Seismic waves, acoustic impedance and reflections

Quick refresher

The P-wave is faster than the S-wave. The S-wave does not pass through liquid or gas. In a loose porous rock, replacing water with gas clearly lowers the P-wave speed and the density. It does not change the resistance to shear, so the S-wave only speeds up a little, because the rock has become lighter.
Impedance is $Z = \rho V$ . The reflection coefficient at normal incidence (when the wave hits the boundary head-on) is $R = (Z_2 - Z_1)/(Z_2 + Z_1)$ . Ordinary boundaries of a sedimentary section give $|R|$ below 0.1, and strong ones 0.2–0.4. The share of energy that is reflected is $R^2$ , not $R$ : for $R = 0.2$ it is 4%.
Both speed and density matter. At the top of a gas sand, density gives more than half of the contrast. For a water sand under shale, a calculation from speed alone can even give the opposite sign.
A trace is the reflection coefficients “convolved” with the wavelet (each coefficient is replaced by a copy of the wavelet, and the copies are added up): $s(t) = \sum R_k\,w(t - \tau_k)$ . A synthetic seismogram from a log ties the reflections to the layers. We must know the polarity of the section. By the SEG (Society of Exploration Geophysicists) convention, an increase of impedance on a zero-phase (symmetric) wavelet is a peak (an upward swing of the trace). In the North Sea and some other areas the opposite is used.
The wavelength is $\lambda = V/f$ . Layers thinner than about $\lambda/4$ are not resolved: we cannot see their top and base separately. The reflections from the top and base merge, the time between the peak and the trough stops getting smaller, and the amplitude falls almost in proportion to the thickness. For a Ricker wavelet the greatest amplitude comes at a thickness in time (two-way travel time) of $0.39/f_p$ .
With depth the speed grows, and high frequencies are absorbed ( $A/A_0 = e^{-\pi f t/Q}$ ), so the wavelength grows and resolution gets worse. The amplitude of a reflection also depends on the angle of incidence, and AVO analysis is based on this.

Key relationships

Z = \rho\,V_p

acoustic impedance; $10^6$ kg/(m²·s) is numerically equal to g/cm³ × km/s

R = \dfrac{Z_2 - Z_1}{Z_2 + Z_1} \approx \tfrac12\,\Delta\ln Z

reflection coefficient at normal incidence; the share of reflected energy is $R^2$

s(t) = \sum_k R_k\,w(t - \tau_k)

convolution model: a trace made of single reflections

w(t) = (1 - 2\pi^2 f_p^2 t^2)\,e^{-\pi^2 f_p^2 t^2}

Ricker wavelet with peak frequency $f_p$

\Delta t_\text{tun} = \dfrac{\sqrt6}{2\pi f_p} \approx \dfrac{0.39}{f_p},\quad h_\text{tun} = \dfrac{V\,\Delta t_\text{tun}}{2}

tuning thickness of a thin layer for a Ricker wavelet (in time and in metres)

A/A_0 = e^{-\pi f t/Q}

absorption: the amplitude factor of frequency $f$ after travel time $t$

Common pitfalls

Calculating the reflection coefficient from speeds alone: in gas sands and water sands, density changes the size and even the sign of the reflection.
Mixing up the reflection coefficient (a ratio of amplitudes) with the share of reflected energy $R^2$ .
Measuring the thickness of a thin layer from the time between the peak and the trough: below the tuning thickness this time stops getting smaller.
Not checking the polarity and phase of the section: the same boundary is a peak on one section and a trough on another.
Taking a bright amplitude for commercial gas, without allowing for tuning (the strengthening of the reflection from a thin layer), lithology and low-saturation gas.

After this lesson you can

explain what the speeds of P-waves and S-waves in sedimentary rocks depend on, and why replacing water with gas changes one of them strongly and the other hardly at all
calculate acoustic impedance and the reflection coefficient at a boundary between rocks, tell the reflection coefficient from the share of reflected energy, and predict the polarity of a reflection (peak or trough) in the accepted convention
describe the Ricker wavelet and zero-phase and minimum-phase wavelets, and build a synthetic trace from a log with the convolution model
estimate the vertical resolution and the tuning thickness of a thin layer, and explain when its thickness can be measured from time and when only from amplitude
explain why resolution falls with depth, and in outline how the amplitude of a reflection changes with the angle of incidence (AVO), and why this is a sign and not a proof of gas

P-waves and S-waves in rocks

An explosion, a blow from a vibrator (a heavy truck that shakes the ground), or a blast from an air gun (a device that fires compressed air into the water) pushes the rock. The push passes from particle to particle. In a P-wave (compressional wave), the particles move back and forth along the direction in which the wave travels. The rock is squeezed and stretched in turn, like air in a sound wave. In an S-wave (shear wave), the particles move sideways, across the direction of travel, and the rock is sheared. Liquid and gas do not resist shear. So the S-wave goes only through the solid framework of the rock, and the P-wave goes through the framework together with the fluid in the pores.

The speeds of both waves are set by the elastic moduli and the density: $V_p = \sqrt{(K + \tfrac43 G)/\rho}$ and $V_s = \sqrt{G/\rho}$ , where $K$ is the bulk modulus (resistance to squeezing) and $G$ is the shear modulus (resistance to shear) (wave speeds in the Earth). The same formulas work in sonic logging. Everything that changes the speed in a rock is covered in the lesson on sonic logging: the mineral framework, porosity, compaction and cement, effective stress, and the fluid in the pores. Three consequences matter for seismic exploration.

Speed grows with depth. Sediments are compacted and cemented. In an ordinary section the P-wave speed grows from 1.6–2 km/s near the surface to 4–6 km/s at a depth of a few kilometres. The wavelength grows with it, and so resolution gets worse (see the section on thin layers).
A fluid acts on the waves in different ways. Gas in a loose, highly porous rock sharply lowers its resistance to squeezing. The P-wave speed falls by 10–20% or more, and the density falls too. Resistance to shear does not depend on the fluid, so the S-wave speed even grows a little, because the rock has become lighter. The ratio $V_p/V_s$ of a gas sand is clearly smaller than that of a water sand (the S-wave and gas). In a stiff, cemented rock the effect of the fluid is small.
Contrasts in a clastic section are weak. Sandstone and shale at one depth often have close speeds and densities, and the sign of the difference can change from one package to the next and with depth. Strong contrasts come from tight carbonates and anhydrite under shales, gas sands, and coals.

