• [3 pt] US exercises Solve denoted exercises.
• [2 pt] US simulation Simulating properties of ultrasound element excitation pulse

Useful 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 $R = \frac{(Z_1 - Z_2)^2}{(Z_1 + Z_2)^2}$

1.1 Signal propagation

1.1.1 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 $1450 m/s$. [0.09425 m]

1.1.2 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? [$1.54 \cdot 10^6 Rayl$]

1.2 Reflectivity

1.2.1 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. [$R_{tr-air}=0.9999$; $R_{tr-gel}=0.8228$]

1.2.2 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? [0.585]

1.3 Miscellaneous

1.3.1 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 $330 m/s$. [axial resolution 0.15m]

1.3.2 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? [0.0045s]

HW 1.1 Ultrasound resolution [1.0 pt] Calculate the minimum frequency of ultrasound that will allow you to see details as small as $0.5 mm$ in human tissue. What is the effective depth to which this sound is effective as a diagnostic probe? Hint: whenever a wave is used as a probe, it is very difficult to detect details smaller than its wavelength $\lambda$ and there is a rule of thumb for effective ultrasound depth of penetration which is $500\lambda$.

HW 1.2 Reflectivity II [1.0 pt] Consider two types of tissue, 1) adipose tissue that has specific weight $911\;\text{kg} \cdot m^{-3}$ and 2) cartilage that has specific weight $1100\;\text{kg} \cdot m^{-3}$, ultrasound propagates through both tissues 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 1) adipose tissue – bone boundary and 2) cartilage – bone boundary?

HW 1.3 Sampling frequency [1.0 pt] Let's assume that the penetration depth of ultrasound in human body is $10 cm$, 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?

In this part we will examine the basic properties of US imaging systems. So far it will be only the properties of the transducer, a more complex simulation of ultrasound with the Field II simulator will follow in the Ultrasound simulation lab.

The impulse response of the transducer is the the function that simulates the effect of exciting the piezoelectric element in the transducer by for example Dirac impulse, which causes it to oscillate for a few periods. This oscillations are transferred to the environment (the examined tissue), propagate through the environment as an ultrasound wave, get reflected, etc.

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:

• Sampling frequency 100 MHz
• 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)

Task 2.1 Visualize the impulse response in time domain and in frequency domain.

fs = ; %TODO sampling frequency
fc = ; % TODO central frequency
bw = ; % TODO fractional bandwidth
bwr = ; % TODO fractional bandwidth reference level
tpe = ; %  TODO drop in trailing pulse envelope for cutoff time estimation
tc = gauspuls('cutoff',fc,bw,bwr,tpe);
t = - tc:1/fs:tc;
y = gauspuls(t,fc,bw);
 
figure()
subplot(2,1,1)
plot(t,y)
title({'Impulse response of an piezoelectric element','in time domain (top) and in frequency domain (bottom)'})
xlabel('t [s]')
axis tight
subplot(2,1,2)
f = linspace(-fs/2,fs/2,512);
plot(f,fftshift(abs(fft([y zeros(1,512 - length(y))]))))
xlabel('f [Hz]')

HW 2.2 [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 in time and frequency domain, preferably in one graph so that the differences stand out clearly. What are the implication for the US imaging?

HW 2.3 [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 in time and frequency domain, preferably in one graph so that the differences stand out clearly. What are the implication for the US imaging?