BE Civil Engineering (IOE, TU) Remote Sensing and GIS (IOE, CE 754) Question Paper 2079 Nepal

Interaction of EM radiation with the atmosphere

As solar/terrestrial radiation passes through the atmosphere it is modified by three processes:

• Absorption — gases (H$_2$O, CO$_2$, O$_3$) convert radiant energy to heat at characteristic wavelengths, removing energy from the beam.
• Scattering — redirection by particles/molecules: Rayleigh (particles ≪ λ, ∝ λ⁻⁴, causes blue sky), Mie (particles ≈ λ, dust/aerosols), and non-selective (particles ≫ λ, clouds appear white).
• Transmission — the fraction that passes through largely unaffected.

Atmospheric windows

Wavelength regions where the atmosphere is highly transmissive (low absorption) are called atmospheric windows. Remote sensing sensors are designed to operate only within these windows. Major windows include:

Strong absorption bands (e.g. around 2.7, 6.3 µm by H$_2$O and 15 µm by CO$_2$) block the intervening regions.

Laws

Wien's displacement law:

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

Stefan-Boltzmann law:

$$M = \sigma T^4, \qquad \sigma = 5.67\times10^{-8}\ \text{W·m}^{-2}\text{·K}^{-4}$$

(a) Peak wavelengths

Sun:

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

This lies in the visible (green) region of the spectrum.

Earth:

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

This lies in the thermal infrared region (8–14 µm window) — hence thermal sensors detect Earth's emitted radiation.

(b) Ratio of total radiant emittance

$$\frac{M_\odot}{M_\oplus} = \left(\frac{T_\odot}{T_\oplus}\right)^4 = \left(\frac{5800}{300}\right)^4 = (19.333)^4$$ $$= 1.397\times10^{5}$$

The Sun emits about $1.40\times10^{5}$ times more energy per unit area than the Earth.

(Check: $M_\odot = 5.67\times10^{-8}\times5800^4 = 6.42\times10^{7}\ \text{W/m}^2$; $M_\oplus = 5.67\times10^{-8}\times300^4 = 459\ \text{W/m}^2$; ratio $= 6.42\times10^{7}/459 = 1.40\times10^{5}$ ✓)

Four types of resolution

• Spatial resolution — the smallest ground area represented by one pixel (ground projection of the IFOV). Example: Landsat-8 OLI multispectral bands = 30 m.
• Spectral resolution — the number and width of wavelength bands a sensor records. Example: Hyperspectral AVIRIS records 224 narrow bands (~10 nm each); a panchromatic band is a single broad band.
• Radiometric resolution — the number of brightness (grey) levels = quantisation bit depth. Example: Landsat-8 = 12-bit (4096 levels) vs older Landsat = 8-bit (256 levels).
• Temporal resolution — the revisit interval over the same area. Example: Landsat = 16 days; MODIS = ~1 day.

Calculations

(a) Ground sample distance at nadir

GSD = IFOV (in radians) × altitude:

$$\text{GSD} = \beta \times H = (42.5\times10^{-6}\ \text{rad})\times(705\times10^{3}\ \text{m})$$ $$= 29.96\ \text{m} \approx \mathbf{30\ m}$$

(b) Swath width

$$SW = 2H\tan\!\left(\frac{FOV}{2}\right) = 2(705\ \text{km})\tan(7.5^\circ)$$ $$\tan(7.5^\circ) = 0.13165$$ $$SW = 1410 \times 0.13165 = 185.6\ \text{km} \approx \mathbf{185.6\ km}$$

(c) Number of grey levels

$$2^{12} = \mathbf{4096\ grey\ levels}\ (\text{values }0\text{–}4095).$$

Enhancement vs classification

• Image enhancement improves the visual interpretability of an image for a human analyst by manipulating pixel brightness/contrast (e.g. contrast stretch, filtering). It does not assign meaning to pixels.
• Image classification assigns each pixel to a thematic class/category (e.g. water, forest, urban) based on its spectral signature, producing a thematic map. It is the analytical/quantitative step.

Linear contrast stretch expands a narrow DN range to fill the full display range using:

$$\text{DN}_{out} = \frac{\text{DN}_{in}-\text{DN}_{min}}{\text{DN}_{max}-\text{DN}_{min}}\times(L-1)$$

where $L=256$ for 8-bit data.

Histogram equalisation is a non-linear stretch that redistributes DN values so the cumulative distribution function (CDF) becomes approximately linear, assigning more output levels to the most frequently occurring (most populated) DN values, thus maximising overall contrast.

