Introducción a la Teledetecciónepsem.upc.edu/~jorge/Bolivia/M3-02.pdf · Fundamentos teóricos de la obtención de datos A Definition: Remote sensing is the practice of deriving

Fundamentos teóricos de la obtención de datos

IntroducciIntroduccióón a la n a la TeledetecciTeledeteccióónn


…desde el espacio

¿Qué es teledetección (“remote sensing”)?Código UNESCO: 2506.16, ámbito de la Geología

Teledetección es la adquisición de información de un sistema desde la distancia (p.e. …?).

…desde el aire...

…en tierra.


A Definition:

Remote sensing is the practice of deriving information about theearth’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.

James P. Campbell, 1996

Another Definition:

The science of remote sensing consists of the interpretation of measurements of electromagnetic energy reflected from or emittedby a target from a vantage point that is distant from the target.

Paul Mather, 19??


Ventajas de la teledetección (espacial)

• Proporciona una visión regional (áreas extensas).

• Proporciona visualización repetida de la misma área.

• Los sensores remotos “ven” en una porción del espectro electromagnético más ancha de la que percibe el ojo humano.

• Los sensores pueden registrar simultáneamente una imagen en varias longitudes de onda (bandas o canales).

• Proporciona datos digitales geo-referenciados.

• Algunos sensores de teledetección operan en todas las estaciones del año, durante la noche, en condiciones meteorológicas adversas,…



Variables medibles de modo directo son función de los datos adquiridos por el sensor se pueden extraer directamente si conocemos los

parámetros de adquisición

• Reflectividad en el espectro solar (VIS, IRC, SWIR)• Temperatura en el térmico• Coeficiente de retro-dispersión (micro-ondas)• Altitud, a partir de la visión estereoscópica en el visible• Medición de distancias con lidar y la interferometría radar• Topografía de la superficie marina, a partir del altímetro de

micro-ondas

Variables en Teledetección


Variables indirectas derivadas a partir de las que son medibles de modo directo,

gracias a que suponen una modificación de la señal recibidapor el sensor

se requiere un modelo que relacione la variable con lainformación contenida en las imágenes• Contenido de clorofila (modifica la reflectividad)• Índice de área foliar• Radiación absorbida por la planta• Humedad del suelo• Humedad de las hojas• Turbidez del agua• Contenido de CO2 en la atmósfera• Evapotranspiración• Productividad vegetal neta


Precision Agriculture

Global ChangeNatural DisastersEnvironmental Management

• Land-use mapping• Forest and agriculture applications• Telecommunication planning• Environmental applications

• Hydrology and coastal mapping• Urban planning• Emergencies and Hazards• Global change• Meteorology


Fundamentos fFundamentos fíísicossicos


Componentes de la Teledetección

A Fuente de energíao Iluminación

B Atmósfera

C Interacción con el objetivo

D Registro de Energía por el Sensor

E Transmisión, Recepción, y Procesamiento

F Interpretación y Análisis

G Aplicación


• La teledetección utiliza la energía radiante que es reflejada y emitida desde la Tierra para diferentes “longitudes de onda” del espectro electromagnético.

• Nuestros ojos sólo son sensibles a la porción de “luz visible” del espectro electromagnético.

El espectro electromagnético

Longitud de onda






4 5 7

1 2

4

3

5 7


Comunicación Submarina< 100 Hz> 3000 km

Radio, onda larga300 kHz – 30 kHz1 – 10 km

Radio, onda media1 MHz300 m

Radio, onda corta30 MHz – 3 Mhz10 – 100 m

Televisión300 MHz 1 m

Teléfonos móviles1 GHz 30 cm

Micro-ondas3 GHz 10 cm

Radar, TV satélite300 GHz – 10 GHz1 mm – 3 cm

Escáneres Térmicos1 THz10 m – 0.3 mm

Escáneres de Infrarrojo4x1014 – 3x1013 Hz800 nm – 10 m

Luz Visible1015 – 4x1014 Hz300 - 800 nm

Solarium (UV), radiografías> 3x1015 Hz< 100 nm

AplicaciónFrecuenciaLongitud de onda

c = f

c : velocidad de la luz,= 3x108 m/s

f : frecuencia (Hz)

: longitud de onda(nm, µm)



Magnitudes físicas en teledetección


Densidad de flujo radiante : cantidad de flujo radiante (Φ) dividido por el área de la superficie sobre la que incide (irradiancia) o de la que es emitida (emitancia radiativa o exitancia); W/m2.

