<< Chapter < Page | Chapter >> Page > |
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 ${}_{k}$ .
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}$
or $${H}_{d}=Wh$$ So
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 $$h(n)=\frac{e^{in}}{M}\sum_{k=0}^{M-1} {H}_{d}({}_{k})e^{i\frac{2\pi nk}{M}}=e^{in}\mathrm{IDFT}({H}_{d}({}_{k}))$$
$h(n)$ 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.
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 $h(n)$ .
$h(n)$ symmetric, odd length, linear phase, real coefficients, and ${}_{k}$ equally spaced: $\forall k, 0\le k\le M-1\colon {}_{k}=\frac{n\pi k}{M}$
To yield real coefficients, $A()$ mus be symmetric $$(A()=A(-))\implies (A(k)=A(M-k))$$
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.
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
$()$∞
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.
Notification Switch
Would you like to follow the 'Digital filter design' conversation and receive update notifications?