(a) Linear contrast stretch

$$\text{DN}_{out} = \frac{\text{DN}_{in}-48}{176-48}\times 255 = \frac{\text{DN}_{in}-48}{128}\times 255$$

For $\text{DN}_{in}=100$:

$$\text{DN}_{out} = \frac{100-48}{128}\times 255 = \frac{52}{128}\times 255 = 0.40625\times255 = 103.6$$ $$\Rightarrow \mathbf{DN_{out}=104}\ (\text{rounded}).$$

(b) 3×3 mean filter on centre pixel

The mean filter replaces the centre value with the average of all nine pixels:

$$\text{mean} = \frac{40+52+60+48+120+64+50+58+70}{9}$$ $$= \frac{562}{9} = 62.44$$ $$\Rightarrow \mathbf{New\ centre\ value = 62}\ (\text{rounded}).$$

Note the high centre value (120) is smoothed down toward its neighbours, illustrating noise suppression by low-pass filtering.

Raster vs Vector

Raster — advantages: simple data structure; efficient for overlay/map-algebra and continuous surfaces (DEM, imagery). Raster — disadvantages: large file sizes at fine resolution; cell-size limits spatial accuracy and produces blocky boundaries.

Vector — advantages: compact storage; precise geometry and topology (good for networks/adjacency). Vector — disadvantages: complex data structure/algorithms; overlay and continuous-surface analysis are computationally harder.

(a) Rows, columns, total cells

$$\text{Columns} = \frac{6000\ \text{m}}{20\ \text{m}} = 300$$ $$\text{Rows} = \frac{4000\ \text{m}}{20\ \text{m}} = 200$$ $$\text{Total cells} = 300\times200 = \mathbf{60{,}000\ cells}$$

(b) Storage size (8-bit = 1 byte/cell)

$$\text{Size} = 60{,}000\ \text{bytes} = \frac{60{,}000}{1024} = \mathbf{58.59\ KB}$$

(c) Effect of halving cell size to 10 m

Halving the cell size doubles both rows and columns, so the number of cells increases by $2\times2 = 4$:

$$\text{New cells} = 600\times400 = 240{,}000\ \text{cells}$$ $$\text{New size} = 240{,}000\ \text{bytes} = 234.4\ \text{KB}$$

The file size increases by a factor of 4 (quadruples).

Working principle of GPS

GPS determines position by measuring the distance (range) from the receiver to several satellites whose positions are precisely known from the broadcast ephemeris. Each satellite transmits a coded signal stamped with the transmission time; the receiver measures the signal travel time $\Delta t$ and computes range $R = c\,\Delta t$.

Trilateration: Knowing the distance to one satellite places the receiver on a sphere; two satellites → intersection circle; three satellites → two candidate points (one rejected as non-physical). Thus three ranges fix a 3-D position if the receiver clock were perfect. A fourth satellite is required to solve for the receiver clock bias as well (4 unknowns: X, Y, Z, clock error).

Major GPS error sources

• Ionospheric and tropospheric delay
• Satellite and receiver clock errors
• Ephemeris (orbit) errors
• Multipath (signal reflection)
• Receiver noise

Dilution of Precision (DOP)

DOP is a dimensionless number describing how satellite geometry amplifies measurement error into position error:

$$\sigma_{position} = \text{DOP}\times\sigma_{measurement}$$

Widely spaced satellites give a low DOP (good geometry); satellites clustered together give a high DOP (poor geometry) and a larger position error. Hence good geometry is essential for accuracy.

(a) Pseudorange

$$R = c\,\Delta t = (3\times10^{8}\ \text{m/s})\times(67.2\times10^{-3}\ \text{s})$$ $$= 2.016\times10^{7}\ \text{m} = \mathbf{20{,}160\ km}$$

(b) Range error from clock bias

A $1\ \mu\text{s}$ clock error produces a range error:

$$\delta R = c\,\delta t = (3\times10^{8})\times(1\times10^{-6}) = \mathbf{300\ m}$$

Because even a tiny clock error causes a large range error, the receiver clock cannot be trusted. The position solution therefore has four unknowns $(X, Y, Z, \Delta t_{clock})$, requiring four independent range equations — hence four satellites — to solve simultaneously for the 3-D position and the clock bias.

NDVI

The Normalised Difference Vegetation Index is:

$$\text{NDVI} = \frac{\text{NIR}-\text{Red}}{\text{NIR}+\text{Red}}$$

Healthy vegetation strongly absorbs red light (chlorophyll) and strongly reflects NIR (leaf mesophyll), so the difference is large and positive. Soil, water and stressed vegetation reflect red and NIR more similarly, giving low values. NDVI ranges from $-1$ to $+1$.

Vegetation pixel

$$\text{NDVI} = \frac{0.52-0.08}{0.52+0.08} = \frac{0.44}{0.60} = \mathbf{0.733}$$

High positive value → dense, healthy vegetation.

Bare-soil pixel

$$\text{NDVI} = \frac{0.28-0.26}{0.28+0.26} = \frac{0.02}{0.54} = \mathbf{0.037}$$

Near-zero value → bare soil / non-vegetated surface.

Interpretation: the large NDVI contrast (0.733 vs 0.037) clearly separates the two cover types, confirming NDVI as an effective vegetation discriminator.