A seismic wave differs from a logging wave first of all in frequency. Seismic exploration works at about 5–100 Hz (hertz, oscillations per second), and sonic logging at 10–20 kHz. The frequency does not change when the wave passes from one layer to another, but the wavelength changes together with the speed:

Wavelength

\lambda = \frac{V}{f}

where:

$\lambda$ wavelength: the distance between neighbouring crests (compressions), m
$V$ speed of the wave in the rock, m/s
$f$ frequency of oscillation, Hz

Units: SI: m, m/s, Hz

Holds when

A wave of one frequency. A real wavelet has many frequencies, and we calculate the wavelength at a characteristic frequency. We must say which one (see the section on the wavelet).
At a boundary the frequency stays the same, and the wavelength grows where the speed grows.

The wavelength is the scale at which seismic data “see” the section: the wave averages everything much smaller than it. A seismic wave in the same rock is hundreds of times longer than a logging wave.

The same shale for a seismic wave and for a logging wave

Wave: Seismic
Frequency: 30 Hz
Wavelength in shale: 101 m

Wave: Sonic logging
Frequency: 15 kHz
Wavelength in shale: 20 cm

Wave	Frequency	Wavelength in shale
Seismic	30 Hz	101 m
Sonic logging	15 kHz	20 cm

The velocity in the shale is 3030 m/s (interval transit time 330 µs/m, as in the examples below). A logging tool tells apart layers tens of centimetres thick, and a seismic wave only layers metres to tens of metres thick. Seismic velocities are also usually a little lower than logging ones, so we calibrate the log time with a check-shot survey, a well survey that measures the travel time from the surface to given depths (tie to seismic).

Acoustic impedance and the reflection coefficient

When a wave reaches a boundary between rocks, part of it is reflected and part goes on. How much is reflected is decided not by the speed or the density alone but by their product, the acoustic impedance. It helps to think of it as the “resistance” of the medium to the wave. To shake the particles of a dense, stiff rock at the same speed, the wave must carry more pressure. If the impedances of two rocks are equal, the boundary is transparent for a P-wave at normal incidence, however different their speeds and densities are when taken separately.

Acoustic impedance

Z = \rho\,V_p

where:

$Z$ acoustic impedance of the rock, kg/(m²·s)
$\rho$ bulk density of the rock, kg/m³
$V_p$ P-wave speed, m/s

Units: SI: kg/(m²·s)

Holds when

The rock is uniform on the scale of the wavelength. A package of layers much thinner than the wavelength is “seen” by the wave through its averaged properties.

The impedances of rocks are millions of kg/(m²·s), so we write them in $10^6$ kg/(m²·s). This number equals the density in g/cm³ times the speed in km/s. A shale with a density of 2.40 g/cm³ and a speed of 3.03 km/s has an impedance of about 7.3, and a tight limestone has 15–17.

In other unit systems (1)

Oilfield logging units

Z\,[10^6\ \text{kg/(m}^2\text{s)}] = \frac{1000\,\rho_b\,[\text{g/cm}^3]}{\Delta t\,[\text{$\mu$s/m}]}

$\rho_b$ is the bulk density from the density log, g/cm³. $\Delta t$ is the interval transit time of the P-wave from the sonic log, µs/m.

$V\,[\text{km/s}] = 1000/\Delta t\,[\mu\text{s/m}]$ : there are $10^6$ µs in a second and 1000 m in a kilometre.

Suppose a compression wave hits a horizontal boundary from above, vertically (normal incidence). The ratio of the pressure in the reflected wave to the pressure in the incident wave is the reflection coefficient:

Reflection coefficient at normal incidence

R = \frac{Z_2 - Z_1}{Z_2 + Z_1}

where:

$R$ reflection coefficient: the ratio of the pressure amplitude in the reflected wave to that in the incident wave, —
$Z_1$ impedance of the rock the wave comes from (above the boundary), kg/(m²·s)
$Z_2$ impedance of the rock beyond the boundary (below it), kg/(m²·s)

Units: dimensionless; impedances in any equal units

Holds when

Normal incidence of a plane P-wave. At oblique incidence, part of the energy goes into S-waves, and $R$ depends on the angle (see the section on AVO).
The boundary is flat and extended, much wider than the zone from which the reflection comes.
The rocks on both sides of the boundary are elastic and uniform.

$|R| \le 1$ . When $R > 0$ (the wave goes into a rock of higher impedance), the reflected wave has the same sign: a compression comes back as a compression. When $R < 0$ , the sign changes: a compression comes back as a rarefaction (a stretching). For small contrasts, $R \approx \tfrac12\ln(Z_2/Z_1) \approx \tfrac12(\Delta V/V + \Delta\rho/\rho)$ : the contributions of the relative changes of speed and density add up.

The reflection coefficient is a ratio of amplitudes, while the energy of a wave is proportional to the square of the amplitude. So a boundary reflects the share $R^2$ of the energy: for $R = 0.1$ it is 1%, and for $R = 0.3$ it is 9%. Almost all the energy goes deeper, and this is exactly why reflections come from many boundaries, one under another. In passing down through a boundary and back up, the wave loses the factor $1 - R^2$ of its amplitude (the derivation is in the advanced section).

Worked exampleWhat reflections a water sand, a gas sand and a tight limestone under shale give

Given (assumed data)

Shale: $\Delta t = 330$ µs/m, $\rho_b = 2.40$ g/cm³
Sand: porosity, interval transit times: 0.326; water-bearing 325, gas-bearing 386 µs/m (the same sand as in the lesson on sonic porosity)
Densities of grains and fluids: quartz 2.65, formation water 1.05, gas in the reservoir 0.15 g/cm³; in the gas-bearing part the water saturation is 0.25
Tight limestone: $\Delta t = 170$ µs/m, $\rho_b = 2.66$ g/cm³

Solution

Density of the water sand: $\rho_b = 2.65 - 0.326\cdot(2.65 - 1.05) = 2.128$ , rounded to 2.13 g/cm³. In the gas-bearing part the fluid is a quarter water and three quarters gas: $0.25\cdot 1.05 + 0.75\cdot 0.15 = 0.375$ g/cm³, and $\rho_b = 2.65 - 0.326\cdot(2.65 - 0.375) = 1.908$ , about 1.91 g/cm³.
Speeds $V = 10^6/\Delta t$ : shale 3030.3 m/s, water sand 3076.9, gas sand 2590.7, limestone 5882.4 m/s.
Impedances $Z = \rho V$ (g/cm³ × km/s): shale $2.40\cdot 3.0303 = 7.273$ ; water sand $2.13\cdot 3.0769 = 6.554$ ; gas sand $1.91\cdot 2.5907 = 4.948$ ; limestone $2.66\cdot 5.8824 = 15.647$ . All are in $10^6$ kg/(m²·s).
Reflection coefficients at the top of each rock under the shale: water sand $\dfrac{6.554 - 7.273}{6.554 + 7.273} = -0.052$ ; gas sand $\dfrac{4.948 - 7.273}{4.948 + 7.273} = -0.190$ ; limestone $\dfrac{15.647 - 7.273}{15.647 + 7.273} = 0.365$ .
Share of reflected energy $R^2$ : 0.27%, 3.6% and 13%.
A calculation from speeds alone, $(V_2 - V_1)/(V_2 + V_1)$ , would give for the gas sand $(2590.7 - 3030.3)/(2590.7 + 3030.3) = -0.078$ , and for the water sand $+0.008$ , a reflection of the other sign.