Irradiancia Eλ = Φ iλ/ A

Exitancia Mλ = Φ iλ/ A

A

A


Magnitudes físicas en teledetección

Radiancia (L)Flujo radiante por unidad de ángulo sólido que abandona una fuente extensa en una dirección dada por unidad de superficie de la fuente proyectada en esa dirección [W · m-2 · sr-1].


,,,,, ,, PtsfL zyx

Remote Sensing Data Collection

The amount of electromagnetic radiance, L (watts m-2 sr-1; watts per meter squared per steradian) recorded within the IFOV of an optical remote sensing system (e.g., a picture element in a digital image) is a function of:

where,

λ = wavelength (spectral response measured in various bands or at specific frequencies). Wavelength (λ) and frequency (f) may be used interchangeably based on their relationship with the speed of light (c) where:


sx,y,z = x, y, z location of the picture element and its size (x, y)

t = temporal information, i.e., when and how often the information was acquired

ө = set of angles that describe the geometric relationships among the radiation source (e.g., the Sun), the terrain target of interest (e.g., a corn field), and the remote sensing system

P = polarization of back-scattered energy recorded by the sensor

= radiometric resolution (precision) at which the data (e.g., reflected, emitted, or back-scattered radiation) are recorded by the remote sensing system.

Remote Sensing Data Collection


Absorción y Transmisión de la Radiación Monocromática

La cantidad de energía, radiancia, que atraviesa un diferencialde área dA en un diferencial de tiempo dt y número de onda vviene dada por la expresión:

L= dE / cos dA dt d d

unidades: W/m2 sr cm-1


Un cuerpo negro es una superficie ideal que emite el máximo posible de energía térmica a una temperatura dada.

El cuerpo negro es un cuerpo ideal de temperatura uniforme T(K) que absorbe perfectamente toda la radiación incidente sobre él, y emite de forma no polarizada e isotrópicamente la radiación térmica de acuerdo a la ley de Planck. Es costumbre indicar como B(T) o B(T) la intensidad específica de dicho cuerpo.

Cuerpo negro


Planck Law:To explain the spectral distribution of radiance emitted by solid bodies, Planck found that the radiance per unit frequency emitted by a blackbody at temperature (T) is given as:

1)exp(

2)(3

TchchTB

Where h is Planck const,k is Boltzmann const.

L = 2 h c2 -5 / exp(hc/KT - 1)

donde h = constante de Plank = 6.6256 x 10-34 J·sy K = constante de Boltzmann = 1.3805 x 10-23 J K-1

Brightness Temperature (or equivalent blackbody temperature) is the temperature estimated by inverting Planck function.


The Planck law describes theoretical spectral distribution for the emissive power of a black body. It can be written as

where C1=3.742x108 (W.mm4/m2) and C2=1.439x104 (mm.K) are two constants. The Planck distribution is shown in the following figure as a function of wavelength for different body temperatures.

The Planck Distribution


Solar Irradiation


Logarithmic representation

The Black Body function in a logarithmic representation.The following properties can be seen:

• The decrease at < max is very steep, the flux radiated at max<max is only 1/4 of the total.

• Below 0.17max there is a fraction of only 10-10 of the total.

• 98% of the energy is radiated between 0.5max and 8 max.

Spectral Radiance of Blackbody

Fundamentos teóricos de la obtención de datosQuestion

1. Which of the following probably has a higher temperature?

a) a red-hot object

b) a white-hot object

c) a blue-hot object

Recall the relation between frequency and wavelength of an electromagnetic wave:

f = c with c = 3 x 108 m/s (speed of Light)

Increasing T Increasing f Decreasing

2.8 kT

Planck Radiation Law

hf = photon energy

U(f)

(Red Blue)


The ratio of emitted radiance by an object to the radiance emitted by a blackbody at the same temperature is calledemissivity ().

