Seismic acquisition and processing

Quick refresher

Field work and processing are one system. Processing cannot restore what is missing from the field record: frequencies, offsets (distances between source and receiver) and density of coverage. And processing errors in velocities, in statics (time shifts for the loose surface layer) or in migration spoil even good field data.
The common midpoint method: for horizontal boundaries, the traces of source–receiver pairs that share a midpoint contain a reflection from one point. We straighten them with the normal moveout correction and add them. The fold $n = N\,\Delta x_\text{rec}/(2\,\Delta x_\text{shot})$ is the number of traces in the sum. It weakens random noise by a factor of $\sqrt n$ compared with the signal.
The traveltime curve of a reflection (its arrival time against offset) is a hyperbola, $t^2 = t_0^2 + x^2/V^2$ . The velocity that straightens the hyperbola is the stacking velocity (the CMP velocity). For gently dipping layers and moderate offsets it is close to the root-mean-square velocity (a kind of average of the layer velocities above the reflection), not to the interval velocity (the velocity inside one layer).
A velocity error of a few per cent weakens the stack clearly. In the lesson's example, a 3% error cuts the amplitude of the reflection by about half. Multiples (waves that bounced several times between boundaries) arrive with a lower stacking velocity and stay curved after the correction, so stacking weakens them.
Without migration, dipping reflections are displaced down-dip and look gentler ( $\mathrm{tan}\,\beta = \sin\theta$ ), and the edges of layers are surrounded by diffractions (waves scattered from a point). Migration returns the reflections to their places. Where the velocity changes sharply sideways, as under salt, we need depth migration.
Before migration, the horizontal resolution is limited by the Fresnel zone (the patch of a boundary that one reflection comes from). Its radius is $(V/2)\sqrt{t_0/f}$ , which is hundreds of metres. 3D migration shrinks it to a fraction of a wavelength. 2D migration shrinks it only along the line. The bin (a cell of the 3D grid) must not be larger than $V/(4 f_{max}\sin\theta)$ .

Key relationships

n = \dfrac{N\,\Delta x_\text{rec}}{2\,\Delta x_\text{shot}}

fold of a regular 2D line

t(x) = \sqrt{t_0^2 + x^2/V_\text{stack}^2}

traveltime curve of a reflection; the normal moveout correction is $\Delta t = t(x) - t_0$

t_\text{st} = \dfrac{h_\text{LVZ}}{V_\text{LVZ}} + \dfrac{E - h_\text{LVZ} - E_0}{V_0}

static correction of a station to the datum $E_0$ (a horizontal reference level)

d = \tfrac{V t_0}{2}\sin\theta,\quad \mathrm{tan}\,\beta = \sin\theta

displacement of the reflection point up-dip, and the dip of the image before migration

r_\text{F} = \dfrac{V}{2}\sqrt{\dfrac{t_0}{f}}

radius of the Fresnel zone before migration

b \le \dfrac{V}{4 f_{max}\sin\theta}

bin size that avoids spatial aliasing (a false overlap of frequencies)

Common pitfalls

Thinking that twice the fold makes the section twice as good. The signal-to-random-noise ratio grows only by a factor of $\sqrt2$ , and fold alone does not remove coherent noise (regular noise, such as surface waves and multiples).
Taking the stacking velocity for the interval velocity of the layer above the reflection, or for the average velocity down to it.
Reading an unmigrated section as a true image. Dipping boundaries on it are displaced and look gentler, anticlines look wider and synclines narrower.
Expecting from a 2D line what 3D gives. 2D migration cannot move reflections that came from the side back into the plane of the line.
Mixing up the sampling interval with resolution. A sampling interval of 2 ms sets only the highest frequency that can be recorded (250 Hz). Resolution is set by the band of frequencies that came back from depth.

After this lesson you can

describe the sources and receivers of seismic surveys on land and at sea, and explain what limits each of them
explain the common midpoint method (offset, spread, common midpoint, fold and bin) and calculate the fold of a 2D line and of a 3D survey
compare 2D lines with 3D surveys by what they give, and estimate the density of a line grid and the cost of a work programme on a block at given rates
list the processing steps in the order they are done and explain what each does: static corrections, noise suppression, deconvolution, velocity analysis, normal moveout correction and stacking, and migration
calculate the traveltime curve of a reflection and the normal moveout correction, and show how a velocity error spoils the stack while stacking weakens multiples
estimate the horizontal resolution (the Fresnel zone before and after migration) and the bin size at which steep boundaries are recorded without distortion

From shot to section: what the receivers record

A seismic survey repeats one experiment thousands or millions of times. At a source point (SP, shot point), the source sends a short pulse of elastic energy into the ground. The waves spread in all directions. They are reflected and refracted (bent) at the boundaries between rocks, and part of the energy returns to the surface. There, receivers at receiver points (RP) pick it up. Each channel (the recording line of one receiver or one group of receivers) records the movement of the ground, or the pressure of the water, for several seconds after the shot. This record is a seismic trace. The traces of all the channels of one shot make a shot gather (a seismogram of one source point).

The record is digital. The movement is measured at equal time steps, usually every 2 ms and sometimes every 1 or 4 ms. The sampling interval $\Delta t$ limits the highest frequency that can still be recorded without distortion. This is the Nyquist frequency, $f_N = 1/(2\Delta t)$ . For 2 ms it is 250 Hz, which is well above the frequencies that return from depth (attenuation). We choose the record length by the deepest target, usually 4 to 8 s.

On a gather we see much more than the reflections that we need. It helps to picture a gather as a set of traveltime curves. A traveltime curve is a line that shows when each wave arrives at each offset (the distance from the source), as in the figure.

CalculatedTraveltime curves of the waves on a gather from one shot (calculated for an assumed section: an upper layer 600 m thick with a velocity of 1900 m/s over a fast layer of 3400 m/s, with two more reflecting boundaries below). Time increases downwards, as on seismic sections. Reflections are hyperbolas, and the deeper the boundary, the flatter the hyperbola. The head wave touches the hyperbola of the reflection from the top of the fast layer.

Direct wave: It travels from the source to the receivers through the top layer. Its traveltime curve is a straight line, $t = x/V_1$ .
Refracted wave, or head wave: It slides along the top of a faster layer and appears beyond the critical offset. Beyond some larger offset it arrives first. We use the first arrivals (the first waves to reach each receiver) of direct and refracted waves to study the upper part of the section. This is needed for static corrections.
Reflected waves (reflections): These are the useful signal. The traveltime curve of each is a hyperbola: the farther the receiver, the longer the path. The deeper the boundary and the higher the velocity, the flatter the hyperbola.
Surface waves (ground roll): They travel slowly along the surface, at hundreds of metres per second. They have low frequencies and are very strong. At short offsets they cover the reflections with a cone of noise.
Air wave: It is a sound wave that travels through the air at about 340 m/s.
Multiples: These waves have bounced several times between strong boundaries on their way, for example between the sea floor and the sea surface. They arrive later than primary reflections (waves reflected once) and look like reflections from boundaries that do not exist.
Random noise: Wind, traffic, working machinery and waves at sea cause vibrations that cannot be followed from trace to trace.

Noise and how to weaken it

Noise: Surface waves
How it differs: slow, low frequency
In the field: receiver groups, charge depth
In processing: filtering by frequency and by apparent velocity (the speed at which a wave moves along the line of receivers)

Noise: Air wave and direct wave
How it differs: straight traveltime curves, known velocity
In the field: a deeper source
In processing: muting (zeroing) the zone, filtering

Noise: Multiples
How it differs: lower stacking velocity, repeat the strong boundaries
In the field: long offsets
In processing: deconvolution, subtraction of multiple models, stacking

Noise: Random noise
How it differs: cannot be followed from trace to trace
In the field: fold, source power
In processing: stacking, filtering

Noise	How it differs	In the field	In processing
Surface waves	slow, low frequency	receiver groups, charge depth	filtering by frequency and by apparent velocity (the speed at which a wave moves along the line of receivers)
Air wave and direct wave	straight traveltime curves, known velocity	a deeper source	muting (zeroing) the zone, filtering
Multiples	lower stacking velocity, repeat the strong boundaries	long offsets	deconvolution, subtraction of multiple models, stacking
Random noise	cannot be followed from trace to trace	fold, source power	stacking, filtering

Coherent (regular) noise, which means surface waves, air waves and multiples, is suppressed by stacking only when its traveltime curves differ markedly from those of reflections. So we weaken it both during the survey and before stacking.

