Remote Sensing

D1

Remote sensing is a quasi-linear estimation problem

Equation of radiative transfer:

o o

L

B B (z)

0

) T e T(z) (z)e −τ−τ° = + α∫

temperature profile

nepers m-1L

z

(z) = (z)∆

τ α∫

∆ τ τo )

L

0

T(z),α(z)

z ° B )

oBT

τ(z)

z

Lec22.5-1 3/27/01

T ( K dz

dz

= (0

T ( K

Radiation from Sky-Illuminated Reflective Surfaces

° = B )

L

zTB

TS

TG 0

R = 1 - ε

2

2− α−τ ∫+ α∫ z

o 0 L

(z)

0

Re T(z) (

reflectivity

emissivity (specular surface)

4

− α∫

−τ

+ α

≅ α τ⎡ ⎤⎣ ⎦

∫

∫

L z

L (z)

0

L (z)

o 0

) (z)e

T(z) (z)e dz for 1 4

+1

1

− τo2 s

sky temperature

3

3

−τ+ε +o GT e

ground temperature

D2 Lec22.5-2 3/27/01

T ( K

dz z)e dz

>>

dz T(z dz

RT e

terms

4

Temperature Weighting Function W(z,f,T)

( )≅ + ∫ L

B o 0

) T T(z)W (z)o

Terms: , ,

D3

↔B ) T(z)�Note: we have ~linear relation:

(not Fourier)

( ) ( )′≅ + −∫ L

B o o o 0

T (f ) T ) ) )

( ) ∂ = ∂

o

B o

)

T(z) )Incremental weighting function:

Alternatively,

Lec22.5-3 3/27/01

T (f z,f,T dz τ >> For 1:

T (f

T(z T (z W ' z,f,T (z dz

T (zW ' z,f,T T(z

1 2 3 4

Atmospheric Temperature T(z) Retrievals from Space

D4

α(ω) (single P)

f

∆ω ∝

∝ 2l l

αo

ωo α ≠o f(P)�

0

altitudes where intrinsic & doppler broadening dominate

z

α(z)

W(z,f) ωo ± ∆

ω ≅ ωo pressure dominates ∆ω

−τ≅ α τ⎡ ⎤⎣ ⎦∫ L

(z)B o

0

T T(z) (z)e 1

ωo ± 2∆ Lec22.5-4 3/27/01

pressure P

area # mo ecu es m

Therefore for P-broadening trace constituents

>> dz for

Atmospheric Temperature T(z) Retrievals from Space z

0 α(z)

W(z,f) ωo ± ∆

ω ≅ ωo

altitudes where intrinsic & doppler broadening dominate

pressure dominates ∆ω

−τ≅ ⎡ ⎤⎣ ⎦∫ L

(z)B o

0

T ) )e 1

W(f,z)

τ ≅ 1

z

ωo ω f

α(ω) increasing P∆ω

D5

ωo ± 2∆

scale height

Lec22.5-5 3/27/01

τ >> T(z d(z dz for

Atmospheric Temperature T(z) Retrievals from Below L

0 α(z)

W(f,z)

ωo

ωo ± ∆

ωo

− α∫= α ω L 0 (z)

oW( ) (z)e i l i

(1/α) >> λ:

0

z

L

transparency, frequency dependentIf α(z) ≅

W(f,z) D6 Lec22.5-6 3/27/01

dz f, z ~decay ng exponentia s, rate s fastest for

Temperature profile retrievals in semi-transparent solids or liquids where

constant:

Atmospheric Composition Profiles

D7

[ ]− α∫ ρ

α ′≅ ρ • = ρρ ⎡ ⎤ ⎢ ⎥⎣ ⎦∫ ∫

L z

o

L L(z)

B B o 0 0

(z)T (z) T( T (z) ( ( )(z)

W(z,f) if viewed from space

Because α(z) and W(z,f) are strong functions of the unknown ρ(z), this retrieval problem is quite non-linear and can be singular (e.g. if T(z) = constant). In this case, good statistics can be helpful. Incremental weighting functions defined relative to a nearly normal ρ(z) can help linearize the problem.

