The Chebychev filter of type II is given by the normalized transfer function

$$H(S)=\epsilon \prod_{k=1}^{n} (S-r_k) \prod_{k=1}^{n} \frac{1}{S-s_k}$$

where n is the order of the filter, ε defines the size of the ripples of the filter's magnitude response in the stop band, the roots of the numerator r_{k} are

$$r_k=\frac{1}{-j \, \cos(\frac{\pi (2k-1)}{2n})}$$

and the roots of the denominator s_{k} are

$$s_k=\frac{1}{-\sinh(\frac{\mathrm{asinh}(\frac{1}{\epsilon})}{n}) \, \sin(\frac{\pi (2k-1)}{2n})+j\,\cosh(\frac{\mathrm{asinh}(\frac{1}{\epsilon})}{n}) \, \cos(\frac{\pi (2k-1)}{2n})}$$

Here sinh and cosh are the hyperbolic sine and cosine functions and asinh and acosh are the inverse hyperbolic sine and cosine functions.

The transfer function above is shown irrespective of the cutoff frequency (i.e., it is normalized for the cutoff frequency 1) and the forms of the Chebychev type II filter require substitutions. Use S = s / ω_{c} to get the Chebychev type II low pass filter, where s = jω, ω_{c} is the cutoff frequency of the filter, and ω is the angular frequency spanning the frequency spectrum between 0 and π. Use S = ω_{c} / s to get the Chebychev type II high pass filter. Use S = (s^{2} + ω_{c}^{2}) / (B s) to get the Chebychev type II band pass filter, where ω_{c} is the midpoint of the pass band and B is the width of the band. Use S = B s / (s^{2} + ω_{c}^{2}) to get the Chebychev type II band stop filter.

This filter is also called the "inverse Chebychev filter". Its poles (the roots of the denominator) are the inverse of the Chebychev filter of type I and as its gain function (shown below) "inverts" the gain function of the Chebychev filter of type I. While the Chebychev filter of type I controls the size of ripples in the pass band, this filter exhibits a flat pass band and controls the size of the ripples in the stop band.

## Example: Fourth order low pass Chebychev type II filter

Set n = 4 and use the substitution S = s / ω_{c}. The transfer function of the fourth order low pass Chebychev type II filter is

$$H(s)=\frac{\epsilon (s-r_1\omega_c) (s-r_2\omega_c) (s-r_3\omega_c) (s-r_4\omega_c)}{(s-s_1\omega_c) (s-s_2\omega_c) (s-s_3\omega_c) (s-s_4\omega_c)}$$

Note that r_{1} and r_{4} are complex conjugates, r_{2} and r_{3} are complex conjugates, s_{1} and s_{4} are complex conjugates, and s_{2} and s_{3} are complex conjugates. Moreover, the roots of the numerator r_{k} have real parts equal to zero, and hence r_{1} + r_{4} = 0, and r_{3} + r_{2} = 0. To make our computations easier, we can rewrite the transfer function as follows.

$$H(s)=\frac{\epsilon (s^2+r_1 r_4 \omega_c^2) (s^2+r_2 r_3 \omega_c^2)}{(s^2-(s_1+s_4)\omega_c+s_1 s_4 \omega_c^2)(s^2-(s_2+s_3)\omega_c+s_2 s_3 \omega_c^2)}$$

or

$$H(s)=\frac{\epsilon (s^4+A s^2 + B)}{s^4+C s^3+D s^2+E s + F}$$ $$A=\omega_c^2 (r_1 r_4 + r_2 r_3)$$ $$B=\omega_c^4 r_1 r_4 r_2 r_3$$ $$C=-\omega_c (s_1+s_4+s_2+s_3)$$ $$D=\omega_c^2 (s_1 s_4+s_2 s_3+(s_1 + s_4)(s_2+s_3))$$ $$E=-\omega_c^3 (s_1 s_4 (s_2+s_3) +s_2 s_3 (s_1 + s_4))$$ $$F=\omega_c^4 s_1 s_4 s_2 s_3$$

After the bilinear transformation s = 2 (z – 1) / (z + 1) the transfer function becomes a Z transform transfer function.

$$H(z)=\frac{a_0+a_1 z^{-1}+a_2 z^{-2}+a_3 z^{-3}+a_4 z^{-4}}{b_0+b_1 z^{-1}+b_2 z^{-2}+b_3 z^{-3}+b_4 z^{-4}}$$ $$a_0=a_4=\epsilon (16+4 A+B)$$ $$a_1=a_3=\epsilon (-64+4B)$$ $$a_2=\epsilon (96-8A+6B)$$ $$b_0=16+8C+4D+2E+F$$ $$b_1=-64-16C+4E+4F$$ $$b_2=96-8D+6F$$ $$b_3=-64+16C-4E+4F$$ $$b_4=16-8C+4D-2E+F$$