The task of the field work is to record the reflections with the best signal-to-noise ratio at a reasonable cost. The task of processing is to remove everything else, add the reflections together and put them in place. Processing gets only what was recorded in the field. No algorithm can restore frequencies that are not in the record, or offsets at which no receivers stood.

Sources and receivers on land and at sea

A source must give the ground enough energy over a wide band of frequencies, roughly from 5–8 Hz to 80–100 Hz. Low frequencies go deep. We need them for deep targets and for inversion (a method that turns seismic data into rock properties). High frequencies give resolution, which is the ability to tell close details apart. A receiver must record this band without distortion, and must not record anything extra.

Explosive source: A small charge in a shallow hole. The hole is usually from a few metres to a few tens of metres deep, and below the loose near-surface zone where possible. One pulse gives a wide spectrum (a wide range of frequencies) and a lot of energy, and the source works in any terrain. But the crew must drill holes, get blasting permits, and keep safe distances from buildings, pipelines and wells.
Vibrator source (Vibroseis): A heavy truck presses a plate against the ground and shakes it according to a set programme, the sweep. Over 8–20 s the frequency changes smoothly, for example from 6 to 90 Hz. The record is correlated with the reference sweep (cross-correlation), and the long sweep “collapses” into a short zero-phase wavelet (the short signal that a reflection leaves on a trace, symmetric in time). A group of several vibrators, and repeated sweeps, build up the energy. Vibrators are easy to control and need no holes. They can work closer to roads and settlements than explosions can. They need ground that a truck can cross, and terrain that is not too steep.
Air gun: A chamber of compressed air at a pressure of about 14 MPa (2000 psi). The air is released into the water in an instant. The air bubble then pulses and gives repeated pulses. So guns of different volumes are combined into arrays, designed so that the first pulses add up and the pulsations cancel each other. The array is towed at a depth of a few metres.
Geophone: A coil on a spring in the field of a permanent magnet. When the ground moves, the coil moves relative to the magnet. The movement induces a voltage in the coil, in proportion to the speed of the movement. The usual natural frequency (the frequency at which the coil swings freely on its spring) is about 10 Hz. Below it, sensitivity falls quickly. Geophones are placed in groups of several per channel, and their signals are added. The group weakens slow surface waves, because they reach its geophones in different phases. Another way is single digital sensors (accelerometers), laid close together, with the noise removed in processing.
Hydrophone: A piezoelectric pressure sensor (it turns pressure into an electric voltage). At sea, hydrophones are built into streamers (seismic cables) several kilometres long. A vessel tows ten or more streamers, with tens to hundreds of metres between them.
Ocean-bottom nodes and cables: Receivers on the sea floor record the pressure and the three components of the ground movement (up–down and two horizontal). They are used to survey shallow water, areas with platforms and pipelines, and repeat surveys over fields that are being produced. In transition zones (shallow water, deltas and the coast), land and marine equipment are combined.

Which sources and receivers are used where

Conditions: Land
Source: explosion in a hole, vibrator
Receivers: geophone groups, digital sensors
What limits the work: terrain, ground, protected zones, surface waves

Conditions: Sea
Source: air gun array
Receivers: streamers with hydrophones, ocean-bottom nodes
What limits the work: rough sea, ghosts, multiples from the sea floor

Conditions: Shallow water and coast
Source: air guns from shallow-draught vessels
Receivers: ocean-bottom nodes, geophones on the shore
What limits the work: water depth, ice, shipping, environmental restrictions

Conditions	Source	Receivers	What limits the work
Land	explosion in a hole, vibrator	geophone groups, digital sensors	terrain, ground, protected zones, surface waves
Sea	air gun array	streamers with hydrophones, ocean-bottom nodes	rough sea, ghosts, multiples from the sea floor
Shallow water and coast	air guns from shallow-draught vessels	ocean-bottom nodes, geophones on the shore	water depth, ice, shipping, environmental restrictions

Marine surveys have a difficulty of their own. The source and the streamer are towed below the water surface, and the sea surface reflects almost everything (its reflection coefficient is close to −1). So after each pulse, with a delay of $2d/V_\text{w}$ , comes an upside-down copy of it, the ghost. Here $d$ is the towing depth, and $V_\text{w} \approx 1500$ m/s is the speed of sound in water. At frequencies where the delay equals a whole number of periods, the pulse and its ghost cancel each other. In the spectrum, the first notch (dip) above zero frequency is at $f = V_\text{w}/(2d)$ . At a depth of 6 m it is 125 Hz, and at 10 m it is 75 Hz, already inside the useful band. Towing shallower is better for high frequencies. Towing deeper is better for low frequencies, and it gives less noise from the waves. Streamers with pressure sensors and particle-motion sensors let us separate the wave from its ghost in processing.

The common midpoint method: offset, spread, fold

Suppose the boundary is horizontal. A wave from the source to the receiver is reflected at a point that lies exactly below the midpoint between them. We call this midpoint the common midpoint (CMP). The reflection point below it is the common depth point (CDP). In Russian-language practice, people usually speak of the common depth point. The distance between the source and the receiver is the offset ( $x$ ). Shots at different source points, with receivers at different offsets, give many source–receiver pairs with the same midpoint. All their traces contain the reflection from the same point of the boundary, but at different times (figure).

SchematicFour pairs of source point and receiver point with a common midpoint. At the top the boundary is horizontal, and all the rays are reflected at one point, the common depth point. At the bottom the boundary dips at 15°. The reflection points spread out up-dip, and the larger the offset, the farther they spread (schematic; the rays are drawn by the laws of reflection).

A set of traces with a common midpoint is a CMP gather. After the offset correction (see the section on velocities), the reflection arrives at the same time on all its traces, and the traces are added. We say that they are stacked. The number of stacked traces is the fold. A reflection that is the same on all traces adds up in phase and grows $n$ times. Random noise adds up on some traces and subtracts on others, so it grows only $\sqrt n$ times. So the signal-to-noise ratio grows by a factor of $\sqrt n$ (advanced section). This is where the name comes from: the common midpoint (CMP) method. We speak of “2D CMP seismic” and “3D CMP seismic”.

The receivers that record one shot form a spread. In a split spread (a symmetric one), the source point is in the middle. In an end-on spread, the source point is at one end. When the shot moves to the next source point, the spread is moved after it.

Fold of a 2D line

n = \frac{N\,\Delta x_\text{rec}}{2\,\Delta x_\text{shot}}

where:

$n$ fold, the number of traces added at one CMP, —
$N$ number of active channels of the spread, —
$\Delta x_\text{rec}$ interval between receiver points (receiver groups); the CMPs are $\Delta x_\text{rec}/2$ apart, m
$\Delta x_\text{shot}$ interval between source points, m

Units: the intervals are in the same units of length

Holds when

A regular layout: all channels work, the shots follow one another at a constant interval, and the spread moves with them.
The formula applies in the middle of the line. At the ends, the fold rises from one to its full value (figure below).
The fold is counted for the nominal layout. Shots skipped because of obstacles, and rejected traces, reduce it.

