Winners First Place: Kay Gemba (University of Hawaiʻi at Mānoa, USA)

Transcription

1 Aoustial Soiety of Ameria International Student Challenge Problem in Aousti Signal Proessing 14 Student Entry Evaluation Report by the Tehnial Committee on Signal Proessing in Aoustis Bakground Bakground information on the international student hallenge problem an be found in the artile by B. G. Ferguson and R. L. Culver, International Student Challenge Problem in Aousti Signal Proessing, Aoustis Today, Volume 1, Issue, pp 6 9, Spring 14 (available online at aoustistoday.org). The artile highlights the pull quote: Students are given the opportunity to distinguish themselves by solving a hallenging problem in aousti signal proessing. Evaluation Proess The solution notes, whih provided guidane for evaluating the student entries, form Appendix A. The entries were assessed by the evaluators to be of a very high standard refleting onsiderable time and effort having been expended by the students and, justifiably, a sense of pride and ahievement permeated their submissions. Students who have learly distinguished themselves by detailing their approah and reasoning in solving the problem and providing good estimates of the parameter set were seleted as Finalists. From the entries of the Finalists, five entries were judged to be Meritorious. The final step required ranking the meritorious entries in order: first, seond and third. Finalists Andrew Aquaviva (Pennsylvania State University, USA) Mihael Biffignani (Pennsylvania State University, USA) Yuta Enomoto (Waseda University, Japan) James Esplin and Matthew Shaw (Pennsylvania State University, USA) Philip Feurtado (Pennsylvania State University, USA) Kay (Kai) Gemba (University of Hawaiʻi at Mānoa, USA) Sahar Hashemgeloogerdi (University of Rohester, USA) Yasuhito Ishihara (Waseda University, Japan) Aaron Lesky (Pennsylvania State University, USA) Mihael Muhlestein (University of Texas at Austin, USA) Bernado Murta, Bruno Knebel, Gil Greo and Sergio Aguirre (Federal University of Santa Maria, Brazil) Ananth Raghavendra (Pennsylvania State University, USA) Rishabh Ranjan and Kaushik Sunder (Nanyang Tehnologial University, Singapore) Leandro Sebastian Rodiño (Universidad Naional de Tres de Febrero, Argentina) Yu Shiduo, Xu Lingji and Yang Long (Northwestern Polytehnial University, China) Sarah Smith and Mihael Heilemann (University of Rohester, USA) R. Troy Taylor (Pennsylvania State University, USA) Kohei Yatabe (Waseda University, Japan) David Zartman (Washington State University, USA) Winners First Plae: Kay Gemba (University of Hawaiʻi at Mānoa, USA) Equal Seond Plae: Yu Shiduo, Xu Lingji and Yang Long (Northwestern Polytehnial University, China); Yuta Enomoto (Waseda University, Japan) Equal Third Plae: James Esplin and Matthew Shaw (Pennsylvania State University, USA); Sahar Hashemgeloogerdi (University of Rohester, USA) 1

2 APPENDIX A. PROBLEM AND SOLUTION NOTES International Student Challenge Problem in Aousti Signal Proessing 14 Bakground. A truk with a 4-stroke diesel engine travels along a straight road with onstant speed. Near the road is a mirophone that senses the radiated aousti noise from the truk during its passage past the mirophone. The output of the mirophone is sampled at the rate of 1, samples/seond and 3 seonds of data are reorded during the truk s transit whih an be found in the attahed wav file. During the reording of the data, the speed of sound propagation in air was a onstant 347 m/s. Problem: Assuming that the truk is a point soure, Plot the spetrogram of the aousti data wav file Given that the strongest spetral line is the engine firing rate, alulate the truk s: (1) engine firing rate (in Hz), () ylinder firing rate (in Hz), (3) number of ylinders, (4) tahometer reading (in revolutions/minute), (5) speedometer reading (speed in km/hour), (6) distane (in meters) of the losest point of approah of the truk to the sensor, (7) time (in seonds) at whih the losest point of approah ours. Right Clik Here to Download file: Your solution should detail your approah and reasoning to solve the problem, as well as your best estimates of the above parameters.

