BE Civil Engineering (IOE, TU) Remote Sensing and GIS (IOE, CE 754) Question Paper 2080 Nepal

(a) EM spectrum regions and atmospheric windows

Remote sensing chiefly uses these regions:

An atmospheric window is a wavelength band where the atmosphere is largely transparent, allowing radiation to pass with little absorption so that a sensor can record useful surface signal. Outside the windows, gases absorb strongly. The dominant absorbers are water vapour (H₂O), carbon dioxide (CO₂) and ozone (O₃). Key windows: 0.4–2.5 µm (visible–SWIR), 3–5 µm and 8–14 µm (thermal), and the microwave region (>1 mm, essentially fully open).

(b) Wien's displacement law

$$\lambda_{max} = \frac{2898\ \mu m\cdot K}{T}$$

(i) Sun, $T = 6000\ \text{K}$:

$$\lambda_{max} = \frac{2898}{6000} = 0.483\ \mu m$$

This falls in the visible (blue-green) region — hence reflected-sunlight (optical) sensing relies on the visible/NIR.

(ii) Earth, $T = 300\ \text{K}$:

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

This falls in the thermal-IR (8–14 µm window) — hence Earth's own emitted energy is sensed thermally.

Significance: Optical sensors are designed around the solar peak (~0.5 µm), while thermal sensors are designed around the terrestrial emission peak (~9.7 µm). The two energy sources dominate different parts of the spectrum, so a single detector cannot optimally serve both.

(c) Spectral reflectance signatures

Reflectance
  ^
  |            Vegetation
  |               ____________
  |        /-----/            \___/\___   (NIR plateau, SWIR water dips)
  |   ____/  (green bump 0.55)
  |  /Soil  ___________________________   (steady rise, dips at 1.4,1.9)
  | /  ____/
  |/__/  Water
  |______\________________________ ___> Wavelength (µm)
  0.4  0.7    1.0     1.5    2.0   2.5

• Healthy vegetation: low blue/red reflectance (chlorophyll absorption), small green peak (~0.55 µm), a sharp red-edge rise near 0.7 µm and a high NIR plateau (leaf cell structure), with SWIR dips at 1.4 and 1.9 µm due to leaf water.
• Water: moderate-low reflectance in visible (highest in blue), and strong absorption beyond ~0.75 µm — near-zero in NIR/SWIR, making water appear very dark in NIR.
• Dry bare soil: smooth, gradually increasing reflectance from visible through NIR/SWIR, with minor moisture/clay absorption dips around 1.4 and 1.9 µm; generally higher than vegetation in red, lower in NIR.

(a) Four resolutions

• Spatial resolution — smallest ground area represented by one pixel; the system's ability to resolve detail. Example: Landsat-8 OLI = 30 m.
• Spectral resolution — number and width of wavelength bands recorded. Example: a sensor with band width 0.07 µm (e.g. 0.63–0.69 µm red) has finer spectral resolution than a single 0.4–0.7 µm panchromatic band.
• Radiometric resolution — number of brightness (grey) levels = $2^n$ for $n$-bit data. Example: 8-bit → 256 levels.
• Temporal resolution — revisit interval over the same area. Example: Landsat = 16 days.

(b) Pushbroom geometry and storage

(i) Ground sampling distance across-track:

$$GSD = \frac{\text{swath width}}{\text{number of detectors}} = \frac{60\ \text{km}}{12000} = \frac{60000\ \text{m}}{12000} = 5\ \text{m}$$

GSD = 5 m.

(ii) Grey levels for 12-bit: $2^{12} = 4096$ levels.

Scene is square with 5 m pixels:

$$\text{pixels per side} = \frac{60000}{5} = 12000$$ $$\text{total pixels} = 12000 \times 12000 = 1.44\times10^{8}$$

12 bits = 1.5 bytes per pixel:

$$\text{bytes} = 1.44\times10^{8} \times 1.5 = 2.16\times10^{8}\ \text{bytes} \approx 206\ \text{MB}$$

4096 grey levels; ≈ 2.16 × 10⁸ bytes (≈ 206 MiB) per band.

