Frequency Response as a Ratio of DTFTs

From Eq.(6.5), we have , so that the frequency response is

and

where

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:

 (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 at to obtain the frequency response , where is continuous radian frequency, we compute instead

where is the desired number of spectral samples around the unit circle in the plane. From Eq.(7.5) we have that this is the same thing as the sampled DTFT of divided by the sampled DTFT of :

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

where

DFT

and , where denotes the sampling rate in Hz.

To avoid undersampling , we must have , and to avoid undersampling , we must have . In general, will be undersampled (when ), because it is the quotient of over . This means, for example, that computing the impulse response from the sampled frequency response will be time aliased in general. I.e.,

IDFT

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 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 , where denotes the maximum pole magnitude. This choice provides more than 60 dB of decay in the impulse response over a duration of samples, which is the time-aliasing block size. (The time to decay 60 dB, or '', is a little less than time constants [84], and the time-constant of decay for a single pole at radius can be approximated by samples, when is close to 1, as derived in §8.6.)

As is well known, when the DFT length is a power of 2, e.g., , 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 is a power of 2).

 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 

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 ('' 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
  plot(w,abs(H));
plot(w,angle(H));

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

 [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): % figure(1); freqplot(w,abs(H),'-k','Amplitude Response',... 'Frequency (rad/sample)', 'Gain'); saveplot('../eps/freqzdemo1.eps'); figure(2); freqplot(w,angle(H),'-k','Phase Response',... 'Frequency (rad/sample)', 'Phase (rad)'); saveplot('../eps/freqzdemo2.eps'); % Plot frequency response in a "multiplot" like Matlab uses: % figure(3); plotfr(H,w/(2*pi)); if exist('OCTAVE_VERSION') disp('Cannot save multiplots to disk in Octave') else saveplot('../eps/freqzdemo3.eps'); end 

 Amplitude Response Phase Response

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Phase and Group Delay
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Polar Form of the Frequency Response