---
title: "Why a point comes out as a blob when echo times become an image"
description: "Turn round-trip times into an image, then use the pulse bandwidth and array aperture to find how much a point blurs and where ghost images come from."
canonical_url: "https://sandboxphysics.com/stories/acoustics-ultrasound/pulse-echo-imaging/"
language: "en"
page_type: "story"
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  zh-CN: "https://sandboxphysics.com/zh/stories/acoustics-ultrasound/pulse-echo-imaging/"
last_modified: "2026-10-29"
---

[Acoustics and ultrasound](https://sandboxphysics.com/stories/acoustics-ultrasound/)

L2 · Chapter 4

# Why a point comes out as a blob when echo times become an image

Sandbox Physics · 2026-10-29

Ultrasound imaging does one thing: send a short pulse, note when and on which element each echo arrives, and turn time into depth. Yet a wire thinner than a hair always shows up as a blob of finite size, and a 2 mm plastic plate looks only 1.3 mm thick. This chapter works out each of these distortions.

## Before you begin

Sound speed and wavelength from Chapter 1, interface echoes from Chapter 2, and the idea behind the Fourier transform: a short signal has a wide band of frequencies. The full width at half maximum of a Gaussian is given where it is used.

- Convert echo times into depth with a single sound speed and explain why a plate with faster sound looks thinner.
- Estimate the depth width, lateral first zero, sidelobes and grating-lobe position of the point spread from the pulse bandwidth, the aperture and the element pitch.

## Time is depth

A point reflector at distance $d$ returns its echo $2d/c$ after transmission: there and back. The scanner inverts this with $z=ct/2$. That step assumes one sound speed, usually that of water or soft tissue.

If the sound crosses a plate with a faster sound speed, it spends less time inside, so the back face appears closer than it is. A plate of thickness $d$ is drawn as $d\,c_{\mathrm{water}}/c_{\mathrm{plate}}$. This is not noise; it is a deterministic error from using the wrong speed.

$$
t=\frac{2d}{c},\qquad z=\frac{ct}{2},\qquad d_{\mathrm{apparent}}=d\,\frac{c_{\mathrm{water}}}{c_{\mathrm{plate}}}
$$

The first two come from the round-trip delay in Eq. 5.4 of Jensen’s notes. The third assumes the image uses the water sound speed everywhere.

A linear array sends one pulse into water. Each wire and each face of the plate sends an echo back; its arrival time $t=2d/c$ becomes depth. Original schematic, not to scale.

Source: J. A. Jensen · R. Barnkob

## Depth blur is set by the pulse length

The pulse has a length, and so does each echo, so a point is stretched along depth. Here the pulse has a Gaussian envelope described by its fractional bandwidth $B$ at −6 dB: the spectrum falls to half its peak at $f_0(1\pm B/2)$.

The shorter the pulse, the wider the band. For a Gaussian pulse the time width is inversely proportional to the bandwidth, and multiplying by $c/2$ turns it into depth. Only the absolute bandwidth $Bf_0$ matters: 5 MHz with $B=0.6$ and 10 MHz with $B=0.3$ give the same depth resolution.

$$
\Delta z_{\mathrm{FWHM}}=\frac{c\,\tau_{\mathrm{FWHM}}}{2}=\frac{2\ln 2}{\pi}\,\frac{c}{Bf_0}\approx0.441\,\frac{\lambda}{B}
$$

This combines $\sigma_f=Bf_0/(2\sqrt{2\ln2})$, $\sigma_t=1/(2\pi\sigma_f)$ and $\tau_{\mathrm{FWHM}}=2\sqrt{2\ln2}\,\sigma_t$.

Source: J. A. Jensen

## Sideways blur is set by the aperture

The array aligns each element’s signal by the round-trip time to every pixel and adds them up: delay-and-sum. At a point reflector all elements add in phase; a little to the side some add in phase and some against. A uniform aperture of width $L$ gives, at depth $z$, a sideways response shaped like a sinc function, with its first zero at $x\approx\lambda z/L$ and its strongest sidelobe about 13.5 dB below the main lobe.

A narrower spot needs a larger aperture or a higher frequency; the ratio $F=z/L$ is called the F-number. Lowering the weight of the edge elements with a Hann taper lowers the sidelobes sharply but widens the main lobe, roughly doubling it.

The elements are not continuous but a row of points with pitch $p$. When $p>\lambda$, all elements also add in phase in a second direction, and a grating lobe appears: a ghost of a target that is not there.

$$
x_{\mathrm{zero}}\approx\frac{\lambda z}{L},\qquad \frac{H_{\mathrm{side}}}{H(0)}=\frac{2}{3\pi}\;(-13.5\,\mathrm{dB}),\qquad \sin\theta_{\mathrm{grating}}=\frac{\lambda}{p}
$$

These are Eqs. 3.10, 3.11 and 3.16 of Jensen’s notes. The first is paraxial and is off by a few per cent when the edge elements see the point at a large angle.

Source: J. A. Jensen

## Worked example: a wire 17 mm deep

The experiment defaults to 5 MHz, $B=0.6$, 64 elements at 0.2 mm pitch, an aperture $L=12.8\,\mathrm{mm}$ and a water wavelength $\lambda=0.299\,\mathrm{mm}$, using the 25 °C sound speed of 1497 m/s from Table II of Barnkob et al. The middle wire is 17 mm deep and its echo arrives after $2\times17\,\mathrm{mm}/1497\,\mathrm{m/s}=22.7\,\mu\mathrm{s}$.

