24.1 Production and use of ultrasound
- Syllabus
- 9702–2028–2029
- Topic
- 24.1
- Level
- A2
Learning objectives
24.1.1A piezo-electric crystal changes shape when a p.d. is applied across it and• understand that a piezo-electric crystal changes shape when a p.d. is applied across it and that the crystal generates an e.m.f. when its shape changes24.1.2Ultrasound waves are generated and detected by a piezoelectric transducer• understand how ultrasound waves are generated and detected by a piezoelectric transducer24.1.3The reflection of pulses of ultrasound at boundaries between tissues can be• understand how the reflection of pulses of ultrasound at boundaries between tissues can be used to obtain diagnostic information about internal structures24.1.4The specific acoustic impedance of a medium as Z = ρc• define the specific acoustic impedance of a medium as Z = ρc, where c is the speed of sound in the medium24.1.5IR / I0 = (Z1 – Z2)2 / (Z1 + Z2)2 for the intensity reflection coefficient• use IR / I0 = (Z1 – Z2)2 / (Z1 + Z2)2 for the intensity reflection coefficient of a boundary between two media24.1.6I = I0e–µx for the attenuation of ultrasound in matter• recall and use I = I0e–µx for the attenuation of ultrasound in matter
A piezoelectric crystal converts energy in both directions
Apply a p.d. across a piezoelectric crystal and it changes shape. Reversing the p.d. reverses the deformation.
Mechanically compress, stretch or vibrate the crystal and charge separation produces an e.m.f. across it.
An alternating p.d. therefore drives alternating deformation; an incoming mechanical vibration produces an alternating electrical signal.
A steady p.d. gives a static deformation, not sustained ultrasound. Generation and detection are reverse conversions in the same material.
A piezoelectric transducer alternates between ultrasound transmitter and receiver
Generation: apply an alternating p.d. to the crystal. It repeatedly changes shape and drives pressure oscillations in the surrounding medium.
Choose the drive frequency equal to the crystal's natural frequency in the ultrasound range (>20 kHz). Resonance gives large-amplitude vibration and a stronger ultrasound pulse.
Detection: a returning ultrasound wave makes the crystal vibrate/change shape; the inverse piezoelectric effect produces an alternating e.m.f. that is amplified and processed.
The transducer does not emit and receive simultaneously in pulse-echo imaging: electronics switch from the drive pulse to listening for the much smaller echo signal.
Pulse-echo ultrasound maps boundary depth and tissue contrast
The probe sends a short ultrasound pulse, then listens. At a boundary between tissues with different acoustic impedances, part of the pulse reflects and returns as an echo.
boundarydepthd=ct/2wheretisthetransmit−to−echotimeandcissoundspeedinthetissue.
For c=1540 m s⁻¹ and echo delay t=80 μs, d=(1540)(80×10⁻⁶)/2=6.16×10⁻² m=6.16 cm.
Echo timing locates boundaries; echo intensity gives information about impedance contrast and therefore boundary type. Repeating along many directions builds a cross-sectional image.
Coupling gel removes the air gap at skin, reducing the severe reflection that an air-tissue impedance mismatch would cause.
Divide by two because the pulse travels to the boundary and back. Pulses create listening intervals so the same probe can distinguish weak echoes from its transmitted signal.
Specific acoustic impedance is density × sound speed
Z=ρcρ=densityofthemedium;c=speedofsoundinthatmedium
With ρ in kg m⁻³ and c in m s⁻¹, Z has unit kg m⁻² s⁻¹.
For tissue with ρ=1060 kg m⁻³ and c=1540 m s⁻¹, Z=(1060)(1540)=1.63×10⁶ kg m⁻² s⁻¹.
It is the contrast between Z values on the two sides of a boundary—not either density alone—that determines the reflected intensity fraction.
Acoustic impedance is not electrical resistance. Use the sound speed and density for the same medium and keep prefix powers consistent.
Acoustic-impedance mismatch sets the reflected intensity fraction
R=IR/I0=((Z1−Z2)/(Z1+Z2))2
If Z₁=Z₂, R=0 and ideally no intensity reflects. If the impedances are very different, R approaches 1 and little intensity transmits.
For water Z₁=1.48×10⁶ and steel Z₂=40.4×10⁶ kg m⁻² s⁻¹, R=[(40.4−1.48)/(40.4+1.48)]²=0.864: about 86.4% reflects.
Attheboundary,neglectingabsorption:T=IT/I0=1−R
R is a dimensionless intensity fraction and the ratio is squared. Multiply by 100 only for a percentage; transmitted percentage is 100(1−R) when boundary absorption is neglected.
Ultrasound intensity attenuates exponentially with distance in matter
I=I0e(−μx)I/I0=e(−μx);μ=−ln(I/I0)/x
I₀ is intensity before a path length x, I is intensity after it, and μ is the linear attenuation coefficient. μ has reciprocal-length units matching x.
If I=0.62I₀ after x=2.1 cm, μ=−ln(0.62)/2.1=0.23 cm⁻¹.
The transmitted fraction is I/I₀; the fraction attenuated is 1−I/I₀. For μ=0.053 cm⁻¹ and x=9.3 cm, 39% is attenuated.
For an echo from depth d in one tissue, propagation covers approximately x=2d before adding boundary-reflection factors; each tissue layer contributes its own μx.
Keep x and μ in reciprocal units. Attenuation through matter and reflection at boundaries are separate losses and may both reduce the detected echo.