Every sixth line of a scene from a sensor with six detectors per band is consistently brighter than the rest. The appropriate correction is

destriping by matching each detector's statistics to the whole image

a second-order polynomial transformation

In a raw scene from a polar-orbiting scanner, successive scan lines are progressively offset toward the west. The cause is

the Earth rotating eastward beneath the sensor during imaging

relief displacement of mountains

Which of the following are properties of nearest-neighbour resampling?

Output values are all original input DNs

It is the fastest of the common methods

It can displace features by up to about half a pixel, giving a blocky look

Output values are all original input DNs

It is the fastest of the common methods

It can displace features by up to about half a pixel, giving a blocky look

It produces the smoothest-looking output

Which of the following statements about geometric correction and registration are correct?

A first-order (affine) transformation can correct translation, rotation, scale and skew

Image-to-image registration is needed before pixel-by-pixel change detection

Relief displacement in hilly terrain requires a DEM (orthorectification) for full removal

A first-order (affine) transformation can correct translation, rotation, scale and skew

Image-to-image registration is needed before pixel-by-pixel change detection

Relief displacement in hilly terrain requires a DEM (orthorectification) for full removal

A higher-order polynomial always gives better accuracy away from the GCPs

Which of the following reduce the brightness difference between a sunlit slope and a shaded slope of the same land cover?

A cosine topographic correction using a DEM

Basic Radiometric and Geometric Corrections: Registration and Resampling

This is a section of Part B2, Image Processing and Analysis, of the Geomatics Engineering (GE) paper. B2's third heading is “Basic Radiometric and Geometric Corrections: Registration and Resampling” — the preprocessing that turns a raw scene into data that can be compared with a map, with another date or with a field measurement. Radiometric correction fixes the numbers: detector calibration from DN to radiance, conversion to top-of-atmosphere reflectance, sun-angle and topographic normalisation, haze removal by dark-object subtraction, and the repair of striping, line drop-outs and bad pixels. Geometric correction fixes the positions: the systematic distortions (Earth rotation, scan skew, panoramic effect) and non-systematic ones (platform altitude and attitude, relief), removed with ground control points and polynomial transformations whose order sets the number of points needed, judged by the RMS error. Registration fits one image to a map or to another image, and resampling assigns values on the new grid by nearest neighbour, bilinear interpolation or cubic convolution. The numericals are calibration, reflectance, GCP counts, RMS errors and bilinear interpolation.

1. Radiometric correction: from DN to reflectance

A DN is proportional to the radiance the detector received, through its calibration: L = gain × DN + offset (or L = L_min + (L_max − L_min) × DN/DN_max). With gain 0.8 and offset −2, DN 150 is L = 118 radiance units. Radiance still depends on how much sunlight arrived, so it is converted to top-of-atmosphere (TOA) reflectance, ρ = π L d²/(ESUN_λ cos θₛ), with d the Earth–Sun distance in astronomical units, ESUN_λ the mean solar exoatmospheric irradiance of the band, and θₛ the solar zenith angle (90° − sun elevation). For L = 100, d = 1, ESUN = 1500 and a sun elevation of 60°: ρ = π × 100/(1500 × cos 30°) = 0.242. Reflectance is comparable between dates and sensors; DN is not.

Radiometric corrections
Correction	Problem	Method
Sun-angle correction	different sun elevations on different dates	divide by the sine of the sun elevation (cos θₛ): DN 60 at 30° elevation normalises to 60/0.5 = 120
Haze (path radiance)	scattered light adds a constant, largest in the blue	dark-object subtraction: subtract the band's histogram minimum (deep water or shadow should be near zero): minimum 32, pixel 85 → 53
Topographic (illumination)	sunlit slopes brighter than shaded ones of the same cover	cosine correction L_H = L_T cos θₛ/cos i (i = local incidence angle, from a DEM); band ratios also cancel it
Striping (banding)	one detector's gain differs from the others	destriping: match each detector's mean and SD (or histogram) to the whole image
Line drop-out	a scan line records zeros	replace by the average of the lines above and below: 84 and 90 → 87
Bad pixels (shot noise)	isolated pixels far from their neighbours	detect by threshold against the neighbourhood; replace by the neighbourhood mean or median

🎯 Why correct before comparing

Change detection and multi-date classification compare numbers. A DN difference caused by a lower sun or a hazier day is indistinguishable from a real change on the ground unless each image is first converted to reflectance and haze-corrected.

