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Frequency Sampling Design Method for FIR filters

Module by: Douglas L. Jones. E-mail the author

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Given a desired frequency response, the frequency sampling design method designs a filter with a frequency response exactly equal to the desired response at a particular set of frequencies ω k ω k .

Procedure

∀k,k=o1…N−1: H d ω k =∑n=0M−1hnⅇ-ⅈ ω k n k k o 1 … N 1 H d ω k n M 1 0 h n ω k n (1)

Note:

Desired Response must incluce linear phase shift (if linear phase is desired)

Exercise 1

What is H d ω H d ω for an ideal lowpass filter, cotoff at ω c ω c ?

Solution

ⅇ-ⅈωM−12if- ω c ≤ω≤ ω c 0if-π≤ω<- ω c ∨ ω c <ω≤π ω M 1 2 ω c ω ω c 0 ω ω c ω c ω

Note:

This set of linear equations can be written in matrix form
H d ω k =∑n=0M−1hnⅇ-ⅈ ω k n H d ω k n M 1 0 h n ω k n (2)
H d ω 0 H d ω 1 ⋮ H d ω N - 1 =ⅇ-ⅈ ω 0 0ⅇ-ⅈ ω 0 1…ⅇ-ⅈ ω 0 M−1ⅇ-ⅈ ω 1 0ⅇ-ⅈ ω 1 1…ⅇ-ⅈ ω 1 M−1⋮⋮⋮⋮ⅇ-ⅈ ω M - 1 0ⅇ-ⅈ ω M - 1 1…ⅇ-ⅈ ω M - 1 M−1h0h1⋮hM−1 H d ω 0 H d ω 1 ⋮ H d ω N - 1 ω 0 0 ω 0 1 … ω 0 M 1 ω 1 0 ω 1 1 … ω 1 M 1 ⋮ ⋮ ⋮ ⋮ ω M - 1 0 ω M - 1 1 … ω M - 1 M 1 h 0 h 1 ⋮ h M 1 (3)

or Hd=Wh H d W h So

h=W-1Hd h W H d (4)

Note:

W W is a square matrix for N=M N M , and invertible as long as ω i ≠ ω j +2πl ω i ω j 2 l , i≠j i j

Important Special Case

What if the frequencies are equally spaced between 00 and 2π 2 , i.e. ω k =2πkM+α ω k 2 k M α

Then H d ω k =∑n=0M−1hnⅇ-ⅈ2πknMⅇ-ⅈαn=∑n=0M−1hnⅇ-ⅈαnⅇ-ⅈ2πknM= DFT! H d ω k n M 1 0 h n 2 k n M α n n M 1 0 h n α n 2 k n M DFT! so hnⅇ-ⅈαn=1M∑k=0M−1 H d ω k ⅇ+ⅈ2πnkM h n α n 1 M k M 1 0 H d ω k 2 n k M or hn=ⅇⅈαnM∑k=0M−1 H d ω k ⅇⅈ2πnkM=ⅇⅈαnIDFT H d ω k h n α n M k M 1 0 H d ω k 2 n k M α n IDFT H d ω k

Important Special Case #2

hn h n symmetric, linear phase, and has real coefficients. Since hn=hM−n h n h M n 1 , there are only M2 M 2 degrees of freedom, and only M2 M 2 linear equations are required.

H ω k =∑n=0M−1hnⅇ-ⅈ ω k n=∑n=0M2−1hnⅇ-ⅈ ω k n+ⅇ-ⅈ ω k M−n−1if M even ∑n=0M−32+hnⅇ-ⅈ ω k n+ⅇ-ⅈ ω k M−n−1hM−12ⅇ-ⅈ ω k M−12if M odd =ⅇ-ⅈ ω k M−122∑n=0M2−1hncos ω k M−12−nif M even ⅇ-ⅈ ω k M−122∑n=0M−32hncos ω k M−12−n+hM−12if M odd H ω k n M 1 0 h n ω k n n M 2 1 0 h n ω k n ω k M n 1 M even n M 3 2 0 h n ω k n ω k M n 1 h M 1 2 ω k M 1 2 M odd ω k M 1 2 2 n M 2 1 0 h n ω k M 1 2 n M even ω k M 1 2 2 n M 3 2 0 h n ω k M 1 2 n h M 1 2 M odd (5)

