1. Bilinear Transformation

2. When the impulse invariance method is used aliasing is encountered.
To reduce the distortion caused by aliasing one can tighten the specifications
on the digital filter.
Aliasing occurs because points in the  axis separated by 2/T are mapped into
the same digital frequency .
In the bilinear transformation method, there is a one-to-one correspondence
between  and . Therefore while transforming the analog filter to a digital filter
aliasing is not encountered.
Since  is limited to [- , ] range but  varies from - to , it becomes clear that
 must be compressed when it is mapped to . In other words bilinear
transformation is a non-linear transformation.

5. Observation:
When <0 , 𝑧 <1.
So the left-plane poles of Ha(s) will be mapped into poles within the unit
circle in the z-plane. In other words, a stable prototype analog filter will
lead to a stable digital filter.
When =0, 𝑧 =1. In other words , z lies on the unit circle and can be
written as:
𝑧= 𝑒 𝑗𝜔
where the digital frequency is
𝜔=𝜃=2𝑎𝑟𝑐𝑡𝑎𝑛 Ω𝑇 2
Alternatively we can express the analog frequency  as:
Ω= 2 𝑇 𝑡𝑎𝑛 𝜔 2

6. Figure below represents the relationship between  and 
Figure below represents the relationship between  and . Clearly compression
occurs in the mapping process.

7. Example:
Design a digital lowpass filter with the following specifications
using the bilinear transformation method and a Butterworth prototype filter.

8. Solution:
For simplicity we assume T =1 . Therefore Ω=2𝑡𝑎𝑛 𝜔 This means that the
digital passband and stopband frequencies below
Are mapped into the analog passband and stopband frequencies:
Therefore the specifications of the analog filter becomes:

9. Next we calculate the Rp and Rs values which denote the passband and
Stopband ripples:
𝑅 𝑝 =20 𝑙𝑜𝑔 =−1
𝑅 𝑠 =20 𝑙𝑜𝑔 =−15
And we know that
Ω 𝑝 = and Ω 𝑠 =1.0191
We can then use the MATLAB function buttord:
[𝑁,Ω 𝑐 ]=𝑏𝑢𝑡𝑡𝑜𝑟𝑑 Ω 𝑝 , Ω 𝑠 , 𝑅 𝑝 , 𝑅 𝑠
MATLAB will return the order of the Butterworth filter as N
and the 3-dB cutoff frequency Ω 𝑐
For this problem N = 6
Ω 𝑐 =

10. To complete the design of the Butterworth filter we then use MATLAB function
butter to obtain the zeros, poles and gains of this filter:
𝑍,𝑃,𝐾 =𝑏𝑢𝑡𝑡𝑒𝑟 𝑁, Ω 𝑐 ,′𝑠′
This will return the zeros of Ha(s) in array Z
and
poles of Ha(s) in array P.
The filter gain which is same as Ω 𝑁 𝑐 , is in the variable K.
For this example there are no zeros.
The gain is
and the poles are as follows:

11. Since the poles are available we can use them to write the transfer
function of the Butterworth filter as:
𝐻 𝑎 𝑠 = 𝐾 𝑘=1 6 𝑠− 𝑝 𝑘
= 𝑠 𝑠 𝑠 𝑠 𝑠 𝑠
Finally the last step in the design exercise is to map the above transfer function into a
Digital filter using the Bilinear Transformation . That is :
𝐻 𝑧 = 𝐻 𝑎 2 1− 𝑧 −1 1+ 𝑧 −1
We can do this by using MATLAB also:
[ 𝑍 𝑑 , 𝑃 𝑑 , 𝐾 𝑑 ]=bilinear 𝑍,𝑃,𝐾,1/𝑇 ; where T is the sampling frequency

12. For this example :
Kd =
Zd = [ -1
-1 ]
and
Pd = [ ]

13. Hence the transfer function of the digital filter can be written as:
𝐻 𝑧 = 𝐾 𝑑 𝑘=1 6 1− 𝑧 𝑘 𝑧 −1 𝑘=1 6 1− 𝑝 𝑘 𝑧 −1
= 𝑧 − − 𝑧 − 𝑧 −2 1− 𝑧 − 𝑧 −2 1− 𝑧 − 𝑧 −2

14. Example #2:
Design a lowpass filter with
Cuttoff frequency c = /5 rad/sample
Transition bandwidth of
  = 0.2  radians/sample = 0.1 cycles/sample
With a ripple of
 = 0.1
Use a Butterworth analog filter
Set T = 1.

16. Map the digital to analog frequencies using
= 2 𝑇 𝑡𝑎𝑛 𝜔 2
𝜔 𝑑1 = 𝜋 5 −0.1𝜋 , 𝜔 𝑑2 = 𝜋 𝜋
Ω 1 = 𝑟𝑎𝑑/𝑠𝑒𝑐 , Ω 2 = 𝑟𝑎𝑑/𝑠𝑒𝑐
We now need to solve for the filter order and 3-dB cutoff frequency
𝑁= 𝑙𝑜𝑔 𝐻 𝑎 Ω 2 −2 −1 / 𝐻 𝑎 Ω 1 −2 −1 𝑙𝑜𝑔 Ω 2 − Ω 1
Ω 𝑐 = Ω 𝐻 𝑎 Ω 2 −2 −1 1 (2𝑁)

17. For this example the order N = 3
and
Ω 𝑐 = −2 −1 1 (2∗3) =
Recall that 𝐻 𝑎 Ω 2 = =0.1
𝐻 𝑎 Ω 1 =1-

18. Knowing N and Ω 𝑐 we can write the transfer function of the analog filter below
after its poles are computed:
Ha(s) Ha(-s) = 𝑠 𝑗 Ω 𝑐 2𝑁
𝑝 𝑘 = Ω 𝑐 𝑒 𝑗 𝜋 2 + 𝜋 2𝑁 + 2𝜋𝑘 2𝑁 , k =0,1,……, (2N-1)
For this example the poles are

19. After writing the analog function Ha(s) we can then transform it to discrete-time
filter using MATLAB as before in Ex#1
𝐻 𝑑 𝑧 = Ω 𝑐 𝑧 − ∗ 1− 𝑧 −1 − 𝑝 1 1− 𝑧 −1 − 𝑝 −𝑧 −1 − 𝑝 3