# 1.2 Frequency sampling design method for fir filters

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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 ${}_{k}$ .

## Procedure

$\forall k, k$
o 1 N 1
H d k n M 1 0 h n k n
Desired Response must incluce linear phase shift (if linear phase is desired)

What is ${H}_{d}()$ for an ideal lowpass filter, cotoff at ${}_{c}$ ?

$\begin{cases}e^{-(i\frac{M-1}{2})} & \text{if -{}_{c}\le \le {}_{c}}\\ 0 & \text{if (-\pi \le < -{}_{c})\lor ({}_{c}< \le \pi )}\end{cases}$

This set of linear equations can be written in matrix form
${H}_{d}({}_{k})=\sum_{n=0}^{M-1} h(n)e^{-(i{}_{k}n)}$
$\begin{pmatrix}{H}_{d}({}_{0})\\ {H}_{d}({}_{1})\\ \\ {H}_{d}({}_{N-1})\\ \end{pmatrix}=\begin{pmatrix}e^{-(i{}_{0}\times 0)} & e^{-(i{}_{0}\times 1)} & & e^{-(i{}_{0}(M-1))}\\ e^{-(i{}_{1}\times 0)} & e^{-(i{}_{1}\times 1)} & & e^{-(i{}_{1}(M-1))}\\ & & & \\ e^{-(i{}_{M-1}\times 0)} & e^{-(i{}_{M-1}\times 1)} & & e^{-(i{}_{M-1}(M-1))}\\ \end{pmatrix}\begin{pmatrix}h(0)\\ h(1)\\ \\ h(M-1)\\ \end{pmatrix}$

or ${H}_{d}=Wh$ So

$h=W^{(-1)}{H}_{d}$
$W$ is a square matrix for $N=M$ , and invertible as long as ${}_{i}\neq {}_{j}+2\pi l$ , $i\neq j$

## Important special case

What if the frequencies are equally spaced between $0$ and $2\pi$ , i.e. ${}_{k}=\frac{2\pi k}{M}+$

Then ${H}_{d}({}_{k})=\sum_{n=0}^{M-1} h(n)e^{-(i\frac{2\pi kn}{M})}e^{-(in)}=\sum_{n=0}^{M-1} h(n)e^{-(in)}e^{-(i\frac{2\pi kn}{M})}=\text{DFT!}$ so $h(n)e^{-(in)}=\frac{1}{M}\sum_{k=0}^{M-1} {H}_{d}({}_{k})e^{i\frac{2\pi nk}{M}}$ or $h(n)=\frac{e^{in}}{M}\sum_{k=0}^{M-1} {H}_{d}({}_{k})e^{i\frac{2\pi nk}{M}}=e^{in}\mathrm{IDFT}({H}_{d}({}_{k}))$

## Important special case #2

$h(n)$ symmetric, linear phase, and has real coefficients. Since $h(n)=h(M-n)$ , there are only $\frac{M}{2}$ degrees of freedom, and only $\frac{M}{2}$ linear equations are required.

$H({}_{k})=\sum_{n=0}^{M-1} h(n)e^{-(i{}_{k}n)}=\begin{cases}\sum_{n=0}^{\frac{M}{2}-1} h(n)(e^{-(i{}_{k}n)}+e^{-(i{}_{k}(M-n-1))}) & \text{if \text{M even}}\\ \sum_{n=0}^{M-\frac{3}{2}} h(n)(e^{-(i{}_{k}n)}+e^{-(i{}_{k}(M-n-1))})h(\frac{M-1}{2})e^{-(i{}_{k}\frac{M-1}{2})} & \text{if \text{M odd}}\end{cases}=\begin{cases}e^{-(i{}_{k}\frac{M-1}{2})}\times 2\sum_{n=0}^{\frac{M}{2}-1} h(n)\cos ({}_{k}(\frac{M-1}{2}-n)) & \text{if \text{M even}}\\ e^{-(i{}_{k}\frac{M-1}{2})}\times 2\sum_{n=0}^{M-\frac{3}{2}} h(n)\cos ({}_{k}(\frac{M-1}{2}-n))+h(\frac{M-1}{2}) & \text{if \text{M odd}}\end{cases}$

