CurveP - J E Patterson

J E Patterson - jepspectro.com - 20231119

Description

The program fits data that may be linear at low x,y amplitudes but y falls off at higher x values. This is a common issue in many physical systems. x1, y1 should preferably be on the linear part of the curve. CurveP is not a regression. It assumes that the data is reasonably precise. Linear and power fit data sets also work well.

The equation used is:

y = a.x^b + c.x.e^-d.x

See Curve Fitting at jepspectro.com.

Originally chart recorders were used to obtain data which occupied the y axis, leaving interpreted results naturally plotted on the x axis, when graphed on the same paper. Here we are using y as the equation output axis.

The parameters a = scale, b = order, c = slope and d = factor are calulated by iteration using 3 standards and a blank. x is first normalised, the equation solved and un-normalised to get y. This limits the magnitude of calculated values in the iteration.

y1, y2 and y3 are known standards with respective readings of x1, x2 and x3.

y1 ENTER, y2 ENTER, y3 GSB A
x1 ENTER, x2 ENTER, x3 GSB B

Do not press ENTER after y3 or x3.
Note that y3>y2>y1 and x3>x2>x1.
After normalisation y3 = x3 = 10 so x3 is not saved. y3 is used instead in the program.

The curve runs through the origin and three points (x1,y1) (x2,y2) (x3,y3). The order of the upper part of the curve is displayed.

y GSB C to get x. This uses SOLVE to run GSB D until zero, the solution x is obtained
x GSB D to get y.

If there is a blank value, enter it as data and subtract the output from all the results.

Test [x,y] data sets [2,2 4,5 5,8] [1,1 2,4 3,9] [1,1 2,2 3,3] [20,3.69 30,8.64 47,22.16]
The last data set relates X-ray fluorescence k line energy in keV to the element atomic number - see Moseley's Law.

Notes:

In this update I have improved the starting values for the iteration and added a refined guess to the stack for SOLVE. A second guess is created by entering the first twice and multiplying by 1.1.

Order is limited to 10.

Because only 3 standards and a blank are used a regression approach is not suitable. The iteration attempts to fit the standards and blank exactly. The standards need to be accurately prepared. The calibration curve can be validated by running other standards as samples.

Program Resources

Labels

Name	Description
A	Enter y1, y2, y3 from stack
B	Enter x1, x2, x3 from stack, normalise, label 1 loop
C	Enter y to get x. Also entered on stack for guess
D	Enter x to get y
1	Iteration loop runs from B to find parameters
4	RCL .5 (temp1) 1/x STO .5 (temp1)
5	Curve function-x = 0 for solve

Storage Registers

Name	Description	Name	Description	Name	Description
1	y1	7	xscale to normalise x data	.4	a - scale parameter
2	y2	8	yscale to normalise y data	.5	temp1
3	y3	9	c - slope parameter = x1/y1	.6	temp2
4	b - order parameter - calculated	.1	x1	.7	temp3
5	d - factor parameter - first guess = 1	.2	x2	.8	xinput
6	count - use to modify convergence rate	.3	b - Order parameter - first guess ln(y3/y2)/ln(x3/x2)	.9	yinput

