EX1-1

Question 1

Active remote sensing

Answer 1

An active sensor system provides its own
energy source. As an example, a radar
sensor sends out radio waves and records
the reflection waves coming back from the
surface. 

Radar; Lidar

Question 2

Passive remote sensing

Answer 2

Passive systems are much more common than active systems

• A passive sensor system needs an external
energy source. In most cases this source is
the sun.
• These sensors generally detect reflected
and emitted energy wave lengths from a
phenomenon

Optical; air photos; multispectral; hyperspectral

Question 3

Spatial Resolution

Answer 3

It is the size of the smallest area that can be separately recorded as an entity on the image.

Spatial resolution depends on:
Property of sensor (e.g. Focal length of camera)
Altitude of sensor (e.g. platform height and stability)

Question 4

Temporal Resolution

Answer 4

minimum time period between two images 
exact same area  
same viewing angle
same platforms using the 
same instruments 
comparable conditions

weather satellites have high temporal but low spatial resolution.

Question 5

Spectral Resolution

Answer 5

observed spectral differences in the energy (wavelength) reflected or emitted from features of interest.

the ability of a sensor to detect small differences in wavelength. The narrower the wavelength range, the finer the spectral resolution.

Question 6

Radiometric Resolution

Answer 6

sensitivity of the sensor to incoming radiance or the differences in the brightness of objects.

(How much change in radiance is required to result in a change in recorded brightness value?)

Question 7

Electromagnetic Radiation (principal spectrums and those that are useful for remote sensing)

Answer 7

Panchromatic – 1 Band (B&W)
Color – 3 Bands (RGB)
Multispectral – 4+ Bands (RGBNIR)
Hyperspectral – 100s of Bands

Limits of the visible spectrum are defined by the sensitivity of the human visual system. It can be divided into 3 segments (“additive primaries”) – B, G and R

Very large spectrum; Categorized into NIR, MIR and Far IR (also known as thermal IR). Along with visible spectrum, NIR and MIR form the “reflective spectrum” (i.e., they are essentially solar radiation reflected from the Earth’s surface). Far IR / thermal IR is radiation emitted by the Earth.

Question 8

Relationship between frequency, wavelength, energy and speed of light

Answer 8

speed of light - c
Frequency - ν
Wavelength – λ

c = λ ν

Frequency and wavelength are inversely proportional

Question 9

Emissivity

Answer 9

ratio between the emittance of a given object and that of a blackbody at the same temperature

•Emissivity ranges from 0-1 and describes an object’s
ability to behave like a black body

Question 10

Blackbody

Answer 10

A blackbody radiator is a hypothetical body
that absorbs and re-radiates all energy
incident upon it with no change in energy

A blackbody emits energy with perfect
efficiency

Question 11

Whitebody

Answer 11

A whitebody is a hypothetical body that
reflects all energy incident upon it with no
absorption or re-radiation of energy.

Question 12

Graybody

Answer 12

Many bodies have a constant emissivity regarding the wave length, but do emit far less radiation than black bodies. They are called grey bodies.

Question 13

Selective radiator

Answer 13

Bodies whos emissivity depends on the temperature and the wave length, such as metals, are called selective radiator.

Question 14

Kirchoff’s Law (understanding of law and its implications for remote sensing)

Answer 14

DEFINES BLACKBODIES:

Ratio of emitted radiation/absorbed radiation

is the same for all blackbodies at the
same temperature.

This law forms the basis for the definition of emissivity.

The emissivity of an object varies at different wavelengths. efficiency of absorption and emission of radiation are material properties.

Question 15

Stefan – Boltzmann Law (understanding of law, the formula and its implications for remote sensing)

Answer 15

This law defines the relationship
between the total emitted radiation (W)
(often expressed in watts per unit area)
and temperature

Stefan - Boltzmann Law:
Total emission radiated from a
blackbody is proportional to the fourth
power of its absolute temperature (T):
W = σT4, 

where: W is the total emission radiatied
T is the absolute temperature in (K)
σ is the Stefan – Boltzmann constant
(5.6697 X 10-8)

Hot blackbodies emit more
energy per unit area than do cool
blackbodies

Question 16

Wien’s Displacement Law (understanding of law, the formula and its implications for remote sensing)