(Check: $2.16\times10^{8}/1024^2 = 206.0$ MiB.)

(c) Orbital period

Orbital radius:

$$r = R_E + h = 6378 + 700 = 7078\ \text{km}$$

Kepler's third law:

$$T = 2\pi\sqrt{\frac{r^3}{GM}}$$ $$r^3 = 7078^3 = 3.546\times10^{11}\ \text{km}^3$$ $$\frac{r^3}{GM} = \frac{3.546\times10^{11}}{3.986\times10^{5}} = 8.896\times10^{5}\ \text{s}^2$$ $$\sqrt{8.896\times10^{5}} = 943.2\ \text{s}$$ $$T = 2\pi \times 943.2 = 5926\ \text{s} = 98.8\ \text{min}$$

Orbits per day:

$$N = \frac{86400\ \text{s}}{5926\ \text{s}} = 14.6\ \text{orbits/day}$$

T ≈ 98.8 minutes; ≈ 14.6 orbits per day (typical of a LEO Sun-synchronous Earth-observation satellite).

(a) Enhancement vs. classification + linear stretch

• Image enhancement improves the visual interpretability of an image (contrast stretching, filtering, ratioing) without assigning meaning to pixels; the output is still an image for human viewing.
• Image classification assigns each pixel to a thematic class (e.g. water, forest, urban) using its spectral values, producing a thematic map; it is the analytical/decision step.

Linear contrast stretch: the narrow input DN range actually present in the data is linearly expanded to fill the full display range (0–255 for 8-bit), increasing visual contrast. Pixels at the input minimum map to 0 and those at the input maximum map to 255.

(b) Linear contrast-stretch transformation

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

Here $\text{DN}_{min}=40$, $\text{DN}_{max}=168$, so $\text{DN}_{max}-\text{DN}_{min}=128$:

$$\text{DN}_{out} = \frac{(\text{DN}_{in}-40)}{128}\times 255 = 1.992\,(\text{DN}_{in}-40)$$

• $\text{DN}_{in}=40$: $\frac{0}{128}\times255 = \mathbf{0}$
• $\text{DN}_{in}=104$: $\frac{64}{128}\times255 = 0.5\times255 = \mathbf{127.5 \approx 128}$
• $\text{DN}_{in}=168$: $\frac{128}{128}\times255 = \mathbf{255}$

The midpoint input (104) maps near mid-grey, confirming a correct linear mapping.

(c) NDVI

$$NDVI = \frac{\rho_{NIR}-\rho_{RED}}{\rho_{NIR}+\rho_{RED}} = \frac{0.52-0.08}{0.52+0.08} = \frac{0.44}{0.60} = 0.733$$

NDVI = 0.73. A value this high indicates dense, healthy green vegetation (strong NIR reflectance, strong red chlorophyll absorption). The theoretical range of NDVI is −1 to +1; bare soil ≈ 0.1–0.2, water and clouds give negative or near-zero values, and dense canopy approaches +1.

(a) Raster vs. vector

(b) Raster size computation

Cell size = 25 m.

$$\text{columns} = \frac{20000\ \text{m}}{25} = 800$$ $$\text{rows} = \frac{15000\ \text{m}}{25} = 600$$ $$\text{total cells} = 800 \times 600 = 480000\ \text{cells}$$

2 bytes per cell:

$$\text{size} = 480000 \times 2 = 960000\ \text{bytes}$$ $$= \frac{960000}{1024\times1024} = 0.9155\ \text{MiB} \approx 0.92\ \text{MB}$$

800 columns × 600 rows = 480,000 cells; ≈ 0.92 MB (960 kB).

(c) Vector overlay

• Union keeps all features and attributes from both input layers; the output extent is the combined area (logical OR of geometry). Useful when no information should be discarded.
• Intersection keeps only the area common to both layers (logical AND), carrying attributes from both.

