Keywords: touch screen panels, touch screen controllers, touch screen, touch panels, N-point calibration, calibration, 3-point calibration, touch systems, resistive
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APPLICATION NOTE 5296

Abstract: Mechanical misalignment and scaling factors lead to a mismatch between the values coming from a touch screen panel (as translated by a touch screen controller) and the display (typically an LCD) on which the touch screen panel is mounted. This tutorial discusses how to calibrate the touch screen panel to match the display.

A similar version of this article appeared in the April 20, 2012 issue of *EE Times* magazine.

Mechanical misalignment and scaling factors can lead to a mismatch between the values from a touch-panel system (in this tutorial, the touch-panel system implies a setup comprised of a touch screen and a touch-screen controller) and the display (typically an LCD) on which the touch screen is mounted. This article discusses the mathematical techniques to calibrate the touch-panel system so you can match the graphics on the display to the output from the touch-screen controller.

Intuition suggests that any coordinate point x, y in an x-y plane that has undergone a transformation should look like:

x_{NEW} = f1(x_{OLD}, y_{OLD}) + constant1 |
(Eq. 1a) |

y_{NEW} = f2(x_{OLD}, y_{OLD}) + constant2 |
(Eq. 1b) |

Where x_{NEW} and y_{NEW} are the transformed coordinates; x_{OLD} and y_{OLD} are the old coordinates; f1() and f2() are functions that transform the old coordinates; constants1 and 2 are just that, constants.

If the transformation is linear, then functions f1() and f2() can be replaced by the following equations:

x_{NEW} = A x_{OLD} + B y_{OLD} + C |
(Eq. 2a) |

y_{NEW} = D x_{OLD} + E y_{OLD} + F |
(Eq. 2b) |

Where A, B, C, D, E, and F are constant coefficients.

Note that f1() = A x_{OLD} +B y_{OLD} and f2() = D x_{OLD} + E y_{OLD}, where constant1 and constant2 are C and F, respectively.

An example, showing a graphic transformation, will demonstrate that the final transformation equations from this exercise will lead to the above equations 2a and 2b and, therefore, our intuition holds true. (The user can bypass this example and jump directly to the **Three-Point Calibration** and **N-Point Calibration** discussions below.)

The illustration below (**Figure 2**) shows a square (Shape A) that has undergone translation, rotation, and scaling (Shape B). In the process it is transformed into a rectangle.

To rotate and scale the square (Shape A), it is important to move its center to the origin first (Figure 3) to ensure that the rotation and scaling happen uniformly.

This move to the center, or translation to the center, changes the value of all the coordinates of the square. For simplicity here, only the transformation of x_{1}, y_{1} in Figure 2 will be explored. Consequently, x_{1a}, y_{1a} translates to x_{1b}, y_{1b} and the new equation representing this new figure is:

x_{1b} = x_{1a} + T_{x0} |
(Eq. 3a) |

y_{1b} = y_{1a} + T_{y0} |
(Eq. 3b) |

Rotate the above figure so that proper scaling can be done in the x and y directions. **Figure 4** shows that the figure has been rotated counterclockwise by angle θ.

The new coordinates (x_{1c}, y_{1c}) after the rotation are shown below. (Please see the Appendix for the derivation of the equation for this rotation.)

x_{1c} = x_{1b} Cosθ - y_{1b} Sinθ |
(Eq. 4a) |

y_{1c} = x_{1b} Sinθ + y_{1b} Cosθ |
(Eq. 4b) |

Substitute Equations 3a and 3b into Equations 4a and 4b, and simplify:

x_{1c} = x_{1a} Cosθ - y_{1a} Sinθ + K_{x} |
(Eq. 5a) |

y_{1c} = x_{1a} Sinθ + y_{1a} Cosθ + K_{y} |
(Eq. 5b) |

Where K_{x} = T_{x0} Cosθ - T_{y0} Sinθ and K_{y} = T_{x0} Sinθ + T_{y0} Cosθ.

Now scale the square in both the x and y directions to make it the same size as the rectangle in Figure 2 (Shape B). If we say that G_{x} is the scaling factor in the x direction and G_{y} is the scaling factor in the y direction, then the square will transform into a rectangle (**Figure 5**).

Multiplying Equations 5a and 5b with the scaling factors G_{x} and G_{y} yields Equations 6a and 6b, which are the coordinates for the rectangle shown in Figure 5.

x_{1d} = x_{1a} G_{x} Cosθ - y_{1a} G_{x} Sinθ + K_{x} G_{x} |
(Eq. 6a) |

y_{1d} = x_{1a} G_{y} Sinθ + y_{1a} G_{y} Cosθ + K_{y} G_{y} |
(Eq. 6b) |

Rotate the rectangle by a so that it matches the rotational orientation of Shape B in Figure 2. This leads to the transformation of (x_{1d}, y_{1d}) into (x_{1e}, y_{1e}), as shown in the **Figure 6**.