Each shot puts one trace into each of $N$ neighbouring CMPs. The next shot is moved by $\Delta x_\text{shot}$ , which is $2\Delta x_\text{shot}/\Delta x_\text{rec}$ CMP intervals. So the fold grows with the number of channels and with the density of shots. We can double it by halving the interval between source points. But the signal-to-random-noise ratio then grows only by a factor of $\sqrt2 \approx 1.4$ , while the number of shots and the cost double.

Worked exampleFold of a 2D line and where it is full

Given (assumed data)

Receiver-point interval: 25 m
Active channels: 320, split spread: 160 on each side of the source point
Source-point interval: 50 m; the source points are halfway between the receiver points
Length of the line: 12 km

Solution

CMP interval: $25/2 = 12.5$ m.
Fold: $n = 320\cdot 25/(2\cdot 50) = 80$ . We can check this by counting shots. Each shot puts one trace into each of 320 neighbouring CMPs. The next shot is 50 m away, which is 4 CMP intervals. So each CMP is “covered” by $320/4 = 80$ shots.
Largest offset: the receivers nearest to the source point are 12.5 m from it, and the farthest are at $12.5 + 159\cdot 25 = 3987.5$ m, about 4 km.
After stacking, the signal-to-random-noise ratio grows by a factor of $\sqrt{80} = 8.94$ .
At the ends of the line the fold is lower. Counting the traces over all the CMPs (figure below) shows that full fold begins 1981 m from the start of the line, which is about half of the largest offset.
If we make the source points denser, 25 m apart, the fold becomes 160. The signal-to-noise ratio then grows by only a further factor of $\sqrt2 \approx 1.41$ , with twice as many shots.

Answer: The fold of the line is 80 (CMPs every 12.5 m). Stacking improves the ratio of signal to random noise by a factor of about 9. Full fold starts about 2.0 km from each end of the line.

What it means. The target must lie where the fold is full. So we extend the line beyond the limits of full fold, by about half of the largest offset and also by the migration aperture (the extra width that migration needs; see the section on resolution). Fold is not a goal in itself. Doubling the fold gives only a 1.4 times better signal-to-random-noise ratio. Against coherent noise, which means surface waves and multiples, we use offsets, receiver groups and processing. The data in this example are assumed for illustration.

CalculatedThe fold along the line of the example, counted trace by trace. At the ends it rises almost linearly, and it reaches the full value of 80 about 2 km from the ends of the line.

We choose the largest offset to suit the target. To find the velocity from the traveltime curve, and to tell multiples from primary reflections, the hyperbolas must bend visibly. For this we need offsets of the order of the depth of the target. Offsets that are too large stretch the wavelet after the correction. Refracted waves arrive there, and the amplitude of the reflection changes more with angle. Small offsets are needed for shallow boundaries, and to follow reflections above the cone of surface waves.

2D lines and 3D surveys

2D seismic is the survey along separate lines, called profiles. Each profile gives a vertical section. The profiles are kilometres apart, and between them the structure is filled in by interpolation. We measure the volume of 2D work in line kilometres (line km). A profile also “sees” what lies to the side. Waves also come from boundaries outside its plane (side reflections, or sideswipe). But 2D processing assumes that everything lies in the plane of the line, so such reflections end up in the wrong place.

3D seismic is the survey over an area. On land the most common layout is orthogonal: the receivers are set in lines, the source points are in lines across them, and each shot records a patch of several receiver lines. The midpoints between sources and receivers cover the area. The area is divided into rectangular cells, the bins, and traces with midpoints in one bin are treated as one CMP. The bin size is half the receiver-point interval by half the source-point interval, usually 12.5–25 m and less often up to 50 m. At sea, a vessel tows several streamers and two sources that fire in turn. With this firing order, the CMP lines are a quarter of the streamer spacing apart, measured across the streamers.

Fold of a 3D survey with an orthogonal layout

n = \frac{N_\text{ch}\,\Delta x_\text{rec}}{2\,D_\text{SL}}\cdot\frac{N_\text{RL}}{2}

where:

$N_\text{ch}$ number of active channels on one receiver line of the patch, —
$\Delta x_\text{rec}$ interval between receiver points along the line, m
$D_\text{SL}$ distance between the lines of source points, m
$N_\text{RL}$ number of active receiver lines in the patch, —

Units: lengths in the same units

Holds when

An orthogonal layout: each source point records a patch that is symmetric about it along and across the lines.
Far from the edges of the survey. Near the edges the fold builds up from low values, as at the ends of a 2D line.

The first factor is the fold along the receiver lines, and the second is the fold across them. For example, take receiver-point and source-point intervals of 50 m, line spacings of 400 m, and a patch of 12 lines of 160 channels each. The fold is $10\cdot 6 = 60$ , and the bin is 25 × 25 m.

The result of a 3D survey is a cube (data volume): traces on a regular grid of bins. From the cube we cut sections along the survey lines (inline), across them (crossline) and in any direction. We also cut time slices, which are horizontal sections on which faults, channels and reefs are visible. The main advantage of 3D is migration in three dimensions. Each reflection returns to its place over the area, not only along a line, and small objects appear where they really are.

We plan a 3D survey with a margin. The fold builds up from the edges, so it is low there, and migration also needs data from outside the target. The margin is wider for deeper targets and steeper slopes. So the area of the survey is clearly larger than the area of the object.

The terms of a subsoil block set the volume of work. For example, take the notice of the electronic auction of 25 December 2026 for 56 blocks. For most exploration and production blocks, the minimum exploration programme includes 2D seismic over three years: from 300 to 1000 line km, most often 300. One block also has in its programme the reprocessing of historical 2D data and a 3D survey. The subsoil user (the company that holds the licence) may choose to do these if a promising object is found (for details, see the “Investors → Subsoil auction” section of this site). The example shows what 300 kilometres give on a large block.

Worked exampleWhat 300 line km of 2D give on a block of 2000 km²

Given (assumed data)

Area of the block: 2000 km² (an assumed block)
Minimum 2D programme: 300 line km
Expected size of traps: about 5 × 3 km
Rates (assumptions of the example): 2D with processing, US$2,000 per line km; 3D, US$30,000 per km²
Margin of the 3D survey area: 2 km on each side of the trap (assumption of the example)

Solution

Suppose the lines form a grid of two perpendicular sets with the same spacing $s$ . Their total length over the area $A$ is about $2A/s$ , so $s = 2\cdot 2000/300 = 13.33$ km.
Put a trap of 5 × 3 km, with its sides parallel to the lines, at a random place. Lines of one direction cross it with a probability of $5/13.33 = 0.375$ , and lines of the other direction with a probability of $3/13.33 = 0.225$ . At least one line will cross it with a probability of $1 - (1 - 0.375)(1 - 0.225) = 0.516$ . Lines of both directions will cross it with a probability of $0.375\cdot 0.225 = 0.084$ .
The 2D programme: $300\cdot 2000 = 0.6$ million US$.
A 3D survey over one trap with a margin of 2 km on each side: $(5 + 4)\cdot(3 + 4) = 63$ km². The cost is $63\cdot 30{,}000 = 1.89$ million US$.

Answer: The grid has a spacing of about 13 km. It crosses a trap of this size with at least one line in about 52% of cases, and with lines of both directions in only 8%. The 2D programme costs about 0.6 million US$, and a 3D survey over one trap costs about 1.89 million US$.

What it means. The minimum programme on a large block is a regional reconnaissance. It ties the block to wells and old lines, shows the structure and outlines leads (areas that may hold a trap but need more study). But to map a trap we need several lines through each structure in two directions, or a 3D survey over the object that was found. So exploration usually goes from a sparse 2D grid to a denser one, and then to 3D. Each step is a decision to buy information (Decision trees and the value of information). The rates here are assumptions, not market prices. The cost depends on the terrain and access, the season, the source, the fold and bin, environmental requirements, and how busy the contractors are. For comparison, the 2026 auction notice estimates the whole minimum programme of a block, with 300 line km of 2D and one well, at US$2–3 million. The probabilities are a rough model for a trap whose sides are parallel to the lines. The data in this example are assumed for illustration.