= M/B

= 1 for blackbody.

< 1 for greybody.

Blackbody radiation

Greybody radiation

M

Emissivity


Stefan Boltzmann Law:

It gives the total radiation of a cavity (blackbody) not spectral distribution of radiation. When Planck function is integrated 0 (zero) to infinity (), S-B is given as:

M(exitance) = T4

Where is stefan Boltzmann constant.

M = T4

donde : = 5.6697 x 10-8 Wm-2K-4

M = exitancia total radiada a partir de la superficie in Watts m-2

T = temperatura en Kelvin del material emisor


Calculate the power radiated by a 10-cm-diameter sphere of aluminum at room temperature (20oC). (Assume it is a perfect radiator.)

Watts13 )m103.14)( W/m(418AJ radiatedPower

m103.14)m10(54r4 A :Area

W/m418

)(293K) K m W 10 5.670(

area)per radiated(power TJ

22-2

22-22-2

2

44-2--8

4SB

Exercise for you: Calculate how much power your body (at T = 310 K) is radiating. If you ignore the inward flux at T = 293 K from the room, the answer is roughly 100 W -- about that of a light bulb!

Thermal Radiation from a Solid

Fundamentos teóricos de la obtención de datosSurface Temperature of the Earth

K5800Tmeters 105.1R

meters 107R

S

11

8S

4S SB S

2R4 sR SB S R

4E SB E

J Sun's flux at its surface T

J 'Sun's flux at earth' T

J Earth's flux at its surface T

Equilibrium Condition:

Power absorbed from sun = Power radiated by earth

(Power = Flux × Area)

Room Temperature!

R Rs

TE Ts

K280TT

T4T

J4J R4JRJ

S

2/1

R2sR

E

4ESB

2

RsR4

SSB

ER

2EE

2ER

Fundamentos teóricos de la obtención de datosComments on: Surface Temperature of the Earth

• Actually we did not account for another important factor: about 30% of the sun’s radiation reflects off our atmosphere!

• 30% less input from the sun means about 8% lower temperature because T P1/4 . This factor reduces our best estimate of the earth’s temperature by about 30 K. Now TE = 250 K, or 0oF. Brrr!

• In fact the average surface temperature of the earth is about 290K, or about 600F, just right for animals and plants.

• The extra 600F of warming is mainly due to the “greenhouse effect”, the fact that some of the radiation from the earth is reflected back to the earth by the atmosphere. What exactly is the “greenhouse effect”?

R Rs

TE Ts

Our homeOur energy source


Remember Molecular vibrations

• Energy levels of an oscillator: En= n.• = hf

Strong adsorption in the infrared. (rotational and vibrational motions)

En= n

Photon

u(f)

2.8 kTS

Planck Radiation Law (not to scale)

hf = photon energy2.8 kTE

earth

sun

Greenhouse gas absorption

CO2 or H2O


The Greenhouse Effect

• Thermal radiation from the earth (infrared) is absorbed by certain gases in our atmosphere (such as CO2) and redirected back to earth.

• These ‘greenhouse gases’ provide additional warming to our planet essential for life as we know it.

• Radiation from the sun is not affected much by the greenhouse gases because it has a much different frequency spectrum.

• Mars has a thin atmosphere with few greenhouse gases: 70oF in the day and -130oF at night. Venus has lots of CO2 : T = 800oF

Earth Sun


Example Problem: Surface Temperature of the Earth (crude estimate)

The energy flux of the sun at the earth’s equator is approximately 1 kilowatt per square meter (1 kW/m2). In equilibrium, the earth radiates an equal flux ME = SBTE

4 . With this information estimate the temperature of the earth’s surface TE.

This temperature (close to boiling water) is way too hot for life as we know it! The main flaw in this estimate is that the earth is radiating energy from all sides, whereas the sun is illuminating only one side.

K364T

1076.1KWm1067.5

m/W10T

Tm/W10

E

11428

234E

4ESB

23


Thermal Radiation from a Solid

Calculate the power radiated by a 10-cm-diameter sphere of aluminum at room temperature (20oC). (Assume it is a perfect radiator.)

