Digital Image Processing Lectures 7 & 8 · Image Quantization Lloyd-Max Quantizer Image Transforms...

Image Quantization Lloyd-Max Quantizer Image Transforms

Digital Image ProcessingLectures 7 & 8

M.R. Azimi, Professor

Department of Electrical and Computer EngineeringColorado State University

February 2011

M.R. Azimi Digital Image Processing


Quantization Process

A quantizer maps the continuous variable x into a discrete xq whichtakes values from a finite set r1, . . . , rL of numbers. This mapping isgenerally a staircase function as shown below.

Quantization Rule:Define tk, k = 1, . . . , L+ 1 as a set of transition or decision levels with

t1 and tL+1 as the Min. and Max. values of x, respectively. If x lies in

the interval [tk, tk+1), then xq = rk, where rk is called the kth

reconstruction level.



Properties:

Quantization mapping is irreversible, i.e. for a given quantizeroutput, the input value cannot be determined uniquely.

Quantization incurs distortion. The design is a trade-offbetween simplicity and performance (minimum distortion).

Quantization is a ”zero-memory” operation, i.e. outputdepends on only one input.

Uniform Quantizer:Simplest and most commonly used (e.g., PCM, differential PCM andtransform coding) quantizer. Let the output of an image sensor takesvalues between 0 to A. If samples are quantized uniformly to L (e.g.L = 256) levels, then transition and reconstruction levels are

tk = A(k−1)L k = 1, . . . , L+ 1

rk = tk + A/2L k = 1, . . . , L

The interval q =∆= tk − tk−1 = rk − rk−1 is constant for different values

of k and is called ”quantization interval”.



Minimum Mean Squared or Lloyd-Max Quantizer

This quantizer minimizes the mean squared error for a given number ofquantization levels. Let x, with 0 ≤ x ≤ A be a real scalar randomvariable with a continuous PDF pX(x). It is desired to find optimum thedecision levels tk and the reconstruction levels rk for an L-level quantizersuch that the mean square error (MSE) (or quantization distortion)

ε = E[(x− xq)2] =

∫ tL+1

t1

(x− xq)2pX(x)dx

is minimized. Note that pX(x) is the PDF of the amplitude of x i.e.∫ A0pX(x)dx = 1. The MSE can be rewritten as

ε =

L∑i=1

∫ ti+1

ti

(x− ri)2pX(x)dx



To minimize ε w.r.t. tk and rk

∂ε

∂tk= 0 ⇒ (tk − rk−1)2pX(tk)− (tk − rk)2pX(tk) = 0

∂ε

∂rk= 0 ⇒ 2

∫ tk+1

tk

(x− rk)pX(x)dx = 0, k ∈ [1, L]

where the first derivative is obtained using the fact thattk ≤ x < tk+1 ⇒ xq = rk, simplification of the above equationsgives

tk =rk + rk−1

2(1)

rk =

∫ tk+1

tkxpX(x)dx∫ tk+1

tkpX(x)dx

(2)

i.e. the optimum transition levels lie halfway between the optimumreconstruction levels, which in turn lie at the center of mass of thePDF in between the optimum transition levels.



These non-linear equations must be solved simultaneously giventhe bounding values t1 and tL+1. In practice, these equations canbe solved by an iterative scheme such as the Newton method.

When the No. of quantization levels is large, an approximatesolution can be obtained by modeling the PDF pX(x) as apiecewise constant function i.e.,

pX(x) ' pX(tj), tj∆=tj + tj+1

2, tj ≤ x ≤ tj+1



Using this approximation and performing the requiredminimization, the equation for optimum transition level becomes

tk+1 ≈(tL+1 − t1)

∫ akt1

[pX(x)]−1/3dx∫ tL+1

t1[pX(x)]−1/3dx

+ t1

with

ak =k(tL+1 − t1)

L+ t1, k ∈ [0, L]

Thus, this together with optimum reconstruction level in (2) i.e.rk =

tk+tk+1

2 are now de-coupled.

