Scaling method by using dual point cubic-like slope control (DPCSC)
View Patent ↗A method using a dual point cubic-like slope (DPCSC) for scaling a source data to a destination data, wherein a function f(x) is determined to describe the destination data, in which x is a deviation from a current reference point 0, and two reference data of f(0) and f(1) are used as reference data. The method comprises setting an initial condition about a slope D with respect to the function f(x) at the point 0, a gain factor G to time the slope D, and f′(0)=f′(1)=DG. The f(x) is a quadratic equation of f(x)=ax 2 +bx+c, which should pass f(0), f(1), and a middle point f(0.5) by a quantity of f(0.5)=[f(0)+f(1)]/2. The coefficients of a, b, and, c, are solved in two ranges of 0≦x<0.5 and 0.5≦x<1, so as to obtain the function f(x), with a joint at the middle point. The foregoing steps are repeated for scaling data in a next source data region. The function preferable is chosen to be symmetric to the middle point.
1. A method for scaling data from source data to destination data used in a scaling device, wherein the scaling device is to determine a function f(x) to describe the destination data, in which x is a deviation from a current reference point 0, and two reference data of f(0) and f(1) are used as reference data, the method comprising:
setting an initial condition about a slope D with respect to the function f (x) at the point 0, a gain factor G to time the slope D, and f′(0)=f′(1)=DG, wherein the initial condition is stored in the scaling device;
setting the f(x) stored in the scaling device as a quadratic equation of f(x)=ax 2 +bx+c, which should pass f(0), f(1), and a middle point f(0.5) by a quantity of f(0.5)=[f(0)+f(1)]/2;
solving coefficients of a, b, and, c by the scaling device, in two ranges of 0≦x<0.5 and 0.5≦x<1, so as to obtain the function f(x), with a joint at the middle point, wherein the solved function f(x) is used to scale the source data to the destination data; and
repeating the foregoing steps for a next source data.
2. The method as claimed in claim 1 , wherein the function f(x) with the range of 0≦x<1 is symmetric to the middle point.
3. The method as claimed in claim 1 , wherein the initial condition and the function f(x) are determined by:
setting the initial condition that f′(0)=f′(1)=DG=(sign)[f(0)−f(1)]G, wherein sign is +1 or −1; and
setting f(x)=2[f(1)−f(0)−DG]x 2 (DG)x+f(0) for the range of 0≦x<0.5; and
setting f(x)=2[DG+f(0)−f(1)]x 2 +[4f(1)−4f(0)−3DG]x+[DG−f(1)+2f(0)] for the range of 0.5≦x<1.
4. The method as claimed in claim 1 , wherein the coefficients a, b, and c of the function f(x) are determined by:
for 0≦x<0.5, setting
f (0)= c
f′ (0)= b=[f (0)−f(1)] G=DG
f (0.5)=0.25 a+ 0.5 b+c= 0.5 [f (0)+ f (1)], to solve the coefficients a, b, and c; and
for 0.5≦x<1, setting
f (1)= a+b+c
f′ (1)=2 a+b=[f (0) −f (1)] G=DG
f (0.5)=0.25 a+ 0.5 b+c= 0.5 [f( 0)+ f (1)]to solve the coefficients a, b, and c.
5. The method as claimed in claim 1 , wherein the function is determined by setting
f (x)=2 [f (1) −f (0)− DG]x 2 ( DG ) x+f (0) for 0 ≦x< 0.5; and
F ( z )=2 [f( 0) −f (1)+ DG]z 2 +( DG ) z+f (1) for −0.5 ≦z< 0, wherein Z -transform of Z ( z )= X ( x )−1 is applied.
6. The method as claimed in claim 5 , further comprising solving the coefficients a, b, and c of the function f(x) based on the following equations:
for 0≦x<0.5:
f (0) =c
f′ (0) =b=[f (0)− f (1)] G =DG
f (0.5)=0.25 a+ 0.5 b+c= 0.5 [f (0)+ f (1)]; and
for −0.5≦z<0
F (0)= c=f (1)
F ′(0)= b=[f( 0)− f (1) ]G=DG
F (−0.5)=0.25 a− 0.5 b+c=M= 0.5[ f (0)+ f (1)].
