Being edited by Prof. Sumana Gupta 1. only symmetric quantizers ie the input and output levels in the 3rd quadrant are negative

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1 Being edited by Prof. Sumana Gupta 1 Quantization This involves representation the sampled data by a finite number of levels based on some criteria such as minimizing of the quantifier distortion. Quantizer design includes input(decision) levels and output (reconstruction) levels as well as number of levels. The decision can be enhanced by psychovisual or psychoacoustic perception. Quantizers can be classified as memoryless(assumes each sample is quant independent) or with memory (takes into account previous sample) Limit our discussion to memoryless quantizers. Another classification is uniform or non-uniform Uniform quantizers Completely defined by the no of levels it has, its step size and whether it is midriser or midtreader. We will consider only symmetric quantizers ie the input and output levels in the 3rd quadrant are negative Non-uniform quantizers Step sizes not constant. Hence non uniform quantization specified by input and output levels (Ist or 3rd quad) Quantizer design Given the numbers of input or output levels quantizer design involves minimizing any meaningful quantizer distortion such as: E[(f ˆf) ]= (f ˆf) p(f)df where input f ranges from to a u and p(f) is probability density function of f above(missing). Mean absolute quantization error (MAQE) a u E[ f ˆf ] = a u f ˆf p(f)df 3. Mean L N norm quantization error E[ f ˆf N ]= f ˆf p(f)df N is any positive integer 4. Wted quantization error a u a u w(f) f ˆf p(f)df

2 Uniform quantizers Symmetric type o/p reconstuction levels (o/p) r 3 r -d 3 r decision levels 1 (i/p levels) -d -d 1 i/p f d1 d d 3 dead zont o/p zero for i/p [- d 1,d 1 ] -r 1 -r mid treader for f: [ d, d 3 ] quantization level is r quantization is lossy -d 3 -d -d 1 r 3 r r 1 -r 1 d 1 d d 3 -r -r 3 mid riser Step size constant

3 Non-Uniform Quantization r r 1 d 1 d d 3 mid treader (has zero o/p level) f (i/p) r 3 r -d -d 1 r 1 d 1 d mid riser (has no zero o/p level) Max Lloyd Quantizer Designed the quantizers minimizing MSQE and developed tables for input governed by standard distribution functions such as gamma, Laplacian, Gaussian, Rayleigh, Uniform. Quantizers can also be designed tailored to histograms. For a well designed predictor, the histogram of predicted errors tend to follow Laplacian distribution. Given the range of input function as from d 0 d 1 d d 3 d 4 d 5 d j-1 d j d j+1 d J r 0 r 1 r r 3 r 4 r j-1 r j r J-1 a u d i =decision level(i/p level) r i =reconst level (O/P level) 3

4 If d j f d j+1 then ˆf = Q(f) =r j and MSQE = ɛ = E[(f ˆf) ]= where p(f)= Probability density function of r.v.f The error ɛ can also be written as J 1 ɛ = (f r) p(f)df r k and are variables, to minimize ɛ set ɛ r k = ɛ =0 ɛ = r k l=0 a u 1 (f r k 1) p(f)df + (f ˆf) p(f)df +1 (f r k ) p(f)df ( r k 1 ) p( ) ( r k )p( )=0 ( r k 1 )=±( r k ) Since ( r k 1 ) > 0and( r k ) < 0 solution that is valid is ( r k 1 )= ( r k ) or =(r k + r k+1 )/ ie I/P level is average of two adjacent O/P levels. Also +1 ɛ (f r k ) df r k Hence = r k = +1 d k 1 (f r k )p(f)df =0 fp(f)df / +1 p(f)df O/P level is centroid of adj i/p levels. This solution is not closed form. To find i/p level one has to find r k and vice versa. However any iterative techs both r k and can be found When number of O/P levels J is large the... design can be approximated as follows: Assuming p(f) is constant over each quantization level p(f) p(d j ) 4