Say that the cutoff frequency of the filter is ω_{c} = 0.6 and that ε = 0.1 (technically, the cutoff frequency of the filter will be ω_{c} = 2 arctan(0.6/2) ≈ 0.583, due to the warping of the frequency domain by the bilinear transformation). The transfer function of this example Chebychev type 1 low pass filter, after scaling to obtain b_{0} = 1 is

$$H(z)=\frac{0.090729-0.190160z^{-1}+0.251568z^{-2}-0.190160z^{-3}+0.090729z^{-4}}{1-2.662409z^{-1}+2.828952z^{-2}-1.378102z^{-3}+0.264003z^{-4}}$$

and the filter itself is

$$y(k) = 0.090729 x(k) – 0.190160 x(k – 1) + 0.251568 x(k – 2)$$ $$– 0.190160 x(k – 3) + 0.090729 x(k – 4) + 2.662409 y(k – 1) $$ $$– 2.828952 y(k-2) + 1.378102 y(k – 3) – 0.264003 y(k – 4)$$

Suppose that the sampling frequency is 2000 Hz. The cutoff frequency then is ω_{c} = (0.6 * 2000) / (2 π) = 191 Hz. The magnitude response of the filter is shown in the graph below.

## Size of the ripples in the stop band

The magnitude function of the Chebychev type II low pass filter can also be written as follows.

$$|H(j\omega)|=\frac{1}{\sqrt{1+\frac{1}{\epsilon^2 T_n^2 (\frac{\omega_c}{\omega})}}}$$

where T_{n} are the Chebychev polynomials

$$T_n(x)=\begin{cases} \cos(n\,\mathrm{acos}(x)) & |x|\le1 \\ \cosh(n\, \mathrm{acosh}(x)) & |x| \gt 1 \end{cases}$$

In the stop band, where ω_{c} / ω < 1, T_{n}(ω_{c} / ω) = \cos(n \mathrm{acos}(ω_{c} / ω)) and 0 ≤ T_{n}^{2}(ω_{c} / ω) ≤ 1. In the stop band

$$0 \lt |H(j\omega)| \le \frac{1}{\sqrt{1+\frac{1}{\epsilon^2}}}$$

Thus, smaller ε implies smaller ripples in the stop band and larger ε implies larger ripples. It should also be expected that |H(jω)| approaches 0 at ω_{c} / ω approaches cos(π / 2n), since then T_{n}(ω_{c} / ω) = 0.

Using the same parameters as above, namely n = 2 and ω_{c} = 0.6, we can plot the |H(jω)| for three different values of ε (0.1, 0.2, and 0.3).

The magnitude response of the three corresponding filters is:

A larger ε creates a filter with a smaller transition band, but with larger ripples in the stop band and worse stop band attenuation. A smaller ε creates a filter with smaller ripples in the stop band and better stop band attenuation, but with larger transition band.

## Example: Second order band stop Chebychev type II filter

Set n = 2 and S = B s / (s^{2} + ω_{c}^{2}). The transfer function of the second order band stop Chebychev type II filter is

$$H(s)=\epsilon \frac{B^2 s^2+r_1 r_2(s^2+\omega_c^2)^2}{B^2 s^2-B(s_1+s_2)(s^2+\omega_c^2)s+s_1 s_2(s^2+\omega_c^2)^2}$$

s_{1}, s_{2}, r_{1} and r_{2} are computed with the same formulae as before. After the bilinear transformation s = 2 (z – 1) / (z + 1), setting, for example, ε = 0.1, B = 0.5 and ω_{c} = 0.8, and scaling to obtain 1 at the beginning of the denominator, we have the following band stop transfer function

$$H(z)= \frac{0.532250-1.506697z^{-1}+2.107234z^{-2}-1.506697z^{-3}+0.532250z^{-4}}{1+1.933150z^{-1}-1.879211z^{-2}+1.065289z^{-3}-0.276782z^{-4}}$$

If we suppose that the sampling frequency is 2000 Hz, then ω_{c} = (0.8 * 2000) / (2 π) = 255 Hz, B = (0.5 * 2000) / (2 π) = 159 Hz, and the magnitude response of the filter is as follows.

## Higher order Chebychev type II filters

Higher order Chebychev type II filters can be designed either using the methods above, or by stacking lower order filters. For example, the transfer function for the fourth order low pass Chebychev type II filter could have also been written as

$$H(s)= \epsilon \frac{(s-r_1\omega_c)(s-r_4\omega_c)}{(s-s_1\omega_c)(s-s_4\omega_c)} \frac{(s-r_2\omega_c)(s-r_3\omega_c)}{(s-s_2\omega_c)(s-s_3\omega_c)}$$

This means that the fourth order low pass Chebychev type II filter can be implemented by stacking two filters of order two (these are not Chebychev type II filters, as the roots of each filter of order two differ from the Chebychev roots). Multiplying two transfer functions means that the impulse response of the total filter is the convolution of the two impulse responses of the lower order filters and the output of the first filter is the input to the second filter. We combine the roots in the manner above, as s_{1} and s_{4}, s_{2} and s_{3}, r_{1} and r_{4}, and r_{2} and r_{3} form complex conjugate pairs. It is not important whether ε is used in the first filter, the second filter, or split between the two.

