Module 1 & 2 Calculations

Question 1

If the speed of a wave is c = 1540 m/s and the frequency is 5 MHz, show that the wavelength is λ = 0.31 mm (2 significant figures are adequate). Also determine the period of the wave.

Answer 1

c = f λ λ=c/f λ = 1540/5x10^6Hz = 0.31x10^-3m = 0.31mm
Period (T) = 1/f = 1/5x10^6Hz = 0.2x10^-6s

Question 2

Sketch the wave front and ray methods of illustrating travelling waves

Answer 2

Draw the concentric circles with ray arrow and wavelength marked. Sketch Huygens wavelets along a front with circles, wavefront and rays.

Question 3

If the speed of a wave in soft tissue is 1540 m/s what is the wavelength for the following frequencies: 1 MHz, 5 MHz, 10 MHz and 15 MHz? Repeat this calculation for a frequency in the audible range. 1000Hz is in the hearing range.

Answer 3

c = f λ λ = c/f
λ = 1540/1 x 10^6 = 1.54 x 10 ^-3m = 1.54mm
λ = 1540/5 x 10^6 = 0.308 10^-3m = 0.31mm
λ = 1540/10 x 10^6 = 0.154 x 10^-3m = 0.15mm
λ = 1540/15 x 10^6 = 0.102 x 10^-3m = 0.10mm
1000Hz is in the hearing range.
c = f λ λ = c/f
λ = 1540/1000 = 1.54m

Question 4

Use the idea of Huygens’ principle and wavelets to extend the wave crest pattern in Figure 1.4.

Answer 4

Just draw more concentric circles, each circle (crest) increased in radius by one
wavelength.

Question 5

Draw two identical sinusoidal waves (like that in Figure 1.1) one above the other and below sketch the time variation of the sum of these two waves. What is the amplitude and relative intensity of the resultant wave compared to the original waves?

Answer 5

The sum of these two original, identical, waves has an amplitude equal to twice that of each original wave, and an intensity equal to four times that of each original wave.

Question 6

The speed of sound in human soft tissue averages 1540 m/s. Using the powers of ten, re-write this speed using distances in m, cm and mm, and times in s, ms and μs (so 8 ‘new’ expressions).

Answer 6

1. 54x10^5cm/s 1.54x10^6mm/s
2. 54m/ms 1.54x10-3m/us
3. 54x10^2cm/ms 0.15cm/us
4. 54x10^3mm/ms 1.54mm/us

Question 7

1 g of muscle occupies a volume of about 0.95 cm3. What is the density of muscle expressed in kg/m3 and g/cm3?

Answer 7

ρ = m/V ρ = 1/0.95 = 1.05g/cm^3 = 1.05 x 10^-6kg/m^3

Question 8

What, very roughly, is the wavelength of a diagnostic ultrasound wave in soft tissue? What organs/tissues/structures in the human body are significantly greater than this value (a couple of examples)? What organs/tissues/structures in the human body are significantly smaller than this value (a couple of examples)?

Answer 8

c=fλ λ= c/f = 1540/5x10^6 = 0.31x10^-3m = 0.31mm
All major organs, bones, tendons and muscles are larger than this.
Things smaller than this might include blood and very small microbubbles of air.

Question 9

Calculate the elapsed time for the echo detection from a structure at a depth of 5cm, assuming the velocity is constant at 1540 m/s.
How would the calculated depth be affected if the overlying tissues had a high fat composition? Provide quantitative estimates to justify your answer.

Answer 9

D = 154000 x ▲t / 2
2D = 154000 x ▲t
▲t = 2D/154000
= 10/154000
= 0.0000649s
= 64.9us
The calculated depth would appear larger than it is in reality as the speed of sound in fat is slower than it is in most other soft tissues. This means the time elapsed is longer and according to D = 154000 x ▲t / 2 D will appear larger. The speed of sound in fat is 1450m/s rather than the assumed speed of sound in soft tissues which is 1540m/s.

Question 10

Water = 1.49 x 10^6 kg/m^2/s
ST = 1.63 x 10^6kg/m^2/s
Bone = 5.7x10^6
Air = 430kg/m^2/s
Using the Z values provided determine the reflection and transmission coefficients for interfaces between (average) soft tissue and:
(i)	air
(ii)	water
(iii)	bone

Answer 10

(i)	air ( I did this back the front)
Air (Z1) = 430kg/m^2/s
ST (Z2) = 1.63 x 10^6kg/m^2/s
R = Ir/Ii = (Z2 – Z1)2/(Z2 + Z1)2
= (1.63 x 10^6 – 430)^2 / (1.63 x 10^6 + 430)2
= 0.9989
T = It/Ii = 4Z2 Z1 /(Z2 + Z1)2
= 4(1.63 x 10^6) x 430 / 1.63 x 10^6 + 430)^2
= 0.00105