Civil application — landfill siting: intersect candidate-land polygons with constraint layers (e.g. slope < 10% AND >500 m from rivers AND outside settlements). The intersection yields only parcels satisfying every criterion simultaneously, giving engineers a defensible set of feasible sites for a sanitary landfill; a union, by contrast, could map the total area touched by any single factor for impact assessment.

(a) GPS segments and trilateration

The three GPS segments are:

1. Space segment — the constellation of ~24+ satellites in ~20,200 km MEO orbits transmitting ranging signals and navigation messages.
2. Control segment — ground monitoring/master control stations that track satellites, compute orbits and clock corrections, and upload them.
3. User segment — the GPS receivers that decode signals and compute position, velocity and time.

Trilateration: each satellite-to-receiver distance defines a sphere of that radius centred on the satellite. The intersection of multiple spheres fixes the receiver's location. Two spheres intersect in a circle, three in (generally) two points, one of which is rejected as non-physical.

Why four satellites: the receiver clock is far less accurate than the satellite atomic clocks, so the measured ranges are pseudo-ranges containing an unknown clock-bias term. There are four unknowns — $X, Y, Z$ and the receiver clock offset $\Delta t$ — therefore at least four range equations (four satellites) are needed to solve them.

(b) Pseudo-range

$$\text{range} = c \times t = 3\times10^{8}\ \text{m/s} \times 0.07\ \text{s} = 2.1\times10^{7}\ \text{m}$$

Pseudo-range = 2.1 × 10⁷ m = 21,000 km (consistent with the GPS orbital altitude of ~20,200 km plus slant geometry).

(c) Horizontal error and accuracy techniques

Horizontal position error:

$$\sigma_H = HDOP \times \sigma_{UERE} = 1.8 \times 4\ \text{m} = 7.2\ \text{m}$$

Expected horizontal error (1σ) = 7.2 m.

• DGPS (Differential GPS): a reference receiver at a known point computes range corrections for common (correlated) errors — satellite clock/ephemeris, ionospheric/tropospheric delay — and broadcasts them to nearby rovers, reducing error to sub-metre/few-metre level. It is code-based and works over moderate baselines.
• RTK / PPP: carrier-phase techniques. RTK uses a base station and resolves integer phase ambiguities in real time to give cm-level accuracy over short baselines; PPP (Precise Point Positioning) uses a single receiver with precise satellite orbit/clock products to reach decimetre–centimetre accuracy globally after convergence, without a local base.

(a) Passive vs. active sensing

• Passive remote sensing records natural energy — reflected sunlight or thermal emission from the surface; it has no own source. Examples: Landsat/Sentinel-2 optical imagers, thermal scanners, aerial photography.
• Active remote sensing supplies its own energy and measures the returned signal. Examples: RADAR / SAR (Sentinel-1), LiDAR.

Key contrasts: passive depends on the Sun/daylight and clear weather; active works day or night and (for radar) through clouds, and can measure precise distance from travel time.

(b) Value of SAR for Nepal

SAR uses microwaves that penetrate clouds, haze and rain and need no sunlight, so it images reliably during the monsoon and at night when optical sensors fail. This all-weather, day-night capability is critical in cloud-prone Nepal. Civil applications: flood-extent mapping during monsoon, landslide/ground-deformation monitoring via InSAR, and infrastructure/subsidence assessment.

(c) Slant range from radar travel time

The pulse travels to the target and back, so the one-way distance uses half the round-trip time:

$$R = \frac{c \times t}{2} = \frac{3\times10^{8}\ \text{m/s} \times 40\times10^{-6}\ \text{s}}{2}$$ $$= \frac{12000\ \text{m}}{2} = 6000\ \text{m}$$

Slant range = 6000 m = 6 km.

(a) Supervised vs. unsupervised classification

• Supervised: the analyst defines training samples for known classes; the algorithm (e.g. maximum likelihood) learns each class signature and labels all pixels. Requires prior ground knowledge; classes are known in advance.
• Unsupervised: the algorithm groups pixels into spectral clusters (e.g. ISODATA, k-means) with no training data; the analyst labels the clusters afterward. Classes emerge from the data.

(b) Accuracy assessment

The diagonal entries are correctly classified pixels.

