Flashcards in Module 1 & 2 Calculations Deck (26)

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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.

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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

2

## Sketch the wave front and ray methods of illustrating travelling waves

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

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.

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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

4

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

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

wavelength.

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?

### 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.

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).

###
1.54x10^5cm/s 1.54x10^6mm/s

1.54m/ms 1.54x10-3m/us

1.54x10^2cm/ms 0.15cm/us

1.54x10^3mm/ms 1.54mm/us

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?

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

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)?

###
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.

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.

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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.

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

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(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

11

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

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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.

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

###
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

13

##
Express the amplitude ratios below in dB:

(a) Ax/Ao = 1/2

(b) Ax/Ao = 1/10

(c) Ax/Ao = 10

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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

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.

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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.

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.

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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.

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.

###
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

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

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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

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

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λ = 2L c = f λ c = f2L L = c/2f

=4000 / 2x10^6 = 0.002m = 2mm

=4000 / 2x10^7 = 0.0002m = 0.2mm

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?

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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%

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?"

### PRP = 1ms

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?"

###
PRF = 1/PRP

= 1/1 x 10^-3

=1000Hz

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?

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Q = f0/B = f0/(fU - fL)

B = 2MHz f0 = 10MHz fU = 12MHz fL = 8MHz

Q = f0/B = 10/2 = 5

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?

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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

24

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

###
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

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.

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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

26