Lec22.5-7 3/27/01

+ ρ −dz z)e dz z) W z, f dz

Optimum Linear Estimates (Linear Regression)

=ˆParameter vector estimate

“determination matrix” data vector

[ ]( )∆

1 Nd =

( ) ( )− −⎡ ⎤⎣ ⎦

tˆ ˆi p p p p

( ) ( ){

( )( )

∂ − − = = ∂

∂= − − = −∂

⎡ ⎤⎣ ⎦

⎡ ⎤ ⎡ ⎤⎣ ⎦⎣ ⎦

t

ij

tt t jj j i

ij

ˆ ˆE p p p p 0D

E p Dd pD

THi

[ ] [ ]

∆

=

= =⎡ ⎤⎣ ⎦

d

i j i j

t tt t t d

= C

;�

F1 Lec22.5-8 3/27/01

p Dd 1,d ,..., d

Choose D to min mize E

Derive D :

d D E 2d D d 2d p

row of D

⎡ ⎤ ⎡ ⎤ ⎣ ⎦ ⎣ ⎦

Therefore D E dd E p d

DE dd E pd C D E dp

Optimum Linear Estimates (Linear Regression)

[ ] [ ]

∆

=

= =⎡ ⎤⎣ ⎦

d

i j i j

t tt t t d

= C

;�

F2

1t t dˆi i l C − ⎡ ⎤= =

⎣ ⎦

Lec22.5-9 3/27/01

⎡ ⎤ ⎡ ⎤ ⎣ ⎦ ⎣ ⎦

Therefore D E dd E p d

DE dd E pd C D E dp

The l near regress on so ution is p Dd where D E dp ⎡ ⎤ ⎣ ⎦

is Least-Square-Error Optimum if:D 1) Jointly Gaussian process (physics + instrument):

( )( )

( ) ( )−⎡ ⎤− ⎢ ⎥⎣ ⎦= π Λ

11 r m r m2 r N 2

1p r e 2

( )( )∆ ⎡ ⎤Λ − −⎣ ⎦

t t= E r m r m

∆ =⎡ ⎤⎣ ⎦ 1 2 Nm =

2) Problem is linear:

= + + odata d n d

parameter vector noise (JGRVZM)

F3 Lec22.5-10 3/27/01

− Λ −

1 2

E r , where r r ,r ,...,r

Mr

Examples of Linear Regression

0 d

D11

p̂

slope = D12

[ ]− = ⎡ ⎤ ⎢ ⎥⎣ ⎦

11 12ˆ D D 1

d

F4

Equivalently:

( )

[ ]( ) = + −12p̂ p D d d

E

slope = D12

p̂

d1

regression plane

D11

note change in slope

d2

slope = D13

Lec22.5-11 3/27/01

1 D : p

• ≡ •

Regression Information

G1

but which is correlated with properties the instrument does see; i

uncorrelated with visible information); it is lost.

Lec22.5-12 3/27/01

1) The instrument alone (via weighting functions)

2) “Uncovered information (to which the instrument is blind,

D retr eves this too.)

3) “Hidden information (invisible in instrument and

Nature of Instrument-Provided Information

Assume linear physics: =

[ ] [ ]

1 2 N

1 2 M

thi

d T le

W ix

W i

=

=

=

=

Where

( )

N ii

i 1 ˆ i l i i=

= =

= =

∑

Claim:

G2 Lec22.5-13 3/27/01

d WT

data vector 1,d ,d ,...,d temperature profi T ,T ,...,T

weighting function matr

row of W

If T a W and noise n 0,

then T Dd T, perfect retr eva f W not s ngular

Proof for Continuous Variables

G3

( )

N ii

i 1 ˆ i l i i=

= =

= =

∑

Claim:

1

2

3

) = b (h)

W (h) = b (h) b (h)

W (h) = b (n) b (h) b (h)

∆

∆

∆

φ

φ + φ

φ + φ + φ

i j ij 0

i ij j j 0

0(h) (

1

;b i i ( i ). d = T( (

∞

∞∆

⎧φ = ⎨⎩

φ

∫

∫

Let:

Lec22.5-14 3/27/01

If T a W and noise n 0,

then T Dd T, perfect retr eva f W not s ngular

11 1

21 1 22 2

31 1 32 2 33 3

W (h

Where: h)dh

are known a pr or from phys cs Then: h)W h)dh

• φ = δ

Proof for Continuous Variables

G4

1

2

) = b (h)

W (h) = b (h) b (h)

