Approximation Methods

Constant delay filter

A constant delay (Bessel–Thomson) filter is a low-pass filter designed so that its group delay $\tau(\omega) = -d\theta/d\omega$ is maximally flat (nearly constant) over the passband. All frequency components are delayed equally, so the waveform shape is preserved (no phase distortion). The Bessel–Thomson response is the standard example.

Third-order Bessel–Thomson transfer function

The ideal constant-delay function (delay normalized to 1 s) is

$$T(s) = e^{-s} = \frac{1}{\cosh s + \sinh s}$$

Write $M = \cosh s$ (even part) and $N = \sinh s$ (odd part). Their ratio has the continued-fraction expansion

$$\coth s = \frac{M}{N} = \frac{1}{s} + \cfrac{1}{\dfrac{3}{s} + \cfrac{1}{\dfrac{5}{s} + \cfrac{1}{\dfrac{7}{s} + \cdots}}}$$

For an $n$th-order approximation, keep the first $n$ terms, write the result as $M/N$ with polynomials, and use $T(s) = K/(M + N)$, with $K$ chosen so that $T(0) = 1$.

Third order (first three terms):

$$\begin{aligned} \frac{M}{N} &= \frac{1}{s} + \cfrac{1}{\dfrac{3}{s} + \dfrac{s}{5}} = \frac{1}{s} + \frac{5s}{15 + s^2} \\ &= \frac{15 + s^2 + 5s^2}{s(s^2 + 15)} = \frac{6s^2 + 15}{s^3 + 15s} \end{aligned}$$

So $M = 6s^2 + 15$, $N = s^3 + 15s$ and

$$T(s) = \frac{15}{s^3 + 6s^2 + 15s + 15}$$

The same polynomial follows from the Bessel recursion $B_n = (2n-1)B_{n-1} + s^2B_{n-2}$ with $B_0 = 1$, $B_1 = s + 1$: $B_2 = s^2 + 3s + 3$ and $B_3 = 5B_2 + s^2B_1 = s^3 + 6s^2 + 15s + 15$.

Poles: $s = -2.322$ and $s = -1.839 \pm j1.754$. The delay is $\tau(0) = 15/15 = 1$ s and stays at 0.996 s at $\omega$ = 1 rad/s and 0.887 s at 2 rad/s.

For an actual delay $\tau_o$, replace $s$ by $s\tau_o$ (frequency scaling by $k_f = 1/\tau_o$).

Order of an inverse Chebyshev low-pass filter

The inverse Chebyshev response is flat (monotonic) in the passband and equiripple in the stopband. Normalized to the stopband edge $\omega_s$ = 1 rad/s:

$$|T(j\omega)|^2 = \frac{\varepsilon^2 C_n^2(1/\omega)}{1 + \varepsilon^2 C_n^2(1/\omega)}, \qquad \alpha(\omega) = 10\log_{10}\left[1 + \frac{1}{\varepsilon^2 C_n^2(1/\omega)}\right]$$

Stopband ($\omega \ge 1$): $1/\omega \le 1$, so $C_n^2(1/\omega)$ swings between 0 and 1. The least attenuation (where $C_n^2 = 1$) must be $\alpha_{min}$:

$$\alpha_{min} = 10\log_{10}\left(1 + \frac{1}{\varepsilon^2}\right) \;\Rightarrow\; \varepsilon^2 = \frac{1}{10^{0.1\alpha_{min}} - 1}$$

Passband: the normalized passband edge is $\omega_p/\omega_s < 1$, where the loss must not exceed $\alpha_{max}$:

$$\begin{aligned} 10\log_{10}\left[1 + \frac{1}{\varepsilon^2 C_n^2(\omega_s/\omega_p)}\right] &\le \alpha_{max} \\ C_n^2\left(\frac{\omega_s}{\omega_p}\right) &\ge \frac{1}{\varepsilon^2(10^{0.1\alpha_{max}} - 1)} = \frac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1} \end{aligned}$$

Since $\omega_s/\omega_p > 1$, $C_n = \cosh(n\cosh^{-1}x)$, so

$$n \ge \frac{\cosh^{-1}\sqrt{\dfrac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1}}}{\cosh^{-1}(\omega_s/\omega_p)}$$

This is the same expression as for the Chebyshev filter.

Order for the given specifications

$\alpha_{max}$ = 0.25 dB, $\alpha_{min}$ = 18 dB, $\omega_p$ = 1000 rad/s, $\omega_s$ = 1400 rad/s.

$$\begin{aligned} 10^{0.1\alpha_{min}} - 1 &= 10^{1.8} - 1 = 62.0957 \\ 10^{0.1\alpha_{max}} - 1 &= 10^{0.025} - 1 = 0.05925 \\ \sqrt{62.0957/0.05925} &= 32.3723, \qquad \frac{\omega_s}{\omega_p} = \frac{1400}{1000} = 1.4 \\ n &\ge \frac{\cosh^{-1}(32.3723)}{\cosh^{-1}(1.4)} = \frac{4.1702}{0.8670} = 4.810 \end{aligned}$$

So n = 5.

