Butterworth filters

The transition between the pass-band and stop-band of a first order filter with cut-off frequency $\omega_c=1/\tau$ is characterized by the the slope of 20 dB per decade of frequency change. To achieve better selectivity, we can cascade a set of $n$ such first order filters to form an nth order filter with a slope of 20n dB per decade.

The FRF of a first-order low-pass filter of unit gain is:

  $$H(j\omega)=\frac{1}{j\omega\tau+1} =\frac{1}{j\omega/\omega_c+1}=\frac{\omega_c}{j\omega+\omega_c}$$ (67)

The FRF of $n$ such filters in series is (assuming they are well buffered with no loading effect):

  $$H(j\omega)=\left(\frac{\omega_c}{\omega_c+j\omega} \right)^n$$ (68)

The cut-off frequency of this nth order filter $\omega_{cn}$ can be found by solving the following equation

  $$\left\vert\frac{\omega_c}{\omega_c+j\omega}\right\vert^n =\left\... ...ega_c}{\sqrt{\omega_c^2+\omega^2}}\right\vert^n =\frac{1}{\sqrt{2}}=2^{-1/2}$$ (69)

to get

  $$\omega=\omega_{cn}=\omega_c\sqrt{2^{1/n}-1}$$ (70)

Example: Design an 4th order LP filter with $\omega_{c4}=2\pi 1000=6.2832\times 10^3$. The cut-off frequency of the first-order LP filter can be found to be

  $$\omega_c=\frac{\omega_{c4}}{\sqrt{2^{1/4}-1}} =\frac{6.2832\times 10^3}{0.435}=1.445\times 10^4$$ (71)

The time constant of the first-order filter is $\tau=RC=1/\omega_c=6.92\times 10^{-5}$. If $C=0.1\;\mu F=10^{-7}\;F$, then

  $$R=\frac{\omega_c}{C}=\frac{6.92\times 10^{-5}}{10^{-7}} =6.92\times 10^2=692\;\Omega$$ (72)

The Butterworth filters have the property that the passing band is flat. The magnitude of the FRF of an nth order low-pass Butterworth filter with cut-off frequency $\omega_c$ is

  $$\left\vert H_{lp}(j\omega)\right\vert=\frac{1}{\sqrt{1+(\omega/\o... ...\omega=0\\ 1/\sqrt{2}&\omega=\omega_c\\ 0 & \omega=\infty\end{array}\right.$$ (73)

where $\omega_c$ is the cut-off frequency at which $\vert H_{lp}(j\omega_c)\vert =1/\sqrt{2}$. The transition between the pass-band and stop-band is controlled by the order $n$. In general, higher order corresponds to more rapid transition. Specially, when $n=0$, $n=1$, and $n=\infty$, we have

• $n=0$, $\vert H_{lp}(j\omega)\vert=1/\sqrt{2}$ is an all-pass filter.
• $n=1$, the Butterworth filter is the regular first-order filter:
    $$\vert H_{lp}(j\omega)\vert=\frac{1}{\sqrt{1+(\omega/\omega_c)^2}}... ...omega^2+\omega_c^2}} =\left\vert\frac{\omega_c}{j\omega+\omega_c}\right\vert$$ (74)
• $n=\infty$, the Butterworth filter becomes an ideal low-pass filter:
    $$\vert H_{lp}(j\omega)\vert=\frac{1}{\sqrt{1+(\omega/\omega_c)^\in... ...ft\{\begin{array}{ll}1&\omega<\omega_c\\ 0&\omega>\omega_c\end{array} \right.$$ (75)

The magnitude of the FRF of an nth order high-pass Butterworth filter with cut-off frequency $\omega_c$ is

  $$\left\vert H_{hp}(j\omega)\right\vert=\frac{1}{\sqrt{1+(\omega_c/... ...\omega=1\\ 1/\sqrt{2}&\omega=\omega_c\\ 1 & \omega=\infty\end{array}\right.$$ (76)

Now we consider the implementation of a Butterworth filter. For simplicity, in the following we assume the frequency is normalized by the cut-off frequency $\omega_c$, i.e., $\omega=\omega/\omega_c$, or $\omega_c=1$. Consider the low-pass case:

  $$\vert H(j\omega)\vert^2=H(j\omega)H(-j\omega) =\frac{1}{1+(\omega/\omega_c)^{2n}} =\frac{1}{1+\omega^{2n}}=\frac{1}{1+(\omega^2)^n}$$ (77)

