And, the (low-pass) Butterworth filter of order m is given in the frequency domain by

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1 Problem Set no.3.a) The ideal low-pass filter is given in the frequeny domain by B ideal ( f ), f f; =, f > f. () And, the (low-pass) Butterworth filter of order m is given in the frequeny domain by B Butterworth ( f ) = f + f m. () For the purposes of numerial evaluation, we onsider the range of frequenies defined by wavenumbers, N,, N. That is, k fk =, N k N, (3) N where N is given; in this ase N =. The ut-off frequeny is defined by the wavenumber k =. The following figure shows the frequeny responses, Bideal ( f k ) and BButterworth ( f k ) ; the latter for different orders: m =, 4, 8.. Ideal Butterworth, m= Butterworth, m=4 Butterworth, m= wavenumber, k Figure : Comparison of Butterworth filter responses for ut-off wavenumber,, and various orders.

2 As the order of the Butterworth filter inreases, its frequeny response approahes that of the ideal filter. Therefore, to improve the suppression of the high wavenumbers (frequenies), one should hoose high order. Another option is to derease the ut-off frequeny, f, as shown in the next figure, where four frequeny responses are plotted, all for the Butterworth filter of order, m =, and for different ut-off wavenumbers, k.. k = k = 8 k = 6 k = wavenumber, k Figure : Comparison of Butterworth filter responses of order,, and various ut-off wavenumbers. Clearly, as the value of the ut-off wavenumber, k, dereases more of the higher frequenies are suppressed. If we do not assoiate the pass band with the ut-off frequeny, we may say that we improve the suppression of high frequenies with some sarifie in retaining the desired passband..b) In -D, the ideal and Butterworth high-pass frequeny responses are given by (, ) high-pass ideal B f f, f f and f f; =, f > f or f > f. (4)

3 ( ) B f, f = high-pass Butterworth m m f f + + f f, (5) where f and f are the ut-off frequenies for the frequenies, f and f, respetively. (To make it even more general, one ould also introdue different orders, m and m, for the two frequenies.) The following figures show these two high-pass frequeny responses: BB Ideal high -pass filter, k = k = Butterworth BB high-pass filter, k = k = Figure 3: Ideal and Butterworth -D high-pass filters. Horizontal axes are for wavenumbers in - D, offset by N = in eah diretion. The order of the Butterworth filter is m = 4..a) Using MathCad, the funtion CFFT(A) returns the Disrete Fourier Transform of a vetor or matrix. The formula is saled by /N (or, ( NN ) in the -D ase), and uses a negative exponent going from the time to the frequeny domain. It is thus similar to our definition exept for the saling. To make it onsistent with our definition (equation 3.34) the result of CFFT(A) must be multiplied by N t (similarly in -D). The MathCad funtion ICFFT returns the inverse Disrete Fourier Transform of a vetor or matrix. The formula is unsaled and uses a positive exponent going from the frequeny to the time domain. Thus to be onsistent with our definition (equation 3.35), the result of ICFFT(A) must be divided by N t (similarly for -D). To verify the use of these funtions, we onsider the profile, k = 7, of gravity anomalies desribed in Problem Set no.. We ompute the Fourier transform diretly using 679 π i k l 68 l Gk = x ge l=, (6) and by CFFT: G = 68 x CFFT( g), where g is the vetor of 68 gravity values in the profile and x = 4 km. Computing the RMS differene:

4 RMS = G G k k, (7) k= we find that RMS = 4 mgal/(y/km), whih is the preision of the omputer, sine the magnitude of the spetral omponents if of the order of - mgal/(y/km) ( ).b) Similarly we ompute ( ) ICFFT NN xcfft g NN x, where g represents the entire 54 by 68 data array. Computing the RMS differene between the result of this operation and 4 the original data array, we find, RMS = 3.5 mgal, whih is the preision of the omputer. 3.a)&b) In this problem we have an N N grid of gravity anomalies, where N = 54 and N = 68 are the dimensions of the data grid. Taking the DFT of this array of data yields a spetrum with wavenumbers, N k N and N k N, or frequenies, f k N x f = k N x, where x = x = 4 km. We note that usually the = ( ) and ( ) wavenumbers are ounted as k N and k N. Therefore, when designing filters to be applied in the spetral domain, we must aommodate this way of arranging the wavenumbers. f = k N x We apply a -D high-pass filter as shown in Problem.b, with parameters, ( ) and f = k ( N x ). In the ase that the ideal filter should pass all frequenies with wavenumbers and higher, we have 3 = ( ) = and f ( ) f 54 4 km 9. y/km 3 = 68 4 km = 7.4 y/km. If the spetrum of the data is given by G k, k, k =, K, N, k =, K, N, then the ideally high-pass filtered spetrum is G k, k, k k and k k ;, k k and N k k N ;, N k k N and k k ; = N k k N N k k N, and ; Gk, k, otherwise. (8) Similarly, the spetrum of the Butterworth high-filtered data is given by where ( ) G = B k, k G, N k N, N k N, (9) k, k k, k

5 ( ) B k, k = m m k k + + k k. () Note that to implement this using the DFT and its inverse with the indies running from to N, we first shift the spetrum, G k, k, so that it is properly defined for negative and positive wavenumbers, N k N and N k N. Then the Butterworth filter an be applied in the normal way as above, equation (9). Finally, the resulting filtered spetrum, G k, k, is shifted bak to non-negative indies (making sure to aount for the inherent periodiities), and subjeted to the inverse DFT. A plot of the original data is shown in the figure below. grid index number for x g grid index number for x data T Figure 4: Original data, arbitrary units. A plot of the ideal high-pass filtered data with ut-off wavenumbers, k = k =, is shown in Figure 5. We see the ripple effet throughout the image due to the sharp frequeny response at the ut-off wavelength. Figure 6 shows the data filtered by a Butterworth filter of order with the same ut-off wavenumbers. The ripples have essentially disappeared. Plots of the data profile, k = 7, are shown in Figure 7, and show that both filters yield approximately the same high frequeny signal (but one, of ourse, has the ripple artifat, whih an best be visualized in the -D images).

6 grid index number for x g grid index number for x Figure 5: Ideally filtered data, ut-off wavenumber is (eah axis). Same units as in Figure 4. grid index number for x g grid index number for x Figure 6: Butterworth-filtered data, ut-off wavenumber is (eah axis); the order is m =. same units as in Figure 4.

7 gravity anomaly [mgal] original data ideal-filtered data Butterworth-filtered data grid interval Figure 7: Profile, k = 7, of original, ideal-filtered, and Butterworth-filtered gravity anomaly. 4. We hoose the k = 7 profile of the gravity anomaly data introdued in Problem Set No.. This profile onsists of data sampled at an interval of 4 km. We re-sample the data at an interval of 8 km (i.e., leave out every other sample), and also at an interval of 6 km. The DFT is omputed in eah ase using the FFT routine disussed in problem.a (above). In the seond ase (interval of 8 km), we multiply CFFT(g) by ( N )( x), where x = 4 km, and N = 68. In effet, we multiply by the same sale as in the first ase. Similarly, in the third ase we sale by the same amount. The three amplitude spetra are shown in Figure 8. At double the original sampling interval, the highest frequeny is half of the original. We see that the highest frequenies of the more oarsely sampled data are orrupted (due to aliasing) more than the lower frequenies. This is learer for the quadruple-interval sampled data, where the highest frequeny is a quarter of the original highest. The plot of the differenes in amplitude spetra (Figure 9) shows also that the greatest aliasing errors generally our loser to the Nyquist frequeny (sine aliasing is a folding of the high-frequeny spetrum about the Nyquist frequeny to the low-frequeny spetrum).

8 . 4 original data double-interval sampled data quadruple-interval sampled data amplitude spetrum [mgal/(y/km)] frequeny [y/km] Figure 8: Amplitude spetrum of original, double-interval, and quadruple-interval sampled data. amplitude spetrum [mgal/(y/km)] 4 double-sample minus original quadruple-sample minus original frequeny [y/km] Figure 9: Differene of amplitude spetra for original, double-interval, and quadruple-interval sampled data.