Answer: The top of the gas sand gives a reflection

R = -0.190

. By amplitude it is 3.7 times stronger than the top of the same sand with water (

-0.052

). The top of the tight limestone is a strong reflection of the opposite sign,

+0.365

What it means. Gas has turned a weak boundary into a strong one, and on the section the gas-bearing part of the sand stands out as a bright negative reflection. This is how “bright spots” appear. Notice how much density decides here. From speed alone, the contrast of the gas sand would come out 2.4 times weaker. And the water sand, which is a little faster than the shale but clearly lighter, would give a reflection of the other sign. So we build the synthetic from two curves, sonic and density (Density log and photoelectric factor (Pe)). Even the strongest boundary here reflects only 13% of the energy. The calculation is for normal incidence. At oblique incidence S-waves come into play (see the section on AVO). The densities of the fluids and the water saturation are assumptions of the example. The data in this example are assumed for illustration.

The seismic wavelet, phase and polarity

The source sends a short push into the ground. But the oscillation that comes back lasts tens of milliseconds. The Earth, the receivers and the processing pass only a band of frequencies, roughly from 5–10 to 60–100 Hz, and the push spreads out into several half-cycles. This oscillation is called the seismic wavelet, or simply the wavelet. Each boundary returns to the surface a copy of the same wavelet, multiplied by its own reflection coefficient.

Zero-phase wavelet: It is symmetric about its centre. The largest extremum (peak or trough) falls exactly at the reflection time, and on both sides are smaller extrema of the opposite sign. This wavelet is the most convenient for interpretation, and processing usually tries to bring the data to it.
Minimum-phase wavelet: It starts at the moment the wave arrives, and it is not symmetric. The energy is concentrated at its start, and the reflection time is marked by the first arrival, not by the main extremum. Wavelets of explosive sources have this shape before processing.

For calculations a simple zero-phase wavelet is convenient: the Ricker wavelet. It is set by one number, the peak frequency $f_p$ , at which the amplitude spectrum of the wavelet (how strong each frequency is) is greatest.

Ricker wavelet

w(t) = \left(1 - 2\pi^2 f_p^2 t^2\right)e^{-\pi^2 f_p^2 t^2}

where:

$w(t)$ amplitude of the wavelet, normalised to its value at the centre, —
$t$ time from the centre of the wavelet, s
$f_p$ peak frequency: the maximum of the amplitude spectrum, Hz

Units: SI: s, Hz

Holds when

A model of a zero-phase wavelet. A real wavelet is found from seismic data near a well, and its shape differs from the Ricker shape.
The shape of the wavelet does not change within the time interval under consideration.

The central peak has a trough on each side, going down to $-0.446$ at $t = \pm0.390/f_p$ . The wavelet crosses zero at $t = \pm 0.225/f_p$ . The time between the troughs, the width of the wavelet, is $\sqrt6/(\pi f_p) \approx 0.78/f_p$ : at 30 Hz this is 26 ms. One divided by this width, $1.28\,f_p$ , is the dominant frequency, measured from the visible period. So “the frequency of a wavelet”, with no more detail, is ambiguous.

CalculatedRicker wavelets with peak frequencies of 30 Hz (solid line) and 60 Hz (dashed line), calculated by the formula above. A frequency twice as high gives a wavelet half as long.

Polarity is the convention about which extremum to show for a reflection with $R > 0$ . By the standard of the Society of Exploration Geophysicists (SEG), an increase of impedance on a zero-phase wavelet is shown as a central peak (a positive amplitude). On a minimum-phase wavelet the same increase starts with a trough. In the North Sea and some other areas, the opposite is used for zero-phase data: an increase of impedance is a trough. On sections, peaks are usually filled in, but colour palettes differ too. So before reading amplitudes, we check the polarity on a reflection with a known sign, such as the sea floor or the top of a tight limestone or anhydrite, and on a synthetic seismogram at a well.

The convolution model and the synthetic seismogram

Picture a section made of horizontal layers, and a wave travelling vertically. Each boundary $k$ returns the wavelet multiplied by its reflection coefficient $R_k$ and shifted by its two-way time $\tau_k$ (two-way time: the time for the wave to go down to the boundary and back up). The trace is the sum of these copies. The sequence of reflection coefficients in time is called the reflectivity, and the summing of shifted copies of the wavelet is its convolution with the wavelet.

Convolution model of a seismic trace

s(t) = \sum_k R_k\,w(t - \tau_k) + n(t)

where:

$s(t)$ seismic trace: amplitude against time, —
$R_k$ reflection coefficient of boundary $k$ , —
$\tau_k$ two-way travel time to boundary $k$ , s
$w(t)$ seismic wavelet, —
$n(t)$ noise, —

Units: time in s; amplitudes as fractions of the amplitude of the incident wavelet

Holds when

The layers are horizontal and the wave falls at normal incidence: a one-dimensional model.
Only single reflections are included. Multiples (waves reflected many times between boundaries), transmission losses (energy lost on passing through boundaries) and spherical divergence (weakening as the wavefront spreads out) are left out, or compensated by processing.
The wavelet is the same over the whole interval.

The trace is the reflection coefficients “blurred” by the wavelet. If boundaries are farther apart than the length of the wavelet, each reflection is seen separately with the amplitude $R_k$ . If they are closer, their wavelets add up, and the amplitudes and times on the trace are no longer equal to $R_k$ and $\tau_k$ .

A synthetic seismogram (synthetic trace) is the same model applied to the log of a well. With it we find out which reflection on the section belongs to which layer, we check the polarity and phase of the data, and we calibrate amplitudes.

How to build a synthetic seismogram at a well

1
Prepare the log
Clean the sonic and density curves of spikes, cycle skips (false jumps of the sonic log) and stretches against washouts (enlarged parts of the hole) (checking the sonic log, when not to trust density). In a deviated well, move to vertical depths.
2
Tie depth to time
Get the two-way time to each depth by adding $2\,\Delta t_i h_i$ over the layers, and fit it to a check-shot survey or to vertical seismic profiling (VSP), where the receivers are in the well (integrated time).
3
Calculate impedance and reflection coefficients
$Z = \rho_b V$ for each layer. Take the reflection coefficients at the layer boundaries, or from the curves at a time step of 1–2 ms.
4
Choose a wavelet
For a first estimate, a Ricker wavelet with the frequency of the data. For an accurate tie, a wavelet extracted from the seismic traces near the well.
5
Convolve and compare
Put the synthetic next to the traces at the well. Shift it as a whole and stretch it slightly until the main reflections match, and assess the similarity. Do not stretch it hard: that way one can “tie” anything to anything.