We need to get its FRF $H(j\omega)$ from its magnitude $\vert H(j\omega)\vert$. To do so, we first consider the transfer function (TF) $H(s)$ in the s-domain corresponding to the FRF, which is the same as $H(j\omega)$ when $s=j\omega$, i.e., $s^2=(j\omega)^2=-\omega^2$. Now the equation above can be written as

  $$\vert H(j\omega)\vert^2=H(j\omega)\,H(-j\omega) =\frac{1}{1+(\om... ...=\frac{1}{1+(-s^2)^n} =\frac{1}{1+(-1)^ns^{2n}}=\vert H(s)\vert^2=H(s) H(-s)$$ (78)

We further find the roots of the denominator, the poles of both $H(s)$ and $H(-s)$, and separate them so that those on the left s-plane are the poles of $H(s)$ (stable and causal), while those on the right s-plane belong to $H(-s)$ (stable and anti-causal).

The roots of the denominator can be found by solving the equation

  $$1+(-1)^ns^{2n}=0,\;\;\;\;\;\;\;\mbox{i.e.}\;\;\;\;\; \left\{\beg... ...0 & \mbox{$n$\ is even}\\ 1-s^{2n}=0 & \mbox{$n$\ is odd}\end{array}\right.$$ (79)

Solving these we get the $2n$ solutions on the unit circle in either of the two different forms depending on whether $n$ is even or odd:

  $$\left\{\begin{array}{ll} s=(-1)^{1/2n}=(e^{j(2k+1)\pi})^{1/2n}=e... ...k\pi/n} & \mbox{$n$\ is odd} \end{array}\right.\;\;\;\;\;\;(k=0,\cdots,2n-1)$$ (80)

• If $n$ is even,
    $$s_k=e^{j(2k+1)\pi/2n},\;\;\;\;\;(k=0,\cdots,2n-1)$$ (81)
  These $2n$ roots form $n$ complex conjugate pairs around the unit circle of the s-plane. Corresponding to each of the $n$ roots $s_k=e^{j(2 k+1)\pi/2n}$ ( $k=0,\cdots,n-1$), there is another root $s_{2n-1-k}$ that is its complex conjugate:
    $$s_{2n-1-k}=e^{j(2(2n-1-k)+1)\pi/2n}=e^{j\pi(4n-2k-1)/2n} =e^{j2\pi} \;e^{-j(2k+1)\pi/2n} =e^{-j(2 k+1)\pi/2n}=s_k^*$$ (82)
  Also, for $s_k=e^{j(2 k+1)\pi/2n}$ to be a pole of $H(s)$, it needs to be on the left s-plane, i.e.,
    $$\frac{(2k+1)\pi}{2n}>\frac{\pi}{2},\;\;\;\;\;\;\mbox{i.e.}\;\;\;\;\;\; k>\frac{n-1}{2}$$ (83)
  Now $H(s)$ can be found in terms of its $n$ poles on the left s-plane:
  

  $$H(s) = \frac{1}{\prod_{k=\left\lceil (n-1)/2 \right\rceil}^{n-1} (s-s_k)... ...})} =\frac{1}{\prod_{k=\left\lceil (n-1)/2 \right\rceil}^{n-1}(s-s_k)(s-s_k^*)}$$

  $$= \frac{1}{\prod_{k=\left\lceil (n-1)/2 \right\rceil}^{n-1}(s^2-(s_... ...}{\prod_{k=\left\lceil (n-1)/2 \right\rceil}^{n-1}(s^2-2\cos((2 k+1)\pi/2n)+1)}$$ (84)

  where $\left\lceil x\right\rceil$ is the ceiling of $x$, and we have used the fact that
    $$s_k+s_k^*=e^{j(2 k+1)\pi/2n}+e^{-j(2 k+1)\pi/2n} =2\cos((2 k+1)\pi/2n), \;\;\;\;\;\; s_k s_k^*=1$$ (85)