2. Geometric distortions and their correction

Sources of geometric distortion
Kind	Examples	Removed by
Systematic (predictable)	Earth rotation during imaging (skew of successive lines toward the west), scan skew, panoramic distortion (pixels grow toward the swath edge), mirror-velocity variation, platform velocity	models from orbit and sensor data, applied at the ground station
Non-systematic (random)	changes in altitude and attitude (roll, pitch, yaw), relief displacement	ground control points and a fitted transformation; a DEM for relief (orthorectification)

GCP-based correction fits a mapping between image coordinates (column, row) and map coordinates (E, N) using ground control points — features sharp on both, such as road intersections. A polynomial of order t per coordinate has (t + 1)(t + 2)/2 coefficients, which is the minimum number of GCPs: first order (affine) — 3 coefficients each, so 3 GCPs (it corrects shift, rotation, scale in two directions and skew); second order — 6 GCPs; third order — 10 GCPs. In practice many more are used and the fit is by least squares. Its quality is the RMS error of the residuals at the GCPs, √(Σ(Δx² + Δy²)/n), usually required to be under about half a pixel; residuals of (0.4, 0.3), (0.0, 0.2), (0.6, 0.8) and (0.3, 0.0) pixels give point errors 0.5, 0.2, 1.0, 0.3 and RMS √(1.38/4) = 0.59 pixel. A point with a large residual is usually a misidentified GCP and is removed. Higher orders fit the GCPs better but can warp badly between and beyond them.

3. Registration

Image-to-map registration (rectification, georeferencing) transforms an image into a map projection and datum so that every pixel has map coordinates. Image-to-image registration fits one image (the slave) to another (the master) so that the same ground falls in the same pixel, whether or not either is on a map projection — essential before change detection, since a misregistration of even a fraction of a pixel creates false change along every edge. Orthorectification additionally removes relief displacement using a DEM and a rigorous or rational-function (RPC) sensor model; without it, hilly terrain cannot be registered to a map by any polynomial.

⚠️ Transformation and resampling are two separate steps

The polynomial says where each output pixel's centre falls in the input image; it almost never lands exactly on an input pixel centre. Deciding what value to put there is resampling, and it is a separate choice — the same transformation can be resampled three different ways.

4. Resampling techniques

The three resampling methods
Method	Input pixels used	Output values	Appearance and use
Nearest neighbour	1	original DNs only, unaltered	blocky, edges stepped, position error up to half a pixel; fastest; preferred before classification and for thematic (class) maps
Bilinear interpolation	4 (2 × 2)	new, distance-weighted averages	smooth, slightly blurred
Cubic convolution	16 (4 × 4)	new values, can overshoot the input range	sharpest and visually best; slowest; for display products

Worked bilinear: neighbours DN 10 at (0, 0), 20 at (1, 0), 30 at (0, 1) and 40 at (1, 1); the output centre falls at (x, y) = (0.25, 0.5). Along the row y = 0: 10 + 0.25 × (20 − 10) = 12.5; along y = 1: 30 + 0.25 × (40 − 30) = 32.5; then between them at y = 0.5: 12.5 + 0.5 × (32.5 − 12.5) = 22.5. Nearest neighbour at (0.25, 0.6) takes the DN of the closest centre, (0, 1): 30.

🧠 Classify first, or resample by nearest neighbour

Bilinear and cubic resampling invent mixed spectra along edges that belong to no class. If an image is to be classified, either classify it in its original geometry and resample the class map by nearest neighbour, or resample the image by nearest neighbour so the spectra stay real.

