BE Civil Engineering (IOE, TU) Remote Sensing and GIS (IOE, CE 754) Question Paper 2078 Nepal

Definition. Remote sensing is the science (and art) of acquiring information about an object, area, or phenomenon through the analysis of data acquired by a device (sensor) that is not in physical contact with the target, typically by measuring reflected or emitted electromagnetic energy.

Role of the EM spectrum. All remote sensing relies on the interaction of EM radiation with matter. The Sun (and the Earth) emit radiation across a continuum of wavelengths; different surface materials reflect, absorb, transmit, and emit this radiation differently as a function of wavelength. By measuring the radiance in selected wavelength bands, the sensor builds a spectral signature that allows discrimination of features (water, vegetation, soil, built-up area, etc.). The usable regions are: visible (0.4-0.7 $\mu$m), near-IR (0.7-1.3 $\mu$m), mid-IR / SWIR (1.3-3 $\mu$m), thermal IR (3-14 $\mu$m), and microwave (1 mm - 1 m).

Atmospheric windows (labelled sketch).

Atmospheric
transmission
 100% |  ___        __        ____            ________________
      | /   \      /  \      /    \          /                \
      |/     \____/    \____/      \________/                  \
   0% +----+----+----+----+----+----+----+----+----+----+----+--->
      0.4  0.7  1.3  3.0      5    8        14   (mm)  microwave  wavelength
        VIS  NIR  SWIR  TIR-1   TIR-2          radio/microwave
  (absorption gaps caused mainly by H2O, CO2, O3)

An atmospheric window is a wavelength region where the atmosphere is largely transparent (high transmittance), so radiation reaches the sensor with little absorption. Outside the windows, gases such as water vapour, carbon dioxide and ozone strongly absorb energy, blocking it from the sensor. The principal windows are: 0.4-2.5 $\mu$m (optical/reflective), 3-5 $\mu$m and 8-14 $\mu$m (thermal), and the microwave region (>1 mm, essentially all-weather). Sensors are designed to operate inside these windows because energy in the absorption bands never reaches the sensor (no signal) and any measurement there would be dominated by atmospheric, not surface, information.

Wien's displacement law.

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

(a) Sun, $T = 5800\,\text{K}$:

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

This lies in the green/visible region, confirming the Sun is the dominant source for reflective (optical) remote sensing.

(b) Terrestrial surface, $T = 300\,\text{K}$:

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

This lies in the 8-14 $\mu$m thermal IR window, confirming that to sense the Earth's own emitted (emissive) energy we use thermal IR sensors.

Comment. Because the Sun peaks at ~0.5 $\mu$m, reflective remote sensing (VIS-SWIR) measures reflected sunlight and needs daylight. Because the Earth peaks at ~9.7 $\mu$m, emissive (thermal) remote sensing measures the surface's own radiation and works day or night. The two regimes therefore use different physical principles and different sensor designs.

Four types of resolution.

1. Spatial resolution - the smallest ground area imaged by one pixel; governs how much spatial detail is visible (e.g., 30 m, 10 m). Determined largely by the IFOV and altitude.
2. Spectral resolution - the number and width of the wavelength bands the sensor records; finer (narrower, more) bands give better material discrimination (multispectral vs. hyperspectral).
3. Radiometric resolution - the sensitivity to differences in energy, expressed as the number of bits (quantisation levels) used to record each pixel; more bits = finer brightness discrimination.
4. Temporal resolution - the revisit interval, i.e., how often the same area is imaged; important for change detection and monitoring dynamic phenomena.

Given: number of detectors $N = 7000$, $\text{IFOV} = 42\,\mu\text{rad} = 42\times10^{-6}\,\text{rad}$, altitude $H = 700\,\text{km} = 700{,}000\,\text{m}$, radiometric resolution = 12 bits.

(a) Ground sample distance (GSD) at nadir.

$$\text{GSD} = \text{IFOV} \times H = (42\times10^{-6}\,\text{rad})(700{,}000\,\text{m})$$ $$= 29.4\,\text{m}$$

GSD = 29.4 m (about a 30 m pixel).

(b) Swath width.

$$\text{Swath} = N \times \text{GSD} = 7000 \times 29.4\,\text{m} = 205{,}800\,\text{m}$$ $$= 205.8\,\text{km}$$

Swath width = 205.8 km.