• If $n$ is odd,
    $$s_k=e^{j2k\pi/2n}=e^{jk\pi/n},\;\;\;\;\;(k=0,\cdots,2n-1)$$ (86)
  These $2n$ roots contain $s_0=e^{j0}=1$ and $s_n=e^{j\pi}=-1$, as well as $n-1$ complex conjugate pairs. Corresponding to each root $s_k=e^{jk\pi/n}$ ( $k=1,\cdots,n-1$), there is another root $s_{2n-k}$ that is its complex conjugate:
    $$s_{2n-k}=e^{j(2n-k)\pi/n}=e^{-jk\pi/n}=s_k^*$$ (87)
  For $s_k$ to be a pole of $H(s)$, it needs to be on the left s-plane, i.e.,
    $$\frac{k\pi}{n}>\frac{\pi}{2},\;\;\;\;\;\;\mbox{i.e.}\;\;\;\;\;\;k>\frac{n}{2}$$ (88)
  Now $H(s)$ can be found in terms of its $n$ poles on the left s-plane:
  

  $$H(s) = \frac{1}{(s-s_n)\prod_{i=\left\lceil n/2 \right\rceil}^{n-1}(s-s_... ...)} =\frac{1}{(s+1)\prod_{i=\left\lceil n/2 \right\rceil}^{n-1}(s-s_k)(s-s_k^*)}$$

  $$= \frac{1}{(s+1)\prod_{i=\left\lceil n/2 \right\rceil}^{n-1}(s^2-(s... ...frac{1}{(s+1)\prod_{i=\left\lceil n/2 \right\rceil}^{n-1}(s^2-2\cos(k\pi/n)+1)}$$ (89)

Specifically, here we find the transfer function $H(s)$ of the nth order Butterworth filter for $n=2,\cdots,6$:

• $n=2$, $1+s^4=0$, $s^4=-1=e^{j(2k+1)\pi}$, the four roots are $s_k=e^{j(2k+1)\pi)/4}$ ($k=0,\cdots,3$):
    $$s_0=e^{j\pi/4}=\frac{1+j}{\sqrt{2}},\;\;\;\; s_1=e^{j3\pi/4}=\fr... ...j5\pi/4}=\frac{-1-j}{\sqrt{2}},\;\;\;\; s_3=e^{j7\pi/4}=\frac{1-j}{\sqrt{2}}$$ (90)
  out of which $s_1$ and $s_2=s_1^*$ on the left s-plane are the roots of $H(s)$:
  

  $$H(s) = \frac{1}{(s-s_1)(s-s_2)}=\frac{1}{(s-(-1+j)/\sqrt{2})(s-(-1-j)/\sqrt{2})}$$

  $$= \frac{1}{s^2+\sqrt{2}s+1}$$ (91)

  Note that the coefficient of the first order term is $-2\cos(3\pi/2)=\sqrt{2}$.
• $n=3$, $1-s^6=0$, $s^6=1=e^{j2k\pi}$, the six roots are $s_k=e^{j k2\pi/6}=e^{j k\pi/3}$ ($k=0,\cdots,5$)
    $$s_0=e^{j0}=1,\;\;\;\; s_1=e^{j\pi/3}=\frac{1+j\sqrt{3}}{2},\;\;\;\;s_2=e^{j2\pi/3}=\frac{-1+j\sqrt{3}}{2}$$ (92)
    $$s_3=e^{j3\pi/3}=e^{j\pi}=-1,\;\;\;\; s_4=e^{j4\pi/3}=\frac{-1-j\sqrt{3}}{2},\;\;\;\;s_5=e^{j5\pi/3}=\frac{1-j\sqrt{3}}{2}$$ (93)
  out of which $s_2$, $s_3=-1$, and $s_4=s_2^*$ on the left s-plane are the roots of $H(s)$:
  

  $$H(s) = \frac{1}{(s-s_2)(s-s_3)(s-s_4)} =\frac{1}{(s+1)(s-(-1+j\sqrt{3})/2)(s-(-1-j\sqrt{3})/2)}$$

  $$= \frac{1}{(s+1)(s^2+s+1)}$$ (94)

