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

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

** 1.1 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 $1450 m/s$

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

** 1.1.2 ** 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?

**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 $$ Z = /rho \cdot c = 1000kg/m^3 \cdot 1540 m/s = 1.54 cdot 10^6 Rayl $$

** 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.

**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.

** 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?

**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 $$R = \frac{(Z_b - Z_t)^2}{(Z_b + Z_t)^2} = \frac{(6 - 0.8)^2}{(6 + 0.8)^2} = 0.585$$

** 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 $330 m/s$.

**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…

** 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?

**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 Ultrasound resolution ** 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 $500\lambda$.

**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 $$\lambda = \frac{c}{f} \rightarrow f = \frac{c}{\lambda} = \frac{1540 m/s}{5 \cdot 10^{-4}m} = 3.08 MHz$$ 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 $0.5 mm$ is $25 cm$

**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?

**Solution**

$$ Z_{adipose} = \rho \cdot c = 911 \cdot 1540 = 1.402 \cdot 10^6 Rayl$$ $$ Z_{cartilage} = \rho \cdot c = 1100 \cdot 1540 = 1.694 \cdot 10^6 Rayl$$ $$ Z_{bone} = 6.75\cdot 10^6 Rayl $$ $$ R_{adipose|bone} = \frac{(Z_{adipose} - Z_{bone})^2}{(Z_{adipose} + Z_{bone})^2} = 0.430$$ $$ R_{cartilage|bone} = \frac{(Z_{cartilage} - Z_{bone})^2}{(Z_{cartilage} + Z_{bone})^2} = 0.359$$

**HW 1.3 Sampling frequency** [1.0 pt] Let's assume that the penetration depth of ultrasound in human body is $8 cm$, 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 512 rays for each image?

**Solution**
It takes $t = \frac {2 \cdot d}{c} = \frac {0.16} {1540} \approx 0.104 ~ms$ for one ultrasound ray to travel to the maximum penetration depth and return to the transducer.
Collecting $n = 512$ rays takes $t_{512} = n \cdot t = 512 \cdot 0.104~ms \approx 0.053~s$ and we cannot acquire the data for one image frame any faster than that the maximum sampling frequency is $$f_s = \frac {c} {2 \cdot d \cdot n} = \frac {1540} {2 \cdot 0.16 \cdot 512} = 18.80~Hz$$

In this part we will examine the basic properties of US imaging systems. So far it will bre only the properties of the transducer, a more complex simulation of ultrasound with the Field II simulator will follow in Lab 07.

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

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 2.1.2** [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 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 2.1.3** [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 in time and frequency domain, preferably in one graph so that the differences stand out clearly. What are the implication for the US imaging?

courses/zsl/labs2024_05_usthe1.txt · Last modified: 2024/05/21 11:34 by anyzjiri