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Lab 11 : Ultrasound theory


  • [1 pt] US exercises Solve denoted exercises.
  • [4 pts] US simulation Simulating various properties of ultrasound with Field II.

US exercises

Used formulas:

  • relationship between the wavelength and frequency in a matter $\lambda = \frac{c}{f}$
  • propagation of acoustic signal in a matter $c = \frac{1}{\sqrt{\rho K}}$ ($K$ is compressibility, the inverse of Young's modulus of elasticity $E$)
  • acoustic impedance $Z_t = \rho c$
  • intensity reflection coefficient on a boundary $R = \frac{(Z_1 - Z_2)^2}{(Z_1 + Z_2)^2}$

1. Signal propagation

1.1 The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound travelling through a piece of fat tissue was $0.13 ms$. At what depth did this reflection occur? The speed of propagation of ultrasound waves in fat is $1450 m/s$

Solution The wave travels $2 \cdot d$ in $0.13 ms$, therefore $$d = \frac {c_{fat} \cdot t_d} {2} = \frac{1450 m/s \cdot 0.00013 s} {2} = 0.09425 m$$

1.2 Calculate the minimum frequency of ultrasound that will allow you to see details as small as $0.250 mm$ in human tissue. What is the effective depth to which this sound is effective as a diagnostic probe?

Solution Whenever a wave is used as a probe, it is very difficult to detect details smaller than its wavelength $\lambda$. The relationship between wavelength and the frequency is following $$\lambda = \frac{c}{f} -> f = \frac{c}{\lambda} = \frac{1540 m/s}{2.5 \cdot 10^{-4}m} = 6.16 MHz$$ There is a rule of thumb for effective ultrasound depth of penetration which is $500\lambda$, therefore the effective depth of penetration for ultrasound waves with wavelength $0.25 mm$ is $12.5 cm$

1.3 The acoustic (ultrasound) waves propagate with a speed $1540 m/s$ through a matter with a specific weight $1000 kg \cdot m^{-3}$. Determine the Young's modulus of elasticity. What is the specific acoustic impedance of the matter?

Solution Young's modulus of elasticity $E$ is the inverse of compressibility $E = \frac{1}{K}$. From the formula for the speed of signal propagation we obtain $$ c = \frac{1}{\sqrt{\rho K}} \quad \to \quad K = \frac{1}{c^2 \rho} \quad \text{therefore} \quad E = c^2 \rho = 1540^2 * 1000 = 2.37 \cdot 10^9 Pa = 2.37 GPa $$ The acoustic impedance is $$ Z = /rho \cdot c = 1000kg/m^3 \cdot 1540 m/s = 1.54 cdot 10^6 Rayl $$

2. Reflectivity

2.1 In the clinical use of ultrasound, transducers are always coupled to the skin by a thin layer of gel or oil, replacing the air that would otherwise exist between the transducer and the skin. Using the values of acoustic impedance for transducer material $30.8 \cdot 10^6 \frac{kg}{m^2 \cdot s}$, air $429 \frac{kg}{m^2 \cdot s}$ and water $1.5 \cdot 10^6 \frac{kg}{m^2 \cdot s}$. Calculate the intensity reflection coefficient between transducer material and air. Calculate the intensity reflection coefficient between transducer material and gel (assuming for this problem that its acoustic impedance is identical to that of water). Based on the results of your calculations, explain why the gel is used.

Solution The intensity reflection coefficient transducer–air $$R_{tr - air} = \frac{(30.8 \cdot 10^6 - 429)^2}{(30.8 \cdot 10^6 + 429)^2}=0.9999$$ The intensity reflection coefficient transducer–water $$R_{tr - gel} = \frac{(30.8 \cdot 10^6 - 1.5 \cdot 10^6)^2}{(30.8 \cdot 10^6 + 1.5 \cdot 10^6)^2}=0.8228$$ In the case of transducer to air boundary, almost all the ultrasound energy is reflected back, in the case of transducer to gel boundary, the intensity reflection coefficient is not great, but not terrible. In actual use, the gel acoustic impedance would be different than the acoustic impedance of water.