The processing flow: from the record to the section

Processing is a long chain of steps. It is called the processing flow (in Russian-language practice, the “processing graph”). The order matters, because each step assumes that the earlier ones are done. Below are the main steps for a CMP survey. We choose the set of steps and their parameters for each survey, by test processing of parts of it.

The main processing steps

1
Data input and geometry
The records are converted to the processing format. Each trace gets the coordinates and elevations of its source and receiver points from the topographic survey. A geometry error will show up everywhere later. We look for it in the first arrivals: with correct geometry they lie on smooth lines.
2
Editing
Dead and noisy channels and failed shots are rejected. The polarity of channels that were connected the wrong way round is corrected.
3
Static corrections
They remove the delays in the uneven near-surface layer and bring all times to one horizontal reference level, the datum (details below).
4
Amplitude recovery
This compensates for the weakening with time that is caused by spherical spreading (the energy thins out as the wavefront grows) and absorption (attenuation). If amplitudes will be used to judge fluids, they are recovered by one law for all traces, and not “equalised” in a sliding window.
5
Noise suppression
We suppress surface waves, air waves, ghosts and some multiples. We do it by filtering by frequency and by apparent velocity, and by subtracting models of the noise.
6
Deconvolution
It compresses the wavelet and suppresses short-period repetitions (below).
7
CMP sorting and velocity analysis
The traces are collected into CMP gathers, and the stacking velocities are chosen (next section).
8
Normal moveout correction, residual statics, stacking
The traveltime curves are straightened. The static corrections are refined using the reflections themselves. The traces of each CMP are added.
9
Migration
It returns the reflections to their places (section on migration). Today it is more often done before stacking.
10
Final processing
Filtering, spectrum balancing, converting the wavelet to zero phase, and scaling. The result is a section or a cube, in time or in depth, in the standard SEG-Y format. It comes with a report on the parameters of all the steps.

Static corrections. Near the surface lies the low-velocity zone (LVZ, or weathering layer). It is made of loose, weathered rocks that are not saturated with water. The velocity of P-waves in it is 200–800 m/s, instead of 1500–2500 m/s below it. Its thickness and velocity change from point to point, and so does the height of the ground. So the wave is delayed differently on its way to neighbouring receivers, and on a gather the reflection breaks into steps. Static corrections (statics) bring all the traces to a horizontal datum, as if the sources and receivers stood on it and there were no low-velocity zone.

Static correction of a station to the datum

t_\text{st} = \frac{h_\text{LVZ}}{V_\text{LVZ}} + \frac{E - h_\text{LVZ} - E_0}{V_0}

where:

$t_\text{st}$ static correction of the station: the time that is subtracted from all traces of this station, s
$h_\text{LVZ}$ thickness of the low-velocity zone beneath the station, m
$V_\text{LVZ}$ velocity of P-waves in the LVZ, m/s
$E$ elevation of the station (height above sea level), m
$E_0$ elevation of the datum, m
$V_0$ velocity of the rocks below the LVZ (the replacement velocity), m/s

Units: SI: m, m/s, s

Holds when

Rays near the surface are vertical. In the slow LVZ the wave really is refracted to nearly vertical.
The datum is below the base of the LVZ, and the LVZ is uniform beneath the station.
The correction of a trace is the sum of the corrections of its source point and its receiver point (the corrections are surface-consistent: a point has the same correction for every trace). If the charge is fired below the LVZ, only the second term remains at the source point, and it is worked out from the depth of the charge.

We find the thickness and velocity of the LVZ from uphole surveys and from the first arrivals of refracted waves. An uphole survey is seismic logging of a shallow hole: we measure the wave travel times between the surface and different depths in the hole. The slow zone counts for more than the relief. At the velocities of the example below, one metre of the LVZ delays the wave by 2 ms, and one metre of rock below it delays the wave by 0.53 ms.

Worked exampleStatic corrections for two neighbouring receiver points

Given (assumed data)

Datum: 100 m above sea level
Velocities: in the LVZ 500 m/s, below it 1900 m/s
RP 101: elevation 182 m, LVZ thickness 8 m (from the uphole survey)
RP 102: elevation 176 m, LVZ thickness 17 m

Solution

RP 101: in the LVZ, $8/500 = 16.0$ ms. From the base of the LVZ to the datum, $(182 - 8 - 100)/1900 = 74/1900 = 38.9$ ms. The correction is $16.0 + 38.9 = 54.9$ ms.
RP 102: $17/500 = 34.0$ ms and $(176 - 17 - 100)/1900 = 59/1900 = 31.1$ ms. The correction is $34.0 + 31.1 = 65.1$ ms.
Corrections for the relief only, as if rock with a velocity of 1900 m/s lay directly beneath the points: $82/1900 = 43.2$ and $76/1900 = 40.0$ ms.
The error of such a correction: for RP 101, $54.9 - 43.2 = 11.7$ ms; for RP 102, $65.1 - 40.0 = 25.1$ ms. The difference between the neighbouring points is 13.4 ms.

Answer: The corrections are 54.9 and 65.1 ms. Point 102 stands 6 m lower, but its correction is 10.2 ms larger, because the LVZ is thicker. A correction for the relief only would be wrong by 13.4 ms between the neighbouring points.

What it means. 13.4 ms is almost the width of the main peak of a Ricker wavelet (a standard shape that is used to model seismic wavelets) with a peak frequency of 30 Hz. Between its zero crossings it is 15 ms wide. The traces of neighbouring points would add up with a shift. The reflections would weaken and lose high frequencies. This is why the LVZ is studied specially, and the rest of the static is refined from the reflections themselves (residual statics). After the normal moveout correction, we pick for each point the shift that lines up the reflections best. Where the LVZ changes quickly and there are few observations, static errors remain. False “structures” a few milliseconds high then appear on the section, for example sags beneath areas of thick sand. The data in this example are assumed for illustration.

Deconvolution. The wavelet on the record is longer than we would like. The “ringing” of the source, and repeated reflections in a near-surface layer or in a water layer, stretch each reflection over tens of milliseconds. In the convolution model, a trace is the reflection coefficients “convolved” with the wavelet: each reflection is replaced by a copy of the wavelet, and the copies are added (convolution model). Deconvolution is the reverse operation: a filter that turns a long wavelet into a short one. We calculate it from the traces themselves. We assume that the reflection coefficients are random and that the wavelet is minimum-phase (its energy comes at the front). By frequency, deconvolution raises the frequencies that are weak in the wavelet and levels the spectrum (makes the energy even across the frequencies). But together with the signal it also amplifies noise. So we level the spectrum only where the signal is stronger than the noise.

Predictive deconvolution suppresses oscillations that repeat with a known period. An example is reverberation, the repeated bouncing of the wave inside a water layer of depth $h$ , with a period $T = 2h/V_\text{w}$ . For 30 m of water this is 40 ms, and for 4 m it is about 5 ms. We cannot remove long-period multiples from a deep sea floor in this way. They are subtracted using a model that is predicted from the data themselves (SRME, surface-related multiple elimination), or they are separated by the curvature of their traveltime curves (the Radon transform).