Layers of the synthetic section

Rock: shale
$h$ , m: 50
$Z$: 7.273
Top, m: 2000
Top, s: 1.6400

Rock: gas sand
$h$ , m: 18
$Z$: 4.948
Top, m: 2050
Top, s: 1.6730

Rock: shale
$h$ , m: 42
$Z$: 7.273
Top, m: 2068
Top, s: 1.6869

Rock: limestone stringer
$h$ , m: 6
$Z$: 14.108
Top, m: 2110
Top, s: 1.7146

Rock: shale
$h$ , m: 44
$Z$: 7.273
Top, m: 2116
Top, s: 1.7168

Rock: water sand
$h$ , m: 40
$Z$: 6.554
Top, m: 2160
Top, s: 1.7459

Rock: shale
$h$ , m: 40
$Z$: 7.273
Top, m: 2200
Top, s: 1.7719

Rock: tight limestone
$h$ , m: below
$Z$: 15.647
Top, m: 2240
Top, s: 1.7983

Rock	$h$ , m	$Z$	Top, m	Top, s
shale	50	7.273	2000	1.6400
gas sand	18	4.948	2050	1.6730
shale	42	7.273	2068	1.6869
limestone stringer	6	14.108	2110	1.7146
shale	44	7.273	2116	1.7168
water sand	40	6.554	2160	1.7459
shale	40	7.273	2200	1.7719
tight limestone	below	15.647	2240	1.7983

$Z$ is the impedance in $10^6$ kg/(m²·s), from the interval transit times and densities of the previous example. The limestone stringer has $\Delta t = 185$ µs/m and $\rho_b = 2.61$ g/cm³. “Top, s” is the two-way time. At a depth of 2000 m it equals 1.640 s from the check-shot survey. Below that, $2\,\Delta t\,h$ of each layer is added to it. The data are assumed for illustration.

CalculatedA synthetic seismogram for the section of the table: impedance, reflection coefficients, and the trace, which is the sum of 30 Hz Ricker wavelets. Polarity by the SEG standard: an increase of impedance is a peak (to the right, filled).

Worked exampleA synthetic trace from a log: what is seen and what is not

Given (assumed data)

Section: layers from the table above; impedances from the previous example
Tie of depth to time: at a depth of 2000 m the two-way time is 1.640 s from the check-shot survey
Wavelet: Ricker, peak frequency 30 Hz; SEG polarity

Solution

The time in each layer is $2\,\Delta t\,h$ . For the shale above the gas sand it is $2\cdot 330\cdot 50 = 33{,}000$ µs, that is 0.0330 s, so the top of the sand arrives at $1.640 + 0.0330 = 1.6730$ s. The sand, 18 m thick, adds $2\cdot 386\cdot 18 = 13{,}896$ µs, and its base is at 1.6869 s. In the same way, the top of the limestone stringer is at 1.7146 s, the top of the water sand at 1.7459 s, and the top of the tight limestone at 1.7983 s.
Reflection coefficients from the impedances: the top and base of the gas sand −0.190 and $+0.190$ , of the stringer $+0.320$ and −0.320, of the water sand −0.052 and $+0.052$ , and the top of the tight limestone $+0.365$ .
The trace is the sum of Ricker wavelets, each multiplied by its $R$ and placed at its time (the figure above). At the top of the gas sand there is a trough −0.274, which is 1.44 times deeper than one boundary with $R = -0.190$ would give. At its base there is a peak $+0.271$ .
At the limestone stringer there is a peak $+0.116$ and a trough −0.126, although at its boundaries $|R| = 0.320$ . The thickness of the stringer in time is $2\cdot 185\cdot 6 = 2220$ µs, or 2.22 ms, six times less than the tuning thickness (see the section on thin layers). For such a thin layer the amplitude is close to $|R|\cdot\Delta t\cdot 6.13\,f_p$ , where $\Delta t$ is now not the interval transit time but the thickness of the layer in time: $0.320\cdot 0.00222\cdot 6.13\cdot 30 = 0.131$ . The peak and the trough are 10.7 ms apart, which is 4.8 times more than the thickness of the stringer in time.
At the top of the water sand there is a trough −0.052, and at its base a peak $+0.047$ , close to the $\mp0.052$ of its boundaries. The sand, 40 m (26 ms) thick, is twice as thick as the tuning thickness, and its boundaries are seen separately. The base is weakened a little by the side trough of the neighbouring reflection. At the top of the tight limestone there is a peak $+0.364$ , almost exactly $R$ .

Answer: The gas sand gives the strongest pair, a trough and a peak (−0.274 and

+0.271

), stronger than its boundaries allow. The limestone stringer, with the most contrasting boundaries, looks like a pale wave of ±0.12–0.13. The water sand and the tight limestone are seen with amplitudes almost equal to their reflection coefficients.

What it means. Three lessons of the synthetic. First, the amplitude on a trace is not always the reflection coefficient. The gas sand is 18 m thick, close to the tuning thickness, and its top and base strengthen each other. Thinner or thicker, the pair would be weaker. Second, a thin layer with very strong boundaries almost “cancels itself”, and the time between its peak and trough is set by the wavelet, not by the thickness. Third, where the boundaries are a wavelet length apart or more, the trace faithfully repeats the reflection coefficients, and such reflections are used to calibrate amplitudes. The model is simplified: only single reflections, no transmission losses, no absorption, and a theoretical wavelet. When tying real data, the wavelet is extracted from the traces near the well, and the time is checked against a check-shot survey. The data in this example are assumed for illustration.

Vertical resolution and thin layers

A layer is seen on a section by the reflections from its top and base. While the layer is thick, these are two separate wavelets, and the time between them gives the thickness: $h = V\,\Delta t/2$ , where $\Delta t$ is the difference of the two-way times. When the layer is thinner than the wavelet, the wavelets overlap. We tell two abilities apart. To resolve a layer is to see its top and base separately and to measure its thickness. To detect it is to notice that it is there. Detection reaches much further than resolution.

For a layer in a uniform rock, such as a gas sand in shale or a limestone stringer in shale, the reflection coefficients at the top and base are equal in size and opposite in sign. As the boundaries come closer, the trough from the top and the peak from the base first strengthen each other. The side trough of the wavelet from the base falls on the main trough from the top. The side peak of the wavelet from the top (for a negative reflection the side extrema are peaks) falls on the main peak from the base. The greatest amplitude is reached at the tuning thickness (tuning). This is when the time distance equals the distance from the centre of the wavelet to its side trough.

