(s 2 + b x s + 1) and s 2 /(s 2 + b^s + 1). These observations are impor- tant because they are true in general. First, the high-pass circuit in Fig. 15.25 was obtained from the low-pass circuit in Fig. 15.21 by interchanging resistors and capacitors. Second, the prototype transfer function of a high- pass filter can be obtained from that of a low-pass filter by replacing s in the low-pass expression with \/s (see Problem 15.48). We can use frequency and magnitude scaling to design a Butterworth high-pass filter with practical component values and a cutoff frequency other than 1 rad/s. Adding an inverting amplifier to the cascade will accommodate designs with nonunity passband gains. The problems at the end of the chapter include several Butterworth high-pass filter designs. Now that we can design both nth-order low-pass and high-pass Butterworth filters with arbitrary cutoff frequencies and passband gains, we can combine these filters in cascade (as we did in Section 15.3) to pro- duce nth-order Butterworth bandpass filters. We can combine these filters in parallel with a summing amplifier (again, as we did in Section 15.3) to produce nth-order Butterworth bandreject filters. This chapter's problems also include Butterworth bandpass and bandreject filter designs. Figure 15.25 A A second-order Butterworth high-pass filter circuit. . circuit whose transfer function has a Butterworth polynomial in its denominator by cascading op amp circuits, each of which provides one of the needed factors. A block diagram of such a cascade