3 frequeny (Hz) ASA Signal Proessing in Aoustis Tehnial Committee Solution Notes to 14 International Student Challenge Problem Two approahes to solving the problem are onsidered below: one is based on time-frequeny analysis and the Doppler Effet, the other on nonlinear least squares estimation of the parameters. A omparison of the parameters estimated using eah of these two approahes (below) are found to be in lose agreement. Solution Method 1. Doppler Effet Approah 1. Generate a spetrogram for the digital time series output of the mirophone during the truk s transit (.wav file) using the fast Fourier transform (FFT) with parameters: FFT size = 56*14 samples, window length = 8*14 samples, frequeny bin width = fs/fftsize.48 Hz, where fs is the sampling rate (1, samples/s) see Fig. 1. The strongest line orresponds to the engine firing rate (EFR), whih is observed at a frequeny of about 1 Hz. The EFR is observed to hange systematially with time due to the aoustial Doppler Effet. There is a harmoni series with a fundamental frequeny of about Hz, whih orresponds to the ylinder firing rate (CFR). The Doppler Effet is more readily observed on the higher frequeny harmonis. The number of ylinders is given by the EFR/CFR=6. The number of ylinders (6) is onfirmed by visual inspetion beause every sixth harmoni of the CFR is a strong spetral line. Spetrogram (fs = 1 khz) time (s) FIG. 1 Spetrogram showing the variation with time of the spetral omponents of the signal. 3

4 The truk s transit past the sensor is represented diagrammatially in Fig. (a), where is the (assumed onstant) veloity of the truk and is the distane of the losest point of approah (CPA) of the truk to the sensor. As noted above, the instantaneous frequeny of the EFR spetral line is observed to hange with time due to the Doppler Effet, whih an be modelled by the urve in Fig. (b). The up-doppler and down-doppler asymptotes are denoted by and, respetively. The time is the time when the soure is at the CPA and is the instantaneous frequeny (IF) of the signal reeived at the sensor at this time. Similarly, is the time when the IF of the reeived signal is equal to the onstant soure (or rest) frequeny. Note that at this time ( ), the soure is past CPA. In other words, the IF of the reeived signal is equal to the soure frequeny at a later time t R, where is the time delay for the sound to propagate at a / onstant speed from the soure to the sensor when the soure is at CPA. The IF of the signal reeived at the sensor is given by 1 where f f ( t) p( t;, s), (1a) v, (1b) v v f, (1) 1 R ( v ) s, (1d) v and t p( t;, s). (1e) 1 s t The asymptoti expressions for f (t) are given by f f ( ), (a) 1 v / f f ( ). (b) 1 v / The frequeny estimates of the two asymptotes an be used to provide estimates of the soure (or rest) frequeny ˆf and the soure speed vˆ : f ˆ (3a) v ˆ (3b) 4

5 where the ˆ symbol denotes estimation of the parameter. FIG.. (a) Geometry of a stationary sensor and a moving soure traveling in a straight line at onstant speed. (b) A typial time-frequeny urve 5

6 db Now Fig. 1 indiates that the variation with time of the IF is negligible during the first and last 5 s of data, so proess these two time frames to estimate the up Doppler frequeny asymptote f ( and the down Doppler frequeny asymptote f (, using the fast Fourier transform (FFT) with parameters: FFT size = samples, window length = samples, frequeny bin width = fs/fftsize.48 Hz, where fs is the sampling rate (1, samples/s) see Fig. 3. Frequeny spetra (fs = 1 khz) 8 6 up Doppler down doppler frequeny (Hz) FIG. 3. Frequeny spetra of the initial and final 5 s of mirophone output data. Zoom the spetra to loalise the asymptotes of the EFR see Fig. 4. Hene, the IF estimates ˆf ( 1.53 Hz and ˆf ( Hz. Substituting these values into (3a) and (3b), where 347 m/s, gives: f ˆ = Hz (4a) v ˆ = 5.49 m/s = km/hr (4b) So the soure (or rest) frequeny of EFR is Hz and the soure frequeny of the CFR is Hz. 6

