ultrasound ii

ultrasound ii - BE101 Foundations of Biomedical Imaging...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
UCSD BE101 Fa06 Lecture 11 Slide 1 BE101 Foundations of Biomedical Imaging Fall 2009 David Hall, Ph.D. Department of Bioengineering University of California–San Diego Ultrasound II
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
UCSD BE101 Fa06 Lecture 11 Slide 2 Generation and Detection of Ultrasound Gray Scale Imaging Doppler Imaging Image Quality Suetens’s: Chapter 7 (pages 158-172) Today’s Outline
Background image of page 2
UCSD BE101 Fa06 Lecture 11 Slide 3 Piezoelectric crystal deforms under electric field (& vice versa) Crystal embedded in the “transducer” serves as transmitter and detector. PZT (lead zirconate titanate) and PVDF (polyvinylidene fluoride) are two common piezoelectric polymer materials. Drive with a sinusoidal electrical signal generates a compression wave at same frequency which propagates through surrounding media. But acoustic impedance mismatch causes reflection at transducer interface back into the crystal… which deforms it and induces additional electric fields which interfere with the driving signal! Optimal compression wave generated when crystal thickness is half the wavelength which causes resonance at the “fundamental resonance frequency” Generation and Detection of Ultrasound
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
UCSD BE101 Fa06 Lecture 11 Slide 4 Want most acoustic energy to enter the tissue. Thus crystal backing material has a different acoustic impedance than the crystal which reflects energy back into the crystal. Acoustic impedance mismatch between crystal and tissue causes reflection of acoustic wave back into the crystal. . Instead of transmitting into the tissue! Therefore a “matching” layer is used with: acoustic impedance=(Z c Z t ) 1/2 where Z c (crystal), Z t (tissue). Optimal when matching layer is odd multiple of ¼-wavelengths. Generation and Detection of Ultrasound US-Ducer: (limited bandwidth)
Background image of page 4
UCSD BE101 Fa06 Lecture 11 Slide 5 Data Acquisition: Pulses (not CW) used to obtain Spatial information: 3 ways (or modes) of Data Acquisition: A-mode (Amplitude Mode) After pulse transmission, transducer used as receiver. Reflected (both specular and scattering) recorded as f(t). Time and depth are equivalent as sound velocity constant in tissue. Simplest form, based on pulse-echo, is A (Amplitude) mode. Detected signal is called RF signal (as MHz range) Gray Scale Imaging
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
UCSD BE101 Fa06 Lecture 11 Slide 6 M-mode (Motion Mode) A-mode signal is constant if repeated on static object and transducer. But if object moves then signal changes and called M (Motion) mode. Create 2D image: amplitude as f( depth, time) Gray Scale Imaging
Background image of page 6
UCSD BE101 Fa06 Lecture 11 Slide 7 M-mode (Motion Mode) Gray Scale Imaging
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
UCSD BE101 Fa06 Lecture 11 Slide 8 B-mode (Brightness Mode) A-mode signal is constant if repeated on static object and transducer. But if transducer moves then signal changes and called B mode.
Background image of page 8
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/30/2010 for the course BENG 101 taught by Professor Silva,g during the Fall '08 term at UCSD.

Page1 / 37

ultrasound ii - BE101 Foundations of Biomedical Imaging...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online