Butterworth Filter Properties 1.1 2008/06/09 14:04:15.638 GMT-5 2008/06/24 00:09:41.599 GMT-5 C. Sidney Burrus csb@rice.edu Daniel Collins Williamson dwilliamson1285@gmail.com C. Sidney Burrus csb@rice.edu This section develops the properties of the Butterworth filter which has as its basic concept a Taylor's series approximation to the desired frequency response. The measure of the approximation is the number of terms in the Taylor's series expansion of the actual frequency response that can be made equal to those of the desired frequency response. The optimal or best solution will have the maximum number of terms equal. The Taylor's series is a power series expansion of a function in the form of F ( ω ) = K 0 + K 1 ω + K 2 ω 2 + K 3 ω 3 + ⋯ where K 0 = F ( 0 ) , K 1 = d F ( ω ) d ω | ω = 0 , K 2 = ( 1 / 2 ) d 2 F ( ω ) d ω 2 | ω = 0 , e t c . , with the coefficients of the Taylor's series being proportional to the various order derivatives of F(ω) evaluated at ω=0. A basic characteristic of this approach is that the approximation is all performed at one point, i.e., at one frequency. The ability of this approach to give good results over a range of frequencies depends on the analytic properties of the response. The general form for the squared-magnitude response is an even function of ω and, therefore, is a function of ω2 expressed as F F ( j ω ) = d 0 + d 2 ω 2 + d 4 ω 4 + . . . + d 2 M ω 2 M c 0 + c 2 ω 2 + c 4 ω 4 + . . . c 2 N ω 2 N In order to obtain a solution that is a lowpass filter, the Taylor's series expansion is performed around ω=0, requiring that FF(0)=1 and that FF(j∞)=0, (i.e., d0=c0, N>M, and c2N≠0). This is written as F F ( j ω ) = 1 + E ( ω ) Combining () and () gives d 0 + d 2 ω 2 + ⋯ + d 2 M w = c 0 + c 2 w + ⋯ + c 2 N ω 2 N + E ( ω ) [ c 0 + c 2 ω + ⋯ ] The best Taylor's approximation requires that FF(jω) and the desired ideal response have as many terms as possible equal in their Taylor's series expansion at a given frequency. For a lowpass filter, the expansion is around ω=0, and this requires E(ω) have as few low-order ω terms as possible. This is achieved by setting c 0 = d 0 , c 2 = d 2 , ⋯ c 2 M = d 2 M , ⋯ c 2 M + 2 = 0 , c 2 N - 2 = 0 , c 2 N ≠ 0 Because the ideal response in the passband is a constant, the Taylor's series approximation is often called “maximally flat". Equation () states that the numerator of the transfer function may be chosen arbitrarily. Then by setting the denominator coefficients of FF(s) equal to the numerator coefficients plus one higher-order term, an optimal Taylor's series approximation is achieved . Since the numerator is arbitrary, its coefficients can be chosen for a Taylor's approximation to zero at ω=∞. This is accomplished by setting d0=1 and all other d's equal zero. The resulting magnitude-squared function is F F ( j ω ) = 1 1 + c 2 N ω 2 N The value of the constant c2N determines at which value of ω the transition of passband to stopband occurs. For this development, it is normalized to c2N=1, which causes the transition to occur at ω=1. This gives the simple form for what is called the Butterworth filter F F ( j ω ) = 1 1 + ω 2 N This approximation is sometimes called “maximally flat" at both ω=0 and ω=∞, since it is simultaneously a Taylor's series approximation to unity at ω=0 and to zero at ω=∞. A graph of the resulting frequency response function is shown in for several N.

Frequency Responses of the Butterworth Lowpass Filter Approximation

The characteristics of the normalized Butterworth filter frequency response are: Very close to the ideal near ω=0 and ω=∞, Very smooth at all frequencies with a monotonic decrease from ω=0 to ∞, and Largest difference between the ideal and actual responses near the transition at ω=1 where |F(j1)|2=1/2. Although not part of the approximation addressed, the phase curve is also very smooth. An important feature of the Butterworth filter is the closed- form formula for the solution, F(s). From (), the expression for FF(s) may be determined as F ( s ) F ( - s ) = 1 1 + ( - s 2 ) N This function has 2N poles evenly spaced around a unit radius circle and 2N zeros at infinity. The determination of F(s) is very simple. In order to have a stable filter, F(s) is selected to have the N left-hand plane poles and N zeros at infinity; F(-s) will necessarily have the right-hand plane poles and the other N zeros at infinity. The location of these poles on the complex s plane for N=1,2,3, and 4 is shown in .