Tuning thickness for a Ricker wavelet

\Delta t_\text{tun} = \frac{\sqrt6}{2\pi f_p} \approx \frac{0.39}{f_p}, \qquad h_\text{tun} = \frac{V\,\Delta t_\text{tun}}{2}

where:

$\Delta t_\text{tun}$ tuning thickness in time: the difference of the two-way times of the top and base, s
$h_\text{tun}$ tuning thickness in metres, m
$f_p$ peak frequency of the Ricker wavelet, Hz
$V$ P-wave speed in the layer, m/s

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

Holds when

A zero-phase Ricker wavelet. For other wavelets the factor is different.
The reflections from the top and base are equal in size and opposite in sign, and the neighbouring reflections are far away.

At the tuning thickness the amplitude is $1 + 2e^{-3/2} = 1.446$ times larger than that of a single boundary. In metres, $h_\text{tun} \approx 0.195\,V/f_p$ . That is about $\lambda/5$ if we take the wavelength at the peak frequency, and exactly $\lambda/4$ if we take it at the dominant frequency $1.28\,f_p$ . This gives the well-known quarter-wavelength rule: layers thinner than about $\lambda/4$ are not resolved.

Below the tuning thickness the picture changes. The time between the peak and the trough hardly gets smaller any more, since it is set by the shape of the wavelet. The amplitude falls roughly in proportion to the thickness: the reflection of a thin layer is like the derivative of the wavelet (its rate of change), and its amplitude is close to $|R|\,\Delta t\,|w'|_{max}$ . For a Ricker wavelet $|w'|_{max} = 6.13\,f_p$ , and with $\Delta t = 2h/V$ this is about $4\pi|R|\,h/\lambda$ , the same result as for a sine wave (the derivation is in the advanced section). A layer is usually called thin when it is thinner than $\lambda/8$ (the Widess criterion): its reflection then hardly differs in shape from the derivative of the wavelet. But the thin-layer approximation gives the amplitude to within about 5% only down to a thickness of about $\lambda/16$ . Near $\lambda/8$ it already overstates the amplitude by 20–25% (the wavelength here is taken at the peak frequency).

CalculatedThe calculated amplitude of the trough at the top of a gas sand in shale (

R = \mp 0.19

V = 2591

m/s, a 30 Hz Ricker wavelet) against the thickness of the sand. This is what a wedge of sand pinching out in shale would show (a wedge model). The horizontal dashed line is the amplitude of a single boundary, and the sloping one is the thin-layer approximation.

CalculatedThe thickness “measured” from the time between the trough and the peak on the calculated trace, compared with the true thickness of the same sand. Below the tuning thickness the apparent thickness hardly changes and is larger than the true one.

Worked exampleA thin gas sand: where resolution ends

Given (assumed data)

Layer: the gas sand of the first example in shale: $R = -0.190$ at the top and $+0.190$ at the base, $V = 2590.7$ m/s
Wavelet: Ricker, $f_p = 30$ Hz
Thicknesses to compare: 5 and 10 m

Solution

The wavelength in the sand is $\lambda = V/f_p = 2590.7/30 = 86.4$ m, and a quarter of the wavelength is 21.6 m.
The tuning thickness in time is $\Delta t_\text{tun} = 0.3898/30 = 13.0$ ms. In metres, $h_\text{tun} = 2590.7\cdot 0.0130/2 = 16.8$ m. This is $\lambda/5.1$ .
The amplitude at tuning is $1.446\cdot 0.190 = 0.275$ , against 0.190 for a single boundary.
Sand of 5 m: $\Delta t = 2\cdot 5/2590.7 = 3.86$ ms. The thin-layer approximation gives $0.190\cdot 0.00386\cdot 6.13\cdot 30 = 0.135$ , and the model calculation gives 0.129. The trough and the peak are 11.26 ms apart, and the “thickness from time” is $2590.7\cdot 0.01126/2 = 14.6$ m.
Sand of 10 m: the amplitude from the model is 0.226 (the thin-layer approximation would give $0.190\cdot 0.00772\cdot 6.13\cdot 30 = 0.270$ ). The trough and the peak are 11.67 ms apart, and the “thickness from time” is 15.1 m.

Answer: The tuning thickness is about 16.8 m, roughly

\lambda/5

. Sands of 5 m and 10 m look the same “by time”, about 14.6–15.1 m, but differ in amplitude: 0.129 and 0.226.

What it means. Below the tuning thickness, we measure thickness on a section not from time but from amplitude. In this range the amplitude is almost proportional to the thickness, and we build a thickness map from a calculated curve calibrated on wells. But the amplitude is changed not only by thickness but also by porosity, fluid and the make-up of the neighbouring rocks. A thin layer with a high amplitude and a thick one with a weak contrast look the same. The thin-layer approximation overstates the amplitude by 5% at 5 m (about $\lambda/17$ ) and already by 19% at 10 m (about $\lambda/9$ ). A sand thicker than about 35–40 m is seen as two separate reflections, and its thickness read from time is correct. Between the tuning thickness and 35 m, the time between the trough and the peak is a little smaller than the true one (the figure above). The data in this example are assumed for illustration.

Thicker than tuning: we measure thickness from the time between the top and the base. Up to about $2.5\,h_\text{tun}$ , interference still distorts both the amplitude and the time a little.
Thinner than tuning: the time between the peak and the trough does not tell the thickness. We estimate thickness from amplitude, if the contrast of the rocks is known.
The limit of detection depends on the signal-to-noise ratio and on the contrast. A layer with strong boundaries is noticeable even at a thickness of a few tenths of tuning, but its thickness cannot be found from the section.
Horizontal resolution is set by something else: the size of the zone on the boundary from which the reflection comes (the Fresnel zone). Migration (a processing step that moves reflections to their true positions) improves it (Seismic acquisition and processing).

Attenuation: why deeper is harder to see

On its way down and back, the amplitude of the wave falls for three reasons. Spherical divergence: the energy spreads over a growing wavefront (the surface that the wave has reached), and the amplitude falls in inverse proportion to the distance travelled. Losses on passing through boundaries: a factor $1 - R^2$ for each thick boundary, down and back. Absorption: part of the energy turns into heat in every oscillation. The first two causes hardly depend on frequency, and processing mostly compensates them. Absorption, however, is stronger for high frequencies: over the same path, a high-frequency wave makes more oscillations.

Absorption at constant quality factor

\frac{A}{A_0} = e^{-\pi f t/Q}

where:

$A/A_0$ share of the amplitude of the component of frequency $f$ left after absorption, —
$f$ frequency, Hz
$t$ travel time along the whole path; for a reflected wave, the two-way time, s
$Q$ quality factor of the rocks: the larger it is, the weaker the absorption; average values for a sedimentary section are tens to hundreds, —

Units: SI: Hz, s; $Q$ is dimensionless

Holds when

The quality factor does not depend on frequency in the seismic range and is the same along the whole path (otherwise we take its average along the path, weighted by time).
Spherical divergence and boundary losses are accounted for separately.

