1. Sebastian Böser sboeser@ifh.de Acoustic sensor and transmitter development Amanda/IceCube Collaboration Meeting Berkeley March 2005

2. Acoustic sensors and transmitters – 2 sboeser@ifh.de Overview Motivation Sensors calibration Methods Results Equivalent noise level Transmitters Ringtransmitter HV signal generator

3. Acoustic sensors and transmitters – 3 sboeser@ifh.de Motivation This talk

4. Acoustic sensors and transmitters – 4 sboeser@ifh.de Calibration Problem interesting frequency ≈ 20 kHz  λ water = 7.5 cm  λ ice = 20 cm Oscillating signal  reflections distort signal  need container with x cont » λ Setup at HSVA water tank 12m × 3m × 70m deep section 12m × 5m × 10m Sensors Reference Hydrophone  Sensortech SA03 -163.3±0.3 dB re 1 V/µPa (5 to 65 kHz) Glass Ball, Iron Ball Transmitter piezoceramic in epoxy  arbitrary signal generator

5. Acoustic sensors and transmitters – 5 sboeser@ifh.de Speed of sound Method compare arrival times of  direct signal  reflection at the surface  reflection at the walls Result v water = 1409.7 ± 4.5 m/s Theory v water = 1411.2 m/s  good agreement

6. Acoustic sensors and transmitters – 6 sboeser@ifh.de Sensitivity: Method Method transmit same signal to  reference  sensor to calibrate compare response  relative calibration Transmitted signals gated burst  precisely measure single frequency  limited by system relaxation time reflections pulse  in one shot measure full spectrum  limited by noise level

7. Acoustic sensors and transmitters – 7 sboeser@ifh.de Sensitivity: Gated burst Time window start: after initial excitation stop: before 1 st reflection Fit A(t) = A 0 sin(2πf·t + φ) + bt +c free phase and amplitude fixed frequency linear offset term  very good χ 2 But: low-f and DC background  large error for small signals  probably overerstimated

8. Acoustic sensors and transmitters – 8 sboeser@ifh.de Sensitivity: pulse method Transmitted signal P ∞ ∂ 2 U in / ∂t 2  “soft” step function Received signal Fourier transform  compare spectral components Errors and noise A(t) = Σ f s(f)e i (2πft + φ s ) + n(f)e i (2πft + φ n ) coherent signal: φ s (f) = const random noise: φ s (f) = random Noise spectrum from average  fourier transform fourier transform  average  define signal dominated regions

9. Acoustic sensors and transmitters – 9 sboeser@ifh.de Comparison of methods Results high sensitivity and S/N Glass ball: factor ≈ 20 Iron ball: factor ≈ 50 very good agreement strongly structured  many different resonance modes only valid for water

10. Acoustic sensors and transmitters – 10 sboeser@ifh.de Equivalent noise level Method fourier transform  scaling, frequency range  backward transform Problem noise recording from water tank lab self noise higher due to EM coupling Equivalent Noise Level [mPa] Frequency range [kHz] 5 - 1205 - 65 Hydrophone50.1± 0.740.3 ± 8.3 Glass Ball17.1 ± 1.715.9 ± 1.7 Iron Ball6.6 ± 0.64.7 ± 0.7

11. Acoustic sensors and transmitters – 11 sboeser@ifh.de How to do it for ice ? Theoretical use formula for transmission in media Problem temperature dependence  resonance modes  amplifier gain× bandwidth solid state vs. liquid Practical use large ice volume (glacier, pole) use small ice block with changing boundary conditions (e.g. air, water)  determine reflections from comparison

12. Acoustic sensors and transmitters – 12 sboeser@ifh.de Transmitters Large absorption length  Need high power transmitter Piezoceramics can be driven with kV signals easy to handle cheap well understood Ring-shaped piezo ceramic azimuthal symmetry larger signals than cylinders more expensive

13. Acoustic sensors and transmitters – 13 sboeser@ifh.de Transmitter: Ringtransmitter Linearity tested from 100 mV to 300 V  perfect linearity Frequency response three resonance modes  width, thickness and diameter  wide resonance at lower frequencies Testing frequency sweep  dominated by reflections  resonance modes of container white noise signal  reflections not in phase  resonance modes of transmitter

14. Acoustic sensors and transmitters – 14 sboeser@ifh.de Power supply Problem build a HV generator for arbitrary signals I max = 2πf C tot U max C ring = 16 nF f = 100 kHz U max = 1kV k 33 = 0.34 I max = 16 A, P ≈ 5.4 kW  too large Solution large capacity at low duty cycles 100 cycle burst  1ms  16 W large inductivity  discharge via capacitance  shortcut after N cycles

15. Acoustic sensors and transmitters – 15 sboeser@ifh.de Next talk