Curriculum · Ultrasound Physics & Instrumentation

Pulsed Ultrasound & Axial Resolution

Pulse duration, spatial pulse length, PRF/PRP, duty factor, and how spatial pulse length sets axial resolution (SPL/2).

~30 min · level: foundation · SPI draft — pending clinical review

Learning objectives

  • Define pulse duration, spatial pulse length, PRF, PRP, and duty factor with their equations.
  • Explain why diagnostic imaging is pulsed and how CW differs (duty factor = 100%).
  • Derive and apply axial resolution = SPL/2 and identify what improves it.

Diagnostic imaging is pulsed: the transducer transmits a brief pulse, then listens for echoes before transmitting again. The listening interval is how the system assigns depth. This introduces parameters absent from continuous-wave sound.

ParameterDefinitionEquationAdjustable?

Pulse duration (PD)

Time the pulse is 'on'

PD=nT=n/fPD = n\,T = n/f

No (damping-set)

Spatial pulse length (SPL)

Physical length of the pulse

SPL=nλSPL = n\,\lambda

No

Pulse repetition frequency (PRF)

Pulses per second

set by depth

Indirectly (depth)

Pulse repetition period (PRP)

Time between pulse starts

PRP=1/PRFPRP = 1/PRF

Indirectly

Duty factor (DF)

Fraction of time transmitting

DF=PD/PRP=PDPRFDF = PD/PRP = PD \cdot PRF

No

Pulsed-ultrasound parameters.

Here nn is the number of cycles in the pulse. Imaging pulses are short (2–3 cycles, heavy damping) so PD0.5PD \approx 0.53μs3\,\mu s. Deeper imaging forces a lower PRF (you must wait longer for distant echoes), which has knock-on effects for frame rate (Lesson 6) and Doppler aliasing.

Axial resolution (also longitudinal / range / depth resolution) is the minimum separation along the beam axis at which two reflectors yield distinct echoes:

Axial resolution=SPL2=nλ2\text{Axial resolution} = \frac{SPL}{2} = \frac{n\,\lambda}{2}
The ÷2 is because the pulse travels down and back (round trip).

Two interfaces closer than half the spatial pulse length produce overlapping echoes that merge. Therefore a shorter pulse → better (smaller-number) axial resolution, achieved by (1) higher frequency (shorter λ\lambda) and (2) fewer cycles / heavier backing damping. Axial resolution is independent of depth — the pulse keeps its length as it travels.

Worked example — axial resolution

A 5 MHz probe emits a 2-cycle pulse in soft tissue. Estimate the axial resolution.

Solution.

λ=1.54/5=0.308 mm\lambda = 1.54/5 = 0.308\text{ mm}; SPL=nλ=2(0.308)=0.616 mmSPL = n\lambda = 2(0.308) = 0.616\text{ mm}; axial resolution =SPL/2=0.308 mm0.31 mm= SPL/2 = 0.308\text{ mm} \approx 0.31\text{ mm}. Raising the frequency or shortening the pulse would improve (reduce) this number.

Key takeaways

  • Axial resolution equals SPL/2 = n*lambda/2, with the divide-by-2 reflecting the pulse's down-and-back round trip.
  • A shorter pulse gives better (smaller-number) axial resolution, achieved by higher frequency (shorter wavelength) and fewer cycles or heavier backing damping; more cycles lengthen the pulse and worsen it.
  • Axial resolution is independent of depth because the pulse keeps its length as it travels.
  • Duty factor = PD/PRP = PD*PRF; B-mode runs at DF < 1% (mostly listening) while CW Doppler has DF = 100%, which is exactly why CW has no depth/range resolution.
  • Deeper imaging forces a lower PRF because the system must wait longer for distant echoes (PRP = 1/PRF).

Check your understanding

Registry-style items with worked rationales.

1Axial resolution is best described as:recall

2Which change would IMPROVE axial resolution?analysis

3Continuous-wave operation is characterized by a duty factor of:recall

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References

  1. Edelman SK. Understanding Ultrasound Physics. 4th ed. ESP Inc.; 2012.
  2. Kremkau FW. Sonography Principles and Instruments. 9th ed. Elsevier; 2016.
  3. Ultrasound Physics and Instrumentation. StatPearls, NCBI Bookshelf.