What to check in a processing report

What: Geometry and static corrections
Why it matters: errors give steps and false structures of small amplitude

What: Stacking velocities
Why it matters: the quality of the stack and the first approximation to depths depend on them

What: Multiple suppression
Why it matters: multiples that remain look like real boundaries

What: Type of migration and the velocities for it
Why it matters: the position of faults, crests and flanks

What: Whether true amplitudes are preserved
Why it matters: whether amplitudes and AVO (the change of amplitude with offset) can be used to judge fluids

What: Phase and polarity of the final wavelet
Why it matters: the tie to wells (wavelet and polarity)

What	Why it matters
Geometry and static corrections	errors give steps and false structures of small amplitude
Stacking velocities	the quality of the stack and the first approximation to depths depend on them
Multiple suppression	multiples that remain look like real boundaries
Type of migration and the velocities for it	the position of faults, crests and flanks
Whether true amplitudes are preserved	whether amplitudes and AVO (the change of amplitude with offset) can be used to judge fluids
Phase and polarity of the final wavelet	the tie to wells (wavelet and polarity)

The traveltime curve, velocities and stacking

A reflection from a horizontal boundary arrives later at offset $x$ than at zero offset, because the path is longer. Suppose the layer above the boundary is uniform. Then each of the two legs of the path is the hypotenuse of a right triangle. The other two sides are $x/2$ and the depth of the boundary. A hyperbola describes the time:

Traveltime curve of a reflection and the normal moveout correction

t(x) = \sqrt{t_0^2 + \frac{x^2}{V_\text{stack}^2}}, \qquad \Delta t(x) = t(x) - t_0

where:

$t(x)$ arrival time of the reflection at offset $x$ , s
$t_0$ two-way time at zero offset, the time $t_0$ below the CMP, s
$x$ offset between source and receiver, m
$V_\text{stack}$ stacking velocity (the CMP velocity), m/s
$\Delta t(x)$ normal moveout correction, the time that is subtracted at offset $x$ , s

Units: SI: s, m, m/s

Holds when

A horizontal boundary below a uniform layer. Then the formula is exact, and $V_\text{stack}$ equals the velocity of the layer.
For a layered overburden, the hyperbola is an approximation at offsets that are not larger than the depth of the boundary, and $V_\text{stack}$ is close to the root-mean-square velocity (below).
For a boundary that dips at an angle $\theta$ in the plane of the line, $V_\text{stack} = V/\cos\theta$ , which is larger than the velocity of the layer.

The correction grows with offset and falls with depth and velocity: deep hyperbolas are flatter. If we subtract it from the times of each trace, the hyperbola is straightened. This is the normal moveout correction (NMO).

We find the velocity at which the hyperbola is straightened best by trial. This is velocity analysis. For each time $t_0$ , we try a series of velocities, apply the correction, add the traces and see at which velocity the sum is largest. A colour map of “time versus velocity” shows a measure of how well the traces agree (semblance). The processor draws a velocity function through its maxima. It is the stacking velocity as a function of time. This is done at analysis points a few hundred metres to a kilometre apart, and the velocities between them are interpolated.

CalculatedA calculated CMP gather with two primary reflections and a multiple (every second trace is shown; offsets 200–3000 m). On the left it is before the normal moveout correction. On the right it is after the correction with the velocities of the primary waves. The primary reflections have been straightened. The multiple has the same time

t_0 = 1.8

s but a lower stacking velocity, and it is still curved. Ricker wavelet, 30 Hz; SEG polarity: positive swings are filled in.

Worked exampleNormal moveout correction and a velocity error

Given (assumed data)

Reflection: $t_0 = 1.800$ s, stacking velocity 2700 m/s
Offsets: 1000, 2000 and 3000 m; for the stack, 30 traces 100 m apart, Ricker wavelet 30 Hz
Multiple: the same $t_0$ , stacking velocity 2200 m/s

Solution

At 1000 m: $t = \sqrt{1.8^2 + (1000/2700)^2} = \sqrt{3.24 + 0.13717} = 1.8377$ s, and the correction is 37.7 ms. In the same way at 2000 and 3000 m the times are 1.9465 and 2.1153 s, and the corrections are 146.5 and 315.3 ms.
Take a velocity 3% lower, 2619 m/s. At 3000 m the correction is $\sqrt{3.24 + (3000/2619)^2} - 1.8 = 333.6$ ms. This is 18.3 ms more than needed. At 2000 m it is 155.3 ms, which is 8.8 ms more. On the far traces the reflection rises above $t_0$ . After the correction it bends upwards.
The sum of 30 traces, in a simplified model: each trace is shifted as a whole by its correction for $t_0$ (more in “What it means”). With the right velocity, the amplitude of the reflection is 1 (in fractions of the amplitude of one trace). With a velocity 3% lower, it is 0.47 (figure below).
The multiple at 3000 m: $t = \sqrt{3.24 + (3000/2200)^2} = 2.2582$ s. We apply the correction for each time separately, along the hyperbola with the velocity of the primary wave. This reading of the trace moves to $\sqrt{2.2582^2 - (3000/2700)^2} = 1.9659$ s, which is 165.9 ms below $t_0$ . The sum of the multiple is 0.14.

Answer: The corrections are 37.7, 146.5 and 315.3 ms. A velocity error of 3% makes the correction on the far traces too large by 18.3 ms. It lowers the amplitude of the stack to 0.47 of its possible value. After the correction the multiple stays curved, and only 0.14 of it is left in the stack.

What it means. At the far offset the correction is more than eight times larger than at the near one. So the far traces are the most sensitive to velocity, and we pick the velocity from them. A 3% error is not rare where reflections are weak and the velocity changes sideways. It costs half of the amplitude and the high frequencies. In velocity analysis we aim for an accuracy of about one per cent. The multiple has the same $t_0$ but a lower stacking velocity. In the stack, the multiple was weakened about 7 times. Stacking suppresses multiples because of this difference. The stack model is simplified: equal amplitudes at all offsets, no noise, and each trace shifted as a whole by the correction for the $t_0$ of the reflection, without stretching the wavelet. An exact recalculation of the times, with stretching of the wavelet, as in the step about the multiple, gives an amplitude of 0.45 at a 3% error. The conclusion is the same. The data in this example are assumed for illustration.

CalculatedThe amplitude of the stack of the 30 traces of the CMP gather, as a function of the velocity used for the normal moveout correction (data of the example). The primary reflection has a sharp maximum at 2700 m/s. The multiple with the same

t_0

has an equally sharp maximum, but at 2200 m/s. We pick velocities from such maxima. At the velocity of the primary reflection, 0.14 of the multiple is left.

What velocity does the analysis give? For a set of horizontal layers, the root-mean-square velocity (RMS velocity) describes the hyperbola roughly. Each layer enters it with the weight of its travel time:

Root-mean-square velocity

V_\text{rms}^2 = \frac{\sum_i V_i^2\,\Delta\tau_i}{\sum_i \Delta\tau_i}

where:

$V_\text{rms}$ root-mean-square velocity down to the base of the $n$ -th layer, m/s
$V_i$ interval velocity of the $i$ -th layer, m/s
$\Delta\tau_i$ vertical two-way time in the $i$ -th layer, s

Units: SI: m/s, s

Holds when

Horizontal, uniform layers.
Offsets that are not larger than the depth of the boundary. At larger offsets the traveltime curve departs from a hyperbola, and the velocity that we pick depends on the range of offsets.
Isotropic rocks. In shales the velocity depends on direction (anisotropy), and the stacking velocity usually comes out higher than the vertical velocity.

The root-mean-square velocity is always a little higher than the average velocity, which is the depth divided by half the time. Fast layers enter with the weight of the square of the velocity. Take a layer of 2100 m/s that is 0.9 s thick and a layer of 3300 m/s that is 0.7 s thick (two-way time). For this set $V_\text{rms} \approx 2692$ m/s, and the average velocity is 2625 m/s. Interval velocities are found from stacking velocities with the Dix formula. The topic Seismic interpretation explains how this is done and what its risks are.

The correction is not the same even within one reflection. At the start of the wavelet it is a little larger than at the end, so after the correction the wavelet is stretched. The stretch is greater at larger offsets and smaller times. We zero the strongly stretched parts of the far traces near the top of the section before stacking (muting).

If the boundary dips, the reflection points of different offsets spread out up-dip (the figure with rays), and the stacking velocity grows as $V/\cos\theta$ . Where a steep and a gentle reflection cross at the same time, no single velocity suits both. The dip moveout correction (DMO, a correction for the dip) or migration before stacking (pre-stack migration) fixes this.