The quantizer MS distortion is

ε ≈ 1

12L2∫ tL+1

t1

[pX(x)]13dx3



Properties of Lloyd-Max Quantizers

1 Unbiased in the mean i.e. E[X] = E[Xq].

Proof: Since Xq is a discrete r.v. we have

E[Xq] =

L∑k=1

rkpXq(xq = rk)

Using the staircase mapping of quantizer, the PMF at rk is

pXq(xq = rk) =

∫ tk+1

tk

pX(x)dx

Thus, using Eq. (2) of Llyod-Max quantizer we get

E[Xq] =

L∑k=1

rk

∫ tk+1

tk

pX(x)dx =

L∑k=1

∫ tk+1

tk

xpX(x)dx

which can be reduced to

E[Xq] =

∫ tL+1

t1

xpX(x)dx = E[X]



2 Orthogonality of error with output i.e. E[(X −Xq)Xq] = 0.Proof: To prove we show E[XXq] = E[X2

q ]. Using the correlationdefinition

E[XXq] =

L∑k=1

∫ tk+1

tk

xxqpX,Xq(x, xq = rk)dx

When x ∈ Ωk i.e. segment k, joint PDF pX,Xq (x, xq = rk) can beexpressed as pXq

(xq = rk|x ∈ Ωk)pX(x ∈ Ωk) = pX(x ∈ Ωk).Thus,

E[XXq] =

L∑k=1

rk

∫ tk+1

tk

xpX(x)dx

Using Eq. (2) of Llyod-Max quantizer and result in property 1,

E[XXq] =

L∑k=1

rk

∫ tk+1

tk

xpX(x)dx =

L∑k=1

r2kpXq

(xq = rk) = E[X2q ]

Byproduct: Eq. (2) can be expressed as conditional mean estimator

rk =

∫ tk+1

tk

xpXq(xq|x ∈ Ωk)dx = E[Xq|X ∈ Ωk]



Remarks

1 Two commonly used PDF’s for quantization of image related dataare:

pX(x) =1√

2πσ2x

exp(− (x− µ)2

2σ2x

) Gaussian

pX(x) =α

2exp(−α|x− µ|) Laplacian

µ is the mean, σ2x is the variance (variance of the Laplacian PDF is

σ2x = 2

α ).

2 For uniform distribution, the Lloyd-Max quantizer become a linearquantizer. Let

pX(x) =

1tL+1−t1 t1 ≤ x ≤ tL+1

0 otherwise

Then from (2) we get rk =(t2k+1−t

2k)

2(tk+1−tk) = tk+1+tk2 . Now combining

this result with (1) yields tk = tk+1+tk−1

2 . Thus,

tk − tk−1 = tk+1 − tk = constant∆= q.



Also

q =tL+1 − t1

L, tk = tk−1 + q, rk = tk + q/2

and we have rk − rk−1 = tk − tk−1 = q.i.e. all transition and reconstruction levels are equally spaced. The

quantization error e∆= x− xq is uniformly distributed over the

interval (−q/2, q/2); hence MSE is

ε =1

q

∫ q/2

−q/2α2dα =

q2

12

The variance σ2x of a uniform random variable (original image)

whose range is A is A2/12. For a uniform quantizer having B bitsL = 2B, we have q = A/2B. This gives

σ2x

ε= 22B ⇒ SNRdB = 10log1022B = 6B dB

i.e. in this special case the SNR achieved is 6dB/bit.M.R. Azimi Digital Image Processing


Image Transforms

2-D unitary transforms have found numerous applications in DIP. Thefollowings are some examples.

1 Data compression using transform coding: reduce the bandwidth bydiscarding or grossly quantizing low magnitude transformcoefficients.

2 Feature extraction: extract salient features of a pattern e.g. in FThigh frequency components correspond to the edge information inthe image.

3 Dimensionality reduction to principal components: reduce thedimensionality of the image space by discarding the smallcomponents.

