chapter 7 - physics

The speed of sound in soft tissue is _______.

1,540 m/s

True or false: The time-of-flight is inversely related to how deep a sound pulse travels.

False; The time-of-flight is directly related to how deep a sound pulse travels. Greater distances prolong the time-of-flight ; lesser distances shorten the time-of-flight.

The 13-microsecond rule

This rule always applies when sound travels through soft tissue. For every 13 µs of go-return time the object creating the reflection is 1 cm deeper in the

Range equation

To create an anatomic image, a sound pulse must travel to a reflector located in the body and return to the transducer. The travel time of this journey allows us to accurately position the reflector.

True or False: When a reflector is located superficially, the time from pulse creation to reception is brief because it isn't very far. Whereas trips to and from deeper reflectors require more time. This is the basis for measuring distances in diagnostic imaging.

True

Go-return time

Ultrasound system's computers are programmed with the average speed of sound in soft tissue. Since the average speed of soft tissue Is known, the time-of-flight and the distance that the pulse travels are directly related.

Time-of-flight or go-return time

Ultrasound systems determine reflector depth by measuring a pulse's time-of-flight with a very accurate stopwatch. is the time needed for a pulse to travel to and from the transducer and the reflector.

Depth (mm) = 1.54mm/µs x go-return-time (µs) / 2

When a reflector is twice as deep as another reflector, the pulse's time-of-flight is doubled for the deeper reflector. In other words, the time of flight will be increased by a factor of 2.

A pulse's time-of-flight is 13µs when a reflector is ______ deep.

1cm deep

A pulse's time-of-flight is 26µs when a reflector is ______ deep.

2cm deep.

PRF (Hz) =

77,000 cm/s / Imaging Depth (cm)

A sound wave is created by a transducer, reflects off an object, and returns to the transducer. The go-return time is 26 µs. What is the depth of the reflector? A.)1cm B.) 2cm C.) 3cm D.) 4cm

B - the reflector depth is 2cm! (This is just TO the reflector!) (Because for 13 microseconds, the object creating the reflection is 1cm deeper in the body!)

A sound wave is created by a transducer, reflects off an object, and returns to the transducer. The go-return time is 26 µs. What is the total distance of the reflector? A.)1cm B.) 2cm C.) 3cm D.) 4cm

D - the total distance traveled (to the reflector and back) is TWICE the depth of the reflector.

The maximum imaging depth (depth of view) during an ultrasound exam is 10cm. The sonographer exam is 10cm. The sonographer adjusts the imaging depth to 20cm. What happens to our PRP? a.) It is unchanged b.) It is halved c.) It is doubled d.) It is 20 times longer.

First off, lets think about our relationship. How are depth of view & PRP related? Directly. So with our depth of view doubling, our PRP is going to double!

A sound wave is created by a transducer, reflects off an object and returns to the transducer. The depth of the reflector is 10cm in soft tissue. What is the go-return time? a.) 13 µs b.) 1.3µs c.) 65µs d.) 130µs

Our problem is giving us the distance, 10cm. We know for every 13 microseconds, the object creating the reflection is 1cm deeper in the body! So we multiply 10 x 13 to get 130!

When the depth of view is set to 7.7cm, PRF is ______.

PRF is 10,000Hz (77,000/7.7 = 10,000)

When the depth of view is set to 10cm, PRP is _____.

PRP is 130 µs (10 x 13µs)

Time-of-flight

This is the elapsed time from pulse creation to pulse reception.

Prp & maximum imaging depth

When we adjust the system's maximum imaging depth, the PRP is altered. When the depth of view is shallow remember, our PRP is short. When the depth of view is deep we have a longer PRP.

PRP (µs) =

imaging depth x 13µs/cm