Answer 16

Wien’s Law:
The wavelength of max radiation is a
function of an object’s temperature (T):
λ max = 2898 mm / T, where:
λ max is the wavelength at which
radiance is at a maximum
T is the absolute temperature in (K)
This law specifies the relationship
between the wavelength of radiation
emitted and the temperature of a
blackbody
As blackbodies become hotter,
the wavelength of maximum
emittance shifts to shorter
wavelengths

Question 17

Reflectance and emittance of electromagnetic radiation

Answer 17

-

Question 18

Atmospheric interactions of EMR – transmission, scattering, absorption

Answer 18

Scattered (i.e. reflected or refracted by atmospheric particles) before it reaches the
Earth’s surface
• Absorbed by atmospheric particles before it reaches Earth’s surface
• Transmitted to the earth’s surface, where it is either reflected, transmitted or
absorbed
• Absorbed by an object on the Earth’s surface, and then emitted by that object

Question 19

transmission (Atmospheric)

Answer 19

• Transmission is what we want the atmosphere to do when we are studying
the earth itself.
• Desirable, but even under the best of conditions, 100% transmission does
not occur
• This is because some radiation is scattered, absorbed or reflected by
atmospheric gases

Question 20

scattering (Atmospheric)

Answer 20

- Scattering is the redirection of radiation by particles suspendedin the atmosphere or by large molecules of atmospheric gases.
 The mount of scattering depends on:
 Size of atmospheric particles (from molecules of gases to dust particles to larger ice and water droplets)
 The wavelength of radiation

Question 21

absorption (Atmospheric)

Answer 21

Absorption occurs when the atmosphere prevents, or strongly attenuates, transmission of radiation through the atmosphere.
 Characteristics of absorption:
 Energy absorbed by the atmosphere
is subsequently reradiated / emitted
at longer wavelengths
 Different gases absorb well at
different λ
 O3 absorbs ……
 CO2 absorbs…….
 WV absorbs …….(role of WV varies
greatly with time and location)

Question 22

3 kinds of scattering

Answer 22

Rayleigh, Mie, and Non-selective

Question 23

Rayleigh scattering

Answer 23

 Occurs when λ (shorter) > particle size;
 Occurs primarily due to molecules of atmospheric gases;
 Scattering inversely proportional to fourth power of λ
 A perfectly clean atmosphere, consisting only of atmospheric gases, causes
scattering such that the amount of scattering increases greatly as λ becomes
shorter
 Violet light scattered 16X Red light; Blue light scattered 4X Red light

Question 24

Mie scattering

Answer 24

Occurs when λ (longer) = particle size;
 Occurs primarily due to dust, pollen, smoke, water droplets and
cloud ice particles (all of which are particles larger than the
atmospheric gas molecules responsible for Rayleigh scattering)
 Tends to affect longer λ in and near the visible spectrum

Question 25

Non-selective scattering

Answer 25

 Occurs when atmosphere is heavily dust laden  Scattering particles include larger water droplets or large particles of airborne dust  “Non-Selective”  scattering not λ dependent  all visible λ scattered equally  appears as whitish or grayish haze

Question 26

Absorption and atmospheric windows

Answer 26

Those wavelengths that are not absorbed and are relatively easily transmitted through the atmosphere are called “atmospheric windows”  Major windows (table 2.4 textbk):  UV and visible  NIR  Thermal  Microwave  We can see in precisely the portion of the spectrum that is not absorbed by the atmosphere (evolution and adaptation of our eyes!!)  Atmospheric windows define those wavelengths that can be used to form images and are therefore effective for RS

Question 27

Surface interactions of EMR

Answer 27

reflection, absorption and transmission

Question 28

reflection (Surface)

Answer 28

Surface roughness < incoming λ = specular reflectance Surface roughness > incoming λ = diffuse reflectance Most surfaces are somewhere between specular and diffuse Important Laws related to Lambertian surface reflectance Lambert’s Cosine Law: The perceived brightness (radiance) of a perfectly diffuse surface does not change with the angle of view Inverse Square Law: The observed brightness decreases according to the square of the distance from the observer to the source

Question 29

absorption (Surface)

Answer 29

• The proportion of energy reflected, absorbed or transmitted varies for different earth features. • Within a given feature, the proportion reflected, absorbed or transmitted will vary with different λ

Question 30

transmission (Surface)

Answer 30

• The proportion of energy reflected, absorbed or transmitted varies for different earth features. • Within a given feature, the proportion reflected, absorbed or transmitted will vary with different λ

Question 31

2 kinds of reflection

Answer 31

Surface roughness < incoming λ = specular reflectance Surface roughness > incoming λ = diffuse reflectance

Question 32

specular reflection

Answer 32

 Angle of incidence = angle of reflection;  reflectors are flat, mirror like surfaces (mirror, smooth metal, calm water body) that redirect radiation in a single direction  The surface is smooth relative to the λ

Question 33

diffuse reflection

Answer 33

reflectors are rough surfaces that reflect uniformly in all directions  The surface is rough relative to the λ  A perfectly diffuse reflector (Lambertian surface) would have equal brightness when observed from any angle.