  Note that the coefficient of the first order term is $-2\cos(2\pi/3)=1$.
• $n=4$, $1+s^8=0$, $s^8=-1=e^{j(2k+1)\pi}$, the eight roots are $s_k=e^{j(2k+1)\pi/8}$ ($k=0,\cdots,7$). Evaluating $-2\cos((2k+1)\pi/2n)$ for $k=2$ and $k=3$, we get the coefficients of the two first order terms $-2\cos(5\pi/8)=0.7654$ and $-2\cos(7\pi/8)=1.8478$.
    $$H(s)=\frac{1}{(s^2+0.765 s+1)(s^2+1.848 s+1)}$$ (95)
• $n=5$, $1-s^{10}=0$, $s^{10}=1=e^{j2k\pi}$, the 10 roots are
    $$s_k=e^{j(2k\pi)/10}=e^{j(k\pi)/5},\;\;\;\;\;\;(k=0,\cdots,9)$$ (96)
  Evaluating $-2\cos(k\pi/n)$ for $k=2$ and $k=3$, we get the coefficients of the two first order terms $-2\cos(2\pi/5)=0.618$ and $-2\cos(3\pi/5)=1.618$, and we get
    $$H(s)=\frac{1}{(s+1)(s^2+0.618 s+1)(s^2+1.618 s+1)}$$ (97)
• $n=6$, $1+s^{12}=0$, $s^{12}=-1=e^{j(2k+1)\pi}$, the 12 roots are
    $$s_k=e^{j(2k+1)\pi)/12},\;\;\;\;\;\;(k=0,\cdots,11)$$ (98)
  Evaluating $-2\cos((2k+1)\pi/2n)$ for $k=3$, $k=4$, and $k=5$, we get the coefficients of the three first order terms $-2\cos(7\pi/12)=0.5176$, $-2\cos(9\pi/12)=1.4142$, and $-2\cos(9\pi/12)=1.319$.
    $$H(s)=\frac{1}{(s^2+0.5176 s+1)(s^2+1.4142 s+1)(s^2+1.9319 s+1)}$$ (99)

In summary, we see that a Butterworth filter can be implemented as a cascade of second order systems in the form of $1/(s^2+a\,s+1)$ if $n$ is even, and an additional first order system in the form of $1/(s+1)$ if $n$ is odd. The block diagrams below are for the 5th and 6th order Butterworth filters:

The first order filter in the cascade of the Butterworth filter can be realized by the first order op-amp low-pass circuit shown above with

  $$H(s)=\frac{1/\tau}{s+1/\tau}=\frac{\omega_c}{s+\omega_c}=\frac{1}{s+1}$$ (100)

where $\omega_c=1/\tau=1/RC$. If we let $R=1$, we get $C=1/\omega_c=1$.

The second order systems in the cascade can be implemented as a Sallen-Key low-pass filter with

  $$H(s)=\frac{1/R_1C_1R_2C_2}{s^2+s(R_1+R_2)/R_1R_2C_1+1/R_1C_1R_2C_2} =\frac{1}{s^2+\Delta\omega s+\omega_n^2}=\frac{1}{s^2+a s+1}$$ (101)

where $\omega_n^2=1/R_1R_2C_1C_2$. If we let $R_1=R_2=1$ for simplicity, we get

  $$2/C_1=a,\;\;\;\;\;1/C_1C_2=\omega_n^2=1$$ (102)

Solving these we get

  $$C_1=2/a,\;\;\;\;\; C_2=1/C_1=a/2$$ (103)

A High-pass Butterworth filter can be similarly implemented with the only difference that all first and second order systems in the cascade are high-pass filters

  $$H(s)=\frac{s}{s+\omega_c}=\frac{s}{s+1},\;\;\;\;\;\;\; H(s)=\frac{s^2}{s^2+\Delta\omega s+\omega_c^2}=\frac{s^2}{s^2+a s+1}$$ (104)

so that the transfer function of the cascade is high-pass filter:

  $$H(s)=\left\{\begin{array}{cc}s^n/(1+s^{2n}) & \mbox{$n$\ is even}\\ s^n/(1-s^{2n}) & \mbox{$n$\ is odd}\end{array}\right.$$ (105)

To convert the results obtained above for normalized cut-off frequency $\omega_n=1$ to unnormalized cut-off frequency $\omega_n\ne 1$, all we need to do is to scale all capacitances $C$ to $C'=C/\omega_c$. The capacitor in the first order filter becomes $C'=C/\omega_c$ so that $1/RC'=\omega_c/C'=\omega_c$; while the two capacitors in the second order filter become $C_1'=C_1/\omega_n$ and $C_2'=C_2/\omega_n$ so that $1/C_1'C_2'=\omega_n^2/C_1C_2=\omega_n^2$.