SPI Review

Question 1

The average rate of attenuation in soft tissue is ______

Answer 1

0.5 dB/cm/MHz

or

0.7 dB/cm/MHz

Question 2

Attenuation

what is half-intensity depth ( or half-value layer thickness)?

Answer 2

This is the point at which the beam is reduced to half its intensity (-3dB)

*The higher the frequency, the less distance the beam can travel before being attenuated

Question 3

Attenuation

Equation for HID (half-intensity depth)

Answer 3

HID = 6/f

*unit: cm

Question 4

Attenuation

medium causing the lowest attenuation ?

vs

medium causing the highest attenuation ?

Answer 4

lowest = water

highest = air

Question 5

What depth has a 10MHz TDR with an intensity of 100mW/cm² experienced 3dB of attenuation?

Answer 5

HID = 6/10 = 0.6 cm

*key word “depth”

ignore intensity

Question 6

attenuation coefficient equation

Answer 6

½f

*note: attenuation coefficient = “toll” TDR has to pay in order to travel inside the medium

Question 7

Attenuation

list the medium in order from the lowest attenuation to the highest.

water, blood, urine, fat soft tissue, muscle, bone, lung, air

Answer 7

Question 8

Increasing the dynamic range range setting affects the ultrasound image by

Answer 8

Decreasing image contrast. As the dynamic range is increased, small changes in echo amplitude result in assignment to different shades of grey

Question 9

When you adjust the TGC wheat component of the ultrasound system implements the changes

Answer 9

Receiver

Question 10

If you increase pulse repetition frequency to a level too great for the depth of field the result will be

Answer 10

Range ambiguity

Question 11

What can improve signal to noise ratio

Answer 11

Frame averaging