Overall accuracy:

$$OA = \frac{\sum \text{diagonal}}{N} = \frac{45 + 70 + 62}{200} = \frac{177}{200} = 0.885 = \mathbf{88.5\%}$$

Kappa coefficient:

$$\kappa = \frac{p_o - p_e}{1 - p_e}$$

where $p_o = OA = 0.885$ (observed agreement).

Expected agreement $p_e$ from marginal totals:

$$p_e = \frac{1}{N^2}\sum_i (\text{row}_i \times \text{col}_i)$$ $$= \frac{(50\times53)+(80\times80)+(70\times67)}{200^2}$$ $$= \frac{2650 + 6400 + 4690}{40000} = \frac{13740}{40000} = 0.3435$$

Therefore:

$$\kappa = \frac{0.885 - 0.3435}{1 - 0.3435} = \frac{0.5415}{0.6565} = 0.825$$

Overall accuracy = 88.5%; Kappa = 0.825, indicating almost perfect agreement (κ > 0.80) beyond what chance alone would produce.

(a) DEM definition and generation

A Digital Elevation Model (DEM) is a raster representation of terrain surface elevation, storing a height (z) value for each grid cell over a regular X–Y grid. (A DTM additionally implies bare-earth surface; a DSM includes objects.)

Two generation methods: (i) stereo-photogrammetry / stereo satellite imagery (e.g. ASTER, SRTM-style); (ii) LiDAR / airborne laser scanning. (Also: contour interpolation from topographic maps, InSAR.)

(b) Slope by central finite difference

Label the window:

 a  b  c     1450 1456 1462
 d  e  f  =  1448 1455 1465
 g  h  i     1446 1452 1460

Cell size $L = 30\ \text{m}$.

East–West gradient (Horn 3rd-order):

$$\frac{dz}{dx} = \frac{(c + 2f + i) - (a + 2d + g)}{8L}$$ $$= \frac{(1462 + 2(1465) + 1460) - (1450 + 2(1448) + 1446)}{8 \times 30}$$ $$= \frac{(1462 + 2930 + 1460) - (1450 + 2896 + 1446)}{240}$$ $$= \frac{5852 - 5792}{240} = \frac{60}{240} = 0.25$$

North–South gradient:

$$\frac{dz}{dy} = \frac{(g + 2h + i) - (a + 2b + c)}{8L}$$ $$= \frac{(1446 + 2(1452) + 1460) - (1450 + 2(1456) + 1462)}{240}$$ $$= \frac{(1446 + 2904 + 1460) - (1450 + 2912 + 1462)}{240}$$ $$= \frac{5810 - 5824}{240} = \frac{-14}{240} = -0.0583$$

Slope magnitude:

$$\text{rise/run} = \sqrt{\left(\tfrac{dz}{dx}\right)^2 + \left(\tfrac{dz}{dy}\right)^2} = \sqrt{0.25^2 + (-0.0583)^2}$$ $$= \sqrt{0.0625 + 0.0034} = \sqrt{0.0659} = 0.2567$$

Slope in percent: $0.2567 \times 100 = \mathbf{25.7\%}$

Slope in degrees: $\arctan(0.2567) = \mathbf{14.4^\circ}$

(a) Buffering and map algebra

• Buffering creates a zone of a specified distance around point, line or polygon features (e.g. a corridor around a road, a protection zone around a river). It answers proximity questions and defines areas of influence or exclusion.
• Map algebra (raster overlay) performs cell-by-cell mathematical/logical operations on one or more aligned raster layers — arithmetic ($+,-,\times$), relational, or Boolean (AND, OR, NOT) — to produce a new raster. It is the basis of weighted suitability modelling.

(b) Buffer area of the road corridor

Buffer on each side = 150 m, so total corridor width:

$$W = 2 \times 150 = 300\ \text{m}$$

Length $= 8\ \text{km} = 8000\ \text{m}$.

$$A = L \times W = 8000 \times 300 = 2{,}400{,}000\ \text{m}^2$$

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

$$A = \frac{2{,}400{,}000}{10{,}000} = 240\ \text{ha}$$

Buffer area = 240 hectares.