Once again the rotation is counterclockwise. The equations for (x_{1e}, y_{1e}) in terms of (x_{1d}, y_{1d}) are:

x_{1e} = x_{1d} Cosα - y_{1d} Sinα + K_{x} |
(Eq. 7a) |

y_{1e} = x_{1d} Sinα + y_{1d} Cosα + K_{y} |
(Eq. 7b) |

Substitute Equations 6a and 6b into Equations 7a and 7b, and simplify:

x_{1e} = x_{1a} A + y_{1a} B + P |
(Eq. 8a) |

y_{1e} = x_{1a} D + y_{1a} E + Q |
(Eq. 8b) |

Where:

A = G_{x} Cosθ Cosα - G_{y} Sinθ Cosα

B = - G_{x} Sinθ Cosα - G_{y} Cosθ Cosα

P = K_{x} G_{x} Cosα - K_{y} G_{y} Sinα

D = G_{x} Cosθ Sinα + G_{y} Sinθ Cosα

E = -G_{x} Sinθ Sinα + G_{y} Cosθ Cosα

Q = K_{x} G_{x} Sinα + K_{y} G_{y} Cosα

A = G

B = - G

P = K

D = G

E = -G

Q = K

Now a translation (**Figure 7**) needs to be done to move the rotated rectangle to where Shape B appears in Figure 2.

Assume that the translation in the x and y direction is T_{x} and T_{y}, respectively. Therefore, Equations 8a and 8b will change to:

x_{1f} = x_{1a} A + y_{1a} B + C |
(Eq. 9a) |

y_{1f} = x_{1a} D + y_{1a} E + F |
(Eq. 9b) |

Where C = P + T_{x} and F = Q + T_{y}.

Equations 9a and 9b are what we want—they match Equations 1a and 1b. Note that the coordinate (x_{1f}, y_{1f}) is in terms of (x_{1a}, y_{1a}). In the touch panel mounted on a display the user will have to determine the A, B, C, D, E, and F during calibration.

We start with a pair of equations, the final pair in our mathematical exercise above.

x_{1f} = x_{1a} A + y_{1a} B + C |
(Eq. 9a) |

y_{1f} = x_{1a} D + y_{1a} E + F |
(Eq. 9b) |

The goal of the touch-panel system calibration is to solve the Equations 2a and 2b (or similarly Equations 9a and 9b) to derive values for A, B, C, D, E, and F.

Looking at these equations we know that there are six unknowns. Therefore, we will need six equations to solve for these unknowns and this can be achieved by doing a three-point calibration for a touch-panel system. The user will generate three pairs of (x, y) coordinates by touching the panel at the three pairs of display coordinates: (x_{1d}, y_{1d}), (x_{2d}, y_{2d}) and (x_{3d}, y_{3d}). If their corresponding touch-screen values (as presented by the touch-screen controller) are (x_{1}, y_{1}), (x_{2}, y_{2}), and (x_{3}, y_{3}), then the six unknowns can be solved by the equations shown below. These points must be independent of each other, as shown in **Figure 8**.

x_{1d} = x_{1} A + y_{1} B + C

x_{2d} = x_{2} A + y_{2} B + C

x_{3d} = x_{3} A + y_{3} B + C

x

x

y_{1d} = x_{1} D + y_{1} E + F

y_{2d} = x_{2} D + y_{2} E + F

y_{3d} = x_{3} D + y_{3} E + F

y

y

Now with six equations for the six unknowns, the above can be written in matrix form as:

Therefore, a little matrix manipulation will yield A, B, C, D, E, and F, as shown below:

Where, **Z**^{-1} is the inverse matrix of **Z**.

We will use the MAX11800 touch-screen controller in this example. Assume that an LCD panel display has a resolution of 256 x 768 and that the three calibration points chosen are (65, 350), (200, 195), and (195, 550). These are the exact points that we want the touch panel to show when it is touched. However, since the resolution for the MAX11800 is 4096 x 4096 (12 bits) and because of mechanical misalignment, the coordinate will come out differently. For this exercise, we assume that these values are: (650, 2000), (2800, 1350), and (2640, 3500), respectively. *Please note: these values are made up and serve as an example only*.

