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Q.79PE

Expert-verifiedFound in: Page 11

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(a) Echo times are measured by diagnostic ultrasound scanners to determine distances to reflecting surfaces in a patient. What is the difference in echo times for tissues that are 3.50 and 3.60cm beneath the surface? (This difference is the minimum resolving time for the scanner to see details as small as 0.100cm, or 1.00mm. Discrimination of smaller time differences is needed to see smaller details.) (b) Discuss whether the period T of this ultrasound must be smaller than the minimum time resolution. If so, what is the minimum frequency of the ultrasound, and is that out of the normal range for a diagnostic ultrasound?**

(a) Difference in echo times will be \[1.29 \times {10^{ - 6}}{\rm{ }}s\]

(b) Minimum frequency will be 0.77MHz and this is out of range of normal range of diagnostic ultrasound.

**A simple speed formula can be given as,**

**\[v = \frac{{\Delta d}}{{\Delta t}}\] ….. (1)**

**Here, v is the speed of sound, \[\Delta d\] which is the distance between the transmission and reflecting surface which is called depth.**

**And \[\Delta t\] represents the time when ultrasound waves just fall on the reflecting surface of the tissue.**

Calculate the difference in echo times that is equivalent to a round trip of the sound wave.

\[\Delta t = 2 \times \left( {\frac{{\Delta d}}{{\Delta v}}} \right)\] ….. (2)

Consider the given data as below.

The speed of sound in tissue, \[v = 1540{\rm{ }}{m \mathord{\left/

{\vphantom {m s}} \right.

\kern-\nulldelimiterspace} s}\]

And the depth is,

\begin{aligned}\Delta d &= (3.6 - 3.5){\rm{ }}cm\\ &= 0.1{\rm{ }}cm\\ &= {10^{ - 3}}{\rm{ }}m \end{aligned}

Substitute known values into equation (2).

\begin{aligned}\Delta t &= 2 \times \left( {\frac{{{{10}^{ - 3}}}}{{1540}}} \right)\\ &= 1.29 \times {10^{ - 6}}{\rm{ }}s\end{aligned}

Hence, a time for echo difference is \[1.29 \times {10^{ - 6}}{\rm{ }}s\].

The wavelength of ultrasound must be larger than that smaller size (0.1cm) then time period is related to wavelength then it will smaller than the minimum time resolution.

Frequency is given by,

\[{f_{\min }} = \frac{1}{{\Delta t}}\] ….. (3)

Here, time is \[\Delta t = 1.29 \times {10^{ - 6}}{\rm{ }}s\]

Substitute this value into equation (3).

\begin{aligned}f &= \frac{1}{{1.29 \times {{10}^{ - 6}}{\rm{ }}s}}\\ &= 0.77{\rm{ }}MHz\end{aligned}

** **

The range of accepted frequencies for diagnostic ultrasound is 2 – 8 MHz.

And 0.77MHz is not in this range so it will be out of the normal range of the diagnostic probe.

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