Frequency Response in Matlab

In practice, we usually work with a sampled frequency axis. That is, instead of evaluating the transfer function $H(z)=B(z)/A(z)$ at $z=e^{j\omega T}$ to obtain the frequency response $H(e^{j\omega T})$ , where $\omega$ is continuous radian frequency, we compute instead

$$H(e^{j\omega_k T}) \eqsp \frac{B(e^{j\omega_k T})}{A(e^{j\omega_k T})},\quad e^{j\omega_k T}\isdefs e^{j2\pi k / N_s},\quad k=0,1,2,\ldots,N_s-1,$$

where $N_s$ is the desired number of spectral samples around the unit circle in the $z$ plane. From Eq.(7.5) we have that this is the same thing as the sampled DTFT of ${\underline{b}}$ divided by the sampled DTFT of ${\underline{a}}$ :

$$H(e^{j\omega_k T}) \eqsp \frac{\mbox{{\sc DTFT}}_{\omega_k T}({\underline{b}})}{\mbox{{\sc DTFT}}_{\omega_k T}({\underline{a}})}$$

The uniformly sampled DTFT has its own name: the discrete Fourier transform (DFT) [84]. Thus, we can write

$$H(e^{j\omega_k T}) \eqsp \frac{\mbox{{\sc DFT}}_{\omega_k T}({\underline{b}})}{\mbox{{\sc DFT}}_{\omega_k T}({\underline{a}})}, \quad k=0,1,2,\ldots,N_s-1$$

where

   DFT$$_{\omega_k T}(x) \isdefs \sum_{n=0}^{N_s-1} x(n) e^{-j\omega_k nT}$$

and $\omega_k \isdef 2\pi f_sk/N_s$ , where $f_s=1/T$ denotes the sampling rate in Hz.

To avoid undersampling $B(e^{j\omega T})$ , we must have $N_s\ge M$ , and to avoid undersampling $A(e^{j\omega T})$ , we must have $N_s\ge N$ . In general, $H(e^{j\omega T})$ will be undersampled (when $N>0$ ), because it is the quotient of $B(e^{j\omega T})$ over $A(e^{j\omega T})$ . This means, for example, that computing the impulse response $h(n)$ from the sampled frequency response $H(e^{j\omega_k T})$ will be time aliased in general. I.e.,

$$h(n) \eqsp$$   IDFT$$_n(H) \isdefs \frac{1}{N_s}\sum_{k=0}^{N_s-1} H(e^{j\omega_k T})e^{j\omega_k nT}$$

will be time-aliased in the IIR case. In other words, an infinitely long impulse response cannot be Fourier transformed using a finite-length DFT, and this corresponds to not being able to sample the frequency response of an IIR filter without some loss of information. In practice, we simply choose $N_s$ sufficiently large so that the sampled frequency response is accurate enough for our needs. A conservative practical rule of thumb when analyzing stable digital filters is to choose $N_s>7/(1-R_{\mbox{max}})$ , where $R_{\mbox{max}}$ denotes the maximum pole magnitude. This choice provides more than 60 dB of decay in the impulse response over a duration of $N_s$ samples, which is the time-aliasing block size. (The time to decay 60 dB, or ``$t_{60}$ '', is a little less than $7$ time constants [84], and the time-constant of decay for a single pole at radius $R$ can be approximated by $1/(1-R)$ samples, when $R$ is close to 1, as derived in §8.6.)

As is well known, when the DFT length $N_s$ is a power of 2, e.g., $N_s=2^{10}=1024$ , the DFT can be computed extremely efficiently using the Fast Fourier Transform (FFT). Figure 7.1 gives an example matlab script for computing the frequency response of an IIR digital filter using two FFTs. The Matlab function freqz also uses this method when possible (e.g., when $N_s$ is a power of 2).

Figure 7.1: Matlab function for computing and optionally plotting the frequency response of an IIR digital filter.

 

function [H,w] = myfreqz(B,A,N,whole,fs)
%MYFREQZ Frequency response of IIR filter B(z)/A(z).
%    N = number of uniform frequency-samples desired
%    H = returned frequency-response samples (length N)
%    w = frequency axis for H (length N) in radians/sec
%    Compatible with simple usages of FREQZ in Matlab.
%    FREQZ(B,A,N,whole) uses N points around the whole 
%    unit circle, where 'whole' is any nonzero value.
%    If whole=0, points go from theta=0 to pi*(N-1)/N.
%    FREQZ(B,A,N,whole,fs) sets the assumed sampling 
%    rate to fs Hz instead of the default value of 1.
%    If there are no output arguments, the amplitude and 
%    phase responses are displayed.  Poles cannot be
%    on the unit circle.

A = A(:).'; na = length(A); % normalize to row vectors
B = B(:).'; nb = length(B);
if nargin < 3, N = 1024; end
if nargin < 4, if isreal(b) & isreal(a), whole=0; 
               else whole=1; end; end
if nargin < 5, fs = 1; end
Nf = 2*N; if whole, Nf = N; end
w = (2*pi*fs*(0:Nf-1)/Nf)';

H = fft([B zeros(1,Nf-nb)]) ./ fft([A zeros(1,Nf-na)]);

if whole==0, w = w(1:N); H = H(1:N); end

if nargout==0 % Display frequency response
  if fs==1, flab = 'Frequency (cyles/sample)';
  else, flab = 'Frequency (Hz)'; end
  subplot(2,1,1); % In octave, labels go before plot:
  plot([0:N-1]*fs/N,20*log10(abs(H)),'-k'); grid('on');
  xlabel(flab'); ylabel('Magnitude (dB)');
  subplot(2,1,2);
  plot([0:N-1]*fs/N,angle(H),'-k'); grid('on');
  xlabel(flab); ylabel('Phase');
end