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Image Restoration and Atmospheric Correction Lecture 3 Prepared by R. Lathrop 10/99 Revised 2/04.

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Presentation on theme: "Image Restoration and Atmospheric Correction Lecture 3 Prepared by R. Lathrop 10/99 Revised 2/04."— Presentation transcript:

1 Image Restoration and Atmospheric Correction Lecture 3 Prepared by R. Lathrop 10/99 Revised 2/04

2 Analog-to-digital conversion process A-to-D conversion transforms continuous analog signal to discrete numerical (digital) representation by sampling that signal at a specified frequency Discrete sampled value Continuous analog signal Radiance, L dt Adapted from Lillesand & Kiefer

3 Analog-to-digital conversion process Sampling rate - must be twice as high as the highest frequency in the signal if that highest frequency is to be resolved (Nyquist frequency) Example: if highest frequency = 4 cycles/sec then the sampling rate should be at least 8/sec dt = 1sec Sweep across 4 line pairs in one second, need to take signal measurement on both line and spacing in between, thus 8 measures pr sec

4 Signal-to-Noise Ratio (SNR) SNR measures the radiometric accuracy of the data Want high SNR Over low reflectance targets (I.e. dark pixels such as clear water) the noise may swamp the actual signal True Signal Noise Observed Signal +

5 Noise Removal Noise: extraneous unwanted signal response Noise removal techniques to restore image to as close an approximation of the original scene as possible Destriping: correct defective sensor Line drop: average lines above and below Bit errors: random pixel to pixel variations, average neighborhood (e.g., 3x3) using a moving window (convolution kernel)

6 Radiometric correction Radiometric correction: to correct for varying factors such as scene illumination, atmospheric conditions, viewing geometry and instrument response Objective is to recover the “true” radiance and/or reflectance of the target of interest

7 Units of EMR measurement Irradiance - radiant flux incident on a receiving surface from all directions, per unit surface area, W m -2 Radiance - radiant flux emitted or scattered by a unit area of surface as measured through a solid angle, W m -2 sr -1 Reflectance - fraction of the incident flux that is reflected by a medium

8 For more info, go to: http://ltpwww.gsfc.nasa.gov/IAS/han dbook/handbook_toc.html

9 Radiometric response function Conversion from radiance (analog signal) to DN follows a calibrated radiometric response function that is unique for channel Inverse relationship permits user to convert from DN back to radiance. Useful in many quantitative applications where you want to know absolute rather than just relative amounts of signal radiance Calibration parameters available from published sources and image header

10 Radiometric response function Radiance to DN conversion DN = G x L + B where G = slope of response function (channel gain) L = spectral radiance B = intercept of response function ( channel offset) DN to Radiance Conversion L = [(LMAX - LMIN)/255] x DN} + LMIN where LMAX = radiance at which channel saturates LMIN = minimum recordable radiance

11 Radiometric response function L DN L Lmin Lmax 0 255 0 Slope = channel gain, G Slope = (Lmax – Lmin) / 255 Spectral Radiance to DNDN to Spectral Radiance Bias = Y intercept

12 Radiometric response function Example: Landsat 5 Band 1 From sensor header, get Lmax & Lmin Lmax = 15.21 mW cm -2 sr -1 um -1 Lmin = -0.15200000 mW cm -2 sr -1 um -1 L = -0.15200000 + ((15.21 - - 0.152)/255) DN L = -0.15200000 + (0.06024314) DN If DN = 125, L = 7.37839 mW cm -2 sr -1 um -1

13 Radiometric response function Example: Landsat 7 Band 1 Note that Landsat Header Record refers to gain and bias, but with different units (W m -2 sr -1 um -1) L = Bias + (Gain* DN) If DN = 125, L = ? Landsat Science Data User’s Handbook ltpwww.gsfc.nasa.gov/IAS/handbook/handbook_htmls/chapter11