(ii)	water (I did this back the front)
Water (Z1) = 1.49 x 10^6 kg/m^2/s
ST (Z2) = 1.63 x 10^6kg/m^2/s
R = Ir/Ii = (Z2 – Z1)2/(Z2 + Z1)2
= (1.63 x 10^6 - 1.49 x 10^6)^2 / (1.63 x 10^6 + 1.49 x 10^6)^2
=0.0020
T = It/Ii = 4Z2 Z1 /(Z2 + Z1)2
 = 4 x 1.63 x 10^6 x 1.49 x 10^6 / (1.63 x 10^6 + 1.49 x 10^6) ^2
= 0.9979
(iii)	bone
ST (Z1) = 1.63 x 10^6
Bone (Z2) = 5.7x10^6
R = Ir/Ii = (Z2 – Z1)2/(Z2 + Z1)2
= (5.7x10^6 – 1.63 x 10^6)^2 / (5.7x10^6 + 1.63 x 10^6)^2
= 0.308
T = It/Ii = 4Z2 Z1 /(Z2 + Z1)2
 = 4x5.7x10^6x1.63 x 10^6 / (5.7x10^6 + 1.63 x 10^6)^2
= 0.6916

Question 11

The amplitude of a CW decreases by a factor of 3. By how much does the intensity of the wave change?

Answer 11

I = const.A^2 A = 3
I = const.3^2
I = const.9
Const. = I/9
Therefore, I has decreased by a factor of 9.

Question 12

Express the intensity ratios below in dB:

(a) Ix/Io = 1/2
(b) Ix/Io = 1/4
(c) Ix/Io = 1/10
(d) Ix/Io = 1/100
(e) Ix/Io = 4

Answer 12

Using dB = 10 log10 (Ix/Io)

(a) Ix/Io = 1/2 dB = -3
(b) Ix/Io = 1/4 dB = -6
(c) Ix/Io = 1/10 dB = -10
(d) Ix/Io = 1/100 dB = -20
(e) Ix/Io = 4 dB = 6

Question 13

Express the amplitude ratios below in dB:

(a) Ax/Ao = 1/2
(b) Ax/Ao = 1/10
(c) Ax/Ao = 10

Answer 13

dB = 10 log10(Ax/Ao)2 = 20 log10 (Ax/Ao)

(a) Ax/Ao = 1/2 dB = -6
(b) Ax/Ao = 1/10 dB = -20
(c) Ax/Ao = 10 dB = 20

Question 14

Estimate the attenuation loss and the corresponding intensity ratio for a 7MHz sound wave with a maximum depth of field of 4 cm.
Repeat this for a maximum depth of field of 6cm and 8 cm.

Answer 14

Attenuation loss = 0.7 x f x 2D dB f is in MHz and D is in cm
= 0.7 x 7 x 2 x 4 =39.2dB
dB = 10 log10 (Ix/I0)
-39.2 = 10 log Ix/I0 Hence Ix/I0 = inverse log (-39.2/10) = 0.00012 = 1.2 x 10^-4
*must remember dB is negative when calculating in terms of attenuation loss

= 0.7 x 7 x 2 x 6 =58.8dB
dB = 10 log10 (Ix/I0)
-58.8 = 10 log Ix/I0 Hence Ix/I0 = inverse log (-58.8/10) = 0.0000013 = 1.3x10^-6

= 0.7 x 7 x 2 x 8 =78.4dB
dB = 10 log10 (Ix/I0)
-78.4 = 10 log Ix/I0 Hence Ix/I0 = inverse log (-78.4/10) = 0.000000014 = 1.4x10^-8
Here you can see that as the distance increases the intensity ratio decreases very rapidly.

Question 15

Estimate the attenuation loss for a sound wave of frequency 1 MHz with a maximum depth of field of 4 cm.
Repeat this for sound wave frequencies of 5MHz and 10MHz.

Answer 15

Attenuation loss  = 0.7 x f x 2D dB 
= 0.7 x 1 x (2x4) dB
 = 5.6dB
= 0.7 x 5 x (2x4) dB
= 28dB
= 0.7 x 10 x (2x4) dB
= 56dB
This shows that the higher the frequency the higher the attenuation, and the lower the signal received back at the transducer.

Question 16

Estimate the maximum depth of field for a transducer of frequency 2 MHz if the maximum system gain is 85dB.
Repeat this for transducer frequencies of 5 MHz and 10 MHz.