5 or J 1 ɛ = p(d j ) d j+1 d j (f r j ) df = 1 p(d j )[(d j+1 r j ) 3 (d j r j ) 3 ] 3 j ɛ =0=p( )[(+1 r k ) ( r k ) ] r k (+1 r k ) =( r k ) (+1 r k )= ( r k )as+1 >r k > r k = +1 + ie, each reconstruction level is mid way between two adjacent decision levels. substituting () in (1) we have [ ɛ = 1 j 1 (dj+1 dj ) ( ) ] 3 dj d j+1 p(d j ) 3 = 1 p(d j )(d j+1 d j ) 3 j () minimise ɛ further wrt d j Set d j+1 d j = d j,then ɛ = 1 p(d j )( d j ) 3 j = j ɛ j where ɛ j = 1 p(d j)( d j ) 3 This is minimum when ɛ j is constant ie independent of j th level where k = constant. Let µ j =[p(dj)] 1/3 d j Then [ thus 1 J 1 µ j = k. Constraint... Set [ 1 ] µ 3 µ j + λ() =0 a u Set 1 J 1 [p(d j )] 1/3 d j 1 p 1/3 (f)df = k (A) J p(d j)( d j ) 3 = ɛ = 1 J 1 ] J 1 µ j k =0 µ 3 j 3 µ l + λ/ = 0 or λ = 3µ l 5

6 µ 0 = µ 1 =...µ J 1 =[p(dj)] 1/3 d j 1 µ lj = k or µ l = k J and or and substitute for k we get [p(d j )] 1/3 d j = µ j = k J k d j = (3) J[p(d j )] 1/3 ɛ = 1 J 1 µ 3 j = () k3 J ɛ = 1 J a u p 1/3 (f)df 3 d 0 = d 1 = +(d 1 d 0 ) d = d 1 +(d d 1 )= +(d 1 d 0 ) Substituting for d m from equ(3) d j = + j (d m d m 1 ) m=1 d j = + d j == + j 1 m=0 j 1 m=0 d m d m j d j = + m d m 1 )= + m=1(d m d j = + k J = + k J d j a u = d J = + k J j 1 m=0 1 [p(d m )] 1/3 [p(d j )] 1/3 df d j [p(d f )] 1/3 df d m [p(r m )] 1/3 [p(r m)] 1/3 6

7 Therefore decision level d j = + k J = a u a u [p(d j )] 1/3 df (a U ) a u +ka/l [p(d j )] 1/3 df [p(d j )] 1/3 df for j = 0,1,...J Quantizers using above equation can be designed for any pdf. when 1 p(f) = = 1 a u A then +1 / +1 r k = fk(f)df p(f)df Since then = r k + r k 1 = 1 (+1 + ) = = dk +1 = constant step size =(a u )/J = q or = 1 + q; r k = +1 + z = q + dk ie, all decision and recent levels are equally spaced(show) Since quantity error is distributed uniformly over each step sixe=q with zero mean, MSQE is MSQE ɛ = 1 q q/ e df = q / q/ Let range of f be A and σ f the var... σ f = 1 A A/ f df = A / A/ A b bit quantizer can rep J o/p levels ie. b = J q = A b ie MSQE= 1 A / b 1. The quantizer o/p is an unbiased estimate of i/p ie E[ ˆf] =E[f] 7

8 . The quantizer error is orthogonal to the quantizer o/p ie, E[(f ˆf) ˆf] =0 quantizer noise is uncorrelated with quantization o/p. 3. The variance of quant o/p is reduced by the factor 1 f(b) wheref(b) denotes the mean square distortion of the B-bit quantizer for unity variance i/p s σ ˆf =(1 f(b))σ ˆf 4. It is sufficient to design mean square quantizers for zero mean and unity variance distributions. Suppose o/p of image sensor takes values from 0 to 10. if samples are quant- uniformly to 56 levels then decision and reconstruction levels are: 1 = q or d m = 10 (k 1) for k =1, and r k = + q/ =10/56(k 1) + 5/ If p k is the probability of r k ie. p k = +1 p f (f z )df then E( ˆf) L p k E[f fɛt k ] L = +1 fp f (f)df d L+1 = fp f (f)df = E(f) d 1. E(f ˆf) =E[( ˆf) ] an interesting model for the quantizer follows from this ie f opt MSQuant fˆ f + η - fˆ The quantizer noise η is uncorrelated with ˆf and we ca write f = ˆf + η σ η = E[(f ˆf) ]=E[(f )] E[( ˆf) ] Since σ η 0 average power of quantizer output is reduced by average power σ η Also quantizer noise is dependent on input since... 8

