Axial and lateral resolution

Two reflectors are resolved as two only if their echoes do not overlap. Along the beam this is set by the length of the pulse: the spatial pulse length is SPL = n · λ for a pulse of n cycles, and because the sound has to travel out and back, the axial resolution is half of it, Δz = n·λ/2. A clean two-cycle pulse therefore resolves reflectors about one wavelength apart.

Across the beam the limit is beam width. Two reflectors at the same depth merge into one indication if they are closer than the −6 dB beam diameter at that depth. Lateral resolution is therefore just the beam width at the range of interest, and it degrades with depth as the beam spreads. This is why a cluster of porosity reads as a single indication and why measured indication length is always longer than the true defect.

The two are pulled in opposite directions by probe selection. Short, broadband, heavily damped pulses give good axial resolution but lower sensitivity and a wider bandwidth; a large aperture gives good lateral resolution but a long near field. Raising frequency improves both, at the cost of attenuation and penetration.

Pulse length also creates the near-surface dead zone: while the transmit pulse is still ringing, nothing can be received. Anything shallower than roughly the spatial pulse length is invisible on a single-crystal probe, which is why twin-crystal (TR) probes are used for thin wall and near-surface work.

Compression or shear

Worked example

Wave modecompression
Material velocity (override)0 m/s
Probe frequency5 MHz
Cycles in the pulse2
Element diameter10 mm
Sound path to point of interest100 mm
Velocity used5900 m/s
Wavelength λ1.18 mm
Spatial pulse length2.36 mm
Pulse duration0.4 µs
Axial (depth) resolution1.18 mm
Near field length N21.19 mm
Lateral resolution at that range12.06 mm
Approximate near-surface dead zone2.36 mm

10 mm, 5 MHz compression probe, two-cycle pulse. λ = 1.18 mm so SPL = 2 × 1.18 = 2.36 mm and axial resolution = 2.36/2 = 1.18 mm — one wavelength. Pulse duration = 2/5 = 0.4 µs. Laterally, sin θ = 0.51 × 1.18/10 = 0.06018 and the −6 dB beam is 2 × 100 × tan(3.4501°) = 12.06 mm wide at a 100 mm path, so two side-by-side reflectors 10 mm apart at that depth would still merge into one indication.

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