Doing the Bilinear Transform for a 2nd order filter again

Well it was fun yesterday learning about MathJax and playing around describing the bilinear transform for a 2nd order filter. As I was looking at a couple of sites last night I noticed that a good bunch of descriptions also shown the same transformation done slightly differently so as I got pencil and paper out last night out of curiosity to see how they came to the same result I thought I'd put another post so that both can be seen:

Starting again with our 2nd-order low-pass transfer function in the s-domain:

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

Yesterday we used the following bilinear transform with frequency warping:

$$s \leftarrow \frac {1}{tan \frac {\omega}{2}} \frac {1-z^{-1}}{1+z^{-1}}$$

What I'd like to do today is to use a slightly different writing of this and see how the math juggling comes out:

$$s \leftarrow \frac {1}{K} \frac {z-1}{z+1}$$

Where $K= tan \frac {\omega}{2}$

This is the same transform, which can be seen by multiplying by $\frac {z^{-1}}{z^{-1}}$. The maths just plays out a little differently.

Substituting this in to the transfer function:

$$H(s) = \frac {1}{s^2 + \frac {s}{Q} + 1} \leftarrow \frac {1}{\frac {1}{K^2} \frac {(z-1)^2}{(z+1)^2}+\frac {1}{KQ} \frac {(z-1)}{(z+1)}+1}$$

Multiplying top and bottom by $K^2(z+1)^2$ gives something slightly easier:

$$H(z) = \frac {K^2(z+1)^2}{(z-1)^2 + \frac {K}{Q}(z-1)(z+1) + K^2(z+1)^2}$$

Expanding:

$$H(z) = \frac {K^2(z^2 + 2z + 1)}{(z^2+2z+1)+ \frac {K}{Q}z^2 - \frac {K}{Q}+K^2(z^2+2z+1)}$$

Collecting together the terms:

$$H(z) = \frac {K^2z^2 + 2K^2z + K^2}{z^2(1+\frac{K}{Q}+K^2) + z(-2+2K^2)+(1-\frac{K}{Q}+K^2)}$$

All very easy, which gives us our coefficients:

• $b_2 = K^2$
• $b_1 = 2K^2$
• $b_0 = K^2$
• $a_2 = K^2+\frac{K}{Q}+1$
• $a_1 = -2(1-K^2)$
• $a_0 = K^2-\frac{K}{Q}+1$

What I wanted to see is how these then can be manipulated to get the same equations as in my previous post.

First step then is to get the top-half functions all looking similar, so we can take $K^2$ and using the trig identity $K^2 = tan^2 \frac {\omega}{2} = \frac {(1-cos \omega)}{(1+cos \omega)}$. We can use the numerator in the top and the denominator in the bottom. This makes the b coefficients:

• $b_2 = 1-cos \omega$
• $b_1 = 2 (1-cos \omega)$
• $b_0 = 1-cos \omega$

Now taking $a_0$ first, with the bottom half $(1+cos \omega)$ and substituting K:

$$a_0 = (1+cos \omega) \left(  tan^2 \frac {\omega}{2} - \frac {1}{Q} tan \frac {\omega}{2} + 1 \right)$$

Using our favourite trig identities $K =  tan \frac {\omega}{2} = \frac {sin \omega}{(1+cos \omega)}$ and $K^2 = tan^2 \frac {\omega}{2} = \frac {(1-cos \omega)}{(1+cos \omega)}$ and substituting $1 = \frac {(1+cos \omega)}{(1+cos \omega)}$ gives:

$$a_0 = (1+cos \omega) \left( tan^2 \frac {omega}{2} - \frac {1}{Q} tan \frac {\omega}{2} + 1 \right) = (1+cos \omega) \left(  \frac {(1-cos \omega)}{(1+cos \omega)} - \frac {1}{Q} \frac {sin \omega}{(1+cos \omega)} + \frac {(1+cos \omega)}{(1+cos \omega)} \right)$$

And crossing out terms gives:

$$a_0 = (1-cos \omega) - \frac {sin \omega}{Q} + (1+cos \omega) = 2 - \frac {sin \omega}{Q}$$

and similarly for $a_2$. Now taking $a_1$:

$$a_1 = -2 (1+cos \omega)\left(1-tan^2 \frac {\omega}{2} \right)= -2 (1+cos \omega) \left( \frac {(1+cos \omega)}{(1+cos \omega)} - \frac {(1-cos \omega)}{(1+cos \omega)} \right) = -2\left((1+cos \omega) - (1-cos \omega)\right) = -2(2cos \omega)$$

and dividing all the terms by 2 gives:

• $b_2 = \frac {(1-cos \omega)}{2}$
• $b_1 = (1-cos \omega)$
• $b_0 = \frac {(1-cos \omega)}{2}2$
• $a_2 = (1+\frac {sin \omega}{2})$
• $a_1 = -2cos \omega$
• $a_0 = (1-\frac {sin \omega}{2})$ 

Which looks just like the coefficients calculated by the previous method. QED