(c) Boolean AND overlay

Output cell = 1 only where both layers are 1:

L1   L2   AND
1 0  1 1  1 0
1 1  0 1  0 1

Result:

 1  0
 0  1

Number of suitable cells = 2 (top-left and bottom-right).

(a) Datum, projection, CRS

• Datum: a reference model of the Earth (an ellipsoid plus its position/orientation) that defines the origin for measuring coordinates, e.g. WGS-84 (geocentric) or Everest 1830 (local). It fixes where zero is.
• Map projection: the mathematical transformation that flattens the curved ellipsoidal surface onto a plane, inevitably introducing some distortion of area, shape, distance or direction (e.g. UTM, Transverse Mercator).
• Coordinate Reference System (CRS): the complete framework = datum + projection + coordinate units, allowing any location to be expressed unambiguously (e.g. UTM Zone 45N on WGS-84).

Nepal: topographic maps (Survey Department) use the Modified Universal Transverse Mercator (MUTM) / Transverse Mercator projection on the Everest 1830 datum (newer products use UTM on WGS-84). Nepal spans UTM zones 44N and 45N.

(b) Georeferencing with GCPs and RMSE

Georeferencing assigns real-world coordinates to an un-referenced scanned map/image:

1. Display the scanned map and identify ground control points (GCPs) — features whose true coordinates are known (road intersections, survey monuments, well-defined corners).
2. For each GCP, link its image (pixel) location to its known map/world coordinate.
3. Collect several well-distributed GCPs (a 1st-order/affine transform needs ≥3; higher-order/polynomial needs more) and fit a transformation (affine/polynomial) that maps pixel coordinates to ground coordinates.
4. Apply the transform and resample (nearest-neighbour, bilinear or cubic) to produce the georeferenced raster.

Role of RMSE: the Root Mean Square Error measures the residual between each GCP's transformed position and its true position:

$$RMSE = \sqrt{\frac{1}{n}\sum_{i=1}^{n}\left[(x_i'-x_i)^2 + (y_i'-y_i)^2\right]}$$

A low RMSE (ideally less than one pixel / acceptable map tolerance) indicates a good fit; high RMSE flags poor or mislocated GCPs, which should be re-measured or removed before accepting the georeferencing.

(a) Flood hazard / watershed management

Remote sensing supplies the spatial inputs and GIS performs the analysis:

• DEM (from satellite stereo/SAR/LiDAR) is processed in GIS to delineate the watershed boundary, stream network, flow direction and accumulation, and to derive slope and drainage density.
• Multispectral imagery maps land use / land cover and impervious surfaces, which feed runoff models (e.g. SCS curve number) by assigning infiltration characteristics.
• During floods, SAR/optical imagery maps inundation extent in near-real time (all-weather with SAR), and GIS overlays this on settlements and infrastructure to estimate exposure.
• GIS map algebra / weighted overlay of slope, rainfall, land cover, drainage proximity and elevation produces a flood-hazard zonation map to guide embankment design, land-use control and early warning.

(b) Two further applications

1. Transportation route alignment: RS imagery and DEM provide terrain, slope, geology and land-cover data along candidate corridors. GIS least-cost-path / suitability analysis weights gradient, cut-and-fill volume, settlement avoidance and hazard zones to identify the optimal, lowest-cost, lowest-impact road/highway alignment.

2. Landslide susceptibility mapping: RS detects slope-failure indicators and land-cover change; GIS overlays causative factors — slope, aspect, lithology, distance to faults/roads, rainfall and land use — using weighted overlay or AHP to produce a susceptibility map. This directs slope-stabilization, drainage works and safe siting of structures — especially relevant in Nepal's mountainous, monsoon terrain.

(Alternatively: urban land-use planning uses time-series imagery for change detection and GIS for zoning, utility planning and growth modelling.)

Together, RS provides timely, wide-area, multi-temporal data while GIS stores, integrates, analyses and visualizes it for engineering decisions.