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RECOVERING SIGNAL FREQUENCIES WITH THE FOURIER TRANSFORM

KENNETH L. HO

Suppose that we are given a signal of the form

f(t) =? isin(2πωit),(1)

i.e.,fis the superposition of fundamental waves of amplitudesAiand frequenciesωi, and that we are only

allowed to samplefat a finite number of equally-spaced time points. How might we determine, on the basis

of the samples that we have collected, the identities of the frequencies composingf? A commonly used tool for this task is the Fourier transform. In the continuous setting, the Fourier transform

1is the mapping

f(ω) =? f(t)e-2πıωtdt?C

offinto the frequency domain, where|ˆf(ω)|gives the "contribution" of the fundamental wave of frequency

ωtof. To see this a little more explicitly, it is instructive to introduce also the inverse Fourier transform

f(t) =? -∞ˆf(ω)e2πıωtdω,

which is just an integral decomposition offinto its fundamental components. Viewed as a generalization of

(1),ˆf(ω) clearly represents the amplitude of the wave of frequencyω.

In the posited situation, however, we do not have access to the entire signalf, but only a discrete number

of samplesf1,...,fn, wherefi=f(ti), withti+1-ti= Δt, where Δtis the intersample time. Consequently,

we will not use the Fourier transform as described above, but instead the discrete Fourier transform (DFT),

defined by j=n-1? k=0f ke-2πıjk/n, j= 0,...,n-1.

Similarly to the continuous case,|Fj|gives the contribution of "frequency"jto the sample signal. But what

physical frequency doesjcorrespond to?

Clearly, the maximum frequencyωmaxthat we can hope to capture is limited by the sampling frequency

1/Δt. But due to something called the Nyquist condition, we can achieve only half of this-intuitively, it

takes two points to represent each wave cycle: one for the "peak" and another for the "trough". Therefore,

max=12Δt.(2) Now what about the minimum frequencyωmin? For real signalsf, there is a symmetry between positive

and negative frequencies, so in factωmin=-ωmax. Since the DFT hasncomponents, equally spaced from

mintoωmax, the resolution between the discrete frequencies is

Δω=ωmax-ωminn

=1nΔt≡1T ,(3) whereTis the total time period over which we samplef.1

There are many definitions of the Fourier transform-the one that we use here is certainly not the most common, though

it is the most convenient for our purposes.

2 KENNETH L. HO

Figure 1.Moduli of Fourier transform coefficients|F|offsampled withT= 10 s and Δt= 0.01 s as a function of the frequencyω(including negative frequencies; Hz). So which components ofFcorrespond to the negative frequencies? To answer this, we consider the periodicity of the DFT kernel: since exp(2πı) = 1, soFn-j=F-jis actually the amplitude for "frequency"-j. Hence, the DFT is redundant, and the first?n/2?components cover all non-negative frequencies captured. Thus, forj=

0,...,?n/2?, the physical frequency corresponding toFjisjΔω. From this perspective, the maximum

frequency that we can recover is

which, of course, is consistent with (2). Observe that (2) and (3) provide guidelines on how to selectTand

Δt.

Let"s do an example now where we actually use this to recover the fundamental frequencies of the signal

f(t) = sin(2π·10·t) + 2sin(2π·20·t)-0.7sin(2π·30·t),

i.e.,fis composed of waves of 10, 20, and 30 Hz, iftis given in seconds. We will additionally corruptf

with white noise to simulate an imperfect sensor. The DFT is computed inMatlabby the commandfft, for the fast Fourier transform-this is how the DFT is almost always computed in practice. To ensure that we can capture the highest frequency of 30 Hz

2, we use (2) to derive the condition

s. The moduli|Fj|of the transformed signal are shown in Figure 1. Strong peaks about 10, 20, and 30

Hz clearly pinpoint the constituent frequencies. However, there are also peaks about 70, 80, and 90 Hz; as

discussed above, these correspond to the negative frequencies-30,-20, and-10 Hz, respectively, as can

be seen from the fact that the amplitudes of the corresponding frequencies are identical-this is the sense in

which we mean that the DFT is redundant.

We can clean this up a bit by dropping off the second half of the DFT data. There are a few details in

doing this cleanly (primarily involving promotingnup to the next power-of-two); these are all given in the

functionfftfreq 3:2 Note that we don"t usually have such knowledgea priori.

3All codes can be downloaded athttp://www.courant.nyu.edu/~ho/teaching/2011/nyu/spring/comput-med-biol/.

RECOVERING SIGNAL FREQUENCIES WITH THE FOURIER TRANSFORM 3 Figure 2.Moduli of Fourier coefficients|F|offsampled withT= 10 s and Δt= 0.01 s

as a function of the frequencyω(Hz).%FFTFREQ Compute a FFT restricted to non-negative frequencies and return% the frequencies associated with each Fourier component .

% X = FFTFREQ(X,T,DT) returns the discrete Fourier transform of X % sampled over the time interval T in time increments of DT, restricted

% to non-negative frequencies only . The time interval T must have the% form [TMIN, . . . , TMAX] ; only the f i r s t and last elements are used .

% [X,W] = FFTFREQ(X,T,DT) also returns the frequencies W associated with % each component of X. % See also FFT. function[X, w] = f f tf r e q (x , t , dt )

n =ceil(( t (end)-t (1)) / dt );nfft = 2ˆnextpow2(n );X =fft(x , nfft ) / n ;X = X(1:ceil( nfft /2));w = 1/(2?dt )?linspace(0 , 1 ,ceil( nfft /2));end

The scriptdemofftfreqbelow demonstrates its use:%% i n i t i a l i z e

f = @( t )(sin(2?pi?10?t ) + . . .% input signal2?sin(2?pi?20?t )-. . .0.7?sin(2?pi?30?t ) ) ;T = 10;% sampling time intervaldt = 0.01;% sampling time period%% compute

n =ceil(T/dt );% number of sample pointst =linspace(0 , T, n );% sampling timesx = f ( t ) +randn(1 ,n );% sampled noisy signal[X, w] = f f tf r e q (x , t , dt );% compute FFT%% plot

figure plot(w,abs(X))% frequency contributionsxlabel( " Frequency(\omega)[Hz] " )

ylabel( " Fourierc o e f f i c i e n tmodulus(|F|) " )Figure 2 shows the Fourier coefficients for the same sampling parameters as Figure 1, but with only the

first half of the DFT data shown. Finally, to show the effects of resolution, we give two more plots: Figure

3 withT= 1 s, and Figure 4 withT= 100 s.

4 KENNETH L. HO

Figure 3.Moduli of Fourier coefficients|F|offsampled withT= 1 s and Δt= 0.01 s as

a function of the frequencyω(Hz).Figure 4.Moduli of Fourier coefficients|F|offsampled withT= 100 s and Δt= 0.01 s

as a function of the frequencyω(Hz).quotesdbs_dbs2.pdfusesText_4