K323 / 2027
General properties of waves overview

Topic 6 of 6

Using ultrasound

Ultrasound is sound above the usual upper limit of human hearing, conventionally above about 20 kHz. Its echoes can locate reflecting boundaries.

1 kHz = 1000 Hz, so 20 kHz is 20000 Hz. Ultrasound is still a mechanical sound wave and needs a material medium. The high frequency does not make it electromagnetic radiation.

Underwater echo sounding

An active echo-sounding system can emit a short ultrasound pulse into water. Its transducer converts an electrical signal into sound and converts a received echo back into an electrical signal. The time between emission and reception indicates the distance along the beam when the sound speed is known.

For a vertical beam reflected from the seabed, the pulse travels down from the transducer and back up. Using an approximately uniform water sound speed, depth below the transducer = vt/2.

Sonar calculation

A vertical return after 0.060 s

Use the supplied speed 1500 m/s. The round-trip interval is 0.060 s.

d = 1500 x 0.060 / 2 = 45 m.

This is the seabed distance below the transducer. It is not automatically the depth below the sea surface if the transducer is submerged. A sloping beam would measure distance along its path rather than the vertical depth.

This example uses an active sonar system with ultrasound. Sonar more broadly uses sound: some systems listen without emitting a ranging pulse, and not every sonar frequency is ultrasonic.

Use reflected pulses to locate a boundary

Active echo sounding with ultrasound

Supplied sound speed 1500 m/s; return delay 0.060 s. A vertical beam locates the seabed 45 m below the transducer.

A sonar pulse travels down from the transducer and returns from the seabedA submerged transducer sends an ultrasound pulse vertically down to a level seabed. A second upward arrow shows the returning echo. The paths are offset to distinguish directions. The supplied 0.060 second delay and uniform speed of 1500 metres per second give a one-way distance of 45 metres. A depth bracket starts at the transducer face, below the water surface, and ends at the seabed. The drawing is schematic and does not assign a depth below the water surface.Water surfaceTransducerOutBack45 mSeabed

A probe receives echoes from tissue boundaries

This supplied uniform-speed model uses 1540 m/s along the path. Depths are measured from the probe face; the coupling layer is included in this simplified path.

A probe sends ultrasound through gel and receives separate echoes from nearer and farther boundariesA probe face contacts a thin gel layer above the skin. It emits a pulse downwards. Part reflects from boundary A, 3.08 centimetres from the probe face; part continues and reflects from boundary B, 6.16 centimetres away. Upward arrows show separate echoes from the two boundaries. The directions are separated sideways for clarity, not literal bent sound paths. In the supplied uniform-speed model, A returns an echo after 40 microseconds and B after 80 microseconds. Their depths have a two-to-one ratio. Many beam measurements, using echo times and strengths, are needed to build an image.ProbeGelSkinAB6.16cmDown: pulse. Up: returning echoes.

A: 3.08 cm, echo after 40 microseconds. B: 6.16 cm, echo after 80 microseconds.

Echo times distinguish the two boundariesA schematic pulse trace has time in microseconds on its horizontal axis and signal in arbitrary units on its vertical axis. The emission reference is at zero. Echo A is at 40 microseconds and echo B at 80 microseconds. The later echo corresponds to the farther boundary under the stated uniform-speed model. Pulse heights and widths are schematic and do not supply a numerical reflectivity, tissue identity or depth scale.Signal / arbitrary unitsEmittedAB020406080100Time / microsecondsPulse heights are schematic
The sonar model measures a return from the seabed. The soft-tissue model receives separate echoes from two boundaries. Under its uniform-speed assumption, the later echo comes from the farther boundary. Echo heights are schematic and do not measure depth or identify tissue.

Locate soft-tissue boundaries

A scanning probe sends ultrasound pulses into the body and receives returning echoes. At a boundary between tissues, some of a pulse can reflect while some continues farther. Later echoes can therefore return from deeper boundaries.

  1. Emit a pulse: record its emission time.
  2. Receive echoes: identify the delays of the returned signals.
  3. Estimate depth: combine each round-trip delay with an appropriate assumed sound speed.
  4. Build an image: combine information from many beam positions. Echo strength as well as timing contributes to the displayed image.

One echo does not give a complete photograph of an organ. The system assembles information from many reflections. In the simple model here, timing locates a boundary along the beam; a taller signal does not by itself mean a deeper boundary.

Why the probe uses coupling gel

Gel fills gaps between the probe and skin so that sound can pass into the body more effectively. An intervening air layer would cause strong reflection at the boundaries and reduce transmission. This does not mean that air cannot carry sound; the change between materials affects how much is transmitted or reflected.

Soft-tissue calculation

An echo after 80.0 microseconds

Use the supplied uniform sound speed 1540 m/s. One microsecond (µs) is 0.000001 s, so 80.0 µs = 0.0000800 s.

  1. Complete path: vt = 1540 x 0.0000800 = 0.1232 m.
  2. Boundary depth: d = 0.1232/2 = 0.0616 m = 6.16 cm.

Under the same model, a 40 µs echo comes from a nearer boundary at 3.08 cm. These depths are measured from the probe along the beam. The supplied speed is a model value, not a claim that every tissue has exactly the same sound speed.

Convert the time and retain the return path. Microseconds are millionths of a second. Multiplying speed by an echo delay gives the complete path; the boundary depth is half of it in this model.

Optional check An ultrasound pulse returns from one tissue boundary after 80.0 microseconds. Using the supplied uniform speed 1540 m/s, what depth does this echo indicate below the probe?
An ultrasound pulse returns from one tissue boundary after 80.0 microseconds. Using the supplied uniform speed 1540 m/s, what depth does this echo indicate below the probe?