Ultrasound Physics Flashcards

Impedance

Impedance determines the amount of an incident beam that is reflected back and the amount that is transmitted to the next medium.
$Z = \rho v$ , where:
- $Z$ = impedance
- $\rho$ = density of the medium
- $v$ = sound propagation speed in that medium
Units:
- Rayls (Z)
- Typical values: 1.25 - 1.75 Mrayls
Characteristic of the medium only.
Impedance is calculated, not measured.
Reflection of an ultrasound wave depends upon a difference in the acoustic impedances at the boundary between the two media.
Because almost all the soft tissues have the same velocity, the acoustic impedance values are similar.
Exceptions to the above statement: bone and air.
An interface is a boundary where two tissues come together.
Several interactions occur at the boundaries: absorption, reflection, scattering, diffraction, divergence, and refraction.
All of them originate attenuation, except for Interference which may increase or decrease intensity of the beam.

Perpendicular Incidence

Sound beam strikes the boundary between two media at 90 degrees.
Reflected sound bounces back to the first medium.
Transmitted sound does not change direction.
Intensities of reflected echo and transmitted sound depend on:
- Incident intensity at the boundary
- The impedances of the media involved
Occurs with specular reflectors, where: $\thetai = \thetar$
- $\theta_i$ = Angle of incidence
- $\theta_r$ = Angle of reflection
- Transducer MUST be perpendicular to the reflector interface (0 to <3° deviation).
The primary factor determining the amount of reflected echoes and transmitted ones is the difference in acoustic impedance of the two media.
The larger the difference, the greater the echo reflected back.
An echo is reflected ONLY if the acoustic impedances on each side are different.
If the impedance is exactly the same for both media, all the incident sound is transmitted with no reflection.
Energy transfer is most effective when media have the same impedances.
Perpendicular incidence + same impedances in both media: $Z1 = Z2$
Perpendicular incidence + different impedances: $Z1 \neq Z2$
For Perpendicular Incidence, the energy or echo reflected at a smooth, large soft tissue depends on the difference between the acoustic impedances of the two media on either side of the interface.
Incident sound beam hits the boundary at 90 degrees (also called perpendicular, orthogonal, right angle).
Echoes are reflected at the boundary regardless of the thickness of the media.
The amount of reflection depends on the surface; it is the difference that matters not the value of "Z".

Incident, Reflected, and Transmitted Sound

Incident Intensity:
- The sound's intensity immediately before it hits the interface.
Reflected Intensity:
- The intensity of the reflected echo.
Transmitted Intensity:
- The intensity of the beam that continues forward in the same direction.
Small reflections occur (less than 1% of the incident beam) because almost all the soft tissues have the same velocity, the acoustic impedance values are similar.

Intensity Reflection Coefficient (IRC)

IRC is the fraction of the incident intensity that is reflected.
$IRC = \frac{Reflected Intensity}{Incident Intensity} = \frac{Ir}{Ii}$
In clinical imaging, less than 1% of the incident beam is reflected.
Reflection and Transmission Coefficients must add to 1 (100%).
Example:
- If $Ii = 10 \frac{mW}{cm^2}$ and $Ir = 1 \frac{mW}{cm^2}$ , then:
 - $IRC = \frac{1}{10} = 0.1$
 - 10% of the incident intensity is reflected.

Intensity Transmission Coefficient (ITC)

ITC is the fraction of the incident intensity that is transmitted.
$ITC = \frac{Transmitted Intensity}{Incident Intensity} = \frac{It}{Ii}$
In clinical imaging, most of the incident beam is transmitted.
Example:
- If $Ii = 10 \frac{mW}{cm^2}$ and $It = 9 \frac{mW}{cm^2}$ , then:
 - $ITC = \frac{9}{10} = 0.9$
 - 90% of the incident intensity is transmitted.

Reflection Coefficient and Perpendicular Incidence

$IRC = \frac{Ir}{Ii} = \frac{(Z2 - Z1)^2}{(Z2 + Z1)^2}$
With NORMAL incidence:
- Reflection occurs only if the two media at the boundary have different acoustic impedances.
- The greater the impedance differences between the two media, the greater the IRC and the greater amount of reflection.
$\% Reflection = \frac{(Z2 - Z1)^2}{(Z2 + Z1)^2}$
The IRC depends on the difference between impedances.