(c) Brightness levels for 12-bit radiometric resolution.

$$\text{levels} = 2^{12} = 4096$$

Each pixel can record 4096 distinct brightness levels (digital numbers 0-4095).

Trade-off (spatial vs. temporal). For a given sensor, increasing spatial resolution (smaller GSD) usually means a narrower swath, so the satellite covers less ground per pass and therefore takes longer to revisit the same area, degrading temporal resolution. Conversely, wide-swath, coarse-resolution sensors (e.g., MODIS) revisit daily. Designers balance these depending on the application: a daily wide view for weather/monitoring, or fine detail with infrequent revisit for mapping.

Enhancement vs. classification.

• Image enhancement improves the visual interpretability of an image for a human analyst (e.g., contrast stretching, filtering, ratioing). It changes how the data look but does not assign meaning/labels to pixels.
• Image classification assigns each pixel to a thematic class (water, forest, urban...) based on its spectral values, producing a categorical (labelled) output map. It is automated interpretation, typically supervised or unsupervised.

Linear contrast stretch. The input data only span DN $= 3$ to DN $= 5$, so the image looks flat (low contrast). A linear stretch maps the input minimum to 0 and the input maximum to 7:

$$DN_{out} = \left(\frac{DN_{in} - \min}{\max - \min}\right)\times(L_{max}) = \left(\frac{DN_{in} - 3}{5 - 3}\right)\times 7$$

with $\min = 3$, $\max = 5$, $L_{max} = 7$.

Lookup table (rounded to nearest integer):

(Values 0, 1, 2, 6, 7 do not occur in the input.)

Stretched output image:

0 4 4 7 0
4 7 7 4 4
0 4 7 7 4
4 4 4 7 0
0 4 4 4 7

The three original grey values (3, 4, 5) are now spread across 0, 4 and 7, dramatically increasing visible contrast while preserving the relative ordering of the data.

Raster vs. vector comparison.

Given: study area $6\,\text{km}\times4\,\text{km}$, cell size $= 20\,\text{m}$, 8 bits = 1 byte per cell.

(a) Rows, columns, total cells.

• Width $= 6\,\text{km} = 6000\,\text{m} \Rightarrow$ columns $= 6000/20 = 300$.
• Height $= 4\,\text{km} = 4000\,\text{m} \Rightarrow$ rows $= 4000/20 = 200$.
• Total cells $= 300 \times 200 = 60{,}000$ cells.

Columns = 300, Rows = 200, Total = 60,000 cells.

(b) Uncompressed storage. Each cell = 8 bits = 1 byte.

$$\text{Storage} = 60{,}000 \text{ cells} \times 1 \text{ byte} = 60{,}000\,\text{bytes}$$ $$= \frac{60{,}000}{1024} = 58.59\,\text{KB}$$

Storage ≈ 58.6 KB (60,000 bytes).

Where vector is preferable. For a road or utility network analysis (e.g., finding the shortest route for an ambulance, or tracing flow through a pipe network), the vector model is clearly better: it stores explicit topology (connectivity of arcs and nodes), allows precise linear referencing, and supports network algorithms that a raster grid cannot represent without large error. Similarly, for cadastral/land-parcel mapping the exact polygon boundaries and attributes are required, which raster cells cannot reproduce sharply.

GPS working principle. GPS determines position by measuring the time EM signals take to travel from satellites (of known position) to the receiver and converting these times to distances (ranges). It has three segments:

• Space segment - the constellation of ~24+ satellites in ~20,200 km medium Earth orbits, each broadcasting a coded signal with its position (ephemeris) and a precise atomic-clock time.
• Control segment - ground monitoring/master stations that track the satellites, compute orbit and clock corrections, and upload them.
• User segment - the receivers that pick up the signals and compute position, velocity and time.

Trilateration. The receiver measures its distance to several satellites. Each distance defines a sphere of that radius centred on the satellite. The receiver lies on the surface of each sphere; the intersection of the spheres fixes the position. With three perfect ranges, three spheres intersect at (essentially) one point in 3-D.

Why four satellites? The receiver's own clock is much cheaper than the satellites' atomic clocks and carries an unknown time bias $\Delta t$. The four unknowns are the position coordinates $(X, Y, Z)$ and the clock bias $\Delta t$. Four equations (one pseudorange per satellite) are needed to solve for these four unknowns. Hence at least four satellites are required for a 3-D fix; the fourth effectively synchronises the receiver clock.