Migration: each reflection to its place

After stacking, each CMP is shown as one trace, as if the source and receiver were at the same point, and the reflection is drawn directly beneath it. For horizontal boundaries this is correct. But a wave with the source and receiver at one point is reflected from a dipping boundary along the normal to it, at a right angle. So the reflection point lies not under the CMP but to the side, up-dip. On an unmigrated section, dipping reflections are therefore displaced down-dip and look gentler. Anticlines look wider and synclines narrower. In narrow synclines the reflections from the two limbs cross each other and make a “bow-tie”.

Displacement of a reflection from a dipping boundary before migration

d = \frac{V t_0}{2}\sin\theta, \qquad z = \frac{V t_0}{2}\cos\theta, \qquad \mathrm{tan}\,\beta = \sin\theta

where:

$d$ horizontal distance from the CMP to the reflection point, up-dip, m
$z$ true depth of the reflection point, m
$V$ velocity in the overburden above the boundary (constant), m/s
$t_0$ two-way time of the reflection on the CMP section, s
$\theta$ true dip of the boundary, the dip angle of the layer (in the lessons on structural geology it is $\delta$ ). Do not mix it up with the angle $\theta$ between the ray and the normal to the boundary in the topic on AVO, °
$\beta$ dip of its image on the unmigrated section when the section is redrawn in depth by $V t_0/2$ , °

Units: SI: m, s, m/s; angles in degrees

Holds when

A constant velocity above the boundary, and a plane boundary.
The line runs in the dip direction of the boundary. On an oblique line, part of the displacement is across the line, and 2D migration will not fix it.

The displacement grows with dip and with the reflection time. The image on an unmigrated section is never steeper than 45°: even at $\theta = 90°$ , $\mathrm{tan}\,\beta = 1$ . Migration moves each image point up-dip and makes the reflections steeper.

CalculatedA dipping boundary (30°) in a uniform overburden, and its image on an unmigrated CMP section redrawn in depth (data of the example below). The reflection for point X came along the normal to the boundary, but it is drawn below X, 1.54 km farther down-dip than the point it came from. The image is gentler than the boundary. The edge of the boundary at the fault gives a diffraction arc, which migration collects back to a point.

Worked exampleWhere migration moves a reflection from a dipping boundary

Given (assumed data)

Velocity: 2800 m/s, constant (assumption of the example)
Boundary: plane, dipping at 30° in the plane of the line
At point X of the line: on the CMP section the reflection arrived at $t_0 = 2.2$ s

Solution

The path along the normal to the boundary: $L = V t_0/2 = 2800\cdot 2.2/2 = 3080$ m. On the unmigrated section, redrawn in depth, the reflection is drawn below point X at a depth of 3080 m.
The true reflection point is up-dip by $L\sin 30° = 3080\cdot 0.5 = 1540$ m, at a depth of $L\cos 30° = 3080\cdot 0.866 = 2667$ m, which is 413 m higher.
Dip of the image: $\mathrm{tan}\,\beta = \sin 30° = 0.5$ , so $\beta = 26.6°$ instead of 30°. On a time section this is 357 ms of two-way time per kilometre of line.

Answer: Migration moves this point 1540 m up-dip and 413 m up, and it restores the dip of the boundary from 26.6° to 30°.

What it means. A vertical well at point X would meet this boundary not at 3.08 km, as drawn on the unmigrated section, but at $L/\cos 30° \approx 3.56$ km. This is a point about a kilometre and a half down-dip of the one that gave the reflection. On the flanks of structures, displacements of hundreds of metres are usual. So we make maps and well plans only from migrated data. The example is idealised: the velocity is constant and the line runs in the dip direction. On an oblique line the boundary leaves the plane of the line, and 2D migration moves the reflection only in part. The data in this example are assumed for illustration.

The edge of a layer that was cut off by a fault is a point that scatters the wave in all directions. On an unmigrated section, the hyperbola of a diffracted wave spreads out from it:

Traveltime curve of a diffracted wave on a CMP section

t(x) = \sqrt{t_0^2 + \frac{4x^2}{V^2}}

where:

$t(x)$ diffraction time on the section at a point that is $x$ away from the apex, s
$t_0$ time of the apex, $2z/V$ , s
$x$ distance along the line from the point above the scatterer, m
$V$ velocity (constant), m/s

Units: SI: s, m, m/s

Holds when

A constant velocity. A CMP section, with the source and receiver at one point. The scatterer lies in the plane of the line.

The diffraction hyperbola is twice as steep as the traveltime curve of a reflection at the same $x$ . Here each of the two legs of the path is shifted horizontally by the whole distance $x$ , but in a CMP gather it is shifted by only $x/2$ . The tails of the hyperbola cross the reflections and hide faults. We can picture migration like this. For each point of the section, we add up the amplitudes along the hyperbola that a diffraction from this point would give (Kirchhoff migration). Where the point really reflects or scatters, the amplitudes add up in phase. Where it does not, they cancel each other.

Migration needs a velocity, because the shape of the hyperbolas depends on it. If the velocity is too low, the section is undermigrated: the diffractions are not fully collapsed. If it is too high, the section is overmigrated: instead of apexes, “smile” arcs appear. We refine the velocities step by step: migration, analysis, a new model.

Time migration: The result stays in time. It assumes that the velocity changes smoothly and that each diffraction is a hyperbola. It is fast and stable, and it fits most calm sections.
Depth migration: It builds the image directly in depth, from a depth–velocity model of interval velocities. It takes into account the bending of rays at sharp velocity boundaries: under salt, under reefs and in thrusts. The model is built step by step (by tomography, which adjusts the model until it fits the recorded travel times), and an error in the model is an error in the image.
Pre-stack and post-stack: Pre-stack migration works with the traces before they are added. It correctly joins steep and gentle reflections that have different offsets. It is now the standard. The abbreviations are PSTM for pre-stack time migration and PSDM for pre-stack depth migration.

Resolution: what the survey and the processing decide

The length of the wavelet sets the vertical resolution. Layers thinner than about a quarter of the wavelength cannot be told apart (thin layers and tuning thickness). The survey and the processing decide which band of frequencies reaches the section. Several things affect it. One is the spectrum of the source, together with ghosts. Another is receiver groups, which cancel not only noise but also the high frequencies of dipping reflections. A third is the accuracy of statics and velocities, because trace shifts of a few milliseconds cut the high frequencies when the traces are stacked. The last is deconvolution.

Horizontally, a reflection comes not from a point but from a patch, the Fresnel zone. This is the set of all points of the boundary for which the path is longer than the shortest one by less than a quarter of a wavelength, counted one way. Their reflections arrive nearly in phase and add up. On an unmigrated section, details smaller than this patch are blurred.

Radius of the first Fresnel zone

r_\text{F} = \frac{V}{2}\sqrt{\frac{t_0}{f}} \approx \sqrt{\frac{z\lambda}{2}}

where:

$r_\text{F}$ radius of the first Fresnel zone on the boundary, m
$V$ average velocity down to the boundary, m/s
$t_0$ two-way time of the reflection, s
$f$ dominant frequency of the reflection, Hz
$z,\ \lambda$ depth of the boundary ( $V t_0/2$ ) and wavelength ( $V/f$ ), m

Units: SI: m, m/s, s, Hz

Holds when

A constant (average) velocity, zero offset, a horizontal boundary, and a depth much larger than the wavelength.
If the velocity grows with depth, the rays bend, and the zone is a little wider.

The zone grows with depth and velocity and falls with frequency. At a depth of 2–4 km it is hundreds of metres. Migration collects the energy of diffractions and shrinks the zone. In the ideal case it shrinks it to the order of $\lambda/(4\sin\theta_{max})$ , where $\theta_{max}$ is the largest angle that the migration captures (its aperture). With the full aperture, the zone shrinks to a quarter of the wavelength, as in the vertical direction. 2D migration shrinks the zone only along the line. Across the line it stays the same.