The exponent is proportional to the number of oscillations on the path, $f t$ . After $Q/\pi$ oscillations the amplitude falls by a factor of $e$ . So with depth the wavelet loses high frequencies, its peak frequency falls, it becomes wider, and resolution gets worse.

Worked exampleResolution at 0.6 s and at 2.8 s

Given (assumed data)

Shallow reflection: two-way time 0.6 s; speed in the layer 2100 m/s; peak frequency 50 Hz
Deep reflection: 2.8 s; 4200 m/s; 20 Hz
Average quality factor of the path: $Q = 80$

Solution

Shallow: $\lambda = 2100/50 = 42.0$ m and $\lambda/4 = 10.5$ m. The tuning thickness for a Ricker wavelet is $2100\cdot 0.3898/50/2 = 8.2$ m.
Deep: $\lambda = 4200/20 = 210$ m and $\lambda/4 = 52.5$ m. The tuning thickness is 40.9 m.
Absorption down to 0.6 s: the 20 Hz component keeps $e^{-\pi\cdot 20\cdot 0.6/80} = 0.624$ of its amplitude, and the 60 Hz component keeps $e^{-\pi\cdot 60\cdot 0.6/80} = 0.243$ . The ratio is 0.389.
Down to 2.8 s: 20 Hz keeps 0.1109, and 60 Hz keeps 0.001364. The ratio is 0.0123.

Answer: At shallow depth, layers are resolved from about 8.2–10.5 m, and deep down only from 40.9–52.5 m. By 2.8 s the 60 Hz component is weakened relative to 20 Hz about 81 times, and by 0.6 s only 2.6 times.

What it means. A deep layer must be five times thicker than a shallow one for its top and base to be seen separately. There are two reasons, and both work the same way: the speed grows with depth, and the frequencies fall, because the high ones are absorbed. Processing restores some high frequencies, but it strengthens noise along with them, so only part can be recovered. This gives a practical rule: the deeper the accumulation, the more of its internal structure must come from wells and not from seismic data. The speeds, frequencies and quality factor are assumptions of the example. The data in this example are assumed for illustration.

CalculatedShare of the amplitude left after absorption against frequency, for reflections at 0.6 and 2.8 s (

Q = 80

, the formula above).

Amplitude and angle of incidence: AVO in brief

Seismic receivers (the sensors that record the waves) stand at different distances (offsets) from the source, and the wave reaches the boundary at different angles. At oblique incidence (when the wave hits the boundary at a slant) part of the energy of the P-wave turns into S-waves at the boundary. Then the reflection coefficient begins to depend on the angle of incidence $\theta$ and on the contrast of S-wave speeds. This dependence is described exactly by the Zoeppritz equations (the exact equations for how a wave splits at a boundary). For small contrasts and angles up to about 30°, a two-term approximation is convenient. The analysis of how amplitude depends on offset or angle is AVO (amplitude versus offset).

Two-term approximation of the reflection coefficient (Shuey)

R(\theta) \approx A + B\sin^2\theta,\quad A = \tfrac12\!\left(\frac{\Delta V_p}{V_p} + \frac{\Delta\rho}{\rho}\right),\quad B = \tfrac12\frac{\Delta V_p}{V_p} - 2\frac{V_s^2}{V_p^2}\!\left(\frac{\Delta\rho}{\rho} + 2\frac{\Delta V_s}{V_s}\right)

where:

$\theta$ angle of incidence of the P-wave on the boundary, °
$A$ intercept: the reflection coefficient at normal incidence in the linear approximation, —
$B$ gradient: the rate of change of $R$ with $\sin^2\theta$ , —
$\Delta V_p,\ \Delta V_s,\ \Delta\rho$ differences of P-wave speed, S-wave speed and density: lower rock minus upper rock, m/s; g/cm³
$V_p,\ V_s,\ \rho$ averages over the two rocks, m/s; g/cm³

Units: dimensionless $R$ , $A$ , $B$ ; speeds and densities in any consistent units

Holds when

Small relative contrasts of properties at the boundary.
Angles of incidence less than about 30°. At larger angles a third term or the exact formulas are needed.
A flat boundary between uniform isotropic rocks.

$A$ is set by the impedance contrast, and $B$ mostly by the contrast of resistance to shear. At the top of a gas sand under shale, $A < 0$ and $B < 0$ , and the reflection gets stronger as the angle grows. Searching for gas on pre-stack gathers (seismic records before summing) is based on this.

Classes of reflections from the top of a gas sand under shale

Class: I
Impedance of the sand: higher than the shale
Reflection at normal incidence: positive
As the angle grows: gets weaker

Class: II
Impedance of the sand: almost the same as the shale
Reflection at normal incidence: close to zero
As the angle grows: grows or falls, and may change sign

Class: III
Impedance of the sand: lower than the shale
Reflection at normal incidence: negative, strong
As the angle grows: gets even stronger

Class: IV
Impedance of the sand: lower than the shale
Reflection at normal incidence: negative, strong
As the angle grows: gets weaker

Class	Impedance of the sand	Reflection at normal incidence	As the angle grows
I	higher than the shale	positive	gets weaker
II	almost the same as the shale	close to zero	grows or falls, and may change sign
III	lower than the shale	negative, strong	gets even stronger
IV	lower than the shale	negative, strong	gets weaker

Classes I–III were proposed for gas sands, and class IV was added later. The class is set by the intercept and the gradient, and these depend on the rocks above and below the boundary. So the same sand in different sections can belong to different classes.

Worked exampleGas sand and water sand at angles up to 30°

Given (assumed data)

Shale: $V_p = 3030.3$ m/s, $V_p/V_s = 2.2$ , $\rho = 2.40$ g/cm³
Water sand: $V_p = 3076.9$ m/s, $V_p/V_s = 1.9$ , $\rho = 2.13$ g/cm³
Gas sand: $V_p = 2590.7$ m/s, $\rho = 1.91$ g/cm³; the shear modulus is the same as for the water sand, so $V_s = V_{s,\text{w}}\sqrt{2.13/1.91}$

Solution

S-wave speeds: shale $3030.3/2.2 = 1377.4$ , water sand $3076.9/1.9 = 1619.4$ , gas sand $1619.4\cdot 1.0560 = 1710.1$ m/s. For the gas sand $V_p/V_s = 1.51$ .
The shale–gas sand boundary: the averages are $V_p = 2810.5$ and $V_s = 1543.75$ m/s, and $\rho = 2.155$ g/cm³. Then $\Delta V_p/V_p = -0.1564$ , $\Delta V_s/V_s = +0.2155$ , $\Delta\rho/\rho = -0.2274$ , and $(V_s/V_p)^2 = 0.3017$ .
The intercept is $A = -0.5\cdot(0.1564 + 0.2274) = -0.1919$ . The gradient is $B = -0.5\cdot 0.1564 - 2\cdot 0.3017\cdot(2\cdot 0.2155 - 0.2274) = -0.2011$ .
At 30° ( $\sin^2 30° = 0.25$ ): $R = -0.1919 - 0.25\cdot 0.2011 = -0.2422$ .
In the same way for the water sand: $A = -0.0520$ , $B = -0.0905$ , $R(30°) = -0.0746$ .

