Microwave Radiometry Ch6 Ulaby & Long INEL 6669 Dr. X-Pol Outline l Introduction l Thermal Radiation l Black body radiation Rayleigh-Jeans l Power-Temperature correspondence l Non-Blackbody radiation, brightness temperature T P, apparent temperature T, antenna temperature l More realistic ntenna Effect of the beam shape Effect of the losses of the antenna Thermal Radiation Thermal Radiation l ll matter (at T>0K radiates electromagnetic energy! l toms radiate at discrete frequencies given by the specific transitions between atomic energy levels. (Quantum theory Incident energy on atom can be absorbed by it to move an e- to a higher level, given that the frequency satisfies the Bohr s equation. f = (E 1 - E /h where, 3 h = Planck s constant = 6.63x10-34 J l absorption => e- moves to higher level l emission => e- moves to lower level (collisions cause emission bsortion Spectra = Emission Spectra l atomic gases have (discrete line spectra according to the allowable transition energy levels. 4 Molecular Radiation Spectra tmospheric Windows l Molecules consist of several atoms. l They are associated to a set of vibrational and rotational motion modes. l Each mode is related to an allowable energy level. l Spectra is due to contributions from; vibrations, rotation and electronic transitions. l Molecular Spectra = many lines clustered together; not discrete but continuous. 5 bsorbed (blue area Transmitted (white 6 tmosphere 1
Radiation by bodies (liquids - solids CommonTemperature -conversion l Liquids and solids consist of many molecules which make radiation spectrum very complex, continuous; all frequencies radiate. l Radiation spectra depends on how hot is the object as given by Planck s radiation law. 7 l 90 o F = 305K = 3 o C l 80 o F = 300K = 7 o C l 70 o F = 94K = 1 o C l 3 o F = 73K=0 o C l 0 o F = 55K = -18 o C l -80 o F = 100K = -173 o C 8 Spectral brightness intensity [Planck s Law] = hf 3 " 1 $ ' c # e hf /kt 1& [W/m sr Hz] Sun 9 10 Solar Radiation T sun = 5,800 K = hf 3 " 1 $ ' c # e hf /kt 1& I λ = hc " 1 $ ' λ 5 # e hc/λkt 1& Properties of Planck s Law l f m = frequency at which the maximum radiation occurs f m = 5.87 x 10 10 T [Hz] where T is in Kelvins l Maximum spectral Brightness B f (f m (f m = c 1 T 3 where c 1 = 1.37 x 10-19 [W/(m srhzk 3 ] 11 1 tmosphere
Problem 4.1 l Solar emission is characterized by a blackbody temperature of 5800 K. Of the total brightness radiated by such a body, what percentage is radiated over the frequency band between f m / and f m, where f m is the frequency at which the spectral brightness B f is maximum? Stefan-Boltzmann Total brightness of body at T l Total brightness is σ T I = 4 df = π http://energy.sdsu.edu/testcenter/testhome/ javaapplets/planckradiation/blackbody.ht 0 where the Stefan-Boltzmann constant is σ= 5.67x10-8 W/m K 4 sr 0M W/m sr 13 67 13M W/m sr 14 Solar power l How much solar power could ideally be captured per square meter for each Steradian? I = ( 0MW/m sr (.35sr = 7M W/m Blackbody Radiation - given by Planck s Law l Measure spectral brightness [Planck] = hf 3 " 1 $ ' [W/m sr Hz] c # e hf /kt 1& l For microwaves, Rayleigh-Jeans Law, condition hf/kt<<1 (low f, then e x -1 x = f kt c = kt [W/m sr Hz] 15 t T<300K, the error < 1 for f<117ghz, and error< 3 for f<300ghz 16 Rayleigh-Jeans pproximation Mie Theory Total power measured due to objects Brightness, Rayleigh-Jeans f<300ghz (λ>.57mm T< 300K T Wien f +Δf = 1 r ( θ,ϕ, f df f I=Brightness=radiance [W/m sr] = spectral brightness (B per unit Hz I λ = spectral brightness (B per unit cm Boltzmann' s k frequency = normalized antenna radiation pattern Ω= solid angle [steradians] r =antenna aperture on receiver 3 17 18 = 1.38 10 J / K tmosphere 3
