Question

Develop a piecewise linear transfer function for the calibration data shown below (and on your handout) Use a (s,T) coordinate of (5.5,0) as your transition point. Report your answer as an equation using the equation editor function. 225 200 175 150 U 125 2 100 O 75 50 - 25 ト + ト + 25 75 89 10 Stimulus (mV)

Answer 1

In piecewise linear transform, we break the non-linear graph onto smaller linear pieces.

Here, it is mentioned in the question that point (5.5, 0) is the transition point. So, the graph needs to be broken in two pieces

(i) Graph before S = 5.5

(ii) Graph after S = 5.5

Case (i):

The equation for a line is given by,

y = mx + c

Here, temperature is on y-axis and stimulus is on x-axis. So, the equation of line can be written as,

T = mS + c

To, derive the equation, two points on line before S = 5.5 needs to be considered. Let us consider point - 1 as (0.7,-40) and point - 2 as (5.5,0).

Slope m = (T₂ - T₁)/(S₂ - S₁)

= (0 - (-40))/(5.5 - 0.7)

= 8.33

To find value of c, put values of T, S and m in equation.

T = mS + c

0 = (8.33 * 5.5) + c

c = -45.83

So, the equation for the first part of graph is given by,

T = 8.33*S - 45.83

Case (ii):

To, derive the equation, two points on line after S = 5.5 needs to be considered. Let us consider point - 1 as (10,178) and point - 2 as (5.5,0).

Slope m = (T₂ - T₁)/(S₂ - S₁)

= (0 - 178)/(5.5 - 10)

= 39.55

To find value of c, put values of T, S and m in equation.

T = mS + c

0 = (39.55 * 5.5) + c

c = -217.53

So, the equation for the first part of graph is given by,

T = 39.55*S - 217.53

--> So, the total piecewise equation is given by,

T = 8.33*S - 45.83 for S <= 5.5

= 39.55*S - 217.53 for S > 5.5