Answer: a 5th-order inverse Chebyshev filter is required, with $\varepsilon^2 = 1/62.096 = 0.01610$.

Characteristics of the elliptic (Cauer) response

The elliptic low-pass response is

$$|T(j\omega)|^2 = \frac{1}{1 + \varepsilon^2 R_n^2(\omega)}$$

where $R_n(\omega)$ is a Chebyshev rational function (Jacobi elliptic function).

• Equiripple in both the passband and the stopband.
• Has finite zeros of transmission on the $j\omega$ axis in the stopband, where the attenuation becomes infinite.
• Sharpest transition (narrowest transition band) of all classical approximations for a given order; so for given $\alpha_{max}$, $\alpha_{min}$, $\omega_s/\omega_p$ it needs the lowest order.
• Stopband attenuation does not keep increasing; it ripples down to $\alpha_{min}$ between the zeros.
• Poorest phase linearity: group delay varies strongly near the band edge, causing pulse distortion.
• Design is complex (elliptic integrals); normally done with tables or software.
• Realization needs both poles and finite zeros (e.g. LC ladders with parallel-resonant series arms, or notch biquads).

 |T|
  1 |~v~v~~.
    |       \        elliptic
    |        \
    |         \  /\    /\
    |          \/  \__/  \___
    +--------------------------> w
          wp  ws (zeros at dips)

Comparison

The elliptic filter is chosen when a very narrow transition band is needed with minimum components and phase is not important; Chebyshev/inverse Chebyshev are simpler alternatives.

Importance of all-pass filters (delay equalization)

An all-pass filter has $|T(j\omega)| = 1$ at all frequencies and changes only the phase. Its zeros are mirror images of its poles about the $j\omega$ axis:

$$T_1(s) = \frac{a - s}{a + s}, \qquad T_2(s) = \frac{s^2 - \dfrac{\omega_o}{Q}s + \omega_o^2}{s^2 + \dfrac{\omega_o}{Q}s + \omega_o^2}$$

The first-order section has phase $\theta = -2\tan^{-1}(\omega/a)$ and delay $\tau = \dfrac{2a}{a^2 + \omega^2}$.

     jw
  x  |  o      x pole, o zero
     |
  x  |  o      |T(jw)| = 1 for all w
 ----+-------> sigma

Importance:

• Sharp filters (Chebyshev, elliptic) have a delay that peaks near the band edge. Cascading an all-pass network adds delay where the filter's delay is small, making the total delay flat without changing the magnitude response. This is delay equalization.
• Used in data transmission, telephone lines, video and radar where pulse shape must be kept.
• Used as phase shifters (e.g. 90° networks for SSB modulation) and as pure delay lines.

Third-order Bessel–Thomson low-pass filter

The ideal constant-delay function (delay normalized to 1 s) is

$$T(s) = e^{-s} = \frac{1}{\cosh s + \sinh s}$$

Write $M = \cosh s$ (even part) and $N = \sinh s$ (odd part). Their ratio has the continued-fraction expansion

$$\coth s = \frac{M}{N} = \frac{1}{s} + \cfrac{1}{\dfrac{3}{s} + \cfrac{1}{\dfrac{5}{s} + \cfrac{1}{\dfrac{7}{s} + \cdots}}}$$

For an $n$th-order approximation, keep the first $n$ terms, write the result as $M/N$ with polynomials, and use $T(s) = K/(M + N)$, with $K$ chosen so that $T(0) = 1$.

Third order (first three terms):

$$\begin{aligned} \frac{M}{N} &= \frac{1}{s} + \cfrac{1}{\dfrac{3}{s} + \dfrac{s}{5}} = \frac{1}{s} + \frac{5s}{15 + s^2} \\ &= \frac{15 + s^2 + 5s^2}{s(s^2 + 15)} = \frac{6s^2 + 15}{s^3 + 15s} \end{aligned}$$

So $M = 6s^2 + 15$, $N = s^3 + 15s$ and

$$T(s) = \frac{15}{s^3 + 6s^2 + 15s + 15}$$

The same polynomial follows from the Bessel recursion $B_n = (2n-1)B_{n-1} + s^2B_{n-2}$ with $B_0 = 1$, $B_1 = s + 1$: $B_2 = s^2 + 3s + 3$ and $B_3 = 5B_2 + s^2B_1 = s^3 + 6s^2 + 15s + 15$.