Power-Temperature correspondence = 1 r! kt$ " # & df f +Δf f if B f is approximately constant over Δf = 1 kt Δf r but the pattern solid angle is Ω p = = ktδf = ktb r 19 nalogy with a resistor noise Direct linear relation power and temperature ntenna Pattern T = ktb nalogous to Nyquist; noise power from R T R P n = ktb *The blackbody can be at any distance from the antenna. 0 Non-blackbody radiation Emissivity, e (θ,φ Isothermal medium at physical temperature T For Blackbody, I bb = B = k But in nature, we find variations with direction, I(θ,φ =>So, define a radiometric temperature (bb equivalent I(θ,ϕ = e(θ,ϕi bb = k B l The brightness temperature of a material relative to that of a blackbody at the same temperature T. (it s always cooler e(θ,ϕ = I(θ,ϕ I = T ( θ,ϕ B bb T where 0 e 1 I bb = Ulaby = kt λ B is related to the self-emitted radiation from the observed object(s. 1 Quartz versus BB at same T Ocean (color visible radiation Emissivity depends also on the frequency. l Pure Water is turquoise blue l The ocean is blue because it absorbs all the other colors. The only color left to reflect out of the ocean is blue. Sunlight shines on the ocean, and all the colors of the rainbow go into the water. Red, yellow, green, and blue all go into the sea. Then, the sea absorbs the red, yellow, and green light, leaving the blue light. Some of the blue light scatters off water molecules, and the scattered blue light comes back out of the sea. This is the blue you see. Robert Stewart, Professor Department of Oceanography, Texas &M University 3 4 tmosphere 4
pparent Temperature, T P ntenna Temperature, T Is the equivalent T in connection with the power incident upon the antenna = T UP + ϒ a (T SE + T SS (θ,φ I inc (θ,ϕ = k θ,ϕ Δf 5 ( P = 1 k λ B r P = 1 P n = kt B T = r r kb r r Noise power received at antenna terminals. 6 ntenna Temperature (cont ntenna Beam Efficiency, η M l Using T ' = r λ = Ω we can rewrite as ( θ,ϕ l for discrete source such as the Sun. Ω sun T = Tsun Ω 7 l ccounts for sidelobes & pattern shape # T = $ T = + & ' *, #.. +. - $ T = η b T +(1- η b T SL & ' * 8,... - Combining both effects Radiation Efficiency, η l Heat loss on the antenna structure produces a noise power proportional to the physical temperature of the antenna, given as T N = (1-ξT o The η l accounts for losses in a real antenna T T = ξt +(1-ξT o T 9 l Combining both effects T = ξ η b T + ξ (1- η b T SL +(1-ξT o T = 1/(ξ η b T + [(1- η M / η b] T SL +(1-ξT o/ ξη B T = at +b *where, T = measured, T = to be estimated a= scaling factor b= bias term 30 tmosphere 5
Ej. Microwave Radiometer l K-band radiometer measures blackbody radiation from object at 00K. The receiver has a bandwidth of 1GHz. What s the maximum power incident on the radiometer antenna? nswer : P bb = P max = (1.38 10 =.8 10 = ktb 1 3 = 115.6 dbw 9 (00K(10 31 Ej. The recibo Observatory measured an antenna temperature of 45K when looking at planet Venus which subtends a planar angle of 0.003 o. The recibo antenna used has an effective diameter is 90m, its physical temperature is 300K, its radiation efficiency is 0.9 and it s operating at 300GHz (λ=1cm. l what is the apparent temperature of the antenna? T =39K l what is the apparent temperature of Venus? T Venus = 130K l If we assumed lossless, what s the error? T Venus = 136K (4 error nswer : T ' = η T + (1 η T Ω = λ / l r Ωvenus = (.003 3 ( steradians l o Demo of Rayleigh-Jeans l Let s assume T=300K, f=5ghz, Δf=00MHz I TOT bb = B = kt = (1.38 10 3 (300 f c ( f = 4.9GHz = 7.4 10 1 [W / m sr] ( f = 5.0GHz = 7.7 10 1 [W / m sr] ( f = 5.1GHz = 8 10 1 [W / m sr] 8.0 7.7 7.4 B f 5.0 f I TOT bb = rea = (00(7.4+ 1 00M(8 7.4 =1.54 10 1 I TOT bb rea = (00(7.7 =1.54 10 1 33 tmosphere 6