Pole Locations for Analog Butterworth Filter Transfer Function on the Complex s Plane

Because of the geometry of the pole positions, simple formulas are easy to derive for the pole locations. If the real and imaginary parts of the pole location are denoted as s = u + j w the locations of the N poles are given by u k = - cos ( k π / 2 N ) ω k = sin ( k π / 2 N ) for N values of k where k = ± 1 , ± 3 , ± 5 , . . . , ± ( N - 1 ) for N even k = 0 , ± 2 , ± 4 , . . . , ± ( N - 1 ) for N odd Because the coefficients of the numerator and denominator polynomials of F(s) are real, the roots occur in complex conjugate pairs. The conjugate pairs in (,) can be combined to be the roots of second-order polynomials so that for N even, F(s) has the partially factored form of F ( s ) = ∏ k 1 s 2 + 2 cos ( k π / 2 N ) s + 1 for k=1,3,5,...,N-1. For N odd, F(s) has a single real pole and, therefore, the form F ( s ) = 1 s + 1 ∏ k 1 s 2 + 2 cos ( k π / 2 N ) s + 1 for k=2,4,6,⋯,N-1 This is a convenient form for the cascade and parallel realizations discussed in elsewhere. A single formula for the pole locations for both even and odd N is u k = - sin ( ( 2 k + 1 ) π / 2 N ) ω k = cos ( ( 2 k + 1 ) π / 2 N ) for N values of k where k=0,1,2,...,N-1 One of the important features of the Butterworth filter design formulas is that the pole locations are found by independent calculations which do not depend on each other or on factoring a polynomial. A FORTRAN program which calculates these values is given in the appendix as Program 8. Mathworks has a powerful command for designing analog and digital Butterworth filters. The classical form of the Butterworth filter given in () is discussed in many books , , , , . The less well-known form given in () also has many useful applications . If the frequency location of unwanted signals is known, the zeros of the transfer function given by the numerator can be set to best reject them. It is then possible to choose the pole locations so as to have a passband as flat as the classical Butterworth filter by using (). Unfortunately, there are no formulas for the pole locations; therefore, the denominator polynomial must be factored. Summary This section has derived design procedures and formulas for a class of filter transfer functions that approximate the ideal desired frequency response by a Taylor's series. If the approximation is made at ω=0 and ω=∞, the resulting filter is called a Butterworth filter and the response is called maximally-flat at zero and infinity. This filter has a very smooth frequency response and, although not explicitly designed for, has a smooth phase response. Simple formulas for the pole locations were derived and are implemented in the design program in the appendix of this book.

Butterworth Filter Design Procedures This section considers the process of going from given specifications to use of the approximation results derived in the previous section. The Butterworth filter is the simplest of the four classical filters in that all the approximation effort is placed at two frequencies: ω=0 and ω=∞. The transition from passband to stopband occurs at a normalized frequency of ω=1. Assuming that this transition frequency or bandedge can later be scaled to any desired frequency, the only parameter to be chosen in the design process is the order N. The filter specifications that are consistent with what is optimized in the Butterworth filter are the degree of “flatness" at ω=0 (DC) and at ω=∞. The higher the order, the flatter the frequency response at these two points. Because of the analytic nature of rational functions, the flatter the response is at ω=0 and ω=∞, the closer it stays to the desired response throughout the whole passband and stopband. An indirect consequence of the filter order is the slope of the response at the transition between pass and stopband. The slope of the squared-magnitude frequency response at ω=1 is s = F F ' ( j 1 ) = - N / 2 The effects of the increased flatness and increased transition slope of the frequency response as N increases are illustrated in . In some cases specifications state the response must stay above or below a certain value over a given frequency band. Although this type of specification is more compatible with a Chebyshev error optimization, it is possible to design a Butterworth filter to meet the requirements. If the magnitude of the frequency response of the filter over the passband of 0<ω<ωP must remain between unity and G, where ωp<1 and G<1, the required order is found by the smallest integer N satisfying N ≥ log ( ( 1 / G ) 2 - 1 ) 1 log ( ω p ) This is illustrated in where |F| must remain above 0.9 for ω up to 0.9, i.e., G = 0.9 and ωp = 0.9. These requirements require an order of at least N=7.

Passband Specifications for Designing a Butterworth Filter

If stopband performance is stated in the form of requiring that the response stay below a certain value for frequency above a certain value, i.e., |F|<G for ω>ωs, the order is determined by the same formula () with ωp replaced by ωs. Note |F(j1)|=1/(2) which is called the “half power" frequency because |F(j1)|2=1/2. This frequency is normalized to one for the theory but can be scaled to any value for applications. Example. Design of a Butterworth Lowpass IIR Filter To illustrate the calculations, a lowpass Butterworth filter is designed. It is desired that the frequency response stay above 0.8 for frequencies up to 0.9. The formula () for determining the order gives a value of 2.73; therefore, the order is three. The analytic function corresponding to the squared-magnitude frequency response in () is | F ( j ω ) | 2 = 1 1 + ω 6 The transfer function corresponding to the left-half-plane poles of F'(s) are calculated from () to give F ( s ) = 1 ( s + 1 ) ( s + 0 . 5 + j 0 . 866 ) ( s + 0 . 5 - j 0 . 866 ) F ( s ) = 1 ( s + 1 ) ( s 2 + s + 1 ) F ( s ) = 1 s 3 + 2 s 2 + 2 s + 1 The frequency response is obtained by setting s=jω which has a plot illustrated in for N=3. The pole locations are the same as shown in c.

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