For migration to collect dipping reflections, they must be recorded without spatial aliasing (a false overlap of frequencies when the traces are too far apart). A high-frequency dipping reflection must shift between neighbouring CMPs by less than half a period. Otherwise we cannot follow it unambiguously from trace to trace, and migration creates false dipping events.

The largest bin without spatial aliasing

b \le \frac{V_\text{int}}{4 f_{max}\sin\theta}

where:

$b$ bin size (distance between CMPs) in the dip direction, m
$V_\text{int}$ interval velocity above the target, m/s
$f_{max}$ highest useful frequency, Hz
$\theta$ largest dip that must be imaged, °

Units: SI: m, m/s, Hz

Holds when

A CMP section (zero offsets). The velocity above the target is constant. The estimate is made for the highest frequency that we still want to keep.

Between neighbouring CMPs, a dipping reflection shifts by $2b\sin\theta/V_\text{int}$ of two-way time. The allowed shift is at most half a period, $1/(2f_{max})$ . A large bin is cheaper, but steep slopes, faults and high frequencies “fall apart” on it. The same formula estimates the horizontal resolution after migration. For this, $\theta$ is the largest angle that the migration uses, and the frequency is the dominant one.

Worked exampleFresnel zone, migration and bin size

Given (assumed data)

Target: $t_0 = 2.4$ s; average velocity down to it 2900 m/s; interval velocity above it 3500 m/s
Frequencies: dominant 30 Hz, highest useful 60 Hz
Dips: steepest slopes 20°; migration uses angles up to 30°

Solution

Radius of the Fresnel zone before migration: $r_\text{F} = (2900/2)\sqrt{2.4/30} = 1450\cdot 0.2828 = 410$ m. The diameter is 820 m.
Wavelength above the target: $\lambda = 3500/30 = 116.7$ m. A quarter of it is 29.2 m.
After 3D migration with angles up to 30°: $\lambda/(4\sin 30°) = 3500/(4\cdot 30\cdot 0.5) = 58.3$ m.
The largest bin for slopes of 20° at 60 Hz: $b = 3500/(4\cdot 60\cdot 0.3420) = 3500/82.08 = 42.6$ m.

Answer: Before migration, the reflection is collected from a patch about 820 m across. After 3D migration, details about 58.3 m in size can be told apart. The bin must be no larger than 42.6 m: the standard bin of 25 m is suitable, and 50 m is not.

What it means. The difference is fourteen times, and it exists only with 3D migration. On a 2D line, migration shrinks the zone along the line, but across it the line still “sees” a strip about 820 m wide. So small reefs, narrow channels and faults with a small throw are mapped reliably only with 3D. The estimates are idealised. Real resolution is limited by noise, by the accuracy of velocities, and by the aperture of the survey at its edges. The data in this example are assumed for illustration.

Terms and symbols

SP, RP: The source point, where the source stands, and the receiver point, where a geophone or a group of receivers stands.
Offset $x$: The distance between the source point and the receiver point.
CMP, CMP gather: The common midpoint of source–receiver pairs, and the set of their traces. For horizontal boundaries the traces contain reflections from one point.
Fold $n$: The number of traces added at one CMP or bin. The signal-to-random-noise ratio grows as $\sqrt n$ .
Bin: A cell of the area of a 3D survey. Traces with midpoints in it are treated as one CMP.
Static correction, LVZ: A shift of the times of a point to the datum. It removes the delays in the loose low-velocity zone and the effect of relief.
Normal moveout correction, stacking velocity: Subtracting $t(x) - t_0$ to straighten the traveltime curve. The stacking velocity is the velocity at which the curve is straightened best.
Root-mean-square velocity $V_\text{rms}$: The velocity that describes the traveltime curve of a set of horizontal layers. It is higher than the average velocity.
Deconvolution: A filter that compresses the wavelet and suppresses short-period repetitions.
Migration: Moving reflections to their true places and collecting diffractions into points. It can be in time or in depth, before or after stacking.
Fresnel zone: The patch on a boundary whose reflections add up in phase. It limits the horizontal resolution before migration.
Spatial aliasing: The distortion of dipping high-frequency reflections that are recorded with too wide a CMP interval.

Common mistakes and misconceptions

Wrong

“A fold of 80 means that the reflections are 80 times stronger than the noise.”

In fact

The reflection in the sum grows 80 times, but the noise grows too, by $\sqrt{80} \approx 9$ times. The signal-to-random-noise ratio improves about 9 times. Fold alone does not suppress coherent noise that has similar traveltime curves.

Wrong

“The CMP velocity is the velocity in the layer above the reflection.”

In fact

It is the velocity that straightens the traveltime curve of the whole overburden above the boundary. For gently dipping layers it is close to the root-mean-square velocity. Interval velocities are found with the Dix formula, and the errors then grow many times over (Seismic interpretation).

Wrong

“A reflection on a section lies below the point where it was recorded.”

In fact

This is true only for horizontal boundaries. Before migration, a dipping reflection is displaced down-dip and looks gentler. In the lesson's example, it is displaced by 1.5 km and 3.4°. We make maps and well plans from migrated data.

Wrong

“The longer the spread, the better.”

In fact

Offsets of the order of the depth of the target are needed for velocities and for suppressing multiples. But at very large offsets the wavelet is stretched after the correction, refracted waves arrive, the amplitude changes more with angle, and the cost grows.

Wrong

“A 2D line that passes near an object does not see it.”

In fact

It does see it, and that is the danger. Reflections also come from the side, from a strip about as wide as the diameter of the Fresnel zone, and 2D migration cannot put them back in place. The object “appears” on the line as if it lay beneath it.

In depth: advanced detail

Where the hyperbola comes from, and why the CMP velocity is the root-mean-square velocity

Take a uniform layer with velocity $V$ above a horizontal boundary at depth $h$ . A ray from the source to the receiver, $x$ apart, is reflected below the midpoint. Each of the two legs is a hypotenuse with the other sides $h$ and $x/2$ . So $(Vt/2)^2 = h^2 + (x/2)^2$ , and with $t_0 = 2h/V$ we get $t^2 = t_0^2 + x^2/V^2$ . This is an exact hyperbola.

In a set of layers, a ray bends at each boundary, and the traveltime curve is no longer a hyperbola. If we expand $t^2$ in a series in powers of $x^2$ , we get $t^2 = t_0^2 + x^2/V_\text{rms}^2 + c\,x^4 + \ldots$ The first term is set only by the root-mean-square velocity. The next terms become noticeable when the offset is comparable with the depth or larger. So at moderate offsets the stacking velocity that we pick is close to $V_\text{rms}$ . At far offsets the traveltime curve “lags” behind the hyperbola, and we straighten it with a higher-order correction.

For a boundary that dips at an angle $\theta$ in the plane of the line, the CMP traveltime curve is also a hyperbola, but now $t^2 = t_0^2 + x^2\cos^2\theta/V^2$ . So the stacking velocity equals $V/\cos\theta$ . At a dip of 30° it is about 15% higher than the velocity of the layer. So steep and gentle reflections with the same $t_0$ have different stacking velocities. The dip moveout correction (DMO) or pre-stack migration removes this confusion.

Why stacking weakens random noise by a factor of √n

Let each of $n$ traces have the same reflection of amplitude $A$ and independent random noise with a root-mean-square value $\sigma$ . When the traces are added, the amplitudes of the reflection add up: $nA$ . Noise adds up differently. The variances (the squares of the noise levels) of independent terms add up, so the variance of the sum is $n\sigma^2$ , and its root-mean-square value is $\sigma\sqrt n$ . The signal-to-noise ratio of the sum is $nA/(\sigma\sqrt n) = \sqrt n\,A/\sigma$ . This is $\sqrt n$ times larger than on one trace.