Solving for A, B, C, D, E, and F using Equations 10a and 10b, we get the following:

A = 0.0635

B = 0.0024

C = 18.9116

D = -0.0227

E = 0.1634

F = 37.8887

B = 0.0024

C = 18.9116

D = -0.0227

E = 0.1634

F = 37.8887

Therefore, the equations that will generate the x and y coordinates for this particular example are:

x_{d} = 0.0635 x + 0.0024 y + 18.9116

y_{d} = -0.0227 x + 0.1634 y + 37.8887

y

Where (x, y) are the coordinates coming from the touch-panel system and (x_{d}, y_{d}) is the adjusted value.

The user may choose to calibrate using more points. A generalized form of the sets of equations will look like the following:

x_{1d} = x_{1} A + y_{1} B + Cx _{2d} = x_{2} A + y_{2} B + Cx _{3d} = x_{3} A + y_{3} B + C... x _{nd} = x_{n} A + y_{n} B + C |
(Eq. set 11a) |

y_{1d} = x_{1} D + y_{1} E + Fy _{2d} = x_{2} D + y_{2} E + Fy _{3d} = x_{3} D + y_{3} E + F... y _{nd} = y_{n} D + y_{3} E + F |
(Eq. set 11b) |

Where, (x_{1d}, y_{1d})...(x_{nd}, y_{nd}) are coordinates generated by the display; (x_{1}, y_{1})...(x_{n}, y_{n}) are the corresponding values (for n points) determined by the MAX11800 from the touch-screen panel. The goal is to determine the coefficients using these values.

In the Equation set 11a there are three unknowns, A, B, and C, but the set of coordinates is more than three. This implies that there are more equations than unknowns. Therefore, in this instance it makes sense to use the least square fit to utilize all the points and derive an average value of the coefficients. This also means that more calibration points would help lower errors. The least square fit is represented in **Figure 9**. The same technique will be applied for determining the unknowns D, E, and F for the y values.

Equation sets 11a and 11b can be written in matrix form, as shown:

By using the least square fit in this matrix form, the coefficients are given by the equations below. (Please refer to a book on regression analysis for the derivation of the matrix form of the least square fit form.)

Once again, assume that an LCD panel display has a resolution of 256 x 768 and the five calibration points are (100, 350), (50, 200), (200, 200), (210, 600), and (65, 600). These are the exact points that we want the touch panel to show when it is touched. However, since the resolution for the MAX11800 is 4096 x 4096 (12 bits) and because of mechanical misalignment, the points may come out differently. Again assume that they are (1700, 2250), (750, 1200), (3000, 1500), (2500, 3400), and (600, 3000), respectively. *Please note: these values are made up and meant to serve as an example only*.

Solving for A, B, C, D, E, and F using Equations 12a and 12b, we get:

A = 0.0677

B = 0.0190

C = -33.7973

D = -0.0347

E = 0.2100

F = -27.4030

B = 0.0190

C = -33.7973

D = -0.0347

E = 0.2100

F = -27.4030

Therefore, the equations to generate the x and y coordinates for this particular example are:

x_{d} = 0.0677 x + 0.0190 y - 33.7973

y_{d} = -0.0347 x + 0.2100 y - 27.4030

y

Where (x, y) are the coordinates coming from the touch-panel controller, and (x_{d}, y_{d}) is the adjusted coordinate value to match what is shown on the display.

If the least squares technique is used on three points, it will yield the same coefficient as the three-point calibration. It is mathematically easier, therefore, to handle the three-point calibration as three simultaneous linear equations with three unknowns instead of the lengthier calculation with the least squares technique.

The original point (x, y) can be written in polar form as:

x = r Cosθ and y = r Sinθ, | (Eq. A) |

Where r is the radial distance from the origin to the (x, y) coordinate.

After the rotation the (x, y) coordinates become (x', y'):

x' = r Cos(θ + Φ) and y' = r Sin(θ + Φ) | (Eq. B) |

Using the trig identities below:

Cos(θ + Φ) = CosΦ Cosθ - SinΦ Sinθ Sin(θ + Φ) = SinΦ Cosθ - CosΦ Sinθ |
(Eq. C) |

And substituting in Equation B, we get:

x' = r CosΦ Cosθ - r SinΦ Sinθ y' = r SinΦ Cosθ - r CosΦ Sinθ |
(Eq. D) |

Substituting Equation A into Equation D:

x' = x CosΦ - y SinΦ y' = x SinΦ - y CosΦ |
(Eq. E) |

The derivation for the clockwise rotation is left as an exercise for the user.

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APP 5296: APPLICATION NOTE 5296,AN5296, AN 5296, APP5296, Appnote5296, Appnote 5296 |