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Next: Estimating camera response function Up: .  Introduction: Variable gain image Previous: Quantimetric imaging

Estimation in the presence of noise

Consider first, a special case in which we know each of the relative exposure values $K_i$. For this special case, images are related to to the actual quantity $q$ as well as to each other, by:

f_i (x) = f( k_i q ( x ) +n_q_i ) +n_f_i where $n_{q_i}$ includes sensor noise, and $n_{f_i}$ includes image noise due to quantization, compression, transmission. (For precise definitions of these two noise sources, see [7].)

In the presence of noise, each picture provides an estimate of the actual quantity of light falling on the image sensor: q_i (x) = 1k_i f^-1(f_i( x)) where $\hat{k}_i$ is an estimate of the actual exposure constant $k_i$, and $\hat{f}$ is an estimate of the true camera response function $f$, assuming $n_{q_i} >> n_{f_i}$ [7].

Multiple estimates of the actual quantity of light falling on the image sensor may be combined as follows: q(x) = _i c_i q_i(x) _i c_i

Photographic film is traditionally characterized by the so-called ``Density versus log Exposure'' characteristic curvewyckoff[9]. Similarly, in the case of electronic imaging, we may also use logarithmic exposure units, $Q=\log(q)$, so that one image will be $K=log(k)$ units darker than the other: (f^-1(f_1(x))) =Q =( f^-1( f_2( x ) ) ) - K The existence of an inverse for $f$ follows from a semimonotonicity assumption. Semimonotonicity follows from the fact that we expect pixel values to either increase or stay the same with increasing quantity of illumination, $q$1. Since the logarithm function is also monotonic, the problem comes down to estimating the semimonotonic function $F()=\log(f^{-1}())$ and the scalar constant $K$.

The unknowns ($F$ and $K$) may be solved in a least squares sense2.


next up previous
Next: Estimating camera response function Up: .  Introduction: Variable gain image Previous: Quantimetric imaging
Steve Mann 2002-05-25