Answer 16

Maximum gain = 0.7 x f x 2 Dmax  
2Dmax = max gain / (0.7f)
Dmax = [max gain/(0.7f)] x ½
Dmax = Max gain/ (2 x 0.7f)
Where Max gain = 85dB and f = 2
Dmax = 85/ (2 x 0.7 x 2)
Dmax = 30.36cm

Question 17

Estimate the total intensity loss for a 7 MHz ultrasound beam producing an echo from a muscle / fat interface at a depth of 3 cm.
(Use the values of acoustic impedance provided in topic 2). Express your answer in dB, and as an intensity ratio.
Air Z = 430 kg/m2/s
Water Z = 1.49 x 106 kg/m2/s
Soft tissue Z = 1.63 x 106 kg/m2/s
Muscle Z = 1.70 x 106 kg/m2/s
Fat Z = 1.41 x 106 kg/m2/s
Bone Z = 5.7 x 106 kg/m2/s

Answer 17

R = Ir/Ii = (Z1 - Z2)^2/(Z1 + Z2)^2
=0.008695
Reflective loss = 10 log (Ir/Ii) = 10 log R
=-20.607dB
Total intensity loss = (0.7 x 2 x D x f) + 10 log R
=-50.007dB
dB = 10log(Ix/Io)
= -50.007/10 = log(Ix/Io)
Ix/Io = Inverse log -5.0007
= 9.8 x 10^-6

Question 18

Calculate the LZT crystal thickness if it is to ring with a natural frequency of 1 MHz. You need cLZT = 4000 m/s. Also calculate the LZT crystal thickness if it is to ring at a natural frequency of 10 MHz

Answer 18

λ = 2L c = f λ c = f2L L = c/2f
=4000 / 2x10^6 = 0.002m = 2mm
=4000 / 2x10^7 = 0.0002m = 0.2mm

Question 19

A voltage pulse produces an ultrasound pulse that lasts for 10 μs. This voltage pulse is repeated every 1 ms. What is the duty cycle?

Answer 19

DF (%) = sound pulse duration (to) / total time (PRP) x 100
DF(%) = (t0/PRP) ´ 100
= 10us / 1ms = 1 x 10^-7 / 1 x 10^-3 x 100 = 0.01%

Question 20

Determine the PRP from “A voltage pulse produces an ultrasound pulse that lasts for 10 μs. This voltage pulse is repeated every 1 ms. What is the duty cycle?”

Answer 20

PRP = 1ms

Question 21

Determine the PRF from “A voltage pulse produces an ultrasound pulse that lasts for 10 μs. This voltage pulse is repeated every 1 ms. What is the duty cycle?”

Answer 21

PRF = 1/PRP
= 1/1 x 10^-3
=1000Hz

Question 22

A LZT transducer produces an ultrasound pulse with a centre frequency of 10 MHz and a bandwidth of 2 MHz. What is the Q-factor and what are the upper and lower half-power frequencies?

Answer 22

Q = f0/B = f0/(fU - fL)
B = 2MHz   f0 = 10MHz  fU = 12MHz fL = 8MHz
Q = f0/B = 10/2 = 5

Question 23

Differentiate between the spatial pulse length and the pulse duration. Provide typical values of these two parameters for a 10 MHz ultrasound pulse, and indicate the acronyms used. How are these two parameters related?

Answer 23

A typical transmit pulse lasts 3-5 wavelengths. The length of these wavelengths combined is the spatial pulse length.
So the pulse duration will be 3-5 times the period of the ultrasound wave. (The period is the duration of a single cycle) The overall time is referred to as the pulse duration.
PD = Period x 3-5 (wavelength)
Period (T) = 1/f = 1/10 000 000 = 0.1us
Assuming 3 wavelengths the PD = 0.3us
c = f λ T = 1/f

Question 24

Explain the relation between echo time and interface depth, and calculate the interface depth if the echo time is 60 μs.

Answer 24

Echo time (Δt) is the total time for a round trip of the ultrasound pulse. By using echo time and the speed of sound in soft tissue the interface depth is able to be calculated.
The pulse will have travelled a total distance, x, with x = 2d, (out and back) so the echo time is
Δt = x/c = 2d/c
d = c Δt/2
= 1540 x 60 x10^-6 /2
= 0.0462m
=4.6cm

Question 25

Show, in fact, that the below statement is true; that for every additional 1 cm depth into the patient, the time delay for the echo is increased by 13 us.

Answer 25

Δt = 2d/c
c = 1540m/s d = 0.01m
2 x 0.01/1540 = 13us
2 x 0.02/1540 = 26us
2 x 0.03/1540 = 39us

Question 26

If there are 40 frames generated and displayed each second, what is the frame time (give your answer in s and ms). Is this frame rate considered real-time?

Answer 26

FR = 1/FT
FT = 1/FR
FT = 1/40
= 0.025s
=25ms
This frame rate is considered real time as it is greater than 20 frames per second.