OSE801 Engineering System Identification. Lecture 05: Fourier Analysis

Transcription

1 OSE81 Engineering System Identification Lecture 5: Fourier Analysis What we will study in this lecture: A short introduction of Fourier analysis Sampling the data Applications Example

2 1 Fourier Analysis and Signal Aliasing Suppose we have a set of measurements on vibrating structures, q(t), f(t),..., where q s can be velocity, displacement or acceleration vectors, and f s are excitations. As most vibrating structural systems exhibit periodic responses, it is natural to approximate the time series of measured data in terms of their frequency contents. The starting point for this data processing is the wellknown Fourier analysis (after J.B.J. Fourier ( )). In Fourier analysis we approximate a function x(t) by x(t) a 2 + (a k cos kt + b k sin kt) (1) k=1 where a k, b k and a are coefficients to be determined. An attractive feature of the Fourier expansion is in the simplicity of determining the expansion coefficients given by 2π a k = 1 x(t) cos ktdt, b k = 1 x(t) sin ktdt (2) π π for which the following results are repetitively employed: 2π 2π 2π 2π k m cos kt cos mt dt = π k = m 2π k = m = sin kt cos mt dt = for all k & m k m sin kt sin mt dt = π k = m (3) There are two essential properties of the Fourier function: 1) The error of a truncated approximation can be assessed in terms of the function rather than some high-order derivatives required in typical polynomial expansions. 2) The rate of convergence can be estimated easily from the discontinuities of the function and its derivatives. Of course, the routine Fourier analysis tool we would like to understand is Fast Fourier Transform (FFT) rather than the above continuous Fourier analysis. First, as sampling profoundly affects a proper usage of FFT, let us begin 2

3 this section by examining the aliasing phenomenon and its impact on Fourier analysis in general. To this end, we introduce a thought experiment: We are watching a Western movie. A stage coach starts up and the wheels start going faster and faster. If we keep our eyes on the wheels, we see the wheels gradually slow down (even though the speed of the coach is not!), then stop, then occasionally go backwards, slow down, stop, go forward, etc. This effect is due solely to the sampling that our eyes perform on the real scene of the rotating wheels. Hence, once we know the sampling rate, we can tell exactly what sampled frequencies correspond to the rotating wheel speeds (frequencies). In other words, signals with different frequencies may appear to be the same signal, depending upon the sampling rate. To illustrate this, let us consider the following two pathological signals, an example given in Hamming: y 1 (t) = cos[π(n + ε)t + φ] y 2 (t) = cos[π(n ε)t φ] (4) Notice that the two signals have the same values at the three sampling points (t =, 1, 2) as shown in Fig This can be verified from the following identity: cos[π(n + ε)5 + φ] cos[π(n ε)t φ] 2 sin(nπt) sin(nεt + φ) =. (5) if both n and t are integers. Therefore, once we made the measurements, there is no way for us to distinguish the two signals; they appear as if they were the same frequency. This is an important concept to remember in all the data processing involving Fourier analysis. 2 Fourier Integral and Discrete Fourier Transform In practice, the evaluation of the Fourier coefficients a k and b k has to be carried out by an efficient discrete algorithm. This involves the Fourier integral of y(t) defined by X(f) = and its inverse x(t), once X(f) is given, is obtained by: x(t) = + x(t)e j2πft dt (6) X(f) e j2πft df (7) 3

4 Fig. 1. Hamming s Aliasing Example where f is the frequency variable in Hz. Before we can derive Fast Fourier Transforms from the above Fourier integrals, we would need two intermediate steps: a sampled representation of Fourier series and discrete Fourier transform. First, we represent the function x(t) in a discrete periodic form: x(t) = a + 2 (a k cos 2πkt k=1 T + b k sin 2πkt T ) a k = 1 T x(t) cos 2πkt dt, k T T b k = 1 T x(t) sin 2πkt dt, 1 k T T (8) Now let us introduce a complex coefficient that absorbs both a k and b k X(k) = (a k j b k ) = 1 T T x(t)e j(2πkt/t ) dt, j = 1, k (9) 4

