Simulated Sine-Wave Analysis in Matlab

In this section, we will find the frequency response of the simplest lowpass filter

$$y(n) = x(n) + x(n-1),\quad n=1,2,\ldots,N.$$

using simulated sine-wave analysis carried out by a matlab program. This numerical approach complements the analytical approach followed in §1.3. Figure 2.3 gives a listing of the main script which invokes the sine-wave analysis function swanal listed in Fig.2.4. The plotting/printing utilities swanalmainplot and swanalplot are listed in Appendix H starting at §H.13.

Figure 2.3: Main matlab program for computing the frequency response of the simplest low-pass filter $y(n) = x(n) + x(n - 1)$ by means of simulated sine-wave analysis.

 

% swanalmain.m - matlab program for simulated sine-wave
%             analysis on the simplest lowpass filter:
%
%                 y(n) = x(n)+x(n-1)}

B = [1,1]; % filter feedforward coefficients 
A = 1;     % filter feedback coefficients (none)

N=10;           % number of sinusoidal test frequencies
fs = 1;         % sampling rate in Hz (arbitrary)

fmax = fs/2;    % highest frequency to look at
df = fmax/(N-1);% spacing between frequencies
f = 0:df:fmax;  % sampled frequency axis
dt = 1/fs;      % sampling interval in seconds
tmax = 10;      % number of seconds to run each sine test
t = 0:dt:tmax;  % sampled time axis
ampin = 1;      % test input signal amplitude
phasein = 0;    % test input signal phase

[gains,phases] = swanal(t,f/fs,B,A); % sine-wave analysis

swanalmainplot;    % final plots and comparison to theory

Figure 2.4: Matlab function for performing simulated sine-wave analysis.

 

function [gains,phases] = swanal(t,f,B,A)
% SWANAL - Perform sine-wave analysis on filter B(z)/A(z)
  
ampin = 1;      % input signal amplitude
phasein = 0;    % input signal phase

N = length(f);       % number of test frequencies
gains = zeros(1,N);  % pre-allocate amp-response array
phases = zeros(1,N); % pre-allocate phase-response array
if length(A)==1
  ntransient=length(B)-1; % no. samples to steady state
else
  error('Need to set transient response duration here');
end

for k=1:length(f)    % loop over analysis frequencies
  s = ampin*cos(2*pi*f(k)*t+phasein); % test sinusoid
  y = filter(B,A,s); % run it through the filter
  yss = y(ntransient+1:length(y)); % chop off transient
  % measure output amplitude as max (SHOULD INTERPOLATE):
  [ampout,peakloc] = max(abs(yss)); % ampl. peak & index
  gains(k) = ampout/ampin;  % amplitude response
  if ampout < eps  % eps returns "machine epsilon"
    phaseout=0;    % phase is arbitrary at zero gain
  else
    sphase = 2*pi*f(k)*(peakloc+ntransient-1);
    % compute phase by inverting sinusoid (BAD METHOD):
    phaseout = acos(yss(peakloc)/ampout) - sphase;
    phaseout = mod2pi(phaseout); % reduce to [-pi,pi)
  end
  phases(k) = phaseout-phasein;
  swanalplot; % signal plotting script
end

In the swanal function (Fig.2.4), test sinusoids are generated by the line

        s = ampin * cos(2*pi*f(k)*t + phasein);

where amplitude, frequency (Hz), and phase (radians) of the sinusoid are given be ampin, f(k), and phasein, respectively. As discussed in §1.3, assuming linearity and time-invariance (LTI) allows us to set

        ampin = 1;      % input signal amplitude
        phasein = 0;    % input signal phase

without loss of generality. (Note that we must also assume the filter is LTI for sine-wave analysis to be a general method for characterizing the filter's response.) The test sinusoid is passed through the digital filter by the line

        y = filter(B,A,s); % run s through the filter

producing the output signal in vector y. For this example (the simplest lowpass filter), the filter coefficients are simply

        B = [1,1]; % filter feedforward coefficients 
        A = 1;     % filter feedback coefficients (none)

The coefficient A(1) is technically a coefficient on the output signal itself, and it should always be normalized to 1. (B and A can be multiplied by the same nonzero constant to carry out this normalization when necessary.)

Figure 2.5 shows one of the intermediate plots produced by the sine-wave analysis routine in Fig.2.4. This figure corresponds to Fig.1.6 in §1.3 (see page ). In Fig.2.5a, we see samples of the input test sinusoid overlaid with the continuous sinusoid represented by those samples. Figure 2.5b shows only the samples of the filter output signal: While we know the output signal becomes a sampled sinusoid after the one-sample transient response, we do not know its amplitude or phase until we measure it; the underlying continuous signal represented by the samples is therefore not plotted. (If we really wanted to see this, we could use software for bandlimited interpolation [91], such as Matlab's interp function.) A plot such as Fig.2.5 is produced for each test frequency, and the relative amplitude and phase are measured between the input and output to form one sample of the measured frequency response, as discussed in §1.3.

Figure 2.5: Input and output signals for the sixth test frequency $f(k)=f_s/4$ in the simulated sine-wave analysis of the simplest lowpass filter.