Intensity Transmission Coefficient & Intensity Reflection Coefficient

With NORMAL incidence: $ITC = \frac{It}{Ii} = 1 - IRC$
If Intensity Reflection Coefficient increases, Intensity Transmission Coefficient decreases.
If the impedances are equal in both media and the incident intensity is perpendicular, there is no reflected echo, and Transmitted Intensity would be same as the Incident Intensity.

Reflection & Transmission With Oblique Incidence

Extremely complex physics regarding transmission & reflection with obliquity.
With oblique incidence, we are uncertain as to whether reflection will occur.
Incident Intensity = Transmitted + Reflected
Reflection Angle = Incident Angle
Oblique Incidence is very common in ultrasound.
Physics and calculations for Oblique Incidence are much more complicated.
Calculations about IRC and TRC are related to Perpendicular Incidence.
With Oblique Incidence, reflection can occur even if the impedances are the same for both media, and the opposite is true too (reflection may not happen even when the impedances are different).
If reflection occurs, the reflected angle will be equal to the incident angle. And, the reflected and transmitted energy should equal 1 or 100%.

Specular vs Diffuse Reflection

Specular:
- Smooth boundary
- One direction
- Organized
Diffuse (Backscatter):
- Irregular boundary
- More than one direction

Refraction

Refraction is the change of direction of the traveling sound wave as it goes from one medium to another.
The sound beam strikes the interface at an angle different than 90 degrees
Change of direction happens when there is different speed of propagation of sound in the two media.
If the velocities of the two media are the same, no refraction occurs even though the impedances may be different.
Refraction will not occur at perpendicular (normal) incidence, regardless of the relative velocities in the two media.

Snell's Law

The Snell's Law calculates the refracted angle in relation to the angle of incidence and to the speeds of sound in the two media.
$\frac{Sinei}{Sinet} = \frac{v1}{v2}$
- $\theta_i$ is the incident angle
- $\theta_t$ is the transmitted angle
- $v_1$ is the velocity of sound in medium 1
- $v_2$ is the velocity of sound in medium 2
Sine varying from "0" to "1" as angle varies from 0 degrees to 90 degrees.
$\frac{Sinei}{c1} = \frac{Sinet}{c2}$
Because in soft tissues the speed of sound in different tissues are about the same, refraction is NOT a major problem; but, may be source of artifacts.
Refraction is advantageous for imaging ultrasound in focusing the beam (internal focusing using curved crystal).

Critical Angle Reflection/Refraction

This happens when a sound beam strikes a smooth acoustical interface at angles other than perpendicular.
In this case, more of the energy is reflected than transmitted.
Under these circumstances, a critical angle will be reached when the reflected echo travels along the interface.
Two conditions should be present:
- The velocity in medium1 being slower than the velocity in medium2; and,
- The angle of incidence beyond the so called critical angle.
If these two elements are present, then the beam travels along the interface and no energy is transmitted.

Range Equation

When know the time-of-flight, we can determine the distance.
Ultrasound systems measure "time-of-flight" and relate that measurement to distance traveled.
Since the average speed of US in soft tissue (1.54 km/sec) is known, the time-of-flight and distance are directly related.
Time-of-flight: The time needed for a pulse to travel from the transducer to the reflector and back to the transducer.
$distance (mm) = \frac{go-return time (\mu s) \times speed (\frac{mm}{\mu s})}{2}$
In soft tissue: $distance (mm) = time (\mu s) \times 0.77 \frac{mm}{\mu s}$
The purpose of the Range Equation is to determine the distance to a reflector or boundary where the echo was produced.
The system cannot measure this distance directly; instead, it measures the time it takes for an echo to leave the transducer and to come back to it.
$d = vt$
The time measured by the system is for the round-trip.
So, the distance to a reflector can be calculated from the propagation speed and the pulse round-trip travel time.
Then the Range Equation is: $d (mm) = \frac{[v (\frac{mm}{\mu s}) \times t(\mu s)]}{2}$
The round-trip time used in the formula is the time used for the pulse from and to the reflector. The system needs only the distance.
$v = 1.54 \frac{mm}{\mu s}$
$\frac{1}{2}v$ = 0.77
$1 \frac{mm}{\frac{mm}{\mu s}} = 0.65 \frac{\mu s}{mm}$
This means: 1 cm takes 6.5 μs
So, for a 1 cm round trip (2 cm) time is: 13 μs
The pulse round trip travel time is 13μs for each centimeter of distance from source to reflector.