Poles: $s = -2.322$ and $s = -1.839 \pm j1.754$. The delay is $\tau(0) = 15/15 = 1$ s and stays at 0.996 s at $\omega$ = 1 rad/s and 0.887 s at 2 rad/s.

For an actual delay $\tau_o$, replace $s$ by $s\tau_o$ (frequency scaling by $k_f = 1/\tau_o$).

All-pass filter

An all-pass filter has $|T(j\omega)| = 1$ at all frequencies and changes only the phase. Its zeros are mirror images of its poles about the $j\omega$ axis:

$$T_1(s) = \frac{a - s}{a + s}, \qquad T_2(s) = \frac{s^2 - \dfrac{\omega_o}{Q}s + \omega_o^2}{s^2 + \dfrac{\omega_o}{Q}s + \omega_o^2}$$

The first-order section has phase $\theta = -2\tan^{-1}(\omega/a)$ and delay $\tau = \dfrac{2a}{a^2 + \omega^2}$.

     jw
  x  |  o      x pole, o zero
     |
  x  |  o      |T(jw)| = 1 for all w
 ----+-------> sigma

Importance

• Delay (phase) equalization: added after a filter to make its total group delay flat without changing its magnitude response.
• Phase shifting networks (e.g. 90° phase splitters for SSB), pure time delay of analog signals.
• Correcting phase distortion in communication channels and audio systems.

Second-order constant delay filter

Let

$$T(s) = \frac{a_0}{s^2 + a_1 s + a_0}$$

The phase is $\theta(\omega) = -\tan^{-1}\dfrac{a_1\omega}{a_0 - \omega^2}$, and the group delay is

$$\tau(\omega) = -\frac{d\theta}{d\omega} = \frac{a_1(a_0 + \omega^2)}{a_0^2 + (a_1^2 - 2a_0)\omega^2 + \omega^4}$$

Condition 1 – normalized delay of 1 s at dc: $\tau(0) = a_1a_0/a_0^2 = a_1/a_0 = 1 \Rightarrow a_1 = a_0$.

Condition 2 – maximally flat delay: $\tau(\omega)$ is flattest when the $\omega^2$ terms of numerator and denominator are in the same ratio as the constant terms:

$$\frac{a_1}{a_1^2 - 2a_0} = \frac{a_1 a_0}{a_0^2} \;\Rightarrow\; a_1^2 = 3a_0$$

With $a_1 = a_0$: $a_0^2 = 3a_0 \Rightarrow a_0 = 3,\ a_1 = 3$.

$$T(s) = \frac{3}{s^2 + 3s + 3}$$

Poles: $s = -1.5 \pm j0.866$ ($\omega_o = \sqrt{3}$, $Q = 0.577$). The delay is $\tau(\omega) = \dfrac{9 + 3\omega^2}{9 + 3\omega^2 + \omega^4}$: exactly 1 s at dc and 0.923 s at 1 rad/s.

The same result comes from truncating $\coth s = \dfrac{1}{s} + \dfrac{1}{3/s} = \dfrac{s^2 + 3}{3s}$, giving $M + N = s^2 + 3s + 3$. For a delay $\tau_o$, replace $s$ by $s\tau_o$.

Constant delay filter

A constant delay (Bessel–Thomson) filter is a low-pass filter designed so that its group delay $\tau(\omega) = -d\theta/d\omega$ is maximally flat (nearly constant) over the passband. All frequency components are delayed equally, so the waveform shape is preserved (no phase distortion).

Significance

• A filter with phase $\theta(\omega) = -\omega\tau_o$ delays every frequency by the same $\tau_o$, so the output is an exact delayed copy of the input within the passband.
• Needed where waveform shape matters: pulse and digital data transmission, video, radar, oscilloscope and ECG signals, and audio crossovers.
• Bessel filters have little overshoot and ringing in their step response, unlike Butterworth or Chebyshev filters.
• The cost is a poor magnitude roll-off (wider transition band).

Second-order constant delay filter

Let

$$T(s) = \frac{a_0}{s^2 + a_1 s + a_0}$$

The phase is $\theta(\omega) = -\tan^{-1}\dfrac{a_1\omega}{a_0 - \omega^2}$, and the group delay is

$$\tau(\omega) = -\frac{d\theta}{d\omega} = \frac{a_1(a_0 + \omega^2)}{a_0^2 + (a_1^2 - 2a_0)\omega^2 + \omega^4}$$

Condition 1 – normalized delay of 1 s at dc: $\tau(0) = a_1a_0/a_0^2 = a_1/a_0 = 1 \Rightarrow a_1 = a_0$.