Depth: $\sigma_f=1.27\,\mathrm{MHz}$, $\tau_{\mathrm{FWHM}}=0.294\,\mu\mathrm{s}$, so $\Delta z=0.220\,\mathrm{mm}$. The simulated image gives 0.224 mm, 1.8 % more. With the bandwidth cut to 0.15 the formula gives 0.881 mm and the simulation 0.885 mm.

Sideways: the first zero is at $\lambda z/L=0.299\times17/12.8\,\mathrm{mm}=0.398\,\mathrm{mm}$. Summing the elements directly at one frequency gives 0.411 mm, 3.5 % more, because the edge elements see the wire at a sizeable angle. With only 16 elements the aperture is a quarter as wide and the zero moves out to 1.59 mm.

Grating lobe: at 10 MHz and 0.4 mm pitch, $\lambda/p=0.374$, and the far-field estimate puts the ghost at $17\,\mathrm{mm}\times\tan(\arcsin0.374)=6.86\,\mathrm{mm}$. In the simulated image it appears at 6.56 mm, 19.4 dB below the real wire.

Plate: the front face of the 2 mm polystyrene plate is at 28 mm and the back face is drawn at $28+2\times1497/2350=29.27\,\mathrm{mm}$, so the plate looks 1.27 mm thick.

Source: J. A. Jensen · R. Barnkob

## Try it in the Lab

### [Pull each blur out on its own](https://sandboxphysics.com/labs/ultrasound-pulse-echo/)

1. In the default preset, read the depth blur and the sideways blur of the wire.
2. Choose “A long pulse smears depth”: the depth blur grows about four times while the sideways blur barely changes. Check it with this chapter’s formula.
3. Choose “A narrow array blurs sideways” and then “Tapering hides the sidelobes”, and watch how the main-lobe width and the brightest sidelobe of the sideways cut change.
4. Choose “Sparse elements make ghosts”, find the ghosts beside the wire in the image and measure how far they are from the real wire.

## Exercises with explained answers

1. How thick does a 2 mm Pyrex plate look in the image?

   Show the answer

   $2\,\mathrm{mm}\times1497/5674=0.528\,\mathrm{mm}$. Sound in glass is 3.8 times faster than in water, so the back-face echo comes back much earlier and the plate looks less than a third of its real thickness.

2. A 10 MHz array has 0.2 mm pitch. Does the grating lobe of a wire 17 mm deep fall inside an image $\pm8\,\mathrm{mm}$ wide?

   Show the answer

   $\lambda/p=0.1497/0.2=0.749$, and the far-field estimate is $17\times\tan(\arcsin0.749)\,\mathrm{mm}=19.2\,\mathrm{mm}$, outside the image. The smaller the pitch relative to the wavelength, the farther out the grating lobe. Jensen’s notes recommend a pitch of half a wavelength so that no grating lobe enters the image even when the beam is steered.

3. At a single frequency, why does a Hann taper move the sideways first zero from 0.40 mm to about 0.78 mm?

   Show the answer

   The taper brings the weights of the edge elements close to zero, which shrinks the effective aperture and roughly doubles the main lobe, to about $2\lambda z/L$. In return the sidelobes drop sharply, so a weak target next to a strong one stays visible.

[Previous: Standing waves on drum skins, on plates and in a liquid-filled channel](https://sandboxphysics.com/stories/acoustics-ultrasound/standing-waves-and-resonators/) [Next: How acoustic radiation force pushes beads to the nodes of a standing wave](https://sandboxphysics.com/stories/acoustics-ultrasound/acoustic-radiation-force/)

## Sources and what we recomputed

1. [R. Barnkob, P. Augustsson, T. Laurell and H. Bruus, Acoustic radiation- and streaming-induced microparticle velocities determined by micro-PIV in an ultrasound symmetry plane, arXiv:1208.6534v1; Physical Review E 86, 056307 (2012)](https://arxiv.org/abs/1208.6534v1)

   Version 1, 31 August 2012, read in full. Eq. 1 (standing-wave field), Eq. 2 (radiation force), Eqs. 5–7 (streaming scale and Rayleigh streaming), Eqs. 9–10 (geometry and thermoviscous corrections), Eqs. 12–14 (velocity amplitude, velocity ratio, critical diameter), Eqs. 16–18 and Table I (wall drag), Table II (material data), Tables IV–V and Figs. 8–9 (measured energy densities, streaming coefficients and velocity ratios). We recompute the theory values; the measured data are quoted, not reproduced.

2. [J. A. Jensen, Notes for the International Summer School on Advanced Ultrasound Imaging, Technical University of Denmark (21 May 2024)](https://home.healthtech.dtu.dk/jaj/aui_2024/reading_material/notes_aui_2024.pdf)

   Chapter 2, Eqs. 2.45–2.51 (linear acoustic equations and wave equation). Chapter 3, Eqs. 3.8–3.11 (line aperture, first zero, first sidelobe), Eqs. 3.15–3.17 (grating lobes) and Eq. 3.24 (F-number). Chapter 5, Eqs. 5.3–5.4 (pulse and round-trip delay). The impedance reflection coefficients are not taken from these notes.