1. 3.4 Frequency-domain Filters
Commonly used filters
* Butterworth filters
* Elliptic filters
* Chebyshev filters
* Bessel filters a type of linear filter with a maximally flat group delay (maximally linear phase response). Bessel filters are often used in audio crossover systems.

2. Maturity in analog lowpass filter design
(1.
(2.
(3.

3. 3.4.1 Removal of high-frequency noise: Butterworth lowpass filters
Properties of Butterworth filters:
1. Most commonly used frequency-domain filters
2. Simplicity
3. A maximally flat magnitude response in the pass-band
4. 2N  1 derivatives of the squared magnitude response at ( = 0) = 0, for Butterworth lowpass filter of order N
5. Monotonic filter response in the pass-band and the stop-band
6. The squared transfer function |Ha(j)|2 =1/[1+( j/jc)2N]
Ha: the frequency response of the analog filter
c: the cutoff frequency (in radians/s)
7. Completely specified by the cutoff frequency c and the order N.
8. N increases  more flat pass-band response; sharper pass-band to stop-band transition

4. Properties of Butterworth filters:
9. |Ha(jc)|2 = 1/2 for all N
10. The squared transfer function Ha(s) Ha(s) =1/[1+( s/jc)2N]
11. The poles of the squared transfer function
sk = c exp{j[1/2 + (2k  1) / 2N]}, k = 1, 2, 3, …, 2N
(a) For the filter coefficients to be real  complex poles must appear in conjugate pairs
(b) For a stable and causal responser  Ha(s) only has poles on the left-hand side of the s-plane
Ha(s) = G/[(s  p1) (s  p2) (s  p3)…(s  pN)]
12. By using the bilinear transform s = (2/T)[(1  z1)/(1 + z1)] to map Ha(s) to the z-domain,
H(z) = G’(1 + z 1)N/(N k = 0 ak zk), k = 0, 1, 2, …, N
a0 = 1
G’ is such that H(z) = 1 at z = 1, i.e., at DC
y(n) = N k = 0 bk x(n-k) - N k = 0 ak y(n-k)
13. H(z) is an IIR filter

7. Step 1 1.

8. Step 2: Pole locations S(1) = -0.5561 + 1.3425i

9. Step 3: Form Ha(s) Choose s(1:4) as the poles for Ha(s)
Ha(s) = 分子 /(s – s(1)) (s – s(2)) (s – s(3)) (s – s(4))
Ha(s) = DC , so the 分子 = * =
Ha(s) = /(s – s(1)) (s – s(2)) (s – s(3)) (s – s(4)) (equation 3.62)

10. Sep 4: Bilinear transform
S = (2/T)(1-z-1)/(1+z-1)
Ha(s) = /(s – s(1)) (s – s(2)) (s – s(3)) (s – s(4))
 H(z) = Equation 3.63
 b = [ , ….
a = [1, , …..
Use [h,w] = freqz(b,a)
fs = 200;
a = [1, , , , ];
b = [ , , , , ];
% a = [ ];
% b= [ ];
[h,w] = freqz(b,a);
subplot(2,1,1);
plot(w*(fs/2)/pi,abs(h));xlabel('Frequency, Hz'); ylabel('Magnitude'); grid;
subplot(2,1,2);
plot(w*(fs/2)/pi,angle(h) * 180 / pi); xlabel('Frequency, Hz'); ylabel('Phase, degree'); grid;
fvtool(b,a);

11. [b,a] = butter(4,0.4);
fvtool(b,a); % The four zeros are at the same location.

12. a = [1, , , , ];
b = [ , , , , ];
fvtool(b,a);

13. a = [ ];
b= [ ];
fvtool(b,a);

14. Step 5
y(n) = N k = 0 bk x(n-k) - N k = 0 ak y(n-k) (Equation 3.59)

15. The poles of Ha(s) Ha(-s)

16. Plot the frequency response

17. Homework due on 2009.11.16 (Monday midnight)
Design a Butterworth LPF with N = 4, fc = 50 Hz. (fs = 200 Hz)
(i) Ha(s) = ?
(ii) H(z) = ?
(iii) Plot the magnitude response and the phase response.
Put the answers in a Word file and turn in on the Black Board System before the midnight of

18. Figure 3.29

19. Figure 3.30

20. Figure 3.31

21. Figure 3.32

22. Figure 3.33

23. Figure 3.34

24. Design by Matlab
B = [ ]
A = [ ]
H(z) = ( z z z z-4)
/ ( z z z z-4)

25. With textbook coefficients

26. By matlab

27. Figure 3.35

28. Bilinear Transform

30. Bilinear Transform

31. Why s = (ln z)/T  (2/T)(1-z^-1)/(1 + z^-1)

35. Sampling theory

37. To compensate for the distortion caused by bilinear transform

39. Figure 3.36

40. Figure 3.37

41. 3.4.2 Butterworth HPF

42. Figure 3.38

43. Figure 3.39