Program

Line	Display	Key Sequence	Line	Display	Key Sequence	Line	Display	Key Sequence
000			055	16	CHS	110	44 .7	STO . 7
001	42,21,11	f LBL A	056	12	eˣ	111	45,20, 4	RCL × 4
002	44 8	STO 8	057	45,20, 3	RCL × 3	112	44 4	STO 4
003	33	R⬇	058	45,20, 9	RCL × 9	113	45 .7	RCL . 7
004	44 2	STO 2	059	30	−	114	1	1
005	33	R⬇	060	45 3	RCL 3	115	30	−
006	44 1	STO 1	061	45 4	RCL 4	116	43 16	g ABS
007	45 8	RCL 8	062	43,30, 2	g TEST x<0	117	1	1
008	1	1	063	45 .3	RCL . 3	118	26	EEX
009	0	0	064	44 4	STO 4	119	16	CHS
010	44 3	STO 3	065	14	yˣ	120	6	6
011	10	÷	066	10	÷	121	43,30, 8	g TEST x<y
012	44,10, 2	STO ÷ 2	067	44 .4	STO . 4	122	22 1	GTO 1
013	44,10, 1	STO ÷ 1	068	45,10, 1	RCL ÷ 1	123	45 4	RCL 4
014	43 32	g RTN	069	45 .1	RCL . 1	124	43 32	g RTN
015	42,21,12	f LBL B	070	45 4	RCL 4	125	42,21,14	f LBL D
016	44 7	STO 7	071	14	yˣ	126	44 .8	STO . 8
017	33	R⬇	072	20	×	127	45,10, 7	RCL ÷ 7
018	44 .2	STO . 2	073	44 .5	STO . 5	128	45 3	RCL 3
019	33	R⬇	074	1	1	129	20	×
020	44 .1	STO . 1	075	43,30, 8	g TEST x<y	130	44 .8	STO . 8
021	45 7	RCL 7	076	32 4	GSB 4	131	45 4	RCL 4
022	45 3	RCL 3	077	1	1	132	14	yˣ
023	10	÷	078	45 .5	RCL . 5	133	45,20, .4	RCL × . 4
024	44,10, .2	STO ÷ . 2	079	30	−	134	45 5	RCL 5
025	44,10, .1	STO ÷ . 1	080	43 12	g LN	135	45,20, .8	RCL × . 8
026	45 3	RCL 3	081	16	CHS	136	16	CHS
027	45 2	RCL 2	082	45,10, .1	RCL ÷ . 1	137	12	eˣ
028	10	÷	083	44 5	STO 5	138	45,20, .8	RCL × . 8
029	43 12	g LN	084	45,20, .2	RCL × . 2	139	45,20, 9	RCL × 9
030	45 3	RCL 3	085	16	CHS	140	40	+
031	45 .2	RCL . 2	086	12	eˣ	141	45,20, 8	RCL × 8
032	10	÷	087	45,20, .2	RCL × . 2	142	45 3	RCL 3
033	43 12	g LN	088	45,20, 9	RCL × 9	143	10	÷
034	10	÷	089	45 .4	RCL . 4	144	43 32	g RTN
035	44 4	STO 4	090	45 .2	RCL . 2	145	42,21,13	f LBL C
036	44 .3	STO . 3	091	45 4	RCL 4	146	44 .9	STO . 9
037	1	1	092	14	yˣ	147	36	ENTER
038	44 5	STO 5	093	20	×	148	36	ENTER
039	0	0	094	40	+	149	1	1
040	44 6	STO 6	095	44 .6	STO . 6	150	48	.
041	45 1	RCL 1	096	45 2	RCL 2	151	1	1
042	45,10, .1	RCL ÷ . 1	097	45,10, 3	RCL ÷ 3	152	20	×
043	44 9	STO 9	098	3	3	153	42,10, 5	f SOLVE 5
044	42,21, 1	f LBL 1	099	0	0	154	43 32	g RTN
045	1	1	100	20	×	155	42,21, 4	f LBL 4
046	44,40, 6	STO + 6	101	45,10, 6	RCL ÷ 6	156	45 .5	RCL . 5
047	45 3	RCL 3	102	11	√x̅	157	15	1/x
048	45 4	RCL 4	103	45,20, .6	RCL × . 6	158	44 .5	STO . 5
049	43,30, 7	g TEST x>y	104	45 .6	RCL . 6	159	43 32	g RTN
050	45 .3	RCL . 3	105	45,30, 2	RCL − 2	160	42,21, 5	f LBL 5
051	44 4	STO 4	106	10	÷	161	32 14	GSB D
052	45 3	RCL 3	107	15	1/x	162	45,30, .9	RCL − . 9
053	45 5	RCL 5	108	1	1	163	43 32	g RTN
054	45,20, 3	RCL × 3	109	40	+