Range Equation Examples

Pulse Round-Trip Travel Times for Various Reflector Depths
- 0. 5 cm: 6.5 μs
- 1 cm: 13 μs
- 2 cm: 26 μs
- 3 cm: 39 μs
If an echo returns 130μs after a pulse was emitted, find the depth of the reflector.
- Using Range Equation: $d (mm) = 0.77 \frac{mm}{\mu s} \times 130 \mu s = 100 mm$
  - d = 10 cm (reflector is located 10 cm from the transducer).
- Using the 13μs rule: $d (cm) = \frac{130 \mu s}{\frac{13 \mu s}{cm}} = 10 cm$
  - Reflector is located 10 cm from the transducer.

PRP & Maximum Imaging Depth

PRP is the time from beginning of one pulse to the beginning of the next one.
When the depth of view is superficial, PRP is SHORT.
When depth is deep, PRP is LONG.
If the depth of view is 20 cm, what is the PRP?
- $PRP (\mu s) = imaging depth(cm) \times \frac{13 \mu s}{cm}$
  - $PRP(\frac{\mu s}{cm}) = 20 cm \times \frac{13 \mu s}{cm}$
  - PRP = 260 μs

PRF & Maximum Imaging Depth

PRF is the number of pulses per second.
When the depth of view is superficial, PRF is HIGH.
When depth is deep, PRF is LOW.
$PRF (Hz) = \frac{77,000 \frac{cm}{s}}{imaging depth (cm)}$
If the depth of imaging is 7.7 cm, then the PRF would be?
- $PRF (Hz) = \frac{77,000 \frac{cm}{s}}{7.7 cm}$
  - PRF (Hz) = 10,000

Axial Resolution

The ability to image accurately (accuracy, not merely quality)
The ability to distinguish two structures that are close to each other front to back, parallel to, or along the beam's main axis.
Synonyms: longitudinal or axial
Units: mm, cm (units of distance)
With shorter pulses, axial resolution is improved.
The shorter the pulse, the smaller the number, the better the picture quality.
No, a new transducer is needed to change axial resolution.
"Short pulse" means a short spatial pulse length or a short pulse duration.
Ultrasound transducers are designed by the manufacturers to have a minimum number of cycles per pulse, so that the numerical value is low and the image quality is superior.
Typical Values: 0.05-0.5mm
Axial Resolution is the minimum reflector separation along the direction of the scan line to produce separate echoes.
The important factor determining axial resolution is the spatial pulse length.
$Axial Resolution = \frac{1}{2} spatial pulse length$
Axial Resolution allows to differentiate between two reflectors close together along the scan line (parallel to the sound beam). It is measure in millimeters.
The smaller the Axial Resolution, the better the detail of the image.
The better the Axial Resolution, the closer two reflectors can be identified as separate echoes.
In order to improve the Axial Resolution we need to shorten the spatial pulse length (increasing the frequency).

Wavelength and Axial Resolution

spl = wavelength \times # cycles in pulse
To improve spl, either wavelength or the number of cycles in the pulse has to be reduced.
Wavelength can be reduced by increasing frequency.
Number of cycles is reduced by damping (already done to a minimum by design).
$Axial Resolution = \frac{1}{2} spatial pulse length$
AR: Typical Values: 0.1 to 1.0 mm
Axial Resolution can be modified only by increasing the frequency.
Since transducer frequency has already been set up, the Sonographer cannot adjust Axial Resolution.
However, the Sonographer can switch to a higher frequency transducer to improve resolution; but, at the same time, this will yield shorter penetration.