optional · CSS study notes
Image Sensing, Acquisition, and Representation
CSS Optional — Paper-II, Section-B · P-II.V.II · Covers two syllabus items: Image Sensing and Acquisition Techniques and Representing Digital Image. Pixel neighbourhoods and distance measures are prime material for short numericals. Sampling and quantization have their own page (next).
1. Image Sensing and Acquisition
An image is formed when an illumination source (sunlight, X-rays, ultrasound…) interacts with a scene and the reflected or transmitted energy is captured by a sensor, which converts it to a voltage. That voltage is then digitized.
| Arrangement | How it works | Examples |
|---|---|---|
| Single sensor | One photodiode (often with a colour filter); the sensor or the object is moved mechanically in x and y to cover the scene | Drum scanners, microdensitometers, laser scanning |
| Sensor strip (line) | A row of sensors captures one line; motion perpendicular to the strip builds the second dimension | Flatbed scanners, satellite push-broom imagers; ring-shaped strips in CT scanners |
| Sensor array (2-D) | A full grid of sensors captures the whole image at once — no motion needed | Digital cameras, phone cameras (CCD / CMOS) |
2. CCD vs CMOS Sensors
| Feature | CCD | CMOS |
|---|---|---|
| Read-out | Charge shifted across the chip to one amplifier | Each pixel has its own amplifier |
| Image quality | Traditionally lower noise, high uniformity | Now comparable or better |
| Power / cost | Higher power, costlier | Low power, cheap, on-chip processing |
| Where used | Scientific, astronomy, older cameras | Almost all phones and modern cameras |
Colour capture: most cameras use one sensor with a Bayer filter mosaic (R-G-G-B) — 50% green because the eye is most sensitive to green. Missing colours at each pixel are interpolated (demosaicing). High-end cameras may use three sensors with a prism.
3. Acquisition Techniques Beyond the Camera
- X-ray / CT: X-rays pass through the body; CT reconstructs cross-sections from many angles.
- MRI: radio waves in a strong magnetic field; excellent soft-tissue contrast.
- Ultrasound: reflected sound waves; real-time, no radiation.
- Thermal infrared: emitted heat — night vision, fever screening, building inspection.
- LiDAR / radar: laser or microwave ranging to build depth maps (autonomous vehicles, terrain mapping).
- Multispectral satellite imaging: several bands (visible + infrared) for vegetation and water analysis.
4. A Simple Image Formation Model
f(x, y) = i(x, y) · r(x, y)
- i(x, y) — illumination: amount of source light falling on the scene, 0 < i < ∞ (set by the source).
- r(x, y) — reflectance: fraction reflected by the object, 0 < r < 1 (0 = total absorption, 1 = total reflection; set by the object).
- Typical reflectance: black velvet ≈ 0.01, stainless steel ≈ 0.65, snow ≈ 0.93.
- For transmission imaging (X-ray), r is replaced by a transmissivity term.
This model is the basis of homomorphic filtering (separating illumination from reflectance using logs).
5. Representing a Digital Image
- A digital image is a 2-D function f(x, y) sampled into an M × N matrix; each element is a pixel (picture element).
- Origin at the top-left (0, 0); x indexes rows (downward), y indexes columns (to the right) in Gonzalez's convention.
- Each pixel holds one of L = 2^k intensity levels.
- Three ways to show it: (a) a 3-D surface plot (x, y, intensity), (b) a visual intensity array (the picture), (c) a numerical matrix (what the computer stores).
- Dynamic range: ratio of max to min detectable intensity; contrast: difference between highest and lowest intensity in the image.
- Image types: binary (1-bit), grayscale (8-bit), colour (3 × 8-bit), indexed, multispectral.
6. Relationships Between Pixels
Neighbourhoods of pixel p at (x, y):
- N4(p) — 4 horizontal/vertical neighbours: (x±1, y), (x, y±1)
- ND(p) — 4 diagonal neighbours: (x±1, y±1)
- N8(p) = N4(p) ∪ ND(p) — all 8 neighbours
Adjacency (V = set of intensity values that count as "connected", e.g. V = {1} in a binary image):
- 4-adjacency: q ∈ N4(p)
- 8-adjacency: q ∈ N8(p)
- m-adjacency (mixed): q ∈ N4(p), OR q ∈ ND(p) and N4(p) ∩ N4(q) contains no pixel with a value in V. Introduced to remove the multiple-path ambiguity of 8-adjacency.
Path, connectivity, region, boundary: a path is a sequence of adjacent pixels; pixels are connected if a path exists between them within a set; a connected set is a region; its boundary is the set of region pixels that have a neighbour outside the region.
7. Distance Measures
For p = (x, y) and q = (s, t):
| Distance | Formula | Pixels at distance ≤ r form a |
|---|---|---|
| Euclidean De | √((x−s)² + (y−t)²) | Disk |
| City-block D4 | x−s | |
| Chessboard D8 | max( | x−s |
📐 Worked Numericals
N1 — Distances: p = (2, 3), q = (5, 7). |Δx| = 3, |Δy| = 4.
- De = √(9 + 16) = 5
- D4 = 3 + 4 = 7
- D8 = max(3, 4) = 4
Note D8 ≤ De ≤ D4 always holds.
N2 — Image formation: sunlight illumination i = 90,000 lm/m² on snow (r = 0.93) vs black velvet (r = 0.01). f(snow) = 83,700; f(velvet) = 900 → ratio ≈ 93:1 — this is the dynamic range the sensor must capture.
