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Frequency Response as a Ratio of DTFTs

From Eq.$ \,$(6.5), we have $ H(z)=B(z)/A(z)$, so that the frequency response is

$\displaystyle H(e^{j\omega T})=\frac{B(e^{j\omega T})}{A(e^{j\omega T})},


B(e^{j\omega T}) &=& \mbox{{\sc DTFT}}_{\omega T}({\underline{...
...^{j\omega T}) &=& \mbox{{\sc DTFT}}_{\omega T}({\underline{a}}),


{\underline{b}}&\isdef & [b_0,b_1,\ldots,b_M,0,\ldots]\\
{\underline{a}}&\isdef & [1,a_1,\ldots,a_N,0,\ldots],

and the DTFT is as defined in Eq.$ \,$(7.1).

From the above relations, we may express the frequency response of any IIR filter as a ratio of two finite DTFTs:

$\displaystyle H(e^{j\omega T}) \eqsp \frac{\mbox{{\sc DTFT}}_{\omega T}({\under...
...^M b_m e^{-j\omega mT}}{\displaystyle\sum_{n=0}^N a_n e^{-j\omega nT}} \protect$ (8.5)

This expression provides a convenient basis for the computation of an IIR frequency response in software, as we pursue further in the next section.

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

$\displaystyle 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}}$:

$\displaystyle 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

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


   DFT$\displaystyle _{\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.,

$\displaystyle h(n) \eqsp$   IDFT$\displaystyle _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)');
  plot([0:N-1]*fs/N,angle(H),'-k'); grid('on');
  xlabel(flab); ylabel('Phase');

Example LPF Frequency Response Using freqz

Figure 7.2 lists a short matlab program illustrating usage of freqz in Octave (as found in the octave-forge package). The same code should also run in Matlab, provided the Signal Processing Toolbox is available. The lines of code not pertaining to plots are the following:

  [B,A] = ellip(4,1,20,0.5); % Design lowpass filter B(z)/A(z)
  [H,w] = freqz(B,A);        % Compute frequency response H(w)
The filter example is a recursive fourth-order elliptic function lowpass filter cutting off at half the Nyquist limit (``$ 0.5\pi$'' in the fourth argument to ellip). The maximum passband ripple8.2is 1 dB (2nd argument), and the maximum stopband ripple is 20 dB (3rd arg). The sampled frequency response is returned in the H array, and the specific radian frequency samples corresponding to H are returned in the w (``omega'') array. An immediate plot can be obtained in either Matlab or Octave by simply typing
However, the example of Fig.7.2 uses more detailed ``compatibility'' functions listed in Appendix J. In particular, the freqplot utility is a simple compatibility wrapper for plot with label and title support (see §J.2 for Octave and Matlab version listings), and saveplot is a trivial compatibility wrapper for the print function, which saves the current plot to a disk file (§J.3). The saved freqplot plots are shown in Fig.7.3(a) and Fig.7.3(b).8.3

Figure: Illustration of using freqz from octave-forge (version 2006-07-09) to produce Fig.7.3 (using Octave version = 2.9.7). Also illustrated are a few plotting and plot-saving utilities from Appendix J.

[B,A] = ellip(4,1,20,0.5); % Design the lowpass filter
[H,w] = freqz(B,A);        % Compute its frequency response

% Plot the frequency response H(w):
freqplot(w,abs(H),'-k','Amplitude Response',...
         'Frequency (rad/sample)', 'Gain');

freqplot(w,angle(H),'-k','Phase Response',...
         'Frequency (rad/sample)', 'Phase (rad)');

% Plot frequency response in a "multiplot" like Matlab uses:
if exist('OCTAVE_VERSION')
  disp('Cannot save multiplots to disk in Octave')

Figure: Frequency response of an order 4 elliptic function lowpass filter computed using the matlab listing in Fig.7.2.

Amplitude Response

Phase Response

Next Section:
Phase and Group Delay
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Polar Form of the Frequency Response