Question 34

Atmospheric correction

Answer 34

•Reduces the effects of haze and absorption • Improves the sharpness of images • Allows normalization of images – Provides for better comparison between different images

Question 35

Spectral properties of objects and how remote sensing uses these properties to distinguish between objects

Answer 35

• The spectral signature of an object is its own unique spectral response pattern (much like a fingerprint) and helps distinguish the object from other objects • It is a record of the unique reflectance properties of an object at different wavelengths

Question 36

Spectral signatures (while you don’t need to memorize spectral signatures of all objects, you should have a general idea of how certain common objects such as water and vegetation behave). You should be able to interpret spectral signature curves and understand the information that they convey

Answer 36

• The proportion of energy reflected, absorbed or transmitted varies for different earth features. • Within a given feature, the proportion reflected, absorbed or transmitted will vary at different λs • Thus, 2 features may be indistinguishable in one spectral range, but they may be very different in another. • Spectral reflectance (the proportion of incident energy that is reflected) of an object can be plotted for different wavelengths to produce a spectral reflectance curve • Comparison of spectral reference curves of different objects can help us figure out the most useful bands for distinguishing between these objects

Question 37

BW Panchromatic imagery

Answer 37

* Panchromatic means “across the colors” * The entire visible spectrum is represented as a single channel / band without distinguishing between R, G and B * This provides a B&W image that records brightness using radiation from the visible spectrum without separating the different colors (a regular B&W photo) * Idea is to use data capacity to capture spatial detail rather than a colored representation * Because Blue light scatters easily and degrades image quality, some instruments capture radiation across G, R and IR regions of the spectrum, thereby providing a sharper, clearer image. * Traditionally, panchromatic referred only to the visible spectrum, but now includes a broader region extending into the NIR

Question 38

Natural Color Imagery (or True Color Composite)

Answer 38

* Natural color imagery is how our eyes / visual system applies band combinations * Totally obvious – we see Blue as Blue, Green as Green and Red as Red (a normal color photo) * It is a familiar representation of scene, but a disadvantage in remote sensing is that ……..

Question 39

False Color Imagery (or False Color Composite)

Answer 39

 Our eyes can perceive only the visible spectrum.  How can we record / represent radiation from outside the visible spectrum?  When recording radiation from outside the visible spectrum, we must make assignments that depart from the natural color model  This creates false color images (things don’t appear the color that we naturally perceive them to be)  Seems nonsensical to represent objects using color other than their natural colors  But an essential task in RS  False color assignment is arbitrary – but certain assignments are commonly accepted

Question 40

Color Infrared Imagery

Answer 40

 NIR very useful for RS  Use of NIR adds a fourth spectrum channel to the natural color model  Because we recognize only three primaries, adding an NIR channel requires omission of one of the visible bands  Which band is easiest to exclude??? Why?  CIR model creates a 3 band color image by excluding … and including NIR  Initially developed in WW II for camouflage detection (to differentiate between actual vegetation and surfaces painted green to resemble vegetation)