Watts13 )m103.14)( W/m(418AJ radiatedPower

m103.14)m10(54r4 A :Area

W/m418

)(293K) K m W 10 5.670(

area)per radiated(power TJ

22-2

22-22-2

2

44-2--8

4SB

Exercise for you: Calculate how much power your body (at T = 310 K) is radiating. If you ignore the inward flux at T = 293 K from the room, the answer is roughly 100 W -- about that of a light bulb!


Wien Law

When Planck equation is differentiated w.r.t wavenumber () or wavelength () and equated to 0 (zero), one can find max for a given temperature (T) called Wien Law.

max =0.2897/T (cm) E

What wavelengthggives maximum energy

max = ( 2.8983 x 10-3 K m ) / T

donde T = temperatura en Kelvin

This relation is known as Wien’s Displacement law, also written:

λmax,1T1 = λmax,2T2


Amount of solar energy hitting Earth’s outer atmosphere is

~1370 watts/m2

Our Sun emits more light in the visible than any other part of the spectrum.


Sun at ~6000K: 600 to 700 nm (0.6 – 0.7 µm)(corresponds with spectral sensitivity of human eye)

Earth at ~300 K: 9600 nm (9.6 µm)

Lava from volcano at ~1000 K: 4200 nm (4.2 µm)

Light bulb at ~3000 K: 1000 nm (1.0 µm)

Wien: max = A/TWavelengths of maximum emittance


The radiative temperature of the sun surface is about 5780 K. After applying Wien law, maximum Planck radiance is obtained at the wavelength (max ) of 0.50 mwhich is the center of the visible region of the spectrum.

On the other hand eath’s atmospheric temperature is about 255 K. Maximun emitted energy takes place around 11 mwhich is infrared region.


Blackbody is a perfect absorber and emitter of all radiation. Theemitted radiation in each wavelength is only a function of T.

Planck's Law : Eλ(T)= C1/λ5(exp(C2/λT)-1) [W/m2μm]

Stefan Boltzmann law: Mλ= σT4 [W/m2]where σ = 5.6697 x 10-8 Wm-2 K-4

T = temp in Kelvin

Wien’s displacement law: λmax = k/Twhere k is a constant = 2898 μm K

Blackbody Radiation

where C1 (=3,74151x108 Wm-2m4) and C2 (=1,43879x104 mK) are constant


A function of the exoatmospheric irradiance, target reflectance and the transmission of the atmosphereB light scattered from the atmosphere, reflects off of the target, and reaches the sensor (downw. irrad)C scattered by the atmosphere and reflected in the direction of the sensor (upwelling irrad)D photons thermally emitted by the target itselfE thermally emitted photons from the atmosphere reflected by the target and reach the sensor (as B in refl)F emission by the atmosphere due to its temperature (as C in refl)G background effects due to other objects. Sometimes negligible.H as G except it applies to thermal emission of photons due to the background's temperature

Target and Path Radiance

Fundamentos teóricos de la obtención de datosRemote Sensing Radiative Transfer

τv

Li

Lt

E0

τ0

0 v

1

λ

atmosphere

Fundamentos teóricos de la obtención de datosIrradiance at Top of Atmosphere (TOA)

E0E0 = Irradiance at TOA

Fundamentos teóricos de la obtención de datosIrradiance at Top of Atmosphere

E0E0 = Irradiance at top of atmosphere

rsun = 6.95e5 kmrdis = 149.6e6 kmTsun = 5784 K

σ = 5.67e-8 Wm-2 K-4

S = solar constant= E0 for average sun-earth distance

Φsun = ΦdisEsunAsun=EdisAdisEdis = Esun(Asun/Adis)Edis = Esun(rsun

2/rdis2)

Edis = σTsun4 (rsun

2/rdis2)

Edis = 1370 W/m2

rsun

rdis

AdisAsun

Φsun Φdis

Documents

Introducción a la Teledetecciónepsem.upc.edu/~jorge/Bolivia/M3-02.pdf · Fundamentos teóricos de la obtención de datos A Definition: Remote sensing is the practice of deriving