Pseudorange error from clock error.

$$\Delta R = c \times \Delta t = (3\times10^{8}\,\text{m/s})(1\times10^{-6}\,\text{s})$$ $$= 300\,\text{m}$$

Pseudorange error = 300 m. (This shows why a 1 $\mu$s bias is huge, and why the clock bias must be solved for, not ignored.)

Differential GPS (DGPS). A reference receiver is placed at a precisely known (surveyed) point. Because it knows its true position, it computes the error in each measured pseudorange (caused by ionospheric/tropospheric delay, satellite clock/orbit error) and broadcasts these corrections to nearby roving receivers. Since these errors are spatially correlated over the local area, the rover applies the corrections and removes most of the common error, improving accuracy from ~5-15 m (standalone) to sub-metre (or centimetre with carrier-phase RTK). In civil surveying this enables accurate setting-out, control densification and topographic survey.

NDVI. The Normalized Difference Vegetation Index exploits the fact that healthy green vegetation strongly absorbs red light (chlorophyll) but strongly reflects near-infrared. It is defined as:

$$\text{NDVI} = \frac{\rho_{NIR} - \rho_{R}}{\rho_{NIR} + \rho_{R}}$$

NDVI ranges from $-1$ to $+1$. Negative/near-zero values indicate water or non-vegetated surfaces; low positive values indicate sparse vegetation/bare soil; high values (≈0.5-0.9) indicate dense, healthy vegetation.

Computations.

P1: $\dfrac{0.10 - 0.30}{0.10 + 0.30} = \dfrac{-0.20}{0.40} = \mathbf{-0.50}$ → NIR < Red, strongly negative → water.

P2: $\dfrac{0.30 - 0.25}{0.30 + 0.25} = \dfrac{0.05}{0.55} = \mathbf{0.091}$ → low positive → bare soil (sparse/no vegetation).

P3: $\dfrac{0.50 - 0.10}{0.50 + 0.10} = \dfrac{0.40}{0.60} = \mathbf{0.667}$ → high positive → healthy vegetation.

Summary table:

Supervised vs. unsupervised.

• Supervised classification: the analyst defines training samples of known classes; the algorithm (e.g., maximum likelihood) learns the spectral signature of each class and labels all pixels accordingly. Requires prior knowledge/ground truth.
• Unsupervised classification: the algorithm (e.g., ISODATA, k-means) groups pixels into spectral clusters automatically; the analyst labels the clusters afterward. Requires little prior knowledge but needs interpretation of the resulting clusters.

Overall accuracy. Sum of the diagonal (correctly classified) divided by total:

$$\text{Diagonal} = 28 + 35 + 24 = 87$$ $$\text{Overall accuracy} = \frac{87}{100} = 0.87 = \mathbf{87\%}$$

Forest class. Diagonal element for Forest = 35.

• Forest column total (reference) = 40.
• Forest row total (classified) = 42.

Producer's accuracy (Forest) = correctly classified / reference (column) total:

$$= \frac{35}{40} = 0.875 = \mathbf{87.5\%}$$

(How well actual forest was mapped; omission error $= 12.5\%$.)

User's accuracy (Forest) = correctly classified / classified (row) total:

$$= \frac{35}{42} = 0.833 = \mathbf{83.3\%}$$

(Reliability that a pixel labelled forest really is forest; commission error $= 16.7\%$.)

DEM. A Digital Elevation Model is a raster (gridded) representation of the continuous terrain surface, where each cell stores the ground elevation (height above a datum). It is the bare-earth surface (cf. DSM which includes objects).

Two civil applications: (1) earthwork/cut-and-fill volume estimation and route alignment design; (2) watershed delineation, drainage and flood/inundation modelling. (Also: line-of-sight, slope-stability and visibility analysis.)