Key takeaways

L = gain × DN + offset; TOA reflectance ρ = πLd²/(ESUN cos θₛ), θₛ = 90° − sun elevation.
Haze: dark-object subtraction; striping: detector matching; line drop-out: average of adjacent lines; terrain: cosine correction or band ratios.
Systematic distortions (Earth rotation, scan skew, panoramic) are modelled; non-systematic ones (attitude, altitude, relief) need GCPs or a DEM.
Polynomial order t needs (t + 1)(t + 2)/2 GCPs: 3, 6, 10 for orders 1, 2, 3; accept an RMS error below about half a pixel.
Resampling: nearest neighbour (1 pixel, original values, for classification), bilinear (4, smooth), cubic convolution (16, sharpest).

Practice questions (14)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A band's calibration is L = 0.8 × DN − 2 (radiance units). The radiance of a pixel with DN 150 is ____.
Numerical answer — type the value.
Show answer
Answer: 118
L = gain × DN + offset = 0.8 × 150 − 2 = 120 − 2 = 118 units. Ignoring the offset gives 120; dividing instead of multiplying by the gain gives an impossible 185.5.
A pixel has at-sensor radiance L = 100 W m⁻² sr⁻¹ µm⁻¹ in a band with ESUN = 1500 W m⁻² µm⁻¹, the Earth–Sun distance is 1 AU and the sun elevation is 60°. Its top-of-atmosphere reflectance (to three decimal places) is ____.
Numerical answer — type the value.
Show answer
Answer: 0.242
The solar zenith angle is 90° − 60° = 30°. ρ = πLd²/(ESUN cos θₛ) = π × 100 × 1/(1500 × 0.8660) = 314.16/1299.04 = 0.242. Using cos 60° (the elevation) gives 0.419, the usual slip.
The minimum DN of the blue band over a scene containing deep clear water is 32. After dark-object subtraction, a pixel with DN 85 becomes ____.
Numerical answer — type the value.
Show answer
Answer: 53
Deep clear water should be nearly black; its DN of 32 is taken as additive path radiance and subtracted from every pixel: 85 − 32 = 53. The correction is band-specific, largest in the blue where Rayleigh scattering is strongest.
A pixel has DN 60 on an image acquired at a sun elevation of 30°. Normalised to an overhead sun by dividing by the sine of the sun elevation, its DN becomes ____.
Numerical answer — type the value.
Show answer
Answer: 120
DN_normalised = DN/sin(elevation) = 60/sin 30° = 60/0.5 = 120. A low sun spreads the same energy over more ground and darkens the scene; the correction removes that seasonal difference before images are compared.
Every sixth line of a scene from a sensor with six detectors per band is consistently brighter than the rest. The appropriate correction is
1. destriping by matching each detector's statistics to the whole image
2. dark-object subtraction
3. a second-order polynomial transformation
4. cubic convolution resampling
Show answer
Answer: A — destriping by matching each detector's statistics to the whole image
A periodic bright line is striping: one of the six detectors is miscalibrated. Adjusting that detector's mean and standard deviation (or histogram) to match the others removes it. Dark-object subtraction corrects haze for the whole band; polynomials and resampling fix geometry, not radiometry.
A dropped scan line is repaired by averaging the lines above and below. For a pixel whose neighbours above and below have DNs 84 and 90, the replacement DN is ____.
Numerical answer — type the value.
Show answer
Answer: 87
(84 + 90)/2 = 87. Adjacent lines are strongly correlated, so their average is a reasonable estimate; it is cosmetic, and the repaired line should not be used for precise measurement.
The minimum number of ground control points needed to compute a second-order polynomial transformation for geometric correction is ____.
Numerical answer — type the value.
Show answer
Answer: 6
A polynomial of order t in x and y has (t + 1)(t + 2)/2 terms: for t = 2, 3 × 4/2 = 6 (1, x, y, x², xy, y²). Each GCP gives one equation for each output coordinate, so 6 points are the minimum; first order needs 3 and third order 10.
After fitting a transformation, four GCPs have residuals (Δx, Δy) of (0.4, 0.3), (0.0, 0.2), (0.6, 0.8) and (0.3, 0.0) pixels. The total RMS error, in pixels (to two decimal places), is ____.
Numerical answer — type the value.
Show answer
Answer: 0.59