7 db 8 7 up Doppler down doppler frequeny (Hz) FIG. 3. Zoom spetra to estimate the frequeny asymptotes of the Doppler shifted EFR. Note FFT size is 56*14 samples, window lengths are 64x14 samples from whih the frequeny asymptotes are 1.53 Hz and Hz. {f1 = 1.59 Hz, f = Hz, v = m/s, f = Hz} When the truk is near the CPA, the IF of the reeived signal dereases rapidly with time see Fig. (b). The maximum magnitude for the rate of hange of f t, whih ours when t, is given by g max df dt t t R f v 3 v. (5) The orresponding IF of the reeived signal at t is f 1 f f, (6) whih is larger than the soure frequeny f. Reiterating, when the soure is at the CPA ( t ), the IF of the reeived signal is reeived signal is equal to f at a later time f, not f. Owing to the propagation delay t R / - see Fig. (b). R /, the IF of the 7

8 frequeny (Hz) The IF of the EFR an be estimated using the fast Fourier transform (FFT) with parameters: FFT size = samples, window length = 14 samples, frequeny bin width = fs/fftsize.48 Hz, where fs is the sampling rate (1, samples/s). Figure 5 shows the variation with time of the raw IF estimates of the EFR as blue irles with a 5 sample smoothed version represented by the red line. Using the smoothed IF estimates, the parameter R an be estimated using either: Rˆ tˆ ˆ 34.3 m (7) where 347 m/s, and from Fig. 5, tˆ s with ˆ s. {tau = s, t = s, R = *(t-tau) = 34.9 m} or: where, from Fig. 5. ˆ ˆ f vˆ R 34.9 m (8) 3 gˆ vˆ max ĝ.95 max {gmax =.95, R = f*(v)^/(^-v^)^1.5/gmax = m} Instantaneous frequeny estimates (fs =1 khz) raw estimates smoothed (K=5) time (s) FIG. 4. Variation with time of the instantaneous frequeny estimates of the EFR 8

9 Finally, by using this approah, the estimates of the parameters are: (1) engine firing rate (in Hz): EFR= ˆf Hz () ylinder firing rate (in Hz): CFR 19.8 Hz (3) number of ylinders: 6 (4) tahometer reading (in revolutions/minute), The tahometer reading is the rank shaft rate, whih for a 4 stroke engine, is equal to CFR 6 37 rpm (5) speedometer reading (speed in km/hour): vˆ 19.8 km/hr (6) distane (in meters) of the losest point of approah of the truk to the sensor: Rˆ tˆ ˆ 34.3 m; ˆ R gˆ max fˆ vˆ vˆ m (7) time (in seonds) at whih the losest point of approah ours: ˆ s The truk driver advised that the speedometer dial reading of the truk was km/hr and the tahometer dial reading was 35 Hz. 9

10 frequeny (Hz) Solution Method. Nonlinear Least Squares (NLS) Estimation of the Parameters R, or equivalently,, s The parameters f, v,,,, an be estimated using a NLS approah, i.e., by minimizing the sum of the squared deviations of the noisy IF estimates from their predited values 1. This approah has been applied to the same truk transit previously but for a 45 s (rather than 3 s) observation interval. Also the data were downsampled to 1 khz. Figure 6 shows the variation with time of the raw IF estimates of the EFR in blue and the time-frequeny urve omputed using (1a) (1e) with the NLS-optimized parameters in red. The estimated values of the soure parameters were: f Hz v 1.9 km/hr.81 s R m fs = 1 khz raw estimates LS fit time (s) FIG. 6. Variation with time of the instantaneous frequeny estimates (blue-filled irles) of the signal reeived by a mirophone during the truk transit, and the nonlinear least squares fit (solid red line) to the observations. {f = Hz, v = m/s, tau =.81 s, R = m} 1

11 frequeny (Hz) Similarly, Fig. 7 shows the variation with time of the smoothed IF estimates of the EFR in blue and the time-frequeny urve omputed using (1a) (1e) with the NLS-optimized parameters in red. The estimated values of the soure parameters were: f Hz v 1.9 km/hr.81 s R m smoothed estimates (K=5) LS fit time (s) FIG. 7. Variation with time of the smoothed instantaneous frequeny estimates (blue-filled irles) of the signal reeived by a mirophone during the truk transit, and the nonlinear least squares fit (solid red line) to the observations. {f = Hz, v = m/s, tau =.898 s, R = m} {For this ase, f(- )=1.697 Hz and f(+ )= Hz, so fmean= Hz} 11