### 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

*sampled DTFT*of divided by the

*sampled DTFT*of :

*discrete Fourier transform (DFT)*[84]. Thus, we can write

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 |

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Example LPF Frequency Response Using freqz

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Software for Partial Fraction Expansion