Condition 2 – maximally flat delay: $\tau(\omega)$ is flattest when the $\omega^2$ terms of numerator and denominator are in the same ratio as the constant terms:

$$\frac{a_1}{a_1^2 - 2a_0} = \frac{a_1 a_0}{a_0^2} \;\Rightarrow\; a_1^2 = 3a_0$$

With $a_1 = a_0$: $a_0^2 = 3a_0 \Rightarrow a_0 = 3,\ a_1 = 3$.

$$T(s) = \frac{3}{s^2 + 3s + 3}$$

Poles: $s = -1.5 \pm j0.866$ ($\omega_o = \sqrt{3}$, $Q = 0.577$). The delay is $\tau(\omega) = \dfrac{9 + 3\omega^2}{9 + 3\omega^2 + \omega^4}$: exactly 1 s at dc and 0.923 s at 1 rad/s.

The same result comes from truncating $\coth s = \dfrac{1}{s} + \dfrac{1}{3/s} = \dfrac{s^2 + 3}{3s}$, giving $M + N = s^2 + 3s + 3$. For a delay $\tau_o$, replace $s$ by $s\tau_o$.

Constant delay filter

A constant delay (Bessel–Thomson) filter is a low-pass filter designed so that its group delay $\tau(\omega) = -d\theta/d\omega$ is maximally flat (nearly constant) over the passband. All frequency components are delayed equally, so the waveform shape is preserved (no phase distortion). Its transfer function is $T(s) = K/B_n(s)$, where $B_n(s)$ is a Bessel polynomial: $B_1 = s + 1$, $B_2 = s^2 + 3s + 3$, $B_3 = s^3 + 6s^2 + 15s + 15$, and $B_n = (2n-1)B_{n-1} + s^2B_{n-2}$.

Design steps

1. Specifications: dc delay $\tau_o$, allowed delay error (e.g. within 1%) up to frequency $\omega_1$, and any attenuation limits.
2. Normalize: take $\tau_o$ = 1 s, so the frequency axis becomes $\omega\tau_o$; the normalized band edge is $\omega_1\tau_o$.
3. Choose the order: from Bessel delay curves/tables, pick the lowest $n$ whose normalized delay stays within the tolerance up to $\omega_1\tau_o$ (also check the loss there).
4. Write $T_n(s) = B_n(0)/B_n(s)$.
5. Denormalize: replace $s$ by $s\tau_o$.
6. Realize the function with a passive ladder or active sections (Sallen–Key/MFB), then apply magnitude scaling for practical values.

Example

Design a filter with dc delay $\tau_o$ = 0.5 ms whose delay is within 1% up to $\omega_1$ = 2000 rad/s.

Normalized frequency: $\omega_1\tau_o = 2000 \times 0.5\times10^{-3} = 1$.

Normalized delay at $\omega$ = 1 rad/s (computed from $\tau = -d\theta/d\omega$):

So $n = 3$: $T(s) = \dfrac{15}{s^3 + 6s^2 + 15s + 15}$. Replacing $s$ by $s\tau_o = s/2000$:

$$\begin{aligned} T(s) &= \frac{15}{(s/2000)^3 + 6(s/2000)^2 + 15(s/2000) + 15} \\ &= \frac{1.2\times10^{11}}{s^3 + 12000s^2 + 6\times10^{7}s + 1.2\times10^{11}} \end{aligned}$$

The magnitude loss at 2000 rad/s is only about 0.9 dB. The function can be realized as a first-order RC section (pole at $-2.322 \times 2000 = -4644$ rad/s) cascaded with a Sallen–Key biquad (poles at $(-1.839 \pm j1.754)\times2000$ rad/s).

Delay (phase) equalization is the process of making the total group delay of a filter or channel nearly constant over the passband, by cascading it with all-pass networks.

Filters with sharp magnitude response (Chebyshev, elliptic, high-order Butterworth) have a delay that peaks near the band edge, which distorts pulses and data signals. Since an all-pass section has $|T| = 1$, it does not change the magnitude response, but it adds its own delay. The total delay is the sum:

$$\tau_{total}(\omega) = \tau_{filter}(\omega) + \tau_{AP}(\omega)$$

How it is done:

1. Plot or compute the delay $\tau_F(\omega)$ of the filter over the passband.
2. Choose first- and/or second-order all-pass sections whose delay is large where $\tau_F$ is small and small where $\tau_F$ is large (a second-order section gives a delay peak near its $\omega_o$, with height set by $Q$).
3. Adjust $a$, $\omega_o$ and $Q$ (and add sections) until $\tau_F + \tau_{AP}$ is flat within the allowed tolerance.
4. Realize the all-pass sections with op-amp circuits and cascade them with the filter.

 tau      filter        equalizer        total
  |        .            .     .
  |      .' `.          `. .'         .------.
  | ...-'     `      ->   `     ->   '        `
  +----------- w      +------ w      +---------- w

First-order op-amp all-pass: an inverting/non-inverting op-amp with equal resistors $R_1$ in the feedback path and an RC network at the + input gives $T(s) = \dfrac{1 - sRC}{1 + sRC}$.