4 Image filtering: Wiener filter and standard 2-D filters.



Discrete Space Fourier Transform

The discovery of FFT for computing the FT of sampled signals made itpossible for the Fourier analysis of signals to be performed efficientlyusing digital computers. The computational saving made possible by FFTare even more important in the 2-D case when a large amount of data isinvolved. In what follows we start with 2-D DSFT and then show therelation between 2-D DSFT and 2-D DFT.

Spatial Sampling of Band-limited Images and DSFTWe saw that the 2-D FT of a continuous space image x(u, v) is

X(ω1, ω2) =

∫ ∫ ∞−∞

x(u, v)e−jω1ue−jω2vdudv

and the inverse FT is

x(u, v) =1

4π2

∫ ∫ ∞−∞

X(ω1, ω2)ej(ω1u+ω2v)dω1dω2



For the sampled image i.e. x(m∆u, n∆v) or simply x(m,n) theFT becomes

X(ω1, ω2) =

∞∑m=−∞

∞∑n=−∞

x(m,n)e−jω1m∆ue−jω2n∆v

∆u,∆v are sampling intervals. Alternatively,

X(Ω1,Ω2) =∑ ∞∑

m,n=−∞x(m,n)e−j(Ω1m+Ω2n)

or 2D DSFTx(m,n) where Ω1∆= ω1∆u,Ω2

∆= ω2∆v, are

discrete frequency variables. Note that since ejΩn = ej(Ω+2kπ)n,we have X(Ω1,Ω2) = X(Ω1 + 2kπ,Ω2 + 2lπ), i.e. the 2-D DSFTis doubly periodic with periods 2π owing to spatial sampling atspacing ∆u and ∆v. The inverse DSFT considering this periodicbehavior becomes

x(m,n) =1

4π2

∫ ∫ π

−πX(Ω1,Ω2)ej(Ω1m+Ω2n)dΩ1dΩ2



Example 1Determine the frequency response of a 2-D system for which the impulseresponse is

h(m,n) =

1 |m| < M, |n| < N0 otherwise

H(Ω1,Ω2) =∑ ∞∑

m,n=−∞h(m,n)e−j(Ω1m+Ω2n)

=

M−1∑m=−M+1

N−1∑n=−N+1

e−jΩ1me−jΩ2n

= (e−jΩ1(−M+1) − e−jΩ1M

1− e−jΩ1)

×(e−jΩ2(−N+1) − e−jΩ2N

1− e−jΩ2)



Then

= (ejΩ1(M− 1

2) − e−jΩ1(M− 1

2)

ejΩ1/2 − e−jΩ1/2)

×(ejΩ2(N− 1

2) − e−jΩ2(N− 1

2)

ejΩ2/2 − e−jΩ2/2)

=sin(Ω1(M − 1/2))

sin(Ω1/2)× sin(Ω2(N − 1/2))

sin(Ω2/2)

Note that this is the aliased version of the 2-D sinc function due tothe sampling in the spatial domain (see Fig. M = 32 and N = 16).

−2

0

2

−2

0

2

0

500

1000

1500

Ω1

Aliased 2−D Sinc

Ω2

H(Ω

1,Ω

2)



Example 2Determine the 2D DSFT of the image x(m,n) = amu(m)v(n) whereu(m) is a 1D unit step function and

v(n) =

1 0 ≤ n ≤ N − 10 elsewhere

X(Ω1,Ω2) =∑ ∞∑

m,n=−∞x(m,n)e−j(Ω1m+Ω2n)

=

∞∑m=−∞

amu(m)e−jΩ1m∞∑

n=−∞e−jΩ2nv(n)

=

∞∑m=0

ame−jΩ1mN−1∑n=0

e−jΩ2n

= (1

1− ae−jΩ1)× e−jΩ2( N−1

2 ) sin(NΩ2/2)

sin(Ω2/2)


Digital Image Processing Lectures 7 & 8 · Image Quantization Lloyd-Max Quantizer Image Transforms...

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