5.1 Pulse Characteristics: PRF, PRP, SPL & Duty Factor

Key Takeaways

  • Pulse repetition frequency (PRF) and pulse repetition period (PRP) are reciprocals: a PRF of 5 kHz corresponds to a PRP of 200 microseconds
  • Each centimeter of imaging depth requires 13 microseconds of round-trip travel time, so deeper imaging forces a lower PRF and a longer PRP
  • Spatial pulse length equals the number of cycles per pulse multiplied by the wavelength; it is set by the transducer and cannot be changed by the sonographer
  • Duty factor is the fraction of time the system is transmitting, roughly 0.1-1% for pulsed imaging but 100% for continuous wave Doppler
  • As PRF increases, PRP and maximum imaging depth decrease while duty factor increases
Last updated: July 2026

Diagnostic ultrasound does not transmit sound continuously. During B-mode imaging, the transducer emits a short burst of sound — a pulse — and then falls silent while it listens for echoes returning from tissue. The characteristics of that transmit-listen cycle determine how deep the system can image, how fine the detail can be, and how much energy the patient receives.

Pulse Repetition Frequency and Pulse Repetition Period

The pulse repetition frequency (PRF) is the number of pulses the system launches each second, reported in hertz (Hz) or kilohertz (kHz). Clinical imaging systems typically operate at PRFs between about 1 kHz and 15 kHz. The pulse repetition period (PRP) is the time from the start of one pulse to the start of the next, reported in microseconds (µs). The two are reciprocals:

  • PRP = 1 / PRF
  • PRF = 1 / PRP

Worked example: if a system fires 5,000 pulses per second (PRF = 5 kHz), then PRP = 1 / 5,000 s = 0.0002 s = 200 µs. If the PRP is 500 µs, the PRF is 1 / 0.0005 s = 2 kHz.

Within each PRP, only a tiny sliver of time is spent transmitting. The rest is listening time — the system waits for echoes to travel to the deepest reflector and back. Because sound travels at 1540 m/s in soft tissue, the round trip takes 13 microseconds per centimeter of depth (6.5 µs down, 6.5 µs back). This drives the most important relationship in this section:

  • Deeper imaging requires a longer PRP and therefore a lower PRF. The sonographer does not set PRF directly; the system lowers it automatically when depth is increased.
  • The maximum depth a system can image without range ambiguity is depth<sub>max</sub> = (1540 m/s × PRP) / 2, or equivalently 77 cm divided by PRF in kHz.
PRFPRPListening time availableApprox. max depth (soft tissue)
10 kHz100 µs~99 µs~7.7 cm
5 kHz200 µs~199 µs~15.4 cm
2 kHz500 µs~499 µs~38.5 cm

Worked example: a transducer images to 15 cm. Round-trip time = 15 cm × 13 µs/cm = 195 µs, so the PRP must be at least ~195 µs and the PRF no more than about 5.1 kHz. Deep abdominal scanning therefore runs at low PRF; superficial thyroid scanning can use a much higher PRF.

Spatial Pulse Length

The spatial pulse length (SPL) is the physical length — in millimeters — that a pulse occupies in space from its first cycle to its last. It equals the number of cycles in the pulse multiplied by the wavelength:

  • SPL = n × λ, and wavelength λ = c / f = 1540 m/s ÷ frequency

Worked example: a 5 MHz transducer has a wavelength of 1540 / 5,000,000 = 0.308 mm. If each pulse contains 3 cycles, SPL = 3 × 0.308 mm = 0.92 mm. In soft tissue, clinical pulses are typically 0.1 to 1 mm long.

Two factors determine SPL, and both are fixed by the transducer manufacturer:

  1. Frequency — higher frequency means shorter wavelength and therefore a shorter SPL
  2. Number of cycles per pulse — set by the damping (backing) material bonded behind the piezoelectric element; heavy damping produces short, few-cycle pulses

Because neither is adjustable from the control panel, the exam loves to ask: the sonographer cannot change the spatial pulse length. A shorter SPL is desirable because it improves axial resolution (Section 5.2).

Pulse Duration and Duty Factor

The pulse duration (PD) is the time a single pulse lasts: PD = n / f (cycles ÷ frequency). For the 5 MHz, 3-cycle pulse above, PD = 3 / 5,000,000 = 0.6 µs.

The duty factor (DF) is the fraction of total time the system spends transmitting:

  • DF = pulse duration / PRP (a unitless ratio, often expressed as a percentage)

Worked example: PD = 0.6 µs and PRP = 200 µs gives DF = 0.6 / 200 = 0.003 = 0.3%. Pulsed imaging systems spend 99% or more of their time listening, with duty factors around 0.1% to 1%. By contrast, continuous wave (CW) Doppler transmits without pause, so its duty factor is exactly 1.0, or 100% — a classic exam contrast. Shallow imaging shortens the PRP, which raises the duty factor even though pulse duration is unchanged; this is why shallow scans deposit slightly more average energy per second.

Keep the cause-and-effect chain straight: increasing depth → longer PRP → lower PRF → lower duty factor. Every reversible form of this chain is fair game on the ARRT exam.

Common Exam Traps

Watch for these frequent confusions:

  • PRF and PRP are reciprocals, not equals — a high PRF always means a short PRP
  • PRF is set by depth, not by the frequency control — changing transducer frequency does not change PRF
  • Spatial pulse length vs. pulse duration — SPL is a distance (mm); pulse duration is a time (µs); both are fixed by the transducer
  • Duty factor has no units — it is a ratio, typically reported as a percentage

Final worked example: a system imaging to 10 cm must wait 10 cm × 13 µs/cm = 130 µs between pulses, so the maximum PRF is 1 / 130 µs ≈ 7.7 kHz. Anything faster risks a range-ambiguity artifact, in which late-arriving deep echoes are misregistered as shallow reflectors on the next scan line.

Test Your Knowledge

A pulsed-wave system is operating at a pulse repetition frequency of 5 kHz. What is the pulse repetition period?

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Test Your Knowledge

Which duty factor is characteristic of continuous wave (CW) Doppler?

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Test Your Knowledge

A sonographer wants to shorten the spatial pulse length to improve detail. Which action will accomplish this?

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