N3 — Neighbourhood count: for a pixel at the image corner, N4 has 2 valid neighbours and N8 has 3; on an edge (not a corner), N4 has 3 and N8 has 5.
N4 — Bayer sampling: a 12-megapixel Bayer sensor has 6 M green, 3 M red and 3 M blue sensing sites — each output pixel's other two colours are interpolated.
🎯 CSS Exam Angle
For "explain image acquisition", use the three sensor arrangements table with a quick sketch of each, then the formation model f = i · r. For "representation", draw a small matrix with the origin marked and explain N4/N8 with a 3 × 3 grid. A distance numerical takes two lines and is a safe inclusion in any representation answer.
📌 Past CSS Paper Frequency
| Year | Question |
|---|---|
| Syllabus | Sensing, acquisition and representation are named explicitly; no verified standalone question found in 2017–2021 papers — expect short notes or an introductory part |
⚠️ MCQ Traps
- Reflectance is bounded 0 < r < 1; illumination is not bounded above.
- Bayer filter is 50% green, 25% red, 25% blue.
- m-adjacency exists to remove ambiguous multiple paths in 8-adjacency — not to add connections.
- D4 is the city-block (diamond) distance; D8 is chessboard (square).
- CT uses X-rays with a ring of sensors; MRI uses radio waves.
✅ Self-Check (with Answers)
Q1. Name the three sensor arrangements used for image acquisition, with one device each.
Ans: Single sensor with mechanical motion in x and y (drum scanner); sensor strip with motion in one direction (flatbed scanner, satellite push-broom, CT ring); 2-D sensor array capturing the whole image at once (CCD/CMOS digital camera).
Q2. Write the image formation model and state the ranges of its components.
Ans: f(x, y) = i(x, y) · r(x, y), where illumination 0 < i < ∞ (set by the source) and reflectance 0 < r < 1 (set by the object: 0 = total absorption, 1 = total reflection).
Q3. Why does a Bayer filter have twice as many green elements as red or blue?
Ans: The human eye is most sensitive to green, and green carries most of the luminance (brightness) detail, so sampling it more densely gives a sharper-looking image. Missing colours at each pixel are filled in by demosaicing.
Q4. Why is m-adjacency needed?
Ans: 8-adjacency can create multiple (ambiguous) paths between two pixels along a diagonal. m-adjacency allows a diagonal link only when the two pixels share no common 4-neighbour in V, removing the ambiguity while keeping connectivity.
Q5. Find De, D4 and D8 between p = (1, 1) and q = (4, 5).
Ans: |Δx| = 3, |Δy| = 4 → De = √(9 + 16) = 5; D4 = 3 + 4 = 7; D8 = max(3, 4) = 4.
Q6. CCD or CMOS for a smartphone — why?
Ans: CMOS: each pixel has its own amplifier, so it uses far less power, is cheaper, and allows on-chip processing — essential for battery-powered phones. Its image quality now matches CCD.
📜 Past-Paper Questions & MCQs (CSS 2016–2026)
2019 · Paper-I · Q5(c) · 5 marks
Q. An image is a representation of information. How does a computer represent an image internally? Name different algorithms used to extract features from images.
Ans.
- Internal representation: an image is a two-dimensional array (matrix) of pixels f(x, y). Each pixel holds an intensity value. A grey-scale pixel uses 8 bits (0–255), and a colour pixel uses three values (R, G, B) of 8 bits each (24 bits). A binary image uses 1 bit. The array is stored in memory row by row, and file formats (BMP, PNG, JPEG) add headers and compression.
- Feature extraction algorithms:
- Edges: Sobel, Prewitt, Canny.
- Corners and interest points: Harris, FAST, SIFT, SURF, ORB.
- Shape and texture: histogram of oriented gradients (HOG), local binary patterns (LBP), Gabor filters, Hough transform.
- Colour: colour histograms.
- Learned features: convolutional neural networks.
2024 · Paper-II · Q7(c) · 6 marks
Q. Compare and contrast various image sensing and acquisition techniques. Discuss the advantages and limitations of methods such as CCD and CMOS.
Ans.
- Image acquisition: a sensor converts light (energy) into an electrical signal, which is sampled and quantised into a digital image. Sensor arrangements are a single sensor, a line (strip) sensor and an array sensor (a camera).
| Basis | CCD | CMOS |
|---|---|---|
| Readout | Charge is shifted pixel by pixel to a single output amplifier | Each pixel has its own amplifier and converter |
| Image quality | Low noise, uniform, high quality (astronomy, medical, scientific) | Was noisier, now close to CCD quality |
| Power | High consumption | Low consumption |
| Speed | Slower readout | Faster, with windowed readout and high frame rates |
| Cost and integration | More expensive, special manufacturing | Cheap and on-chip processing (system on a chip) |
| Used in | Scientific and high-end cameras | Phones, webcams and most modern cameras |
- Other acquisition methods: scanners (line sensor), X-ray, CT, MRI, ultrasound and satellite and radar imaging.
MCQs
2025 P-II · Q6. For a continuous image f(x, y), quantization is defined as:
- (A) Digitizing the coordinate values
- (B) Digitizing the amplitude values ✅
- (C) Both (A) and (B)
- (D) None of these
- Answer: (B).
- Why: quantisation converts the continuous intensity (amplitude) into a finite number of grey levels. Digitising the coordinates is sampling, so (A) and (C) are wrong.