The following is a comparison of the magnitude responses of the fourth order and second order Chebychef type II filters with ε = 0.1, sampling frequency of 2000 Hz, and ω_{c} = 0.6.

## Odd order Chebychev type II filters

When the Chebychev type II filter is of odd order n, one of the roots r_{k} of the numerator cannot be computed, as the computation produces a division by zero. At k = (n – 1) / 2 we have r_{k} = 1 / (-j cos(π / 2)). Thus, when n is odd, the middle root of the numerator is omitted from the transfer function, and the numerator becomes a polynomial of s of a lower degree than the denominator. Take, for example, the low pass Chebychev type II filter of order n = 3. Its transfer function, after omitting the middle root r_{2}, is as follows.

$$H(s)= \epsilon \frac{(s-r_1\omega_c)(s-r_3\omega_c)}{(s-s_1\omega_c)(s-s_2\omega_c)(s-s_3\omega_c)}$$

Since r_{1} and r_{3} are complex conjugates, s_{1} and s_{3} are complex conjugates, and s_{2} is a real number, we can rewrite the transfer function above as a function with real coefficients in the following way.

$$H(s)= \frac{\epsilon (s^2+r_1 r_3 \omega_c^2)}{s^3-(s_1+s_3+s_2)\omega_c s^2+(s_1 s_3+(s_1+s_3)s_2)\omega_c^2 s-s_1 s_3 s_2 \omega_c^3}$$

After the bilinear transformation this filter has the transfer function

$$H(z)= \frac{a_0+a_1 z^{-1}+a_2 z^{-2}+a_3 z^{-3}}{b_0+b_1 z^{-1}+b_2 z^{-2}+b_3 z^{-3}}$$ $$a_0=a_3=\epsilon(4+r_1 r_3 \omega_c^2)$$ $$a_1=a_2=\epsilon(-4+3 r_1 r_3 \omega_c^2)$$ $$b_0=8-4(s_1+s_3+s_2)\omega_c+2(s_1 s_3+(s_1+s_3)s_2)\omega_c^2-s_1 s_3 s_2\omega_c^3$$ $$b_1=-24+4(s_1+s_3+s_2)\omega_c+2(s_1 s_3+(s_1+s_3)s_2)\omega_c^2-3 s_1 s_3 s_2\omega_c^3$$ $$b_2=24+4(s_1+s_3+s_2)\omega_c-2(s_1 s_3+(s_1+s_3)s_2)\omega_c^2-3 s_1 s_3 s_2\omega_c^3$$ $$b_3=-8-4(s_1+s_3+s_2)\omega_c-2(s_1 s_3+(s_1+s_3)s_2)\omega_c^2-s_1 s_3 s_2\omega_c^3$$

For example, when ω_{c} = 0.6 and after scaling to obtain b_{0} = 1 in the denominator, the filter would be

$$H(z)=\frac{0.0369-0.0211z^{-1}-0.0211z^{-2}+0.0369z^{-3}}{1-2.179z^{-1}+1.666z^{-2}-0.431z^{-3}}$$

This filter, however, produces a magnitude response in the pass band of about -5 dB. To ensure that the filter does not attenuate the pass band, we add some gain by scaling the coefficients in the numerator, so that H(s), evaluated at zero, is equal to 1. The gain is

$$\frac{1}{|H(0)|}=\frac{s_1 s_3 s_2 \omega_c^3}{\epsilon r_1 r_3 \omega_c^2}$$

In the example above the gain is 1.8 and the filter itself changes to

$$H(z)=\frac{0.0665-0.0380z^{-1}-0.0380z^{-2}+0.0665z^{-3}}{1-2.179z^{-1}+1.666z^{-2}-0.431z^{-3}}$$

The magnitude response of this third order filter is shown below with the magnitude responses of two Chebychev type II filter so order 2 and 4.

## Comments

## Older comments

admin: First posted on 2011 12 01

Werner Van Belle, 2011 12 01: Very nice summary. I found it remarkably hard to find the correct transfer function for the type 2 filter. So, everything is working out quite well now, except for the odd orders. What does one do with the middle zero ? This gives an expression (s+infinity*j)

mic, 2011 12 03: Yes. When the order of the filter is odd, one of the zeroes of the numerator becomes "infinite" (results in a division by zero). From what I have seen, this zero is simply removed from the numerator. This means that the (s + j * infinity) that you mentioned is not included in the numerator. The numerator becomes a polynomial of lower order than the polynomial in the denominator.

mic, 2011 12 17: We added an example of an odd order Chebychev type II computations. When the middle root of the numerator is dropped, the gain of the filter should be adjusted as described above.

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