Removing linear phase from both sides yields A ω k =2∑n=0M2−1hncos ω k M−12−nif M even 2∑n=0M−32hncos ω k M−12−n+hM−12if M odd A ω k 2 n M 2 1 0 h n ω k M 1 2 n M even 2 n M 3 2 0 h n ω k M 1 2 n h M 1 2 M odd Due to symmetry of response for real coefficients, only M2 M 2 ω k ω k on ω∈0π ω 0 need be specified, with the frequencies - ω k ω k thereby being implicitly defined also. Thus we have M2 M 2 real-valued simultaneous linear equations to solve for hn h n .

Special Case 2a

hn h n symmetric, odd length, linear phase, real coefficients, and ω k ω k equally spaced: ∀k,0≤k≤M−1: ω k =nπkM k 0 k M 1 ω k n k M

hn=IDFT H d ω k =1M∑k=0M−1A ω k ⅇ-ⅈ2πkMM−12ⅇⅈ2πnkM=1M∑k=0M−1Akⅇⅈ2πkMn−M−12 h n IDFT H d ω k 1 M k M 1 0 A ω k 2 k M M 1 2 2 n k M 1 M k M 1 0 A k 2 k M n M 1 2 (6)

To yield real coefficients, Aω A ω mus be symmetric Aω=A-ω⇒Ak=AM−k A ω A ω A k A M k

hn=1MA0+∑k=1M−12Akⅇⅈ2πkMn−M−12+ⅇ-ⅈ2πkn−M−12=1MA0+2∑k=1M−12Akcos2πkMn−M−12=1MA0+2∑k=1M−12Ak-1kcos2πkMn+12 h n 1 M A 0 k M 1 2 1 A k 2 k M n M 1 2 2 k n M 1 2 1 M A 0 2 k M 1 2 1 A k 2 k M n M 1 2 1 M A 0 2 k M 1 2 1 A k 1 k 2 k M n 1 2 (7)

Simlar equations exist for even lengths, anti-symmetric, and α=12 α 1 2 filter forms.

Comments on frequency-sampled design

This method is simple conceptually and very efficient for equally spaced samples, since hn h n can be computed using the IDFT.

Hω H ω for a frequency sampled design goes exactly through the sample points, but it may be very far off from the desired response for ω≠ ω k ω ω k . This is the main problem with frequency sampled design.

Possible solution to this problem: specify more frequency samples than degrees of freedom, and minimize the total error in the frequency response at all of these samples.

Extended frequency sample design

For the samples H ω k H ω k where 0≤k≤M−1 0 k M 1 and N>M N M , find hn h n , where 0≤n≤M−1 0 n M 1 minimizing ∥ H d ω k −H ω k ∥ H d ω k H ω k

For ∥l∥∞ l norm, this becomes a linear programming problem (standard packages availble!)

Here we will consider the ∥l∥2 2 l norm.

To minimize the ∥l∥2 2 l norm; that is, ∑n=0N−1| H d ω k −H ω k | n N 1 0 H d ω k H ω k , we have an overdetermined set of linear equations: ⅇ-ⅈ ω 0 0…ⅇ-ⅈ ω 0 M−1⋮⋮⋮ⅇ-ⅈ ω N - 1 0…ⅇ-ⅈ ω N - 1 M−1h= H d ω 0 H d ω 1 ⋮ H d ω N - 1 ω 0 0 … ω 0 M 1 ⋮ ⋮ ⋮ ω N - 1 0 … ω N - 1 M 1 h H d ω 0 H d ω 1 ⋮ H d ω N - 1 or Wh=Hd W h H d

The minimum error norm solution is well known to be h=W¯W-1W¯Hd h W W W H d ; W¯W-1W¯ W W W is well known as the pseudo-inverse matrix.

Note:

Extended frequency sampled design discourages radical behavior of the frequency response between samples for sufficiently closely spaced samples. However, the actual frequency response may no longer pass exactly through any of the H d ω k H d ω k .

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