5 A crucial step here is to discretize the above complex coefficient X(k) by X(k) = 1 T N 1 s= = t (N t) x(s) e j(2πks t/t ) t N 1 s= x(s) e j(2πks/n) since T = N t (1) The quantity in the denominator (N t) should be interpreted for periodic functions as the time duration or the period of the fundamental frequency content of the signal. Clearly, X(k) in the above equation is the discrete counterpart to the continuous Fourier integral (6). Following a similar step, the discrete inverse Fourier transform that corresponds to the continuous case can be derived as x(s) = N 1 k= X(k) e j(2πks/n) (11) Remark: It should be noted that in MATLAB the discrete Fourier transform pairs are defined by X(k) = N 1 s= x(s) e j(2πks/n) x(s) = 1 N N 1 k= X(k) e j(2πks/n) (12) Hence, for plotting or sometimes scaling purposes, MATLAB Fourier transform X(k) needs to be scaled by t/(n t). However, these MATHLAB pairs need no scaling when using them eventually to obtain an approximate x(t) from {x(s), s =,..., N 1}, viz., x(s) ifft (fft ( x(s) ) ) (13) 3 Sampling Theorem and Aliasing Revisited At the start of this section a physical view of aliasing was presented. Let us now revisit aliasing in conjunction with sampling rules. Suppose x(t) is sampled at discrete time intervals t = t, 2 t, 3 t,..., N t. The sampled discrete 5

6 Fig. 2. Box Wave Transform Pairs values of x(t) denoted as ˆx(t) at those sampling instants can be represented by ˆx(t) = k= where δ(t k t) is the Dirac delta functions. x(k t) δ(t k t) (14) Now suppose that x(t) is zero for frequencies that are greater than ω max = 2πf c. One such function is a box function in frequency domain by {X(f) = 1; f < f c } x(t) = 2f c sin (2πf ct) 2πf c t whose theoretical transform pairs are shown in Fig. 2. (15) To examine the effect of sampling, three sampling stepsizes ( t =.1,.5,, 85) are chosen and their respective Fourier transforms are shown in Fig It should be noted that only half of the symmetric part of the frequency vs. amplitude ( f 1) is plotted. Comparing the discrete transform results with the theoretical transform given in Fig. 5.2, we see that the case with the sampling rate t =.1 gives both the amplitude and frequency transform correctly, except small oscillations around the cut-off frequency f = 1 due to 6

7 Dt=.85 Dt= Frequency(Hz) Dt = Fig. 3. Discrete Fourier Transform of sin(2πt) 2πt the well-known Gibbs phenomenon. Second, the case of t =.5 not only gives the correct magnitude and frequency, but more importantly there is no oscillation around the cut-off frequency. The third case with the sampling rate t =.1 shows the discrete case fails to approximate the theoretical result. The reconstructed three time signals by performing the inverse discrete Fourier transforms of the three discrete Fourier transform cases are shown in Fig

8 D t=.1.2 Dt= Dt= Fig. 4. Reconstructed sin(2πt) 2πt Function Notice that the case of t =.1 reconstructs the original signal reasonably accurately. The case with t =.5 captures the correct magnitude at time t =, and it is zero everywhere else. The case with sampling rate t =.85 are again completely in error. This dependency of the foregoing discrete transform accuracies on the sampling rate can be understood as follows. Since the period of the example problem is T c = 1/f c = 1, the sampling rate 8

9 1.8 o o o o o :Sampling Points o o o Fig. 5. Nyquist Sampling of cos(2πt) Function that picks the absolute maximum function values is t N = T c /2 = 1/2f c. The sinusoidal function is then approximated by a saw-tooth function. Notice that the saw-tooth approximation preserves both the peak values and the frequency. This is illustrated in Fig Therefore, if the sampling rate is large than this value, aliasing will occur, meaning the frequency will be lower 9