          R1        R1
 Vi o--+-/\/\-+---/\/\---+
       |      |  |\      |
       |      +--|-\     |
       R         |  >----+--o Vo
       |      +--|+/
       +------+  |/
              C
              |
             GND

Applications: data modems, telephone lines, video and pulse transmission, where waveform shape must be preserved.

Order of a Butterworth low-pass filter

For the normalized Butterworth response (passband edge $\omega_p$ = 1 rad/s)

$$|T(j\omega)|^2 = \frac{1}{1+\varepsilon^2\omega^{2n}}, \qquad \alpha(\omega) = 10\log_{10}(1 + \varepsilon^2\omega^{2n})\ \text{dB}$$

Passband condition ($\omega = 1$, $\alpha = \alpha_{max}$):

$$\alpha_{max} = 10\log_{10}(1+\varepsilon^2) \;\Rightarrow\; \varepsilon^2 = 10^{0.1\alpha_{max}} - 1$$

Stopband condition (at $\omega = \omega_s/\omega_p$ the attenuation must be at least $\alpha_{min}$):

$$\begin{aligned} 10\log_{10}\left[1 + \varepsilon^2\left(\frac{\omega_s}{\omega_p}\right)^{2n}\right] &\ge \alpha_{min} \\ \left(\frac{\omega_s}{\omega_p}\right)^{2n} &\ge \frac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1} \end{aligned}$$

Taking logarithms:

$$n \ge \frac{\log_{10}\left[\dfrac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1}\right]}{2\log_{10}(\omega_s/\omega_p)}$$

$n$ is rounded up to the next integer.

Order for the given specifications

$\alpha_p$ = 0.5 dB, $\alpha_s$ = 20 dB, $\omega_p$ = 1000 rad/s, $\omega_s$ = 2000 rad/s.

$$\begin{aligned} 10^{0.1\alpha_{min}} - 1 &= 10^{2} - 1 = 99.0000 \\ 10^{0.1\alpha_{max}} - 1 &= 10^{0.05} - 1 = 0.12202 \\ \frac{\omega_s}{\omega_p} = \frac{2000}{1000} = 2 \\ n &\ge \frac{\log_{10}(99.0000/0.12202)}{2\log_{10}(2)} = \frac{2.9092}{0.6021} = 4.832 \end{aligned}$$

The order must be an integer, so n = 5.

Check: with $n = 5$, $\varepsilon^2 = 0.12202$, attenuation at $\omega_s$ is $10\log_{10}(1 + 0.12202 \times 2^{10}) = 21.00$ dB $\ge 20$ dB.

Pole locations

$\varepsilon = \sqrt{0.12202} = 0.34931$. Choosing the passband edge exactly at $\omega_p$ (this leaves a margin in the stopband), the poles lie on a circle of radius

$$\omega_o = \frac{\omega_p}{\varepsilon^{1/n}} = \frac{1000}{0.34931^{1/5}} = 1234.12\ \text{rad/s}$$

at angles $\theta_k = \dfrac{(2k-1)\pi}{2n}$ measured from the $j\omega$ axis: $p_k = -\omega_o\sin\theta_k \pm j\omega_o\cos\theta_k$.

           jw
       x   |       radius 1234 rad/s
    x      |       poles 36 deg apart
 x---------+------ sigma
    x      |
       x   |

Transfer function

Normalized 5th-order Butterworth: $(s+1)(s^2+0.618s+1)(s^2+1.618s+1)$. Replacing $s$ by $s/\omega_o$:

$$T(s) = \frac{\omega_o^5}{(s + 1234.12)(s^2 + 762.73s + 1.5231\times10^6)(s^2 + 1996.85s + 1.5231\times10^6)}$$

with $\omega_o^5 = 2.8628\times10^{15}$, $\omega_o^2 = 1.5231\times10^6$, so $T(0) = 1$.

Characteristics of the Butterworth response

The Butterworth (maximally flat) low-pass response has magnitude

$$|T(j\omega)|^2 = \frac{1}{1 + \varepsilon^2\omega^{2n}}$$

(normalized so that $\omega_p$ = 1 rad/s; with $\varepsilon = 1$ the half-power frequency is 1 rad/s). Its characteristics are:

• Maximally flat at $\omega = 0$: the first $2n-1$ derivatives of $|T(j\omega)|^2$ are zero at $\omega = 0$, so the passband is as flat as possible.
• Monotonic: no ripple in the passband or the stopband; the gain falls continuously with frequency.
• $|T(j0)| = 1$; at $\omega = 1$ (when $\varepsilon = 1$) $|T| = 1/\sqrt{2}$ (−3 dB) for every order.
• Roll-off of $20n$ dB/decade far into the stopband; the transition is wider than Chebyshev or elliptic for the same $n$.
• All-pole function: all transmission zeros are at $\omega = \infty$.
• Poles lie on a circle of radius $\omega_o = \varepsilon^{-1/n}$ in the left-half s-plane, equally spaced by $\pi/n$, symmetric about the real axis.
• Phase is fairly linear (better than Chebyshev), with a moderate delay peak near the cut-off.
• As $n \to \infty$ the response approaches the ideal brick-wall.

 |T|
  1 |----.__
    |       `.   n = 2
0.707 - - - - :\.  n = 4
    |         : \ \.
    |         :  `. `-._
    +---------+---------`----> w
              1

Order of a Butterworth low-pass filter

For the normalized Butterworth response (passband edge $\omega_p$ = 1 rad/s)

$$|T(j\omega)|^2 = \frac{1}{1+\varepsilon^2\omega^{2n}}, \qquad \alpha(\omega) = 10\log_{10}(1 + \varepsilon^2\omega^{2n})\ \text{dB}$$

Passband condition ($\omega = 1$, $\alpha = \alpha_{max}$):

$$\alpha_{max} = 10\log_{10}(1+\varepsilon^2) \;\Rightarrow\; \varepsilon^2 = 10^{0.1\alpha_{max}} - 1$$

Stopband condition (at $\omega = \omega_s/\omega_p$ the attenuation must be at least $\alpha_{min}$):

$$\begin{aligned} 10\log_{10}\left[1 + \varepsilon^2\left(\frac{\omega_s}{\omega_p}\right)^{2n}\right] &\ge \alpha_{min} \\ \left(\frac{\omega_s}{\omega_p}\right)^{2n} &\ge \frac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1} \end{aligned}$$

Taking logarithms:

$$n \ge \frac{\log_{10}\left[\dfrac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1}\right]}{2\log_{10}(\omega_s/\omega_p)}$$

$n$ is rounded up to the next integer.

Order for the given specifications

$\omega_p$ = 1000 rad/s, $\alpha_{max}$ = 0.5 dB, $\omega_s$ = 2000 rad/s, $\alpha_{min}$ = 20 dB.

$$\begin{aligned} 10^{0.1\alpha_{min}} - 1 &= 10^{2} - 1 = 99.0000 \\ 10^{0.1\alpha_{max}} - 1 &= 10^{0.05} - 1 = 0.12202 \\ \frac{\omega_s}{\omega_p} = \frac{2000}{1000} = 2 \\ n &\ge \frac{\log_{10}(99.0000/0.12202)}{2\log_{10}(2)} = \frac{2.9092}{0.6021} = 4.832 \end{aligned}$$

The order must be an integer, so n = 5.

Check: with $n = 5$, $\varepsilon^2 = 0.12202$, attenuation at $\omega_s$ is $10\log_{10}(1 + 0.12202 \times 2^{10}) = 21.00$ dB $\ge 20$ dB.

Answer: order $n = 5$.

Characteristics of the Butterworth response

The Butterworth (maximally flat) low-pass response has magnitude

$$|T(j\omega)|^2 = \frac{1}{1 + \varepsilon^2\omega^{2n}}$$

(normalized so that $\omega_p$ = 1 rad/s; with $\varepsilon = 1$ the half-power frequency is 1 rad/s). Its characteristics are:

• Maximally flat at $\omega = 0$: the first $2n-1$ derivatives of $|T(j\omega)|^2$ are zero at $\omega = 0$, so the passband is as flat as possible.
• Monotonic: no ripple in the passband or the stopband; the gain falls continuously with frequency.
• $|T(j0)| = 1$; at $\omega = 1$ (when $\varepsilon = 1$) $|T| = 1/\sqrt{2}$ (−3 dB) for every order.
• Roll-off of $20n$ dB/decade far into the stopband; the transition is wider than Chebyshev or elliptic for the same $n$.
• All-pole function: all transmission zeros are at $\omega = \infty$.
• Poles lie on a circle of radius $\omega_o = \varepsilon^{-1/n}$ in the left-half s-plane, equally spaced by $\pi/n$, symmetric about the real axis.
• Phase is fairly linear (better than Chebyshev), with a moderate delay peak near the cut-off.
• As $n \to \infty$ the response approaches the ideal brick-wall.

 |T|
  1 |----.__
    |       `.   n = 2
0.707 - - - - :\.  n = 4
    |         : \ \.
    |         :  `. `-._
    +---------+---------`----> w
              1

Order of a Butterworth low-pass filter

For the normalized Butterworth response (passband edge $\omega_p$ = 1 rad/s)

$$|T(j\omega)|^2 = \frac{1}{1+\varepsilon^2\omega^{2n}}, \qquad \alpha(\omega) = 10\log_{10}(1 + \varepsilon^2\omega^{2n})\ \text{dB}$$

