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Lecture 4: March 13, 2007

Lecture 4: March 13, 2007. Topic:. 1. Uniform Frequency-Sampling Methods (cont.). Lecture 4: March 13, 2007. Topics:. 1. Nonrecursive FIR Filter Design by DFT Applications (cont.). 2. Nonrecursive FIR Filter Design by Direct Computation of Unit Sample Response.

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Lecture 4: March 13, 2007

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  1. Lecture 4: March 13, 2007 Topic: 1. Uniform Frequency-Sampling Methods (cont.)

  2. Lecture 4: March 13, 2007 Topics: 1. Nonrecursive FIR Filter Design by DFT Applications (cont.) 2. Nonrecursive FIR Filter Design by Direct Computation of Unit Sample Response 3. Recursive FIR Filter Design

  3. 4.2.3. Uniform Frequency-Sampling Method B. Nonrecursive FIR Filter Design by DFT Applications

  4. 4.2.3. Uniform Frequency-Sampling Method B. Nonrecursive FIR Filter Design by DFT Applications • specification of the desired frequency response at a set of equally spaced frequencies, • solving for the unit sample response h(n) of the linear phase FIR filter from these equally spaced specification based on DFT application.

  5. Suppose that we specify the frequency response of the filter at the frequencies Then, we obtain:

  6. This expression allows to compute h(k) from H(n), where frequency response is specified at the set of equally spaced frequencies. Notice:H(n)has to be specified correctly! It is a simple matter to identify the previous expression as DFT of h(k): Thus, we obtain

  7. Important property of H(n) : for n= 1, 2, …, M-1 Proof:

  8. Frequency response of the linear phase FIR digital filters: Proof:

  9. for n= 1, 2, …, M-1 for k= 0, 1, 2, …, M-1 Uniform Frequency-Sampling Method B. : A Review 1. Specification of H(n) by: 2. Computation of h(k) by:

  10. for n= 1, 2, …, M-1 Example: Determine the coefficients h(k) of a linear phase FIR filter of length M=15, frequency response of which satisfies the condition: Solution:

  11. for k= 0, 1, 2, …, M-1

  12. By using MATLAB (function ifft.m) the following filter coefficients can be obtained: h(0)=h(14)= -0.0498 h(1)=h(13)= 0.0412 h(2)=h(12)= 0.0667 h(3)=h(11)= -0.0365 h(4)=h(10)= -0.1079 h(5)=h(9)= 0.0341 h(6)=h(8)= 0.3189 h(7)= 0.4667

  13. Example Solution: Magnitude Response

  14. 4.2.4. Uniform Frequency-Sampling Method C. Nonrecursive FIR Filter Design by Direct Computation of Unit Sample Response

  15. 4.2.4. Uniform Frequency-Sampling Method C. Nonrecursive FIR Filter Design by Direct Computation of Unit Sample Response

  16. 4.2.4. Uniform Frequency-Sampling Method C. Nonrecursive FIR Filter Design by Direct Computation of Unit Sample Response We have shown that It is more convenient, however, to define the samples H(n) of the frequency response for the FIR filter as where

  17. for n= 1, 2, …, M-1 for n= 1, 2, …, M-1 M : even

  18. for n= 1, 2, …, M-1 Proof:

  19. In the view of the symmetry for the real-valued frequency samples G(n), it is a simple matter to determine an expression for h(k) in terms of G(n). If M is even, we have:

  20. In the view of the symmetry for the real-valued frequency samples G(n), it is a simple matter to determine an expression for h(k) in terms of G(n). If M is even, we have:

  21. Because:

  22. for k= 0, 1, 2, …, M-1 We can get:

  23. Thus we can compute h(k) directly from the specified frequency samples G(n) or equivalently, H(n) (method 3). This approach eliminates the need for a matrix inversion (method 1) or for direct DFT computation (method 2) for determining the unit sample response of the linear phase FIR filter from its frequency-domain specification.

  24. Thus we can compute h(k) directly from the specified frequency samples G(n) or equivalently, H(n) (method 3). This approach eliminates the need for a matrix inversion (method 1) or for direct DFT computation (method 2) for determining the unit sample response of the linear phase FIR filter from its frequency-domain specification.

  25. Thus we can compute h(k) directly from the specified frequency samples G(n) or equivalently, H(n) (method 3). This approach eliminates the need for a matrix inversion (method 1) or for direct DFT computation (method 2)for determining the unit sample response of the linear phase FIR filter from its frequency-domain specification.

  26. Summary on the Uniform Frequency-Sampling Method 3.

  27. 1. Unit Sample Response:Symmetrical,

  28. 2. Unit Sample Response: Symmetrical,

  29. 3. Unit Sample Response: Antisymmetrical,

  30. 4. Unit Sample Response: Antisymmetrical,

  31. Example: Determine the coefficients h(n) of a linear phase FIR filter of length M=15 with symmetrical impulse response, frequency response of which satisfies the condition:

  32. Solution: Unit Sample Response:Symmetrical,

  33. h(4)=h(10)= -0.1079 h(5)=h(9)= 0.0341 h(6)=h(8)= 0.3189 h(2)=h(12)= 0.0412 h(7)= 0.4667 h(3)=h(11)= -0.0365 End. Etc. h(0)=h(14)= -0.0498 h(1)=h(13)= 0.0412

  34. Example Solution: Magnitude Response

  35. 4.2.5. Uniform Frequency-Sampling Method D. Recursive FIR Filter Design

  36. 4.2.5. Uniform Frequency-Sampling Method D. Recursive FIR Filter Design

  37. n=0, 1, …, M-1 k=0, 1, …, M-1 4.2.5. Uniform Frequency-Sampling Method D. Recursive FIR Filter Design An FIR filter can be uniquely specified by giving either its impulse response (coefficients h(k))or, equivalently, by the DFT coefficients H(n). These sequences are related by the following expressions:

  38. It has also been shown that H(n) can be regarded as Z-transform of h(k), evaluated at N points equally spaced around the unit circle:

  39. Thus, the transfer function of an FIR filter can easily be expressed in terms of DFT coefficients of h(k):

  40. H1(z) H2(z)

  41. zeros poles ! The transfer function H(z) can be expressed in the form: where

  42. H(z) can be interpreted as a cascade of a comb filteranda parallel bank of single-pole filters The zeros of the comb filtercancel the poles of theparallel bank of filters, so that in effect the final transfer function contains no poles, but it is in fact FIR filter.

  43. Alternative form to can be obtained by combining a pair of complex-conjugate poles of H(z) or two real-valued poles to form two-pole filters with real valued coefficients and taking into account . Then, we can obtain in the forms:

  44. A(k) and B(k) are the real-valued coefficients given by

  45. Proof:

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