BE Civil Engineering (IOE, TU) Remote Sensing and GIS (IOE, CE 754) Question Paper 2077 Nepal

Principle of EM remote sensing

Remote sensing is the acquisition of information about an object or phenomenon without physical contact, by detecting and recording reflected or emitted electromagnetic energy. The Sun (or an active sensor) illuminates a target; the target reflects/emits energy that travels back through the atmosphere to a sensor, which records it as an image. The data are then transmitted, processed, interpreted and applied.

Energy-flow diagram (components A–G)

        (A) Energy Source (Sun)
             |  incident radiation
             v
   (B) Atmosphere  <-- scattering/absorption
             |
             v
   (C) Target / Earth surface  --(D) reflected/emitted energy-->
             |                                |
             v                                v
   (B) Atmosphere again              (E) Sensor / Platform (records)
                                            |
                                            v
                                   (F) Transmission, Reception, Processing
                                            |
                                            v
                                   (G) Interpretation & Application

• A Energy source, B Radiation & the atmosphere, C Interaction with the target, D Recording of energy by the sensor, E Transmission/reception/processing, F Interpretation & analysis, G Application.

Major EM spectral regions in remote sensing

Atmospheric windows

The atmosphere absorbs EM energy strongly in certain bands (mainly due to $H_2O$, $CO_2$, $O_3$). Wavelength ranges where transmission is high are called atmospheric windows (e.g. 0.4–2.5 µm, 3–5 µm, 8–14 µm, and the microwave region). Sensors are designed to operate within these windows.

Wien's displacement law

$$\lambda_{max} = \frac{b}{T}, \qquad b = 2898\ \mu\text{m}\cdot\text{K}$$

Sun ($T = 5800$ K):

$$\lambda_{max} = \frac{2898}{5800} = 0.4997\ \mu\text{m} \approx \mathbf{0.50\ \mu m}$$

This lies in the visible/reflective region, so solar energy is sensed in the reflective (visible–SWIR) bands.

Earth ($T = 300$ K):

$$\lambda_{max} = \frac{2898}{300} = \mathbf{9.66\ \mu m}$$

This lies in the thermal-IR (8–14 µm) region, so the Earth's emitted energy is best sensed by thermal sensors.

Conclusion: Hot sources (Sun) peak in the reflective region; ambient terrestrial sources peak in the thermal region — this is why remote sensing distinguishes reflective bands from emissive (thermal) bands.

Four types of resolution

• Spatial resolution — smallest ground feature distinguishable; pixel size on the ground (e.g. Landsat-8 OLI = 30 m).
• Spectral resolution — number and width of wavelength bands recorded (e.g. hyperspectral Hyperion = 220 narrow bands).
• Radiometric resolution — number of brightness levels (bit depth) the sensor records (e.g. Landsat-8 = 12-bit = 4096 levels).
• Temporal resolution — revisit interval over the same area (e.g. Landsat = 16 days; MODIS ≈ 1 day).

(a) Swath width

For a symmetric field of view $\theta$ at altitude $H$ (flat-Earth approximation):

$$SW = 2 H \tan\!\left(\frac{\theta}{2}\right) = 2 \times 705 \times \tan(7.5^\circ)$$ $$\tan(7.5^\circ) = 0.13165$$ $$SW = 2 \times 705 \times 0.13165 = 185.6\ \text{km} \approx \mathbf{185.6\ km}$$

(b) Ground sample distance (GSD)

$$GSD = \frac{SW}{N} = \frac{185.6\ \text{km}}{6000} = \frac{185{,}600\ \text{m}}{6000} = 30.93\ \text{m} \approx \mathbf{30.9\ m}$$

(At nadir the GSD is slightly finer than the cross-track average; this is the nominal value.)

(c) Radiometric quantization

• Grey levels $= 2^8 = \mathbf{256}$ levels (0–255).
• Dynamic range in dB:

$$DR = 20\log_{10}(2^8) = 20 \times 8 \times \log_{10}2 = 20 \times 8 \times 0.30103 = \mathbf{48.16\ dB}$$

Summary: Swath ≈ 185.6 km, GSD ≈ 30.9 m, 256 grey levels, ~48 dB dynamic range.

Enhancement vs classification

• Image enhancement improves the visual interpretability of an image (contrast stretch, filtering, ratioing). It does not assign meaning to pixels — output is still an image.
• Image classification assigns each pixel to a thematic class (water, forest, urban…) using spectral signatures (supervised/unsupervised). Output is a thematic map.