9 ... = E(f ) E( ˆf ) = E(η ) 3. Since for any mean square quantizer we get σ η = σ ff(b) σ ˆf =(1 f(b))σ f Although uniform quantization is straight forward and appears to be a natural approach it may not be optimal. Suppose f(or u) is much more likely to be in one region than in others. It is reasonable to assign more reconstruction levels to that region. If u(or f) falls rarely between t 1 (ord 1 )andt (ord ) the reconstruction level r 11 is rarely used.. u 7/8 r 4 5/8 r 3 3/8 r 1/8 r 1 u 0 (d 0 ) 1/4 t 1 /4 t 3/4 t 3 t 4 Rearranging reconstruction levels r 1,r,r 3,r 4 that they all lie between t 1,t 4 and d 1,d 4 makes more sense. Quantizer in which reconstruction and transition levels do not have even spacing is called non-uniform quantization. The notion that uniform quantizer is the optimal MMSE when p u (u 0 )orf is uniform suggests another approach. Specifically we can... uniform: quantizer g with a uniform quantization and then perform the inverse nonlinearity. f u Nonlinear g f (u) u g Uniform quantization ĝ g ĝ (Nonlinear) -1 û ĝ The non linearity is called companding one choice of nonlinearity or companding is given by g = C[u] = u x 0 = p u (x)dx 1 9

10 The resulting p g (g) is uniform in the int(-1/,1/). The nonuniform quantization companding minimizes the distortion D = E[(ĝ g) ] In case of unique coding, one has to quantize many scalars. One approach is to quantize each scalar independently. This approach is called scalar quantization of a vector source. Suppose we have N scalars...and each scalar fi is quantized to L i reconstruction levels. If L i canbeexpressedas a power of and each reconstruction level is coded with an equal number of bits, L i will be related to required number of bits B i by L i = B i or B i =log L i Total number of bits B required in coding N scalars is B = N i=1 B i The total number of reconstruction levels L is L = N L i = c B i=1 If we have a fixed number of bits B to code all N scalars using scalar quantization of a vector source, B has to be divided among N scalars. The optimal strategy for this used and the probability density functions of the scalar. The optimal strategy type involves assigning more bits to scalars with larger variance and fewer bits to scalars with small variance. As an example suppose we minimize the mean square error E[( ˆf f i ) ]wrtb i for 1 i N, where ˆf i is the result of quantization fi. If the probability density functions are the same for all scalar except for their variance, and we use the same quantization method for their variance and we use the same quantization method for each of the scalars, an approximate solution to the bit allocation problem is B i = B N + 1 log σ i [ N σj ]1/N j=1 L i = Bi = B/N σ i /(πσ i ) 1/N, 1 i N ie number of reconstruction levels for fi is... Although the above equation is an approximate solution obtained under certain specific conditions, it is useful as a reference in other bit allocation problems. As B i in the equation can be -ve and is not in general an integer, a constraint has to be imposed in solving the bit allocation problem s.t B i is a non -ve integer To prove B i = B N + 1 log σ fi [ N σfi ]1/N j=1 1 i N 10

11 We find an expression for the distortion and then find the bit allocation that minimizes the distortion. We perform the minimization using the method of Lagrange. Let average number of bits/sample to be used by the vector source be R = B/N and average number of bits/sample used by the k th scalar be R k,then R = 1 N R k (1) N where N is number of scalars in vector source. The MSQE ie reconstruction error variance is given by σ L k = α k Rk σ f k () α k is a factor that depends on the input σ f k variance of scalar f k... The objective of bit allocation procedure is to find R k that minimizes equ(3) subject given by eqn(1) Let us assume that alpha k is a constant α for... We can set up the minimization problem in terms of Lagrange s multipliers as J = α N R k σf k λ(r 1 N Taking the derivative of J wrt R k and setting it equal to zero, we obtain Substituting equ(5) in (1) we get a value for λ as: Substituting (6) in (5) we get N R k ) (4) R k = 1 log (α ln σ f k ) 1 log λ (5) λ = N (α ln σf k ) 1/N R (6) R k = R +... This will increase the average bit rate above the constraint. the non zero R k s are uniformly reduced until the average rate is equal to R. 11