Sebastian Böser Acoustic sensor and transmitter development Amanda/IceCube Collaboration Meeting Berkeley March 2005.

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Presentation transcript:

Sebastian Böser Acoustic sensor and transmitter development Amanda/IceCube Collaboration Meeting Berkeley March 2005

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

Acoustic sensors and transmitters – 3 Motivation This talk

Acoustic sensors and transmitters – 4 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 SA ±0.3 dB re 1 V/µPa (5 to 65 kHz) Glass Ball, Iron Ball Transmitter piezoceramic in epoxy  arbitrary signal generator

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

Acoustic sensors and transmitters – 6 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

Acoustic sensors and transmitters – 7 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

Acoustic sensors and transmitters – 8 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

Acoustic sensors and transmitters – 9 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

Acoustic sensors and transmitters – 10 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] Hydrophone50.1± ± 8.3 Glass Ball17.1 ± ± 1.7 Iron Ball6.6 ± ± 0.7

Acoustic sensors and transmitters – 11 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

Acoustic sensors and transmitters – 12 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

Acoustic sensors and transmitters – 13 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

Acoustic sensors and transmitters – 14 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

Acoustic sensors and transmitters – 15 Next talk