14 DN-to-Radiance conversion Example: Landsat ETM BandGainBias 10.7756863-6.1999969 20.7956862-6.3999939 30.6192157-5.0000000 40.6372549-5.1000061 50.1257255-0.9999981 60.0437255-0.3500004 Note that Landsat Header Record refers to gain and bias, but with different units (W m-2 sr-1 um-1)

15 Radiometric response function Example: Landsat 7 Band 1 Note that Landsat Header Record refers to gain and bias, but with different units (W m -2 sr -1 um -1) Gain = 0.7756863 mW cm -2 sr -1 um -1 Bias = -6.1999969 mW cm -2 sr -1 um -1 L = -6.1999969 + (0.7756863) DN If DN = 125, L = 90.76079 W m-2 sr-1 um-1 Same 9.076079 mW cm-2 sr-1 um-1 Landsat Science Data User’s Handbook ltpwww.gsfc.nasa.gov/IAS/handbook/handbook_htmls/chapter11

16 Radiometric response function Example: Landsat 5 Thermal IR Gain = 0.005632 mW cm -2 sr -1 um -1 Bias = 0.1238 mW cm -2 sr -1 um -1 L = 0.1238 + (0.005632) DN To convert to at-satellite temperature ( o K): T = 1260.56 / log e [(60.776/L) + 1] Remember 0 o C = 273.1K For more details see Markham & Barker. 1986. EOSAT Landsat Technical Notes v.1, pp.3-8.

17 At-Satellite Reflectance To further correct for scene-to-scene differences in solar illumination, it is useful to convert to at-satellite reflectance. The term “at-satellite” refers to the fact that this conversion does not account for atmospheric influences. At-Satellite Reflectance, p = (  L  d 2 ) /  (ESUN  cos  ) Where L  = spectral radiance measured for the specific waveband  = solar zenith angle ESUN = mean solar exoatmospheric irradiance (W m -2 um -1 ), specific to the particular wavelength interval for each waveband, consult the sensor documentation d = Earth-sun distance in astronomical units, ranges from approx. 0.9832 to 1.0167, consult an astronomical handbook for the earth-sun distance for the imagery acquisition date

18 Solar Zenith angle   = solar zenith angle    cos  o = 1 As  o cos  o  o = 0  o = 60 Solar elevation angle = 90 - zenith angle

19 At-Satellite Reflectance Example: Landsat 7 Band 1 If Acquisition Date = Dec. 1, 2001 At-Satellite Reflectance = ?

20 http://aa.usno.navy.mil/data/docs/AltAz.html

21 Table 11.4 Earth-Sun Distance in Astronomical Units Julian Day Dista nce Julian Day Dista nce Julian Day Dista nce Julian Day Dista nce Julian Day Distanc e 1.983274.9945152 1.014 0 227 1.012 8 305.9925 15.983691.9993166 1.015 8 242 1.009 2 319.9892 32.9853106 1.003 3 182 1.016 7 258 1.005 7 335.9860 46.9878121 1.007 6 196 1.016 5 274 1.001 1 349.9843 60.9909135 1.010 9 213 1.014 9 288.9972365.9833 Landsat Science Data User’s Handbook ltpwww.gsfc.nasa.gov/IAS/handbook/handbook_htmls/chapter11

22 Solar Spectral Irradiances: Landsat ETM Watts m-2 um-1 Band 11969.0 Band 21840.0 Band 31551.0 Band 41044.0 Band 5225.70 Band 782.07 Band 81368.0 Landsat Science Data User’s Handbook ltpwww.gsfc.nasa.gov/IAS/handbook/handbook_htmls/chapter11

23 At-Satellite Reflectance Example: Landsat 7 Band 1 p = (  L d 2 ) / (ESUN cos  ) Dec. 1, 2001  Julian Day = 335 Earth-Sun d = 0.986 ESUN = 1969.0 Cos  Cos(63.54) = 0.44558 L   90.76079 W m-2 sr-1 um-1 p = (3.14159*90.76079*0.986 2 )/(1969.0*0.44558) p = 277.20558/877.34702 = 0.31596