Answer: The reflection from the top of the gas sand gets stronger from −0.1919 to −0.2422 by 30°. This is class III. The reflection from the top of the water sand goes from −0.0520 to −0.0746.

What it means. Both reflections get stronger as the angle grows, but the gas one noticeably more steeply: for gas, both the intercept and the gradient are larger. On a plot of $B$ against $A$ , gas sands move away from the line along which water sands and shales lie. This is what AVO analysis looks for. The intercept $A$ here (−0.1919) is close to the exact reflection coefficient −0.190 of the first example: the contrasts are moderate. The $V_p/V_s$ ratios of the shale and the water sand are assumptions of the example. The data in this example are assumed for illustration.

CalculatedReflection coefficient from the top of a gas sand and of a water sand under shale against the angle of incidence, by the two-term approximation (data of the example).

Terms and symbols

Acoustic impedance $Z$: The bulk density of the rock multiplied by the P-wave speed. A contrast of impedance creates a reflection.
Reflection coefficient $R$: The ratio of the amplitude of the reflected wave to that of the incident wave. At normal incidence it is $(Z_2 - Z_1)/(Z_2 + Z_1)$ . The share of reflected energy is $R^2$ .
Polarity: The convention about which extremum shows an increase of impedance. By the SEG standard, on a zero-phase wavelet it is a peak.
Seismic wavelet: The short oscillation that every boundary returns. A zero-phase wavelet is symmetric about the reflection time, and a minimum-phase wavelet starts at it.
Ricker wavelet, $f_p$: A simple zero-phase wavelet set by its peak frequency. The dominant frequency, from the visible period, is 1.28 times the peak frequency.
Synthetic seismogram: A trace calculated from the sonic and density logs by convolving the reflection coefficients with a wavelet. It ties the reflections to the layers.
Tuning thickness: The thickness of a layer at which the reflections from its top and base add up to the greatest amplitude. It is about $\lambda/4$ at the dominant frequency.
Resolution and detection: The ability to see the top and base of a layer separately, and the ability only to notice that a layer is there.
Quality factor $Q$: A measure of absorption: the amplitude of frequency $f$ after time $t$ falls by a factor of $e^{\pi f t/Q}$ .
AVO: Analysis of how the amplitude of a reflection depends on offset or angle of incidence: the intercept $A$ and the gradient $B$ .

Common mistakes and misconceptions

Wrong

“Impedance is set by speed, so density can be ignored in the reflection calculation.”

In fact

Density matters as much as speed. In the example of the lesson, at the top of the gas sand it gives more than half of the contrast. For the water sand under shale, a calculation from speed alone even gives the opposite sign of the reflection.

Wrong

“A reflection coefficient of 0.2 means that the boundary reflects 20% of the energy.”

In fact

The reflection coefficient is a ratio of amplitudes. Energy is proportional to the square of the amplitude, so $0.2^2 = 4$ % of the energy is reflected.

Wrong

“The time between a trough and a peak is the thickness of the layer.”

In fact

Only for layers thicker than the tuning thickness. For thinner ones the time is set by the wavelet. In the example of the lesson, sands of 5 and 10 m look the same by time, about 15 m, but they differ in amplitude.

Wrong

“A peak on a section is always an increase of impedance.”

In fact

It depends on the convention of polarity and on the phase of the wavelet. By the SEG standard, zero-phase data show an increase of impedance as a peak, while in the North Sea and some other areas it is a trough. We check the polarity on a known reflection and on a synthetic.

Wrong

“A bright spot at the top of a sand is a gas accumulation.”

In fact

A similar anomaly can come from low-saturation gas, from the tuning of a thin layer, from coals and from processing errors. A bright spot is a reason to check AVO, the closure of the structure and a flat reflection from a contact, and the decision is made from a well.

In depth: advanced detail

Where the formula for the reflection coefficient comes from, and where the energy goes

In a plane P-wave, the pressure $p$ and the particle velocity $u$ are linked by the impedance: $p = Z u$ for a wave travelling forward. At the boundary between two rocks, the pressure and the normal particle velocity must match on both sides. Otherwise the rock would tear apart or “press into” itself. Let the incident wave carry pressure $p_i$ , the reflected wave $p_r$ , and the transmitted wave $p_t$ . The reflected wave travels back, so its particle velocity enters with a minus sign: $u_r = -p_r/Z_1$ .

The conditions at the boundary are $p_i + p_r = p_t$ and $p_i/Z_1 - p_r/Z_1 = p_t/Z_2$ . Substituting the first into the second, we get $p_r/p_i = (Z_2 - Z_1)/(Z_2 + Z_1) = R$ and $p_t/p_i = 2Z_2/(Z_1 + Z_2) = 1 + R$ . The transmitted pressure can be larger than the incident one, when the wave passes into a stiff rock, but energy is conserved: the energy flux of a wave is proportional to $p^2/Z$ , and $R^2 + (Z_1/Z_2)(1 + R)^2 = 1$ .

Weakening on passing a boundary down and back

T_\downarrow T_\uparrow = \frac{2Z_2}{Z_1 + Z_2}\cdot\frac{2Z_1}{Z_1 + Z_2} = 1 - R^2

where:

$T_\downarrow,\ T_\uparrow$ ratios of the pressures of the transmitted and incident waves when passing the boundary downwards and upwards, —
$R$ reflection coefficient of the boundary for a wave going down, —

Units: dimensionless

Holds when

Normal incidence, and thick layers, much thicker than the wavelength. In a package of thin layers, multiple reflections inside the layers partly return energy to the transmitted wave, and a simple product of factors overstates the losses.

The reflection from a deep boundary is weakened by the product of the factors $1 - R^2$ of all the boundaries above it. For ordinary boundaries with $|R| < 0.1$ each factor is greater than 0.99. But under several strong boundaries, such as the top of salt, tight carbonates or coal beds, the losses are noticeable, and the amplitudes of deep reflections are compared with care.

