Ultrasound: Introduction to B-mode scanning Flashcards
If the distance to a target in a tissue mimicking phantom is 10 cm, how long does it take to receive an echo from that target?
130 us
If a B-mode image has a depth of 10 cm and is made up of 100 lines, what is the maximum achievable frame rate?
About 80 Hz
In which clinical application is A-mode widely used?
ophthalmology
In which clinical application is M-mode widely used?
cardiac
Which parameters are displayed along the vertical and horizontal axes of an M-mode trace?
Depth and ‘physiological time’, measured in seconds
Which image formats are used by phased and curved array transducers, respectively?
Sector and fan-shaped
What is the typical duration of an ultrasonic pulse used for B-mode imaging?
About two cycles
Which component of an ultrasound scanner limits its dynamic range most severely?
Digital video display
Typically, how many elements do modern 1D linear array transducers have?
Between 192 and 256
Typically, how wide are the elements in modern 1D linear array transducers?
Between 0.15 and 0.20 mm
Which parameters determine the axial and lateral resolution of an ultrasound scanner, respectively?
Half the pulse duration and the beam width in the scan plane
An L8–4 transducer has three frequency settings, called ‘gen’, ‘res’ and ‘pen’. What respective transmit frequencies might these correspond to?
6, 7 and 5 MHz
If the voltage amplitude of the largest echoes in a signal is 1 V and the dynamic range of the signal is 100 dB, calculate the the voltage amplitude of the smallest echoes. (Answer in μV)
10
The dynamic range is 100 = 5 × 20 dB, so the voltage amplitude of the smallest signals will be 105 times smaller, i.e. 1/105 V or 10 μV
The smallest echoes are likely to come from blood or other low-scatter liquids inside the body.
If a tissue-mimicking phantom has an ‘attenuation’ of 0.8 dB cm−1 MHz−1, what is the amplitude back at the surface of a 4 MHz plane wave that has been reflected from a plane interface at a depth of 10 cm? Assume that the initial amplitude of the wave is 1 MPa, and that the amplitude reflection coefficient of the interface is 20 %. (Answer in Pa)
125
The attenuation loss, 2afbz = 2 × 0.8 × 41.0 × 10 = 64 dB = (2 × 20) + (4 × 6) dB
Now, 20 dB corresponds to an acoustic pressure ratio of 10, and 6 dB corresponds to an acoustic pressure ratio of 2, so that 64 dB corresponds to an acoustic pressure ratio of 102 × 24 = 1600 (i.e. 402)
In addition, the amplitude reflection coefficient at the interface is 0.2, so the final acoustic pressure ratio is 1600 × 5 = 8000
Hence, the amplitude of the wave is p = 1.0/8000 MPa = 125 Pa