24 Basic interactions between EMR and the atmosphere Scattering, S Absorption, A Transmission, T Incident E = S + A + T Within atmosphere, determined by molecular constituents, aerosol particles, water vapor

25 Satellite Received Radiance Total radiance, L s = path radiance L p + target radiance L t Target radiance, L t = 1/  RT  (E 0 delta T  o cos  o delta + E d ) Where R = average target reflectance  o = solar zenith angle   = nadir view angle T  o = atmospheric transmittance at angle  to zenith E 0 = spectral solar irradiance at top of atmosphere E d = diffuse sky irradiance (W m -2 ) Delta   band width,  – 

26 Atmospheric correction Atmospheric correction procedures are designed to minimize scattering & absorption effect due to the atmosphere Scattering increases brightness. Shorter wavelength visible region strongly influenced by scattering due to Rayleigh, Mie and nonselective scattering Absorption decreases brightness. Longer wavelength infrared region strongly influenced by water vapor absorption.

27 Atmospheric correction techniques Absolute vs. relative correction Absolute removal of all atmospheric influences is difficult and requires a number of assumptions, additional ground and/or meteorological reference data and sophisticated software (beyond the scope of this introductory course) Relative correction takes one band and/or image as a baseline and transforms the other bands and/or images to match

28 Atmospheric correction techniques: Histogram adjustment Histogram adjustment: visible bands, esp. blue have a higher MIN brightness value. Band histograms are adjusted by subtracting the bias for each histogram, so that each histogram starts at zero. This method assumes that the darkest pixels should have zero reflectance and a BV = 0.

29 Atmospheric correction techniques: Dark pixel regression adjustment Select dark pixels, either deep clear water or shadowed areas where it is assumed that there is zero reflectance. Thus the observed BV in the VIS bands is assumed to be due to atmospheric scattering (skylight). Regress the NIR vs. the VIS. X-intercept represents the bias to be scattered from the VIS band.

30 Atmospheric correction techniques: Scene-to-scene normalization Technique useful for multi-temporal data sets by normalizing (correcting) for scene-to-scene differences in solar illumination and atmospheric effects Select one date as a baseline. Select dark, medium and bright features that are relatively time-invariant (I.e., not vegetation). Measure DN for each date and regress. DB b1, t2 = a + b DN b1, t1

31 Scene-to-Scene Normalization Example: Landsat 5 vs Landsat 7 Landsat 7: Sept 01 Landsat 5: Sept 95

32 Scene-to-Scene Normalization Example: Landsat 5 vs Landsat 7 Landsat 5: Sept 95 Landsat 7: Sept 99 & 01 99 R 2 = 0.971 01 R 2 = 0.968 99 R 2 = 0.932 01 R 2 = 0.963

33 Terrain Shadowing USGS DEM Landsat ETM Dec 01 Solar elevation = 26.46 Sun Azimuth = 158.78

34 Terrain correction To account for the seasonal position of the sun relative to the pixel’s position on the earth (I.e., slope and aspect) Normalizes to zenith (sun directly overhead) L c = L o cos (  o ) / cos(i) where L c = slope-aspect corrected radiance L o = original uncorrected radiance cos (  o ) = sun’s zenith angle cos(i) = sun’s incidence angle in relation to the normal on a pixel (i = O o - slope)

35 Cosine Terrain correction L c = L o cos (  o ) / cos(i) oo i Terrain: assumed to be a Lambertian surface Sun Sensor 90 o Adapted from Jensen

36 Terrain correction Terrain Correction algorithms aren’t just a black box as they don’t always work well, may introduce artifacts to the image Example: see results on right from ERDAS IMAGINE terrain correction function appears to “overcorrect” shadowed area


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