Squared point errors: 0.25, 0.04, 1.00, 0.09; sum 1.38. RMS = √(1.38/4) = √0.345 = 0.587 ≈ 0.59 pixel. The third point contributes most and would be the first to check for misidentification.
In a raw scene from a polar-orbiting scanner, successive scan lines are progressively offset toward the west. The cause is
1. the Earth rotating eastward beneath the sensor during imaging
2. relief displacement of mountains
3. atmospheric haze
4. a miscalibrated detector
Show answer
Answer: A — the Earth rotating eastward beneath the sensor during imaging
While the sensor moves along its orbit, the Earth turns eastward, so each later line images ground that has moved east; placed one under another, the lines appear skewed westward. It is systematic and removed by a model of Earth rotation. Relief displacement is radial, and haze and detectors affect brightness, not position.
An output pixel centre maps to input position (x, y) = (0.25, 0.5) between four input pixels: DN 10 at (0, 0), 20 at (1, 0), 30 at (0, 1) and 40 at (1, 1). Its value by bilinear interpolation is ____.
Numerical answer — type the value.
Show answer
Answer: 22.5
Along y = 0: 10 + 0.25 × 10 = 12.5; along y = 1: 30 + 0.25 × 10 = 32.5; between them at y = 0.5: 12.5 + 0.5 × 20 = 22.5. Interpolating in the other order gives the same result. Nearest neighbour would return one of the four originals.
Which resampling method uses a 4 × 4 neighbourhood of 16 input pixels?
1. Cubic convolution
2. Bilinear interpolation
3. Nearest neighbour
4. Dark-object subtraction
Show answer
Answer: A — Cubic convolution
Cubic convolution fits cubic weights over 16 pixels; bilinear uses the 4 nearest, nearest neighbour just 1. Dark-object subtraction is a radiometric correction, not a resampling method.
Which of the following are properties of nearest-neighbour resampling?
1. Output values are all original input DNs
2. It is the fastest of the common methods
3. It can displace features by up to about half a pixel, giving a blocky look
4. It produces the smoothest-looking output
Show answer
Answer: A — Output values are all original input DNs; B — It is the fastest of the common methods; C — It can displace features by up to about half a pixel, giving a blocky look
Nearest neighbour copies the closest input value, so no new DNs appear — which is why it is chosen before classification — and it needs no arithmetic. Copying the nearest value shifts features by up to half a pixel and gives stepped edges. Bilinear and cubic convolution are smoother.
Which of the following statements about geometric correction and registration are correct?
1. A first-order (affine) transformation can correct translation, rotation, scale and skew
2. Image-to-image registration is needed before pixel-by-pixel change detection
3. Relief displacement in hilly terrain requires a DEM (orthorectification) for full removal
4. A higher-order polynomial always gives better accuracy away from the GCPs
Show answer
Answer: A — A first-order (affine) transformation can correct translation, rotation, scale and skew; B — Image-to-image registration is needed before pixel-by-pixel change detection; C — Relief displacement in hilly terrain requires a DEM (orthorectification) for full removal
The six affine parameters cover shifts, rotation, two scales and skew; change detection compares the same ground pixel by pixel, so the images must coincide; and relief displacement depends on each point's height, which only a DEM supplies. Higher-order polynomials fit the GCPs more closely but can oscillate between and beyond them — a low residual is not proof of accuracy elsewhere.
Which of the following reduce the brightness difference between a sunlit slope and a shaded slope of the same land cover?
1. A cosine topographic correction using a DEM
2. Taking a ratio of two bands
3. Computing NDVI
4. Nearest-neighbour resampling
Show answer
Answer: A — A cosine topographic correction using a DEM; B — Taking a ratio of two bands; C — Computing NDVI
The cosine correction rescales each pixel by its local illumination angle; and because illumination multiplies all bands by about the same factor, a band ratio — and NDVI, which is built from a ratio of differences — largely cancels it. Resampling changes the grid, not the illumination.