2.2 A soft tissue has the acoustic impedance $800\;\text{kRayl}$ and and a bone $6\;\text{MRayl}$. What is the proportion of reflected energy of the ultrasound wave reflected from the boundary of the soft tissue and the bone?

Solution For simplicity we will consider total reflection ( the wave propagates perpendicularly to the interface). We know the impedance: $Z_t = 0.8 \cdot 10^6 Rayl, Z_b = 6 \cdot 10^6 Rayl$. The coefficient of reflectivity (ratio of the energy of incoming and reflected waves) is therefore $$R = \frac{(Z_b - Z_t)^2}{(Z_b + Z_t)^2} = \frac{(6 - 0.8)^2}{(6 + 0.8)^2} = 0.585$$

3. Miscellaneous

3.1 A bat uses ultrasound to find its way among trees. If this bat can detect echoes $1.00 ms$ apart, what minimum distance between objects can it detect? Could this distance explain the difficulty that bats have finding an open door when they accidentally get into a house? Speed of ultrasound in air is $330 m/s$.

Solution The ability detecting echoes $1ms$ apart implies wavelength $\lambda = \frac{ c \cdot t}{2} = \frac{330 m/s \cdot 10^{-3}s}{2} = 0.15 m$. However this is axial resolution (alongside the ultrasound beam). It would apply if the bat for example circled in the room. The lateral resolution which seems to be more reasonable for this exercise is proportional to the ultrasound frequency, focusing and in medical ultrasound devices to the size of the transducer and distance of the object from the transducer. But who knows how bats are doing these things…

3.2 A dolphin is able to tell in the dark that the ultrasound echoes received from two sharks come from two different objects only if the sharks are separated by $3.50 m$, one being that much farther away than the other. If the ultrasound has a frequency of $100 kHz$, show this ability is not limited by its wavelength. If this ability is due to the dolphin’s ability to detect the arrival times of echoes, what is the minimum time difference the dolphin can perceive?

Solution The wavelength of ultrasound with frequency $100 kHz$ in water, where the waves travel with speed $1540 m/s$, is $ \lambda = \frac{v}{f} = \frac{1540 m/s}{10^5 1/s} = 0.0154m < 3.5m$ So the inability distinguish sharks less than $3.5m$ is not due to the parameters of the dolphin's sonar. The $3.5m$ length travels the wave in $t = \frac{2d}{v} = \frac{2\cdot3.5m}{1540 m/s} = 0.0045s$

HW 1.1 Reflectivity II [0.5 pt] A tissue has specific weight $900\;\text{kg} \cdot m^{-3}$, ultrasound propagates through the tissue with the speed $1540 m \cdot s^{-1}$ and a bone with acoustic impedance $6.75 \cdot 10^6 Rayl$. What is the reflection coefficient of the tissue – bone boundary?

HW 1.2 Sampling frequency [0.5 pt] Let's assume that the penetration depth of ultrasound in human body is $20 cm$, speed of ultrasound is $1540m\cdot s^{-1}$. What is the highest possible sampling frequency (frame rate) we can use for ultrasound imaging if we need 200 rays for each image?

2. US simulation

In this part we will examine the basic properties of US imaging systems using the Field II simulator. The Field II simulator is an environment for the numeric simulation of ultrasound signals. The simulator is capable of numerically simulating emitted and received acoustic fields for a broad range of acoustic transducers. With the simulator it is for example possible to acquire data and reconstruct images similarly to a conventional ultrasound imaging device. However we will only cover simple properties of the acoustic signals and fields generated by an array of transducers. The lab consists of a few steps:

  1. Setting up the acoustic environment
  2. Modelling the impulse response of the transducers
  3. Visualizing the acoustic field


Field II is distributed as a bundle of Matlab functions for various operating system (Win, Unix, …). To use the simulator , do the following:

  1. Download the appropriate version of the Field II simulator for your operating system ( Field II for Windows, Field II for Linux, Field II for Mac OS ) or from project webpage Field II download.
  2. Extract the archive into an appropriate project folder.
  3. Start Matlab and add the Field II simulator to the search path using addpath('path_to_fieldii')
  4. Start the simulator field_init;, simulate signals and fields and then end the simulator field_end;

Description of some functions from the Field II simulator

This file contains the description of several functions. Complete description of the simulator Field II is in the user's guide. Alternatively, instead of writing your simulation from scratch, you can use this prepared m-file.