Passband condition ($\omega = 1$, $\alpha = \alpha_{max}$):

$$\alpha_{max} = 10\log_{10}(1+\varepsilon^2) \;\Rightarrow\; \varepsilon^2 = 10^{0.1\alpha_{max}} - 1$$

Stopband condition (at $\omega = \omega_s/\omega_p$ the attenuation must be at least $\alpha_{min}$):

$$\begin{aligned} 10\log_{10}\left[1 + \varepsilon^2\left(\frac{\omega_s}{\omega_p}\right)^{2n}\right] &\ge \alpha_{min} \\ \left(\frac{\omega_s}{\omega_p}\right)^{2n} &\ge \frac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1} \end{aligned}$$

Taking logarithms:

$$n \ge \frac{\log_{10}\left[\dfrac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1}\right]}{2\log_{10}(\omega_s/\omega_p)}$$

$n$ is rounded up to the next integer.

Order for the given specifications

$\omega_p$ = 2000 rad/s, $\alpha_{max}$ = 1 dB, $\omega_s$ = 3000 rad/s, $\alpha_{min}$ = 12 dB.

$$\begin{aligned} 10^{0.1\alpha_{min}} - 1 &= 10^{1.2} - 1 = 14.8489 \\ 10^{0.1\alpha_{max}} - 1 &= 10^{0.1} - 1 = 0.25893 \\ \frac{\omega_s}{\omega_p} = \frac{3000}{2000} = 1.5 \\ n &\ge \frac{\log_{10}(14.8489/0.25893)}{2\log_{10}(1.5)} = \frac{1.7585}{0.3522} = 4.993 \end{aligned}$$

The order must be an integer, so n = 5.

Check: with $n = 5$, $\varepsilon^2 = 0.25893$, attenuation at $\omega_s$ is $10\log_{10}(1 + 0.25893 \times 1.5^{10}) = 12.02$ dB $\ge 12$ dB.

Answer: order $n = 5$.

The calculated value 4.993 is very close to 5, so a 5th-order filter only just meets the stopband requirement (12.02 dB).

Characteristics of the Butterworth response

The Butterworth (maximally flat) low-pass response has magnitude

$$|T(j\omega)|^2 = \frac{1}{1 + \varepsilon^2\omega^{2n}}$$

(normalized so that $\omega_p$ = 1 rad/s; with $\varepsilon = 1$ the half-power frequency is 1 rad/s). Its characteristics are:

• Maximally flat at $\omega = 0$: the first $2n-1$ derivatives of $|T(j\omega)|^2$ are zero at $\omega = 0$, so the passband is as flat as possible.
• Monotonic: no ripple in the passband or the stopband; the gain falls continuously with frequency.
• $|T(j0)| = 1$; at $\omega = 1$ (when $\varepsilon = 1$) $|T| = 1/\sqrt{2}$ (−3 dB) for every order.
• Roll-off of $20n$ dB/decade far into the stopband; the transition is wider than Chebyshev or elliptic for the same $n$.
• All-pole function: all transmission zeros are at $\omega = \infty$.
• Poles lie on a circle of radius $\omega_o = \varepsilon^{-1/n}$ in the left-half s-plane, equally spaced by $\pi/n$, symmetric about the real axis.
• Phase is fairly linear (better than Chebyshev), with a moderate delay peak near the cut-off.
• As $n \to \infty$ the response approaches the ideal brick-wall.

 |T|
  1 |----.__
    |       `.   n = 2
0.707 - - - - :\.  n = 4
    |         : \ \.
    |         :  `. `-._
    +---------+---------`----> w
              1

Order of a Butterworth low-pass filter

For the normalized Butterworth response (passband edge $\omega_p$ = 1 rad/s)

$$|T(j\omega)|^2 = \frac{1}{1+\varepsilon^2\omega^{2n}}, \qquad \alpha(\omega) = 10\log_{10}(1 + \varepsilon^2\omega^{2n})\ \text{dB}$$

Passband condition ($\omega = 1$, $\alpha = \alpha_{max}$):

$$\alpha_{max} = 10\log_{10}(1+\varepsilon^2) \;\Rightarrow\; \varepsilon^2 = 10^{0.1\alpha_{max}} - 1$$

Stopband condition (at $\omega = \omega_s/\omega_p$ the attenuation must be at least $\alpha_{min}$):

$$\begin{aligned} 10\log_{10}\left[1 + \varepsilon^2\left(\frac{\omega_s}{\omega_p}\right)^{2n}\right] &\ge \alpha_{min} \\ \left(\frac{\omega_s}{\omega_p}\right)^{2n} &\ge \frac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1} \end{aligned}$$

Taking logarithms:

$$n \ge \frac{\log_{10}\left[\dfrac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1}\right]}{2\log_{10}(\omega_s/\omega_p)}$$

$n$ is rounded up to the next integer.