7. A scaling apparatus, configured with a scaling function f(x) to scale source data into output data, wherein x is a deviation from a current reference point 0, and two reference data of f(0) and f(1) are used as reference data, the apparatus comprising:
an initial oneration part to set an initial condition about a slope D with respect to the function f(x) at the point 0, a gain factor G to time the slope D, and f′(0)=f′(1)=DG, wherein the f(x) is a quadratic equation of f(x)=ax 2 +bx+c, which should pass f(0), f(1), and a middle point f(0.5) by a quantity of f(0.5)=[f(0)+f(1)]/2;
a calculation operation part to solve the coefficients of a, b, and, c, in two ranges of 0≦x<0.5 and 0.5≦x<1, so as to obtain the function f(x), with a joint at the middle point; and
a scaling operation part to scale the source data based on the solved function f(x).
8. The apparatus as claimed in claim 7 , wherein the function f(x) with the range of 0≦x<1 is symmetric to the middle point.
9. The apparatus as claimed in claim 7 , wherein the initial condition is f′(0)=f′(1)=DG=(sign)[f(0)−f(1)]G, wherein sign is +1 or −1; and
the function f(x) is
f ( x )=2 [f (1)− f (0)− DG]x 2 +( DG ) x+f (0) for the range of 0 ≦x< 0.5; and
f ( x )=2 [DG+f (0)− f (1) ]x 2 +[4 f (1)−4 f (0)−3 DG]x+[DG−f (1)−2 f (0)]for the range of 0.5 ≦x< 1.
10. The apparatus as claimed in claim 7 , wherein the coefficients a, b, and c of the function f(x) axe a solution of the following equations:
for 0≦x<0.5, setting
f (0) =c
f ′(0) =b=[f (0)− f (1)] G=DG
f (0.5)=0.25 a+ 0.5 b+c= 0.5 [f (0)+ f (1)], to solve the coefficients a, b, and c; and
for 0.5≦x<1, setting
f (1) =a+b+c
f′ (1)=2 a+b =[f (0)− f (1)] G=DG
f (0.5)=0.25 a+ 0.5 b+c= 0.5 [f (0)+ f (1)]to solve the coefficients a,b, and c.
11. The apparatus as claimed in claim 7 , wherein the function f(x) is:
f ( x )=2 [f (1)− f (0)− DG]x 2 +( DG ) x+f (0) for 0 ≦x< 0.5; and
F ( z )=2 [f (0)− f (1)+ DG]z 2 +( DG ) z+f (1) for −0.5≦ z< 0, wherein Z -transform of Z ( z )= X ( x )−1 is applied.
12. The apparatus as claimed in claim 11 , wherein the coefficients a, b, and c of the function f(x) are a solution of the following equations:
for 0≦x<0.5:
f (0) =c
f′ (0)= b=[f (0)− f (1)] G=DG
f (0.5)=0.25 a+ 0.5 b+c= 0.5[ f (0)+ f (1)]; and
for −0.5≦z<0
F (0)= c=f (1)
F′ (0)= b=[f (0)− f (1)] G=DG
F (−0.5)=0.25 a− 0.5 b+c=M= 0.5 [f (0)+ f (1)].
13. A method used in a scaling device for generating destination data samples f(x) in response to two source data samples f(0) and f(1), wherein f(x) is generated for x in a range of 0≦x<1, said method comprising the steps of:
(a) generating a middle point sample f(0.5) by defining f(0.5)=[f(0)+f(1)]/2, wherein step (a) is implemented in the scaling device;
(b) fitting a quadratic equation of f(x)=ax 2 +bx+c to said source data samples f(0) and f(1) and said middle point sample f(0.5), wherein the step (b) is implemented in the scaling device; and
(c) generating a resulting equation f(x)=2[f(1)−f(0)−DG]x 2 +(DG)x+f(0) for one subregion of 0≦x<0.5 and f(x)=2[DG+f(0)−f(1)]x 2 +[4f(1)−4f(0)−3DG]x+[DG−f(1)+2f(0)] for another subregion of 0.5≦x<1, wherein DG designates a slope at said. source data samples f(0) and f(1), wherein the step (c) is implemented in the scaling device.
14. The method as claimed in claim 13 , further comprising:
(d) converting f(x)=2[DG+f(0)−f(1)]x 2 +[4f(1)−4f(0)−3DG]x+[DG−f(1)+2f(0)] for another subregion of 0.5≦x<1 into F(z)=2[(0)−f(1)+DG]z 2 +(DG)z+f(1) for −0.5≦z<0, wherein Z-transform of Z(z)=X(x)−1 is applied, wherein the step (d) is implemented in the scaling device.
15. The method as claimed in claim 13 , wherein D is defined to be [f(0)−f(1)] and G is a gain factor.