Range Equations
Transducers
Sound Beam
Resolution
Display modes
100

This term describes the elapsed time from pulse transmission until the echo returns to the transducer.

What is time of flight, or go-return time?

100

During reception, the PZT converts sound energy into this form of energy.

What is electrical energy?

100

Also called the Fresnel zone, this region extends from the transducer face to the focus.

What is the near zone?

100

This type of resolution distinguishes two reflectors positioned along the sound beam’s axis.

What is axial resolution?

100

This display mode represents returning echoes as spikes whose heights indicate echo amplitude.

What is A-mode?

200

An echo returns after 78 µs. Using the 13-microsecond rule, the reflector is approximately this deep.

What is 6 cm?
78 ÷ 13 = 6 cm.

200

This transducer component absorbs sound behind the active element, reducing ringing and shortening the pulse

What is backing material, or the damping element?

200

At this location, the beam is narrowest and its spatial peak intensity is highest.

What is the focus?

200

This beam characteristic determines lateral resolution at a particular depth.

What is beam width, or beam diameter?

200

In B-mode, brighter dots represent a greater value of this echo characteristic.

What is echo amplitude?

300

When maximum imaging depth doubles, these are the changes in the minimum PRP and maximum PRF.

What are PRP doubles and PRF is halved?

300

These are the approximate thicknesses of the PZT and matching layer, expressed in wavelengths within their respective materials.

What are one-half wavelength for the PZT and one-quarter wavelength for the matching layer?

300

An active element is 18 mm in diameter. The beam is approximately this wide at the focus.

What is 9 mm?

300

A pulse has a spatial pulse length of 3 mm. Its axial resolution is this distance.

What is 1.5 mm?
Axial resolution = SPL ÷ 2.

300

These two quantities appear on the horizontal and vertical axes of an M-mode display, respectively.

What are time and depth?

400

For an imaging depth of 14 cm in soft tissue, the approximate maximum PRF is this value.

What is 5,500 Hz, or 5.5 kHz?
77,000 ÷ 14 = 5,500 Hz.

400

A transducer has a center frequency of 6 MHz and a bandwidth of 4 MHz. Its quality factor is this value.

What is 1.5?
Q = 6 ÷ 4. Q has no units.

400

Two transducers have the same diameter but operate at 3 MHz and 6 MHz. This transducer produces greater far field divergence.

What is the 3 MHz transducer?


Lower frequency produces greater divergence.

400

Transducer A produces 2 cycles per pulse at 3 MHz. Transducer B produces 4 cycles per pulse at 6 MHz. In the same medium, their axial resolutions have this relationship.

What is equal axial resolution?
B has half the wavelength but twice the cycles, giving the same SPL.

400

An A-mode display shows a tall spike near the left side and a short spike farther right. Compared with the second reflector, the first reflector has these two characteristics.

What are shallower depths and a stronger returning echo?

500

An echo has a go-return time of 104 µs. Name both the reflector depth and the total distance traveled by the pulse.

 What are 8 cm deep and 16 cm total travel?
104 ÷ 13 = 8 cm; round-trip distance = 8 × 2.

500

Two pulsed transducers use the same PZT material. Transducer A’s crystal is twice as thick as Transducer B’s. If B operates at 8 MHz, A operates at this frequency.

What is 4 MHz?
Doubling crystal thickness halves the resonant frequency.

500

An unfocused transducer has a near zone length of 12 cm and an 8 cm focal zone centered on the focus. Name the depth where the focal zone begins and the depth where the beam returns to its original diameter.

What are 8 cm and 24 cm?
Focal zone begins at 12 − 4 = 8 cm. Original beam diameter returns at 2 × 12 = 24 cm.

500

A 4 MHz transducer produces 4 cycles per pulse in soft tissue. Using a propagation speed of 1.54 mm/µs, calculate its axial resolution.

What is 0.77 mm?
Wavelength = 1.54 ÷ 4 = 0.385 mm.
SPL = 4 × 0.385 = 1.54 mm.
Axial resolution = 1.54 ÷ 2 = 0.77 mm.

500

A stationary reflector produces a strong echo. Describe how it would appear in A-mode and how its depth would appear over time in M-mode.

What are a tall A-mode spike and a straight horizontal M-mode line?

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