This argument works only for noise that is independent from trace to trace. Coherent noise with a traveltime curve like that of a reflection adds up nearly in phase after the correction, and it is not weakened. A multiple is weakened only when its remaining curvature on the far traces (the time that is left after the correction) is larger than the period of the wavelet. Static errors also weaken the reflection itself in stacking: it is shifted from trace to trace, so it adds up out of phase.

Why the unmigrated image is gentler and is never steeper than 45°

Let a plane boundary dip at $\theta$ , and let the source and receiver be at one point of the line. The shortest path to the boundary is along the normal to it. Its length $L$ is the distance from the point to the plane, and $t_0 = 2L/V$ . If we move the observation point by $\delta x$ down-dip, the distance to the plane grows by $\delta x\sin\theta$ . The unmigrated section draws the reflection at depth $L$ directly below the point. So the dip of the image is $\mathrm{tan}\,\beta = \delta L/\delta x = \sin\theta$ , and $\beta < \theta$ for any dip. The time changes by $2\sin\theta/V$ per metre of line. This is the slope of the reflection on a time section.

The reflection point itself, which is the foot of the normal (where the normal meets the boundary), lies up-dip by $L\sin\theta$ at a depth of $L\cos\theta$ . This is where migration moves the image. For a point scatterer at depth $z$ , the path from a point of the line at a distance $x$ is $\sqrt{z^2 + x^2}$ each way. This gives the hyperbola $t^2 = t_0^2 + 4x^2/V^2$ . It is convenient to picture it as a reflection from an “exploding” point, whose wave travels up at the velocity $V/2$ . This is how many migration algorithms are derived.

Deconvolution in terms of frequencies

Convolution in time is multiplication of spectra: the spectrum of a trace equals the spectrum of the reflection coefficients multiplied by the spectrum of the wavelet. If the reflection coefficients are random, their spectrum is flat on average, and the shape of the amplitude spectrum of the trace is the shape of the spectrum of the wavelet. Deconvolution divides by it. Where the wavelet has little energy at some frequencies, the filter boosts them. The spectrum becomes flatter, and the wavelet becomes shorter.

Division is dangerous where the signal is weaker than the noise, because the filter will amplify the noise. So we add a small amount of “white noise” to the spectrum of the wavelet (usually from a tenth of a per cent to one per cent of its maximum). And we level the spectrum only in the useful band. We cannot learn the phase of the wavelet from the amplitude spectrum. Statistical deconvolution assumes a minimum-phase wavelet. At the end, the section is turned to zero phase, using the wavelet that is found at a well (wavelet and polarity).

Connections to other topics

Seismic waves, acoustic impedance and reflections — reflection, the wavelet, the convolution model, vertical resolution and attenuation: what a survey records.
Geophysical exploration methods — seismic among the other methods, two-way time and false uplift beneath salt.
Seismic interpretation — what is done with a migrated section: tying it to wells, following horizons, converting time to depth.
Decision trees and the value of information — a survey as the purchase of information: when a denser grid or 3D pays off.
Folds, salt diapirs and their relationship to faults — salt domes are the main difficulty in imaging subsalt horizons.
Geological risk and chance of success — how the quality of seismic data affects the probability that a trap exists.

Check yourself

1
A 2D line has 240 channels 25 m apart, and source points every 25 m. What is the fold, and by how much is the signal-to-random-noise ratio better than with source points every 50 m?
1. A.120; about 1.4 times
2. B.60; 2 times
3. C.240; 4 times
4. D.120; 2 times
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Correct answer: A. 120; about 1.4 times
$n = 240\cdot 25/(2\cdot 25) = 120$ . With an interval of 50 m it is 60. The signal-to-random-noise ratio grows as $\sqrt n$ , so the gain is $\sqrt{120/60} = \sqrt2 \approx 1.41$ times, with twice as many shots.
2
A reflection arrived at $t_0 = 1.2$ s, and the stacking velocity is 2400 m/s. Calculate the normal moveout correction for an offset of 1800 m. What happens to the reflection on this trace if the correction is applied with a velocity of 2300 m/s?
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$t = \sqrt{1.44 + (1800/2400)^2} = \sqrt{1.44 + 0.5625} = 1.4151$ s, so the correction is 215.1 ms. With 2300 m/s: $\sqrt{1.44 + 0.6125} = 1.4326$ s, and the correction is 232.6 ms, which is 17.5 ms more than needed. The reflection on this trace will rise above $t_0$ by 17.5 ms if the trace is shifted as a whole by the correction for $t_0$ . It will rise by about 21 ms if each time gets its own correction. The traveltime curve will bend upwards after the correction, and in the stack the far traces will cancel the near ones.
3
Why does a dipping reflection on an unmigrated section look gentler than the boundary really is?
1. A.The opposite: on an unmigrated section the dips are steeper than the true ones
2. B.Because of the absorption of high frequencies
3. C.Because of static corrections
4. D.It is drawn below the CMP, although it came from a point higher up-dip, and this displacement grows down-dip
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Correct answer: D. It is drawn below the CMP, although it came from a point higher up-dip, and this displacement grows down-dip
The reflection comes along the normal to the boundary from a point that is up-dip by $L\sin\theta$ . But it is drawn below the observation point, at depth $L$ . As we move down-dip, $L$ grows by $\sin\theta$ per metre. So $\mathrm{tan}\,\beta = \sin\theta$ : a boundary with a dip of 40° looks as if it dips at $\arctan 0.643 \approx 32.7°$ .
4
The sea is 30 m deep. At what period does reverberation repeat in the water layer, and what step suppresses it? And at a depth of 600 m?
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The period is $T = 2h/V_\text{w} = 2\cdot 30/1500 = 0.04$ s, that is 40 ms. These are short repetitions, and predictive deconvolution suppresses them. At 600 m, the multiple from the sea floor arrives after 0.8 s. This is a long-period multiple. It is subtracted using a model that is predicted from the data. It is also separated by the curvature of its traveltime curve (it is slower than primary waves with the same time), and it is weakened by stacking.
5
A reef is 400 m across at a depth of about 3 km. The two-way time is 2.2 s, the average velocity is 2700 m/s and the dominant frequency is 28 Hz. What will a 2D line show that passes 300 m from the centre of the reef?
1. A.After 2D migration the reef will be exactly in its place
2. B.Nothing: the line does not cross the reef
3. C.The reef will be visible only if the bin is not larger than 50 m
4. D.Reflections from the reef will appear on the line as if the reef lay beneath it: the radius of the Fresnel zone is about 380 m, and across the line 2D migration does not shrink it
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Correct answer: D. Reflections from the reef will appear on the line as if the reef lay beneath it: the radius of the Fresnel zone is about 380 m, and across the line 2D migration does not shrink it
$r_\text{F} = (2700/2)\sqrt{2.2/28} = 1350\cdot 0.2803 \approx 378$ m. This is more than the distance from the line to the centre of the reef. The line collects reflections from a strip of this width, and a side reflection from the reef will lie on the section beneath the line. 2D migration moves reflections only along the line, so the reef “appears” in the wrong place. Only a 3D survey will show it reliably.
6
Which will distort the static correction of a receiver point more: an error of 5 m in its elevation, or an error of 5 m in the thickness of the LVZ? The velocity in the LVZ is 450 m/s, and below it the velocity is 2000 m/s.
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An error in the elevation changes the column of rock beneath the LVZ: $5/2000 = 2.5$ ms. An error in the thickness of the LVZ makes us count 5 m at 450 m/s instead of 2000 m/s: $5/450 - 5/2000 = 11.1 - 2.5 = 8.6$ ms, which is about 3.4 times more. So we measure the thickness and velocity of the LVZ specially, by uphole surveys and from first arrivals.

Updated: 19 October 2026