Question 41

Pixel values / Brightness values / Digital numbers

Answer 41

``` Ls = (Gain x DN) + offset (or “bias”) DN = (Ls – offset)/Gain ```

Question 42

Digital numbers and binary numbers (you need to understand the concept but you won’t get any quantitative problems requiring conversion from one to the other)

Answer 42

-

Question 43

Bits, Bytes, Kilobytes, Megabytes, Gigabytes

Answer 43

 1 byte = 8 bits  1 kilobyte = 1024 bytes (= 2^10)  1 megabyte = 1,048,576 bytes (= 2^20)  1 gigabyte = 1,024 megabytes (= 2^30

Question 44

How to calculate radiometric resolution of a raster image if the range of DN values are known

Answer 44

know your base 2 calculations

Question 45

How to calculate file size of a raster image if number of rows, number of columns, number of bands and bit depth are given

Answer 45

L X W X BITDEPTH/8

Question 46

Image enhancement

Answer 46

• Image enhancement is the process of improving the visual appearance of digital images • An arbitrary exercise • Integrity of original data altered • original BVs are altered in order to improve visual qualities • BVs lose their relationship to original brightness on the ground • Therefore, for further analysis, original and not enhanced image values should be used as input

Question 47

Common Image enhancement techniques

Answer 47

(Linear stretch, Histogram Equalization, Density Slicing, Edge Enhancement)

Question 48

Linear stretch

Answer 48

Linear stretch converts the original digital values into a new distribution, using new min and max values.  This stretch simply takes a range of the data and rescales it to a preset range. For example, 50 to 190 becomes 0 to 255  This technique is useful for datasets that are concentrated in the middle of the grey scale  Since it simply rescales data, it is useless if there is even 1 pixel at either end of the greyscale. In this case, the rescaling has no effect on the data

Question 49

Histogram Equalization,

Answer 49

 Histogram Equalization is a stretch that favorably expands some parts of the DN range at the expense of others by dividing the histogram into classes containing equal numbers of pixels.  For instance, if most of the radiance variation occurs in the lower range of brightness, those DN values may be selectively extended in greater proportion to higher (brighter) values ****Pixels at peak are spread apart -> Contrast gained Pixels at tail are grouped -> Contrast lost

Question 50

Density Slicing

Answer 50

Density slicing is accomplished by arbitrarily dividing the range of brightness in a single band into intervals, then assigning each interval a color. Density slicing emphasizes certain features that may be represented in vivid colors, but it does not convey any more information than does the single image used as the source.

Question 51

Edge Enhancement

Answer 51

• Techniques discussed so far involve moving pixels around on the grey scale. • Each operation affects all pixels with the same grey scale in exactly the same way, without regard for context • Edge enhancement is different. • Edge enhancement makes use of spatial filters to enhance boundaries / transitions between features • What are spatial filters?

Question 52

The concept of spatial filters, how they work and how they change images

Answer 52

(low pass filters, high pass filters, edge detection / enhancement filters)

Question 53

low pass filters

Answer 53

 Low pass filters are called smoothers because they suppress detail, and emphasize continuous tones  They create regions of homogeneous BVs  A “mean” filter is a low pass filter.  It multiplies the data in the window by 1, sums the result, and divides the summation by 9 (the total # of pixels in the filter)  The result (which is the mean) is placed in the central pixel of the output  Notice the output  the magnitude of change is reduced!

Question 54

high pass filters

Answer 54

 High pass filters enhance detail by emphasizing frequent changes in BVs as the filter moves across the image  Useful for urban images

Question 55

edge detection / enhancement filters

Answer 55

 Edge detectors enhance tone changes between two pixels  Used to sharpen roads, field boundaries, other linear features

Question 56

Image scale and distance measurements (map scale, distance on map and distance on ground – how to calculate one unknown when 2 knowns are given)

Answer 56

DISTANCE BETWEEN A & B is 4.6 inches: MAP SCALE = 1:25,000 1/25,000 = MD/GD ``` MD = 4.6 GD = 4.6* 25,000 = 115,000 inches ``` NOW: We divide 115,000 by #of inches in a mile. 115,000/63,360 = 1.82 Miles _____________________________________ PD/GD = 2.2 inches/115,000 inches = 1/52,273

Question 57

Concept of remote sensing

Answer 57

• Remote sensing is the science and art of obtaining information about an object, area, or phenomenon through the analysis of data acquired by a device (sensor) that is not in contact with the object, area or phenomenon under investigation • Because there is no direct contact between the sensor and object of study, some remote interaction must be introduced to carry information from the object to the sensor. • The interaction between electromagnetic radiation and the object is the most common interaction used in modern remote sensing

Question 58

definition of remote sensing

Answer 58

Remote sensing is the practice of deriving information about the Earth’s land and water surfaces using images acquired from an overhead perspective, using electromagnetic radiation in one or more regions of the electromagnetic spectrum, reflected or emitted from the Earth’s surface.

Question 59

overview of remote sensing

Answer 59

• Physical Object:1) Objects that need to be examined • Sensor Data: 2) The instrument (camera, radar) that views the physical objects by recording electromagnetic radiation emitted or reflected from the object • Extracted Information: 3) Transformation / analysis / interpretation of sensor data (which is often abstract due to perspective, resolution and use of spectral signatures outside the visible spectrum) • Applications: 4) The analyzed remote sensing data is combined with other data to address a specific practical problem