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This page will cover the derivation of the transfer functions of low-pass and high-pass Butterworth filters. Butterworth filters are designed to have a very flat frequency response in the passband.


Normalized Butterworth filters are defined in the frequency domain as follows: (1)|Hn(jω)|11+ω2n In order to determine the transfer function, we'll start from the frequency response squared. We'll assume that the transfer function Hn(s) is a rational function with real coefficients. Therefore, Hn(s)=Hn(s). |Hn(jω)|2=Hn(jω)Hn(jω)=Hn(jω)Hn(jω)=Hn(jω)Hn(jω)=11+ω2n We're looking for the transfer function Hn(s), so we'll use the identity s=jωω=sj. Hn(s)Hn(s)=11+(sj)2n

Poles of Hn(s)Hn(s)

The poles of this transfer function are given by: (sj)2n=1s2n=1(j)2ns2n=1(1)ns2n=(1)n+1s2n=ejπ(n+1) Keep in mind that this is a polynomial of order 2n, so it has 2n complex roots. sk=ej2π2k+n+14nk{0,1,2n1} For example, for n=3, the poles are: s0=ej2π0+3+112=ej2π26s1=ej2π2+3+112=ej2π36s2=ej2π4+3+112=ej2π46s3=ej2π6+3+112=ej2π56s4=ej2π8+3+112=ej2π66s5=ej2π10+3+112=ej2π16 These are all points on the unit circle, π/3=60° apart. The poles are stable if they are in the left half plane, if their complex argument is between 90° and 270°: 2π2k+n+14n(π2,3π2)2k+n+1(n,3n)k(12,n12)k(12,n12){0,1,2n1}k{0,1,n1} (2)sk,stable=ej2π2k+n+14nk{0,1,n1}

Poles of Hn(s)

We want our filter Hn(s) to be stable, so we pick the poles in the left half plane to be the poles of Hn(s). The unstable poles, for k{n,n+1,2n1} are the poles of Hn(s). They are the opposites of the poles of Hn(s): sk,unstable=ej2π2k+n+14nk{n,n+1,2n1}lkn=ej2π2(l+n)+n+14nl{0,1,n1}=ej(2π2l+n+14n+π)=ejπej2π2l+n+14n=1ej2π2l+n+14n=sl,stable

Butterworth Polynomials

We'll define the normalized Butterworth polynomial as follows: (3)Bn(s)k=0n1(sej2π2k+n+14n) We'll rearrange the product to group each pole with its complex conjugate. Then, using the identity ejθ+ejθ=2cosθ, we can further simplify this expression:
Even order n: Bn(s)=k=0n1(sej2π2k+n+14n)=k=0n21(sej2π2k+n+14n)l=n2n1(sej2π2l+n+14n)l=nk1=k=0n21(sej2π2k+n+14n)(sej2π2(nk1)+n+14n)=k=0n21(sej2π2k+n+14n)(sej2π2k+3n14n1)=k=0n21(sej2π2k+n+14n)(sej2π2k+3n14nej2π)=k=0n21(sej2π2k+n+14n)(sej2π(2k+3n14n1))=k=0n21(sej2π2k+n+14n)(sej2π2k+3n4n14n)=k=0n21(sej2π2k+n+14n)(sej2π2kn14n)=k=0n21(sej2π2k+n+14n)(sej2π2k+n+14n)=k=0n21(s2sej2π2k+n+14nsej2π2k+n+14n+1)=k=0n21(s22cos(2π2k+n+14n)s+1)

Odd order n:
In this case, n1 is even, and you get a special pole for k=n12: sn12=ej2π2n12+n+14n=ej2π2n4n=ejπ=1 After isolating this pole, we're left with an even number of complex conjugate poles, just like in the case where n was even.

In conclusion, the normalized Butterworth polynomial of degree n is given by: (4)Bn(s)={k=0n21(s22cos(2π2k+n+14n)s+1)even n(s+1)k=0n121(s22cos(2π2k+n+14n)s+1)odd n

Butterworth Transfer Function Hn(s)

The transfer function Hn(s) has no zeros, so the numerator is a constant. The poles of Hn(s) are given by Equation (2), so the denominator is given by Equation (3). Hn(s)=cBn(s)

We wanted a DC gain of 1 (=0dB) for ω=0: |Hn(0j)|=1|cBn(0)|=1|ck=0n1(0ej2π2k+n+14n)|=1|c|k=0n1|ej2π2k+n+14n|=1|c|1=1 If we want no phase offset for low frequencies, we can postulate that Hn(0j)=0: Hn(0j)=0(cBn(0))=0c(k=0n21(022cos(2π2k+n+14n)0+1))=0c1=0 The derivation is analogous for odd n.
Therefore, c=1, and we've eliminated all unknown parameters from the transfer function: (5)Hn(s)=1Bn(s)

High-Pass Butterworth filters

Up until now, we only looked at the low-pass Butterworth filter. There's also a high-pass version: (6)|Hn,hp(jω)|11+ω2n

We can just multiply the numerator and the denominator by ωn to get a more familiar form: |Hn,hp(jω)|=ωn1+ω2n As you can see, the poles will be the same as for the low-pass version. On top of that, there now are n zeros for s=0.
So the transfer function becomes: (7)Hn,hp(s)=snBn(s)

Non-normalized Butterworth Filters

Up until now, we only looked at normalized Butterworth filters, that have a corner frequency of 1 rad/s. To get a specific corner frequency ωc, we can just scale ω, so the definitions become: (8)|Hn,lp(jω)|11+(ωωc)2n (9)|Hn,hp(jω)|11+(ωcω)2n If you start recalculating the transfer functions, you'll quickly realize that this just scales everything by a factor of ωc. The poles no longer lie on the unit circle, but on a circle with radius |sk|=ωc.
This results in the following transfer functions: (10)Hn,lp(s)=1Bn(sωc) (11)Hn,hp(s)=snωcnBn(sωc)

The gain at the corner frequency can easily be determined from the definitions: |Hn,lp(jωc)|=|Hn,hp(jωc)|=11+(ωcωc)2n=12=220.70720log10|Hn(jωc)|=20log10(22)=10log10(12)3.01 dB This is often called the 3 dB-point or the half-power point, because a sinusoidal input signal at that frequency will result in an output signal that has only half of the power of the input signal: |Hn(jωc)|2=12.