Buffer vs Overlay

• Buffer — generates a new polygon enclosing all area within a specified distance of a feature (point/line/polygon). Civil example: a 50 m no-construction zone around a transmission line or river.
• Overlay — combines two or more layers spatially to create a new layer carrying attributes from all inputs (union, intersect, etc.). Civil example: intersecting a slope layer with a land-use layer to find buildable parcels on gentle slopes.

Buffer area

For a line buffered by distance $d$ on both sides (ignoring end caps), the strip is a rectangle of length $L$ and width $2d$:

$$A = L \times 2d = 3200\ \text{m} \times (2\times50\ \text{m})$$ $$= 3200 \times 100 = 320{,}000\ \text{m}^2$$

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

$$A = \frac{320{,}000}{10{,}000} = \mathbf{32\ ha}$$

DEM

A Digital Elevation Model is a raster representation of the continuous terrain (bare-earth) surface, storing an elevation value in each cell. (A DSM includes surface objects such as buildings/trees; a DTM/DEM represents the ground.)

Civil applications: (1) watershed delineation and hydrological/flood modelling; (2) earthwork cut-and-fill and road/canal alignment design; (3) slope-stability and landslide-susceptibility mapping (also viewshed, line-of-sight).

Slope calculation

Horizontal distance between adjacent (east) cells = cell size = 30 m. Rise = $1465 - 1450 = 15\ \text{m}$.

Slope in percent:

$$S\% = \frac{\text{rise}}{\text{run}}\times100 = \frac{15}{30}\times100 = \mathbf{50\%}$$

Slope in degrees:

$$\theta = \tan^{-1}\!\left(\frac{15}{30}\right) = \tan^{-1}(0.5) = \mathbf{26.57^\circ}$$

Supervised vs Unsupervised

• Supervised classification — the analyst defines training samples for known classes; the algorithm (e.g. maximum likelihood) learns each class's spectral signature and labels all pixels. Requires prior ground knowledge.
• Unsupervised classification — the algorithm (e.g. K-means, ISODATA) groups pixels into spectral clusters automatically; the analyst assigns class labels afterwards. Requires no prior training data.

Overall accuracy

Overall accuracy = (sum of correctly classified pixels on the diagonal) ÷ (total pixels):

$$\text{Diagonal} = 45 + 60 + 52 = 157$$ $$\text{Total} = 180$$ $$\text{OA} = \frac{157}{180} = 0.8722 = \mathbf{87.22\%}$$

(For reference, the off-diagonal values represent omission/commission errors, e.g. 4 water pixels misclassified as forest.)

Elements of visual image interpretation

• Tone/colour — relative brightness or hue.
• Shape — outline/form of objects (e.g. rectangular buildings).
• Size — relative or absolute dimensions.
• Pattern — spatial arrangement (e.g. orchard rows).
• Texture — smoothness/roughness of tonal variation.
• Shadow — indicates height/profile, aids identification.
• Association — related features occurring together (e.g. school near playground).
• Site/location — topographic/environmental setting.

Photo scale

For a vertical photograph:

$$\text{Scale} = \frac{f}{H} = \frac{0.152\ \text{m}}{2400\ \text{m}} = \frac{1}{15{,}789}$$ $$\Rightarrow \mathbf{Scale \approx 1{:}15{,}790}$$

Ground length

Ground distance = photo distance × scale denominator:

$$D_{ground} = 90\ \text{mm} \times 15{,}789 = 1{,}421{,}010\ \text{mm}$$ $$= 1421\ \text{m} \approx \mathbf{1.42\ km}$$

(Check via similar triangles: $D = \dfrac{d\,H}{f} = \dfrac{0.090\times2400}{0.152} = 1421\ \text{m}$ ✓)

Active vs Passive remote sensing

• Passive sensors detect natural radiation — reflected sunlight or emitted thermal energy. They depend on an external source (Sun) and weather. Examples: Landsat OLI, MODIS, aerial cameras, thermal scanners.
• Active sensors supply their own energy source, transmit a pulse and record the backscattered return. They are independent of sunlight. Examples: RADAR (SAR), LiDAR.

Advantages of microwave (radar) RS for Nepal

1. All-weather, day-and-night capability — microwaves penetrate clouds, haze and rain, which is invaluable in Nepal's monsoon-clouded, mountainous terrain where optical imagery is frequently obscured. Useful for flood and landslide monitoring during the rainy season.
2. Sensitivity to surface roughness, moisture and structure — radar backscatter responds to soil moisture, surface geometry and ground deformation; SAR interferometry (InSAR) can measure millimetre-scale ground/infrastructure displacement, aiding landslide, subsidence and dam/structure monitoring.

Radar wavelength bands

Radar systems operate in different microwave bands distinguished by wavelength/frequency; longer wavelengths penetrate vegetation/soil more:

Longer-wavelength bands (L) penetrate canopy and soil more deeply than shorter (X) bands.