10 or the period will be artificially longer. The sampling rate given by t N = 1/(2 period of the signal) (16) is called the Nyquist sampling rate. Hence, if the sampling rate is smaller than the Nyquist rate, no aliasing will occur. Finally, it turns out that the frequency band-limited box wave form pairs provides an important role in constructing the continuous signal from a set of sampled discrete signals, which is called the sampling theorem: x(t) = t s= x(s) sin 2πf c(t s t) π(t s t) (17) The above sampling theorem is applicable to the time domain signal reconstruction. There exists the frequency-domain counterpart, which states that if a function x(t) is time-limited as given by x(t) =, t > T c (18) then the continuous Fourier transform can be constructed from a set of sampled Fourier signals by X(f) = 1 2T c k= X(k) sin 2πT c(f k/2t c ) π(f k/2t c ) (19) In practice, rarely one has the luxury of working with frequency band-limited signals. For example, signals measured on the vibrating structures contain a wide range of frequency spectrum so that it is inevitable to truncate the frequency band-width. Such truncations of either time-domain signals or frequencydomain signals do introduce errors called leakage. This will be discussed below. 4 Leakage due to Signal Truncations Let us consider the following example problem: η 1 (t) = sin(2πf t), f = (2) For subsequent discussions, we introduce without proof the well-known theoretical Fourier transform pairs: 1

11 cos(2πf t) 1 2 δ(f f ) δ(f + f ) sin(2πf t) j 1 2 δ(f + f ) j 1 2 δ(f f ) (21) In truncating time-series signals, it is recommended that the sampling time interval be equal to a multiple of the signal period, while satisfying the Nyquist sampling rate criterion to avoid aliasing. If the sampling time interval is not equal to the integer multiplicity of the signal period, then leakage phenomenon occurs. Since for this example problem we have its period T = 1/f = , in order for the sampling rate to be a fraction of the period, we must have t = /N where N is the number of samples per period. Figure 5.6 illustrates the leakage phenomenon for the example problem. Even though the sampling rate is smaller for the leaking FRF plot designated by +, because the sampling interval is 12.8 * period, that is, the sampling duration is not a multiple of its period, its Fourier transform results in significant leakage. However, this is realistic in that one must usually deal with data sets whose period information is not known a priori. The leakage problem has motivated a host of researchers to devise a plethora of filtering schemes that would compensate the fractional-periodic nature of signals. For periodic signals (fortunately vibration records are periodic!), perhaps the most widely used anti-leakage filter is due to Hanning given by x(t) = 1 2πn (1 cos ) n =, 1, 2,..., N 1. (22) 2 N 1 where N is the total number of samples. Observe that the above Hanning function forces the sampled data to be zero at the start and at the end of the sample interval. Figure 7 shows the effect of Hanning filtering on minimizing leakage when the sample interval does not correspond to a multiple of signal period. The significant improvement over the case without filtering is obvious. However, it should be noted that the peak amplitude is considerably reduced (from.5 to about.25) after applying the Hanning filter. As the magnitude of FRF curves is related to the structural damping levels, one must be careful in interpreting the Hanning-filtered results. Finally, the errors in the magnitudes of the case with the sampling rate N t = 12.8 T can be corrected if one judiciously utilizes the sampling formula. For 11

12 .6.5 Magnitude Frequency(Hz) Fig. 6. Leakage in fft(cos 2πf t) due to Sample Truncation. (o plot: T / t = 32 and 16T duration ), (+ plot: T / t = 4 and 12.8T duration ) example, as the closest multiple of the period is 12, the correct magnitudepreserving formula should read: X(k) = t (12 T ) fft (sin 2πf t) (23) 12

13 Magnitude Frequency(Hz) Fig. 7. Effect of Hanning Filter on Leaky fft(cos 2πf t) FRF Result. (o plot: T / t = 4 and 12.8T duration ), (x plot: after Hanning filtering) which should give its peak magnitude.5, a correct one as can be seen from (21). Finally, the following elementary MatLab code and its variations were used to plot several figures used in the preceding sections. 13

14 % example fft of sine (2*pi*f *t); f = ; f = ; T = 1./f; dt = T/4; N=512; time =(:(N-1))*dt; arg = 2*pi*f*time; y = sin(arg); yfft = fft(y); freq=(:(n/2-1))/(n*dt); plot(freq, abs(yfft(1:n/2))/n, + ); axis([ 1.3.6]); %end 14