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Next: Higher order systems Up: Chapter 6: Active Filter Previous: Wien bridge

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.

Cascade1stOrder.png

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

\begin{displaymath}
H(j\omega)=\frac{1}{j\omega\tau+1}
=\frac{1}{j\omega/\omega_c+1}=\frac{\omega_c}{j\omega+\omega_c}
\end{displaymath}

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

\begin{displaymath}
H(j\omega)=\left(\frac{\omega_c}{\omega_c+j\omega} \right)^n
\end{displaymath}

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

\begin{displaymath}
\left\vert H(j\omega) \right\vert=\left\vert\frac{\omega_c}...
...frac{1}{ (1+(\omega/\omega_c)^2)^{n/2} }
=\frac{1}{\sqrt{2}}
\end{displaymath}

to get

\begin{displaymath}
\omega=\omega_{cn}=\omega_c\sqrt{2^{1/n}-1}
\end{displaymath}

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

\begin{displaymath}
\omega_c=\frac{\omega_{c4}}{\sqrt{2^{1/4}-1}}
=\frac{6.2832\times 10^3}{0.435}=1.445\times 10^4
\end{displaymath}

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

\begin{displaymath}
R=\frac{\omega_c}{C}=\frac{6.92\times 10^{-5}}{10^{-7}}
=6.92\times 10^2=692\;\Omega
\end{displaymath}

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

\begin{displaymath}
\left\vert H_{lp}(j\omega)\right\vert=\frac{1}{\sqrt{1+(\om...
...rt{2}&\omega=\omega_c\\
0 & \omega=\infty\end{array}\right.
\end{displaymath}

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

Given the nth-order low-pass Butterworth filter defined above, we can find the corresponding high-pass Butterworth filter by considering

$\displaystyle \vert H_{hp}(j\omega)\vert^2$ $\textstyle =$ $\displaystyle 1-\vert H_{lp}(j\omega)\vert^2
=1-\frac{1}{1+(\omega/\omega_c)^{2n}}
=\frac{(\omega/\omega_c)^{2n}}{1+(\omega/\omega_c)^{2n}}$  
  $\textstyle =$ $\displaystyle \frac{1}{1+(\omega/\omega_c)^{-2n}}
=\frac{1}{1+(\omega_c/\omega)^{2n}}$  

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

\begin{displaymath}
\left\vert H_{hp}(j\omega)\right\vert=\frac{1}{\sqrt{1+(\om...
...rt{2}&\omega=\omega_c\\
1 & \omega=\infty\end{array}\right.
\end{displaymath}

ButterworthFilters1.png

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, i.e., $\omega_c=1$ so that $\omega/\omega_c=\omega$. Consider the low-pass case:

\begin{displaymath}
\vert H(j\omega)\vert^2=H(j\omega)H(-j\omega)
=\frac{1}{1+...
...a_c)^{2n}}
=\frac{1}{1+\omega^{2n}}=\frac{1}{1+(\omega^2)^n}
\end{displaymath}

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

\begin{displaymath}
\vert H(j\omega)\vert^2=H(j\omega) H(-j\omega)
=\frac{1}{...
...2)^n}
=\frac{1}{1+(-1)^ns^{2n}}=\vert H(s)\vert^2=H(s) H(-s)
\end{displaymath}

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

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

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:

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

ButterworthInSPlane.png

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

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:

ButterworthDiagram.png

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

\begin{displaymath}
H(s)=\frac{1/\tau}{s+1/\tau}=\frac{\omega_c}{s+\omega_c}=\frac{1}{s+1}
\end{displaymath}

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

\begin{displaymath}
H(s)=\frac{1/R_1C_1R_2C_2}{s^2+s(R_1+R_2)/R_1R_2C_1+1/R_1C_...
... =\frac{1}{s^2+\Delta\omega s+\omega_n^2}=\frac{1}{s^2+a s+1}
\end{displaymath}

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

\begin{displaymath}
2/C_1=a,\;\;\;\;\;1/C_1C_2=\omega_n^2=1
\end{displaymath}

Solving these we get

\begin{displaymath}
C_1=2/a,\;\;\;\;\; C_2=1/C_1=a/2
\end{displaymath}

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

\begin{displaymath}
H(s)=\frac{s}{s+\omega_c}=\frac{s}{s+1},\;\;\;\;\;\;\;
H(s...
...rac{s^2}{s^2+\Delta\omega s+\omega_c^2}=\frac{s^2}{s^2+a s+1}
\end{displaymath}

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

\begin{displaymath}
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.
\end{displaymath}

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 $1/C_1'C_2'=\omega_n^2/C_1C_2=\omega_n^2$.


next up previous
Next: Higher order systems Up: Chapter 6: Active Filter Previous: Wien bridge
Ruye Wang 2019-05-07