Removing linear phase from both sides yields $A({}_{k})=\begin{cases}2\sum_{n=0}^{\frac{M}{2}-1} h(n)\cos ({}_{k}(\frac{M-1}{2}-n)) & \text{if \text{M even}}\\ 2\sum_{n=0}^{M-\frac{3}{2}} h(n)\cos ({}_{k}(\frac{M-1}{2}-n))+h(\frac{M-1}{2}) & \text{if \text{M odd}}\end{cases}$ Due to symmetry of response for real coefficients, only $\frac{M}{2}$ ${}_{k}$ on $\in \left[0 , \pi \right)$ need be specified, with the frequencies $-{}_{k}$ thereby being implicitly defined also. Thus we have $\frac{M}{2}$ real-valued simultaneous linear equations to solve for $h(n)$ .

## Special case 2a

$h(n)$ symmetric, odd length, linear phase, real coefficients, and ${}_{k}$ equally spaced: $\forall k, 0\le k\le M-1\colon {}_{k}=\frac{n\pi k}{M}$

$h(n)=\mathrm{IDFT}({H}_{d}({}_{k}))=\frac{1}{M}\sum_{k=0}^{M-1} A({}_{k})e^{-(i\frac{2\pi k}{M})}\frac{M-1}{2}e^{i\frac{2\pi nk}{M}}=\frac{1}{M}\sum_{k=0}^{M-1} A(k)e^{i\frac{2\pi k}{M}(n-\frac{M-1}{2})}$

To yield real coefficients, $A()$ mus be symmetric $(A()=A(-))\implies (A(k)=A(M-k))$

$h(n)=\frac{1}{M}(A(0)+\sum_{k=1}^{\frac{M-1}{2}} A(k)(e^{i\frac{2\pi k}{M}(n-\frac{M-1}{2})}+e^{-(i\times 2\pi k(n-\frac{M-1}{2}))}))=\frac{1}{M}(A(0)+2\sum_{k=1}^{\frac{M-1}{2}} A(k)\cos (\frac{2\pi k}{M}(n-\frac{M-1}{2})))=\frac{1}{M}(A(0)+2\sum_{k=1}^{\frac{M-1}{2}} A(k)-1^{k}\cos (\frac{2\pi k}{M}(n+\frac{1}{2})))$

Simlar equations exist for even lengths, anti-symmetric, and $=\frac{1}{2}$ filter forms.

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

$H()$ for a frequency sampled design goes exactly through the sample points, but it may be very far off from the desired response for $\neq {}_{k}$ . 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 errorin the frequency response at all of these samples.

## Extended frequency sample design

For the samples $H({}_{k})$ where $0\le k\le M-1$ and $N> M$ , find $h(n)$ , where $0\le n\le M-1$ minimizing $({H}_{d}({}_{k})-H({}_{k}))$

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

Here we will consider the $(, l)$ norm.

To minimize the $(, l)$ norm; that is, $\sum_{n=0}^{N-1} \left|{H}_{d}({}_{k})-H({}_{k})\right|$ , we have an overdetermined set of linear equations: $\begin{pmatrix}e^{-(i{}_{0}\times 0)} & & e^{-(i{}_{0}(M-1))}\\ & & \\ e^{-(i{}_{N-1}\times 0)} & & e^{-(i{}_{N-1}(M-1))}\\ \end{pmatrix}h=\begin{pmatrix}{H}_{d}({}_{0})\\ {H}_{d}({}_{1})\\ \\ {H}_{d}({}_{N-1})\\ \end{pmatrix}$ or $Wh={H}_{d}$

The minimum error norm solution is well known to be $h=\overline{W}W^{(-1)}\overline{W}{H}_{d}$ ; $\overline{W}W^{(-1)}\overline{W}$ is well known as the pseudo-inverse matrix.

Extended frequency sampled design discourages radical behavior of the frequency response between samples forsufficiently closely spaced samples. However, the actual frequency response may no longer pass exactly through any of the ${H}_{d}({}_{k})$ .

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