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Chapter 2: Q2.7-22Q (page 93)

Exercises 21 and 22 concern the way in which color is specified for display in computer graphics. A color on a computer screen is encoded by three numbers (R, G, B) that list the amount of energy an electron gun must transmit to red, green, and blue phosphor dots on the computer screen. (A fourth number specifies the luminance or intensity of the color.)

22. The signal broadcast by commercial television describes each color by a vector (Y, I, Q). If the screen is black and white, only the Y-coordinate is used. (This gives a better monochrome picture than using CIE data for colors.) The correspondence between YIQand a 鈥渟tandard鈥� RGBcolor is given by

\[\left[ {\begin{array}{*{20}{c}}Y\\I\\Q\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{.299}&{.587}&{.114}\\{.596}&{ - .275}&{ - .321}\\{.212}&{ - .528}&{.311}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}R\\G\\B\end{array}} \right]\]

(A screen manufacturer would change the matrix entries to work for its RGBscreens.) Find the equation that converts the YIQdata transmitted by the television station to the RGB data needed for the television screen.

Short Answer

Expert verified

The equation is\(\left[ {\begin{array}{*{20}{c}}R\\G\\B\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{1.00308}&{0.95484}&{0.61785}\\{0.99677}&{ - 0.27070}&{ - 0.64478}\\{1.00849}&{ - 1.11048}&{1.6995}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}Y\\I\\Q\end{array}} \right]\).

Step by step solution

01

Write the matrix using the MATLAB command

Use the MATLAB command to write the matrix equation\[\left[ {\begin{array}{*{20}{c}}Y\\I\\Q\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{.299}&{.587}&{.114}\\{.596}&{ - .275}&{ - .321}\\{.212}&{ - .528}&{.311}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}R\\G\\B\end{array}} \right]\].

\( > > {\rm{A}} = \left[ {{\rm{.299 }}{\rm{.587 }}{\rm{.114; }}{\rm{.596 }} - {\rm{.275 }} - {\rm{.321; }}{\rm{.212 }} - {\rm{.528 }}{\rm{.311}}} \right]{\rm{;}}\)

\(A = \left[ {\begin{array}{*{20}{c}}{.299}&{.587}&{.114}\\{.596}&{ - .275}&{ - .321}\\{.212}&{ - .528}&{.311}\end{array}} \right]\)

02

Obtain the inverse of matrix A

Compute theinverse of matrix A by using the MATLAB command shown below:

\(\begin{array}{l} > > {\rm{A}} = \left[ {{\rm{.299 }}{\rm{.587 }}{\rm{.114; }}{\rm{.596 }} - {\rm{.275 }} - {\rm{.321; }}{\rm{.212 }} - {\rm{.528 }}{\rm{.311}}} \right]{\rm{;}}\\ > > B = {\rm{A}}\^ - 1\end{array}\)

\(B = {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{1.00308}&{0.95484}&{0.61785}\\{0.99677}&{ - 0.27070}&{ - 0.64478}\\{1.00849}&{ - 1.11048}&{1.6995}\end{array}} \right]\)

Thus, the required equation is \(\left[ {\begin{array}{*{20}{c}}R\\G\\B\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{1.00308}&{0.95484}&{0.61785}\\{0.99677}&{ - 0.27070}&{ - 0.64478}\\{1.00849}&{ - 1.11048}&{1.6995}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}Y\\I\\Q\end{array}} \right]\).

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Most popular questions from this chapter

Suppose Tand Ssatisfy the invertibility equations (1) and (2), where T is a linear transformation. Show directly that Sis a linear transformation. [Hint: Given u, v in \({\mathbb{R}^n}\), let \[{\mathop{\rm x}\nolimits} = S\left( {\mathop{\rm u}\nolimits} \right),{\mathop{\rm y}\nolimits} = S\left( {\mathop{\rm v}\nolimits} \right)\]. Then \(T\left( {\mathop{\rm x}\nolimits} \right) = {\mathop{\rm u}\nolimits} \), \[T\left( {\mathop{\rm y}\nolimits} \right) = {\mathop{\rm v}\nolimits} \]. Why? Apply Sto both sides of the equation \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right) = T\left( {{\mathop{\rm x}\nolimits} + y} \right)\). Also, consider \(T\left( {cx} \right) = cT\left( x \right)\).]

Suppose A, B, and Care \(n \times n\) matrices with A, X, and \(A - AX\) invertible, and suppose

\({\left( {A - AX} \right)^{ - 1}} = {X^{ - 1}}B\) 鈥�(3)

  1. Explain why B is invertible.
  2. Solve (3) for X. If you need to invert a matrix, explain why that matrix is invertible.

Solve the equation \(AB = BC\) for A, assuming that A, B, and C are square and Bis invertible.

If \(A = \left( {\begin{aligned}{*{20}{c}}1&{ - 2}\\{ - 2}&5\end{aligned}} \right)\) and \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2&{ - 1}\\6&{ - 9}&3\end{aligned}} \right)\), determine the first and second column of B.

Suppose \({A_{{\bf{11}}}}\) is invertible. Find \(X\) and \(Y\) such that

\[\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{\bf{0}}\\{\bf{0}}&S\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&Y\\{\bf{0}}&I\end{array}} \right]\]

Where \(S = {A_{{\bf{22}}}} - {A_{21}}A_{{\bf{11}}}^{ - {\bf{1}}}{A_{{\bf{12}}}}\). The matrix \(S\) is called the Schur complement of \({A_{{\bf{11}}}}\). Likewise, if \({A_{{\bf{22}}}}\) is invertible, the matrix \({A_{{\bf{11}}}} - {A_{{\bf{12}}}}A_{{\bf{22}}}^{ - {\bf{1}}}{A_{{\bf{21}}}}\) is called the Schur complement of \({A_{{\bf{22}}}}\). Such expressions occur frequently in the theory of systems engineering, and elsewhere.
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