Slope at the centre cell. Label the centre cell C in the middle of the window. Its immediate four neighbours are:

• North = 1455, South = 1450 (middle column: top and bottom)
• East = 1450, West = 1465

Using the given finite differences with $\Delta x = \Delta y = 30\,\text{m}$:

$$\frac{dz}{dx} = \frac{z_{east} - z_{west}}{2\Delta x} = \frac{1450 - 1465}{2(30)} = \frac{-15}{60} = -0.25$$ $$\frac{dz}{dy} = \frac{z_{north} - z_{south}}{2\Delta y} = \frac{1455 - 1450}{2(30)} = \frac{5}{60} = 0.0833$$

Slope gradient magnitude:

$$\tan\theta = \sqrt{\left(\frac{dz}{dx}\right)^2 + \left(\frac{dz}{dy}\right)^2} = \sqrt{(-0.25)^2 + (0.0833)^2}$$ $$= \sqrt{0.0625 + 0.00694} = \sqrt{0.06944} = 0.2635$$

Slope in percent:

$$\text{Slope} = 0.2635 \times 100 = \mathbf{26.4\%}$$

Slope in degrees:

$$\theta = \tan^{-1}(0.2635) = \mathbf{14.8^{\circ}}$$

Buffer operation. A buffer creates a new polygon (zone) of specified distance around point, line or polygon features. Example: a 150 m buffer around a river to define a no-build setback, or around a hospital to find affected zones.

Overlay operation. Overlay combines two or more spatial layers to create a new layer carrying the combined attributes/geometry. Common types are union, intersect, and clip. Example: intersecting a slope-class layer with a land-use layer to find steep agricultural land suitable for terracing, or overlaying a flood zone on a parcel map to identify at-risk plots.

Buffer area of the highway corridor. The buffer extends 150 m on each side, so the total corridor width is:

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

Length of road $L = 5\,\text{km} = 5000\,\text{m}$ (treated as a straight line, so end caps neglected).

$$\text{Area} = L \times W = 5000\,\text{m} \times 300\,\text{m} = 1{,}500{,}000\,\text{m}^2$$

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

$$= \frac{1{,}500{,}000}{10{,}000} = 150\,\text{ha}$$

Total buffered area ≈ 150 hectares (excluding the small semicircular end caps).

Elements of visual image interpretation.

1. Tone/colour - relative brightness or hue; the most basic clue (water is dark, sand is bright).
2. Shape - the geometric form/outline of an object (rectangular fields, curving roads).
3. Size - both absolute and relative dimensions; helps separate, e.g., a footpath from a highway.
4. Pattern - the spatial arrangement of objects (orchard rows, street grids).
5. Texture - the frequency of tonal change / roughness-smoothness (smooth water vs. coarse forest canopy).
6. Shadow - reveals height/profile of vertical objects (towers, buildings) and is essential for relief.
7. Association - the relationship of a feature to its surroundings (a school near a playground).
8. Site/location - topographic position and environment (a wetland in a low-lying floodplain).

Application-specific elements.

• (a) River vs. canal → Shape (and pattern). A natural river has an irregular, sinuous (meandering) shape, whereas a canal is straight or gently/regularly curved with a uniform width - shape best separates them.
• (b) Residential area → Pattern (with texture). The regular grid-like arrangement of similar small buildings/streets is a distinctive spatial pattern not produced by natural features.
• (c) Landslide scar → Tone (with texture/site). A fresh landslide shows a bright, bare, high-tone exposed-soil scar contrasting sharply with the surrounding darker vegetation; its position on a steep slope (site) confirms it.

Geographic vs. projected coordinate system.

• A geographic coordinate system (GCS) locates positions on the curved Earth using angular units - latitude and longitude (degrees) - referenced to a datum/ellipsoid (e.g., WGS84). Distances and areas are not directly measurable in consistent linear units.
• A projected coordinate system (PCS) transforms the curved surface onto a flat plane using a map projection, giving linear coordinates (metres - easting and northing). This allows direct measurement of distance, area and direction, with some unavoidable distortion.

UTM for Nepal. Nepal spans UTM zones 44 and 45 (North) on the Universal Transverse Mercator projection (a Transverse Mercator projection), commonly on the WGS84 / Everest datum. The Modified Universal Transverse Mercator (MUTM) is also used by the Survey Department of Nepal with central meridians at 81°, 84° and 87° E.

Georeferencing. Georeferencing is the process of assigning real-world map coordinates to an image (or scanned map) so that every pixel has a known ground location. It is done by identifying ground control points (GCPs) of known coordinates, fitting a transformation (e.g., polynomial), and resampling the image so it aligns correctly with other spatial layers in a chosen coordinate system.