Why the Ricker wavelet “tunes” at 0.39/f_p

Let $u = \pi f_p t$ . Then $w = (1 - 2u^2)e^{-u^2}$ . The derivative $dw/du = -2u(3 - 2u^2)e^{-u^2}$ is zero at $u = 0$ (the central peak) and at $u^2 = 3/2$ . These are the side troughs at $t = \pm\sqrt{3/2}/(\pi f_p) = \pm0.3898/f_p$ , of depth $w = -2e^{-3/2} = -0.4463$ . The zeros of the wavelet are at $u^2 = 1/2$ , that is $t = \pm 1/(\pi\sqrt2\,f_p) = \pm 0.2251/f_p$ .

A layer with equal and opposite reflections gives the trace $c(t) = w(t) - w(t - \Delta t)$ (amplitudes as fractions of $|R|$ ). At the moment of reflection from the base, $t = \Delta t$ , it equals $w(\Delta t) - 1$ . In size this is greatest when $w(\Delta t)$ is smallest, that is when $\Delta t$ coincides with the time of the side trough. Then $|c| = 1 + 2e^{-3/2} = 1.446$ . A numerical search confirms that this sum is never larger for any $\Delta t$ .

For a very thin layer, $w(t) - w(t - \Delta t) \approx \Delta t\,w'(t)$ : the trace is the derivative of the wavelet multiplied by the thickness in time. The steepest slope of the Ricker wavelet is at $u^2 = (3 - \sqrt6)/2$ , where $|w'| = 2\sqrt6\,u\,e^{-u^2}\,\pi f_p = 6.132\,f_p$ . With $\Delta t = 2h/V$ and $\lambda = V/f_p$ , the amplitude (as a fraction of $|R|$ ) is $12.26\,h/\lambda$ , almost $4\pi h/\lambda = 12.57\,h/\lambda$ , as for a sine wave. The extrema of the derivative are $2\cdot 0.1670/f_p = 0.334/f_p$ apart. At 30 Hz this is 11.1 ms, and this is the time between trough and peak towards which thin layers tend.

Connections to other topics

Geophysical exploration methods — seismic exploration among other methods, two-way travel time, and converting time to depth.
Sonic log — interval transit time, P-wave and S-wave speeds, and integrated time for the tie.
Density log and photoelectric factor (Pe) — bulk density, the second half of impedance.
Structure and composition of the Earth — elastic moduli and wave speeds, and seismic waves inside the Earth.
Porosity from the sonic log — the same sand with water and with gas: how the fluid changes the interval transit time.
Seismic acquisition and processing — how seismic traces are made: sources, receivers, stacking and migration.
Seismic interpretation — following reflections, converting time to depth, and direct hydrocarbon indicators.
Reservoir quality: geological controls on porosity and permeability — how impedance and seismic attributes help to predict reservoir quality between wells.

Check yourself

1
Under a shale ( $\rho = 2.45$ g/cm³, $V_p = 2900$ m/s) lies a sandstone ( $\rho = 2.25$ g/cm³, $V_p = 3300$ m/s). What will the reflection from its top be on a zero-phase section in SEG polarity?
1. A.A weak peak: the impedance of the sandstone is only a little larger
2. B.A trough: the sandstone is lighter than the shale
3. C.There will be no reflection
4. D.A strong peak: the sandstone is faster than the shale
Show the answer
Correct answer: A. A weak peak: the impedance of the sandstone is only a little larger
$Z_1 = 2.45\cdot 2.9 = 7.105$ , $Z_2 = 2.25\cdot 3.3 = 7.425$ ; $R = 0.320/14.53 \approx +0.022$ , a weak positive reflection, that is a peak. From speed alone it would be $+0.065$ , three times stronger: the lower density of the sandstone has almost eaten the speed contrast.
2
The reflection coefficient of a boundary is $R = -0.25$ . What share of the energy is reflected, and by what factor does the amplitude of the wave from a deeper boundary weaken after passing this one down and back?
Show the answer
The reflected share is $R^2 = 0.0625$ , that is 6.25% of the energy. The minus sign only says that the reflected wave has changed polarity. Passing down and up gives the factor $1 - R^2 = 0.9375$ : the amplitude of the deep reflection falls by about 6%.
3
A Ricker wavelet with a peak frequency of 35 Hz, a sandstone layer with a speed of 3200 m/s in shales. Estimate the tuning thickness and compare it with a quarter of the wavelength.
Show the answer
$\Delta t_\text{tun} = 0.39/35 = 0.0111$ s; $h_\text{tun} = 3200\cdot 0.0111/2 \approx 17.8$ m. The wavelength at the peak frequency is $3200/35 = 91.4$ m, and a quarter of it is 22.9 m. The quarter-wavelength rule with the peak frequency overstates the tuning thickness by almost 30%. With the dominant frequency $1.28\cdot 35 = 44.8$ Hz it gives $3200/44.8/4 \approx 17.9$ m, which matches.
4
A layer thinner than the tuning thickness becomes half as thin. What happens to the reflection?
1. A.The time between the trough and the peak halves, and the amplitude does not change
2. B.The time between the trough and the peak hardly changes, and the amplitude falls by about half
3. C.Nothing changes
4. D.The amplitude grows because of tuning
Show the answer
Correct answer: B. The time between the trough and the peak hardly changes, and the amplitude falls by about half
Below tuning, the reflection of a thin layer is like the derivative of the wavelet. Its shape and the time between the extrema are set by the wavelet, and the amplitude is proportional to the thickness in time. So thin layers are mapped by amplitude, not by time.
5
Why does the same reflection at a two-way time of 3 s look “lower in frequency” on a section than at 1 s? Estimate, for $Q = 100$ , how many times more the 50 Hz frequency is weakened relative to 25 Hz by 1 s and by 3 s.
Show the answer
Absorption is proportional to the number of oscillations on the path. The ratio of the factors of the two frequencies is $e^{-\pi(50 - 25)t/Q}$ . By 1 s: $e^{-0.785} = 0.46$ , so 50 Hz is weakened relative to 25 Hz 2.2 times. By 3 s: $e^{-2.356} = 0.095$ , which is 10.5 times. High frequencies drop out, the wavelet widens, and resolution gets worse with it.
6
The top of a sand under shale gives a negative reflection, and it gets stronger and stronger as the angle of incidence grows. Which AVO class is this, and what else must be checked before we speak of gas?
1. A.Class IV: there cannot be gas
2. B.Class I: gas is proved
3. C.Class II: only a calculation from speeds is needed
4. D.Class III: check tuning, low-saturation gas, lithology and the closure of the structure
Show the answer
Correct answer: D. Class III: check tuning, low-saturation gas, lithology and the closure of the structure
A negative intercept and a reflection that grows with angle is class III, typical of soft gas sands. But such a picture can also come from low-saturation gas, from the tuning of a thin layer, and from coal. What decides is whether the anomaly agrees with the structure, a flat reflection from a contact, and a well.

Updated: 17 October 2026