Next, the one-sample start-up transient is removed from the filter output signal y to form the ``cropped'' signal yss (``$y$ steady state''). The final task is to measure the amplitude and phase of the yss. Output amplitude estimation is done in swanal by the line

        [ampout,peakloc] = max(abs(yss));

Note that the peak amplitude found in this way is approximate, since the true peak of the output sinusoid generally occurs between samples. We will find the output amplitude much more accurately in the next two sections. We store the index of the amplitude peak in peakloc so it can be used to estimate phase in the next step. Given the output amplitude ampout, the amplitude response of the filter at frequency f(k) is given by

        gains(k) = ampout/ampin;

The last step of swanal in Fig.2.4 is to estimate the phase of the cropped filter output signal yss. Since we will have better ways to accomplish this later, we use a simplistic method here based on inverting the sinusoid analytically:

        phaseout = acos(yss(peakloc)/ampout) ...
                     - 2*pi*f(k)*(peakloc+ntransient-1);
        phaseout = mod2pi(phaseout); % reduce to [-pi,pi)

Again, this is only an approximation since peakloc is only accurate to the nearest sample. The mod2pi utility reduces its scalar argument to the range $[-\pi ,\pi )$, and is listed in Fig.2.6.

Figure 2.6: Utility (matlab) for reducing a radian phase angle to $[-\pi ,\pi )$.

 

function [y] = mod2pi(x)
% MOD2PI - Reduce x to the range [-pi,pi)
  y=x;
  twopi = 2*pi;
  while y >= pi, y = y - twopi; end
  while y < -pi, y = y + twopi; end

In summary, the sine-wave analysis measures experimentally the gain and phase-shift of the digital filter at selected frequencies, thus measuring the frequency response of the filter at those frequencies. It is interesting to compare these experimental results with the closed-form expressions for the frequency response derived in §1.3.2. From Equations (1.6-1.7) we have

$$\begin{eqnarray*} G(f) &=& 2\cos(\pi f /f_s) \\ % \qquad \hbox{(Amplitude Response)}\\ \Theta(f) &=& - \pi f /f_s% \qquad \hbox{(Phase Response)} \end{eqnarray*}$$

where $G(f)$ denotes the amplitude response (filter gain versus frequency), $\Theta(f)$ denotes the phase response (filter phase-shift versus frequency), $f$ is frequency in Hz, and $f_s=1/T$ denotes the sampling rate. Both the amplitude response and phase response are real-valued functions of (real) frequency, while the frequency response is the complex function of frequency given by $H(f) = G(f)e^{j\Theta(f)}$.

Figure 2.7 shows overlays of the measured and theoretical results. While there is good general agreement, there are noticeable errors in the measured amplitude and phase-response curves. Also, the phase-response error tends to be large when there is an amplitude response error, since the phase calculation used depends on knowledge of the amplitude response.

Figure 2.7: Overlay of measured and theoretical frequency response when $f_{\hbox {max}}=f_s/2$.

It is important to understand the source(s) of deviation between the measured and theoretical values. Our simulated sine-wave analysis deviates from an ideal sine-wave analysis in the following ways (listed in ascending order of importance):

1. The test sinusoids are sampled instead of being continuous in time: It turns out this is a problem only for frequencies at half the sampling rate and above. Below half the sampling rate, sampled sinusoids contain exactly the same information as continuous sinusoids, and there is no penalty whatsoever associated with discrete-time sampling itself.

2. The test sinusoid samples are rounded to a finite precision: Digitally sampled sinusoids do suffer from a certain amount of round-off error, but Matlab and Octave use double-precision floating-point numbers by default (64 bits). As a result, our samples are far more precise than could be measured acoustically in the physical world. This is not a visible source of error in Fig.2.7.

3. Our test sinusoids are finite duration, while the ideal sinusoid is infinitely long: This can be a real practical limitation. However, we worked around it completely by removing the start-up transient. For the simplest lowpass filter, the start-up transient is only one sample long. More generally, for digital filters expressible as a weighted sum of $M$ successive samples (any nonrecursive LTI digital filter), the start-up transient is $M-1$ samples long. When we consider recursive digital filters, which employ output feedback, we will no longer be able to remove the start-up transient exactly, because it generally decays exponentially instead of being finite in length. However, even in that case, as long as the recursive filter is stable, we can define the start-up transient as some number of time-constants of exponential decay, thereby making the approximation error as small as we wish, such as less than the round-off error.

4. We measured the output amplitude and phase at a signal peak measured only to the nearest sample: This is the major source of error in our simulation. The program of Fig.2.3 measures the filter output amplitude very crudely as the maximum magnitude. In general, even for this simple filter, the maximum output signal amplitude occurs between samples. To measure this, we would need to use what is called bandlimited interpolation [91]. It is possible and practical to make the error arbitrarily small by increasing the sampling rate by some factor and finishing with quadratic interpolation of the three samples about the peak magnitude. Similar remarks apply to the measured output phase.

   The need for interpolation is lessened greatly if the sampling rate is chosen to be unrelated to the test frequencies (ideally so that the number of samples in each sinusoidal period is an irrational number). Figure 2.8 shows the measured and theoretical results obtained by changing the highest test frequency fmax from $f_s/2$ to $f_s/2.34567$, and the number of samples in each test sinusoid tmax from $10$ to $100$. For these parameters, at least one sample falls very close to a true peak of the output sinusoid at each test frequency.

   It should also be pointed out that one never estimates signal phase in practice by inverting the closed-form functional form assumed for the signal. Instead, we should estimate the delay of each output sinusoid relative to the corresponding input sinusoid. This leads to the general topic of time delay estimation [12]. Under the assumption that the round-off error can be modeled as ``white noise'' (typically this is an accurate assumption), the optimal time-delay estimator is obtained by finding the (interpolated) peak of the cross-correlation between the input and output signals. For further details on cross-correlation, a topic in statistical signal processing, see, e.g., [74].

   Using the theory presented in later chapters, we will be able to compute very precisely the frequency response of any LTI digital filter without having to resort to bandlimited interpolation (for measuring amplitude response) or time-delay estimation (for measuring phase).

Figure 2.8: Overlay of measured and theoretical frequency response when $f_{\hbox {max}}=f_s/2.34567$.