∆

∆

φ

φ + φ

j j 0

d = T( (∞∆ ∫

N j j

i 1 = (h)∆

= ∑

( ) N tt t t

j i i ji 0 0 N i N N

i im m jn n i jii 1 m 1 n 1 i 10

d ( ( d

k b (h) b (h) ( ) =

∞

=

∞ ∆

= = = =

= ⇒ = =

⎛ ⎞ = φ φ⎜ ⎟

⎝ ⎠

∑ ∫

∑ ∑ ∑ ∑∫

1 d i ix

k̂ Q d k i i−

=

= = Lec22.5-15 3/27/01

11 1

21 1 22 2

W (hThen: h)W h)dh

If we force T(h) k W

Then: k W h) W h)dh k W W k Q

dh k Q ⎞⎛ ⎟⎜ ⎠⎝

Therefore Qk where Q s a known square matr

So let where Q s non-s ngular

Proof for Continuous Variables

G5

1k̂ Q d k i i− = =

( )

N ii

i 1 ˆ i l i i=

= =

= =

∑Claim:

N N i i i i

i 1 i 1 ˆ ˆT(h) = ) (h) T(h)

∆

= = = =∑ ∑Then:

Equivalently: ( ) ( )∆ − − = = =

1 1d d

So: ∆− =

= =

1D = "mini i l

i 0 Lec22.5-16 3/27/01

So let where Q s non-s ngular

If T a W and noise n 0,

then T Dd T, perfect retr eva f W not s ngular

k W (h k W (exact) Q.E.D.

T = Wk W Q WQ Dd

WQ mum nformation" so ution

Wh ch is exact if T Wk, n

To what is an instrument blind?

1

2

) = b (h)

W (h) = b (h) b (h)

∆

∆

φ

φ + φ

An instrument is blind to T(h) components outside the space spanned by φ1,φ2,…,φN or, equivalently, by its W1, W2,…,Wn.

By definition, the instrument is blind to any φj ⊥ Wi, for all i.

H1 Lec22.5-17 3/27/01

11 1

21 1 22 2

W (h

Statistical Methods Can Reveal “Hidden” Components ∞

= + = + φ∑ ∑

N i i i i

i 1 N 1 (h) a (h)

seen by N

instrument channels all hidden

components

+φ φ1 1(h) (h).2

Thus hidden components can be “uncovered” if correlated with visible ones.

= = = = φ∑ ∑

N N i i i i

i 1 i 1 ˆl T(h) (h) a (h)

( )+ =

= + φ∑ N

1 1 i i i 2

1T̂(h) a a2 shrinks with decorrelation

H2 Lec22.5-18 3/27/01

In general, T(h) k W ��� ���

N 1 Extreme case: suppose always accompanied by

Then our present so ution: k W

φ + φ N 1 Would become:

General Linear Estimate

H3

∆β

∞−

= +

⎡ ⎤= = + φ⎢ ⎥⎣ ⎦ ∑

i=

1 i ij j

i T̂ a

�

� � minimum

information “uncovered” information

Thus retrieval can be drawn only from the space spanned by

=

⎛ ⎞φ φ β =⎜ ⎟

⎝ ⎠ ∑ N

i1 2 N 1 2 N i i 1

ˆ, ; , ionali i ; T

That is, N channels contribute N orthogonal basis functions to the minimum-information solution, plus N more basis functions which are orthogonal but correlated with the first N.

Lec22.5-19 3/27/01

j N 1 Dd where D WQ

���

�� ��

φ β β ,..., ,..., dimens ty s N d D

General Linear Estimate

H4

As N increases, the fraction of the hidden space which is “uncovered” by statistics is therefore likely to increase, even as the hidden space shrinks.

In general:

a priori variance = observed + uncovered + variance lost

AMSU and MSU are passive microwave spectrometers in earth orbit sounding atmospheric temperature profiles from above with ~10-km wide weighting functions peaking at altitudes from 3 to 26 km. Note the larger ratio of uncovered/lost power for AMSU.

Example: 8 channels of AMSU versus 4 channels of MSU

Lec22.5-20 3/27/01

due to noise and decorrelation.

Example: 8-Channel AMSU vs 4-Channel MSU

H5

MID-LATITUDES (TOTAL POWER = 1222 K2, 15 LEVELS)

TROPICS (TOTAL POWER = 184 K2)

0 PERCENT POWER

LOST

3 km WEIGTHING FUNCTION 1610

(WINDOW)

OBSERVED POWER LOST POWER

Lec22.5-21 3/27/01

10 20 30 40 50 60 70 80 90 100

MSU – NADIR, LAND

AMSU - NADIR

UNCOVERED

MSU - 55° INCIDENCE ANGLE, LAND

MSU - 0° ANGLE, NADIR

AMSU - NADIR

AMSU - NADIR 26 22 12 6

UNCOVERED POWER