2.1 Acoustic environment

We set up the acoustic system wit the following parameters:

  • Speed of propagation in the environment 1540 m/s
  • Sampling frequency 100 MHz
  • Transducer – transmitter (function xdc_linear_array)
    • Number of elements 128
    • Space between elements 0.02 mm
    • Width of the elements 0.2 mm
    • Height of the elements 5 mm
    • Coordinates of the fixed focus [0 0 30] mm
    • Subdividion of elements in x 1
    • Subdivision of elements in y 4

2.2 Impulse response

The function xdc_impulse modifies the impulse response of the transducer. We will use the impulse response in shape of Gaussian-modulated sinusoidal pulse. Use the gauspuls function for the generation of the impulse response.

tc = gauspuls('cutoff',fc,bw,bwr,tpe)

In this mode the function returns the cutoff time tc at which the trailing pulse envelope falls below tpe dB with respect to the peak envelope amplitude. After estimating an appropriate values for the cutoff time we obtain the impulse response in the form of Gaussian-modulated sinusoidal pulse with a following call to the gauspuls function.

yi = gauspuls(t,fc,bw,bwr)

For our simulation, we will use the following parameters of the impulse response:

  • Central frequency (fc) 7.5 MHz
  • Fractional bandwidth (bw) 50% with the reference level (bwr) at -6 dB
  • The length of the impulse response will be such that the cutoff time will when the amplitude drops by 40 dB (tpe)
  • The impulse response has generated in time interval from - cutoff time to + cutoff time with correct time steps (sampling frequency)

Let the excitation of the transducer be the Dirac impulse.

Task 2.1.1 Visualize the impulse response in time domain and in frequency domain.

HW 2.1.2 [1 pt] Examine the effect of bandwidth on the impulse response of the transducer. Compare the time domain and frequency domain properties of the original impulse response to the impulse response with lower and greater bandwidth (for example 25% and 75%). Visualize the different impulse responses. What are the implication for the US imaging?

HW 2.1.3 [1 pt] Examine the effect of central frequency on the impulse response of the transducer. Compare the time domain and frequency domain properties of the original impulse response to the impulse response with lower and greater central frequency (for example 5 MHz and 10 MHz). Visualize the different impulse responses. What are the implication for the US imaging?

Acoustic field

When excited by the excitation signal the piezoelectric elements start to oscillate and the transducer emits an acoustic pulse into the environment. The Field II simulator simulates the pressure field for the given transducer and excitation signal. In the simulation we can measure the time dependent changes in pressure at a certain point in the environment.

The emitter which we set up in the acoustic system part emits an ultrasound ray alongside the z-axis. The ray is focused to the distance 30 mm.

Simulate the time dependent changes on a line of points evenly distributed from -1 mm to 1 mm with the step 0.02 mm that runs parallel to the x-axis and goes through the fixed focus (x = -1:0.02:1, y = 0, z = 30 mm). Use function calculate_hp to calculate the pressure field (see the help page for the format of the coordinates).

Task 2.2.1 Visualize the time dependent changes in the simulated pressure field.

HW 2.2.2 [1 pt] Repeat the pressure field simulation, but measure the pressure changes also on lines 25 mm and 35 mm from the transducer. Visualize the maximum values of pressure in dB (20*log10(Pm/max(Pm))) in each point along the three lines. Compare the shape of the normalized pressure maxima, notice especially the side lobes. How do the side lobes affect the US imaging (the lateral resolution)?

HW 2.2.3 [1 pt] Examine the effect of the central frequency on the side lobes. Repeat the steps 2.2.1 and 2.2.2 for two different central frequencies (for example 5 MHz and 10 MHz) and compare the results of the different central frequencies. Discuss the results and their practical implications.

BONUS 2.2.4 [1 pt] Try to eliminate the side lobes by apodization (function xdc_apodization, choose the profil hanning or blackmann), visualize the shape of the normalized pressure similarly to task 2.2.2, discuss the effect of apodization on the lateral resolution.

courses/zsl/labs2020_07_ultrasound.txt · Last modified: 2022/05/13 10:05 by anyzjiri