Matlab Analysis of the Simplest Lowpass Filter

Simulated Sine-Wave Analysis in Matlab

**Search Introduction to Digital Filters**

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

% 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 |

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 = 1; % input signal amplitude phasein = 0; % input signal phasewithout 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 filterproducing the output signal in vector

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

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.

Next, the one-sample start-up transient is removed from the filter
output signal `y` to form the ``cropped'' signal `yss`
(`` 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

gains(k) = ampout/ampin;The last step of

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

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

where denotes the *amplitude response* (filter gain versus
frequency), denotes the *phase response* (filter
phase-shift versus frequency), is frequency in Hz, and
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
.

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.

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):

**The test sinusoids are**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.*sampled*instead of being continuous in time:**The test sinusoid samples are**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.*rounded*to a*finite precision:***Our test sinusoids are**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 successive samples (any*finite duration*, while the ideal sinusoid is infinitely long:*nonrecursive*LTI digital filter), the start-up transient is 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.**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 to , and the number of samples in each test sinusoid`tmax`from to . 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.*, [77,87].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).

Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.

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