Order for the given specifications

$\omega_p$ = 1000 rad/s, $\alpha_{max}$ = 1 dB, $\omega_s$ = 2000 rad/s, $\alpha_{min}$ = 20 dB.

$$\begin{aligned} 10^{0.1\alpha_{min}} - 1 &= 10^{2} - 1 = 99.0000 \\ 10^{0.1\alpha_{max}} - 1 &= 10^{0.1} - 1 = 0.25893 \\ \frac{\omega_s}{\omega_p} = \frac{2000}{1000} = 2 \\ n &\ge \frac{\log_{10}(99.0000/0.25893)}{2\log_{10}(2)} = \frac{2.5825}{0.6021} = 4.289 \end{aligned}$$

The order must be an integer, so n = 5.

Check: with $n = 5$, $\varepsilon^2 = 0.25893$, attenuation at $\omega_s$ is $10\log_{10}(1 + 0.25893 \times 2^{10}) = 24.25$ dB $\ge 20$ dB.

Answer: order $n = 5$.

Order of a Butterworth low-pass filter

For the normalized Butterworth response (passband edge $\omega_p$ = 1 rad/s)

$$|T(j\omega)|^2 = \frac{1}{1+\varepsilon^2\omega^{2n}}, \qquad \alpha(\omega) = 10\log_{10}(1 + \varepsilon^2\omega^{2n})\ \text{dB}$$

Passband condition ($\omega = 1$, $\alpha = \alpha_{max}$):

$$\alpha_{max} = 10\log_{10}(1+\varepsilon^2) \;\Rightarrow\; \varepsilon^2 = 10^{0.1\alpha_{max}} - 1$$

Stopband condition (at $\omega = \omega_s/\omega_p$ the attenuation must be at least $\alpha_{min}$):

$$\begin{aligned} 10\log_{10}\left[1 + \varepsilon^2\left(\frac{\omega_s}{\omega_p}\right)^{2n}\right] &\ge \alpha_{min} \\ \left(\frac{\omega_s}{\omega_p}\right)^{2n} &\ge \frac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1} \end{aligned}$$

Taking logarithms:

$$n \ge \frac{\log_{10}\left[\dfrac{10^{0.1\alpha_{min}} - 1}{10^{0.1\alpha_{max}} - 1}\right]}{2\log_{10}(\omega_s/\omega_p)}$$

$n$ is rounded up to the next integer.

Order for $\alpha_{max}$ = 1 dB, $\alpha_{min}$ = 20 dB, $\omega_s/\omega_p$ = 1.5

$$\begin{aligned} 10^{0.1\alpha_{min}} - 1 &= 10^{2} - 1 = 99.0000 \\ 10^{0.1\alpha_{max}} - 1 &= 10^{0.1} - 1 = 0.25893 \\ \frac{\omega_s}{\omega_p} = 1.5 \\ n &\ge \frac{\log_{10}(99.0000/0.25893)}{2\log_{10}(1.5)} = \frac{2.5825}{0.3522} = 7.333 \end{aligned}$$

The order must be an integer, so n = 8.

Check: with $n = 8$, $\varepsilon^2 = 0.25893$, attenuation at $\omega_s$ is $10\log_{10}(1 + 0.25893 \times 1.5^{16}) = 22.33$ dB $\ge 20$ dB.

Pole locations

Work with the passband edge normalized to $\omega_p$ = 1 rad/s. $\varepsilon = \sqrt{0.25893} = 0.50885$. With the passband edge met exactly, the poles lie on a circle of radius

$$\omega_o = \varepsilon^{-1/n} = 0.50885^{-1/8} = 1.0881\ \text{rad/s}$$

at $p_k = -\omega_o\sin\theta_k \pm j\omega_o\cos\theta_k$, $\theta_k = (2k-1)\pi/16$ (spacing 22.5°):

Transfer function

Each conjugate pair gives a factor $s^2 + 2\omega_o\sin\theta_k\, s + \omega_o^2$ with $\omega_o^2 = 1.1840$:

$$T(s) = \frac{1.9652}{(s^2 + 0.4246s + 1.1840)(s^2 + 1.2091s + 1.1840)(s^2 + 1.8095s + 1.1840)(s^2 + 2.1344s + 1.1840)}$$

where the numerator $\omega_o^8 = 1.9652$ makes $T(0) = 1$. For an actual $\omega_p$, replace $s$ by $s/\omega_p$. (Check: loss at $\omega$ = 1 is 1.00 dB and at $\omega$ = 1.5 is 22.33 dB.)