Linear contrast stretching (concept)

Raw sensor data often occupy a narrow DN range, giving low-contrast images. A linear stretch maps the input range $[DN_{min}, DN_{max}]$ to the full display range $[0, 255]$:

$$DN_{out} = \left(\frac{DN_{in} - DN_{min}}{DN_{max} - DN_{min}}\right)\times 255$$

(a) NDVI

$$NDVI = \frac{NIR - RED}{NIR + RED} = \frac{180 - 45}{180 + 45} = \frac{135}{225} = \mathbf{0.60}$$

Interpretation: NDVI ranges from −1 to +1. A value of 0.60 indicates dense, healthy green vegetation (high NIR reflectance, strong red absorption by chlorophyll).

(b) Min–max linear stretch

Given $DN_{min} = 40$, $DN_{max} = 168$, range $= 168 - 40 = 128$.

$$DN_{out} = \frac{DN_{in} - 40}{128}\times 255$$

For $DN_{in} = 96$:

$$DN_{out} = \frac{96 - 40}{128}\times 255 = \frac{56}{128}\times 255 = 0.4375 \times 255 = 111.6 \approx \mathbf{112}$$

Check: input 40 → 0, input 168 → 255, confirming the full range is used.

Raster vs vector

(a) Rows, columns, cells

Cell size = 20 m.

• Columns (along 4 km = 4000 m): $4000 / 20 = 200$ columns.
• Rows (along 3 km = 3000 m): $3000 / 20 = 150$ rows.
• Total cells $= 200 \times 150 = \mathbf{30{,}000\ cells}$.

(b) Raw storage size

16-bit = 2 bytes per cell.

$$\text{Bytes} = 30{,}000 \times 2 = 60{,}000\ \text{bytes}$$ $$\text{Size} = \frac{60{,}000}{1{,}048{,}576} = \mathbf{0.0572\ MB}\ (\approx 58.6\ KB)$$

(c) Run-length encoding (RLE)

RLE compresses a raster by storing each run of identical adjacent cell values as a single (value, count) pair instead of repeating the value. For thematic rasters with large homogeneous regions (e.g. land-cover blocks), a row like 2 2 2 2 5 5 5 is stored as (2,4)(5,3), sharply reducing file size. Compression is greatest when the data have low spatial variability; it is ineffective for noisy continuous data.

Three GPS segments

• Space segment — the constellation of ≥24 satellites in ~20,200 km MEO orbits broadcasting timing/ephemeris signals.
• Control segment — master and monitor ground stations that track satellites, compute clock/orbit corrections and upload them.
• User segment — receivers that capture signals and compute position, velocity and time (PVT).

Trilateration principle

Each satellite signal gives the distance (range) from receiver to that satellite. One range defines a sphere of possible positions; two ranges intersect in a circle; three ranges intersect at two points (one rejected as non-physical). Thus three ranges fix 3-D position if the receiver clock were perfect.

Why four satellites?

The receiver clock is cheap and not synchronized to GPS time, introducing an unknown clock bias $\Delta t$. There are therefore four unknowns: $x, y, z, \Delta t$. Four pseudorange equations are needed to solve four unknowns — hence a minimum of four satellites for a 3-D fix.

(a) True geometric range

Clock bias = +2 ms means the receiver over-measures the range by:

$$\Delta R = c \cdot \Delta t = 3\times10^{8} \times 2\times10^{-3} = 6\times10^{5}\ \text{m} = 600\ \text{km}$$

True range:

$$R = \rho - \Delta R = 21{,}300 - 600 = \mathbf{20{,}700\ km}$$

(b) Expected 3-D position error

$$\sigma_{pos} = PDOP \times \sigma_{UERE} = 2.5 \times 6.0\ \text{m} = \mathbf{15.0\ m}$$

A lower PDOP (good satellite geometry) gives smaller position error for the same ranging error.

Elements of visual image interpretation

1. Tone/colour — relative brightness or hue of a feature.
2. Shape — geometric outline (regular = man-made; irregular = natural).
3. Size — absolute or relative dimensions.
4. Pattern — spatial arrangement (e.g. orchard grid, drainage network).
5. Texture — frequency of tonal change (smooth water vs rough forest).
6. Shadow — reveals height/profile of objects.
7. Association — related features occurring together (school + playground).
8. Site/location — topographic position (valley, ridge).

River vs road

• Shape: a river meanders with smooth, irregular curves; a road follows straighter, engineered alignments with constant width.
• Tone/texture: water typically appears dark and smooth; roads are lighter/grey.
• Association/pattern: rivers connect to tributaries and floodplains; roads connect to junctions, bridges and buildings. Thus shape, tone and association are most diagnostic.

(a) Average photo scale

Scale of a vertical photo above terrain of elevation $h$:

$$S = \frac{f}{H - h} = \frac{0.152\ \text{m}}{3040 - 760} = \frac{0.152}{2280}$$ $$S = \frac{1}{15{,}000} \Rightarrow \mathbf{1 : 15{,}000}$$

(Check: $2280 / 0.152 = 15{,}000$.)

(b) Ground length of road

Ground distance = photo distance ÷ scale fraction = photo distance × scale denominator:

$$L = 84\ \text{mm} \times 15{,}000 = 1{,}260{,}000\ \text{mm} = 1{,}260\ \text{m} = \mathbf{1.26\ km}$$

(c) Scale–height relation

Since $S = f/(H-h)$, for a fixed focal length the scale is inversely proportional to the flying height above ground: flying higher gives a smaller-scale (more area, less detail) photograph, and flying lower gives a larger-scale photograph.

Buffering

A buffer creates a new polygon enclosing all locations within a specified distance of a feature (point, line or polygon). Civil application: delineating the right-of-way / land-acquisition zone along a road or the protection zone around a water-supply intake.

Overlay

Overlay combines two or more spatial layers to produce a new layer whose attributes derive from the inputs (union, intersection, identity). Civil application: site suitability for a landfill by intersecting slope, land-use and distance-from-river layers.

Buffer corridor area

Buffer width = 150 m on each side ⇒ total corridor width $W = 2 \times 150 = 300\ \text{m}$. Length $L = 12\ \text{km} = 12{,}000\ \text{m}$.

$$A = L \times W = 12{,}000 \times 300 = 3{,}600{,}000\ \text{m}^2$$

Convert to hectares ($1\ \text{ha} = 10{,}000\ \text{m}^2$):

$$A = \frac{3{,}600{,}000}{10{,}000} = \mathbf{360\ ha}$$

Purpose of the error matrix

An error/confusion matrix cross-tabulates the classified result against ground-reference (truth) data, cell-by-cell. The diagonal holds correctly classified pixels; off-diagonal cells show misclassification. From it we derive overall accuracy, producer's accuracy (omission error) and user's accuracy (commission error), and the Kappa coefficient.

Overall accuracy

Sum of the diagonal ÷ total pixels:

$$OA = \frac{45 + 70 + 65}{200} = \frac{180}{200} = 0.90 = \mathbf{90\%}$$

Producer's accuracy for Water

Producer's accuracy = correctly classified pixels of the class ÷ column (reference) total for that class:

$$PA_{Water} = \frac{45}{53} = 0.849 = \mathbf{84.9\%}$$

This means 84.9% of the actual Water pixels on the ground were correctly mapped as Water (omission error = 15.1%).

DEM

A Digital Elevation Model is a raster representation of the continuous terrain surface, storing the ground elevation $z$ at each grid cell. Derived civil products: slope and aspect maps, contour generation, watershed/drainage delineation, viewshed analysis, cut-and-fill/earthwork volume estimation, flood-inundation mapping. (Any two.)

Slope estimate

Gradient components (rise over run, run = cell size = 30 m):

• East–west: $\dfrac{dz}{dx} = \dfrac{9}{30} = 0.30$
• North–south: $\dfrac{dz}{dy} = \dfrac{6}{30} = 0.20$ (magnitude; sign irrelevant for slope steepness)

Resultant gradient magnitude:

$$G = \sqrt{\left(\tfrac{dz}{dx}\right)^2 + \left(\tfrac{dz}{dy}\right)^2} = \sqrt{0.30^2 + 0.20^2} = \sqrt{0.09 + 0.04} = \sqrt{0.13} = 0.3606$$

Slope angle:

$$\theta = \tan^{-1}(0.3606) = \mathbf{19.8^\circ}$$

(Equivalently a slope of about 36.1%.)

Active vs passive remote sensing

Passive sensors record naturally available reflected/emitted radiation; active sensors transmit a pulse and record the backscattered return.

Advantages of imaging radar (microwave) for Nepal

1. All-weather, day-night capability: microwaves penetrate the dense monsoon cloud cover that frequently blocks optical imaging over the Himalaya, so SAR can monitor floods, landslides and glacial lakes (GLOF risk) during the very monsoon season when hazards peak.
2. Sensitivity to structure and moisture / interferometry: radar backscatter responds to surface roughness, geometry and dielectric (moisture) properties, and SAR interferometry (InSAR) can detect millimetre-scale ground deformation — invaluable for monitoring landslide creep, subsidence and structural/embankment stability. (Additional: signal can penetrate light vegetation and some soil, and provides independent geometric information.)