Post on 03-Jan-2016
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5. Math 4
Matlab
Matrix basedPowerful analytical toolHandles transforms wellEasy to program
1. Tools
5. Math 5
Mathcad
Mathematical toolEvolving into handling transfer functionsHas special programming languageDocumentation closer to real math
1. Tools
5. Math 6
Labview
Powerful analysis toolUses graphical language to translate
concepts into C-code and then execute
1. Tools
5. Math 7
2. Matrices (1 of 2)
AdditionSubtractionMultiplicationVector, dot product, & outer productTransposeDeterminant of a 2x2 matrixCofactor and adjoint matricesDeterminantInverse matrix
2. Matrices
5. Math 9
Addition (1 of 2)
cIJ = aIJ + bIJcIJ = aIJ + bIJ
1 -1 0-2 1 -3 2 0 2
1 -1 -1 0 4 2-1 0 1
A= B=
2 -2 -1 -2 5 -1 1 0 3
C=
C=A+B
2. Matrices
5. Math 10
Addition (2 of 2)
Matrix addition using ExcelMatrix addition using Excel
2. Matrices
A + B = C1 -1 0 1 -1 -1 2 -2 -1-2 1 -3 0 4 2 -2 5 -12 0 2 -1 0 1 1 0 3
1. Highlight area for answer2. Type "="3. Highlight area of first matrix4. Type "+"5. Highlight area for second matrix6. Type CTL+SHIFT+ENTER
5. Math 11
Subtraction (1 of 2)
cIJ = aIJ - bIJcIJ = aIJ - bIJ
1 -1 0-2 1 -3 2 0 2
1 -1 -1 0 4 2-1 0 1
A= B=
0 0 1 -2 -3 -5 3 0 1
C=
C=A-B
2. Matrices
5. Math 12
Subtraction (2 of 2)
Matrix subtraction using ExcelMatrix subtraction using Excel
2. Matrices
A - B = C1 -1 0 1 -1 -1 0 0 1-2 1 -3 0 4 2 -2 -3 -52 0 2 -1 0 1 3 0 1
1. Highlight area for answer2. Type "="3. Highlight area of first matrix4. Type "-"5. Highlight area for second matrix6. Type CTL+SHIFT+ENTER
5. Math 13
Multiplication (1 of 2)
cIJ = aI1 * b1J + aI2 * b2J + aI3 * b3J cIJ = aI1 * b1J + aI2 * b2J + aI3 * b3J
1 -1 0-2 1 -3 2 0 2
1 -1 -1 0 4 2-1 0 1
A= B=
1 -5 -3 1 6 1 0 -2 0
C=
C=A*B
2. Matrices
5. Math 14
Multiplication (2 of 2)
Matrix multiplication using ExcelMatrix multiplication using Excel
2. Matrices
A * B = C1 -1 0 1 -1 -1 1 -5 -3-2 1 -3 0 4 2 1 6 12 0 2 -1 0 1 0 -2 0
1. Highlight area for answer2. Type "= MMULT(", or use INSERT FUNCTION3. Highlight area of first matrix4. Type ","5. Highlight area for second matrix6. Type CTL+SHIFT+ENTER
5. Math 15
Transpose (1 of 3)
bIJ = aJIbIJ = aJI
1 -1 0-2 1 -3 2 0 2
1 -2 2 -1 1 0 0 -3 2
A= B=
B=AT
2. Matrices
5. Math 16
Transpose (2 of 3)
Matrix transpose using ExcelMatrix transpose using Excel
2. Matrices
A A-transpose1 -1 0 1 -2 2-2 1 -3 -1 1 02 0 2 0 -3 2
1. Highlight area for answer2. Type "= TRANSPOSE(", or use INSERT FUNCTION3. Highlight area of matrix4. Type CTL+SHIFT+ENTER
5. Math 17
Transpose (3 of 3)
(AB)T = BT AT
1 -1 0-2 1 -3 2 0 2
1 -1 -1 0 4 2-1 0 1
A= B= 1 1 0 -5 6 -2 -3 1 0
(AB)T =
1 -2 2-1 1 0 0 -3 2
1 0 -1 -1 4 0 -1 2 1
AT = BT = BTAT = 1 1 0 -5 6 -2 -3 1 0
2. Matrices
5. Math 18
Vector, dot & outer products (1 of 2)
A vector v is an N x 1 matrixDot product = inner product = vT x v = a
scalarOuter product = v x vT = N x N matrix
2. Matrices
5. Math 19
Vector, dot & outer products (2 of 2)
Matrix inner and outer products using ExcelMatrix inner and outer products using Excel
2. Matrices
inner outerv v' v'*v v*v'1 1 2 3 14 1 2 32 2 4 63 3 6 9
5. Math 20
Determinant of a 2x2 matrix
2x2 determinant = b11 * b22 - bI2 * b212x2 determinant = b11 * b22 - bI2 * b21
B = 1 -1-2 1
= -1
2. Matrices
5. Math 21
Cofactor and adjoint matrices
1 -1 0-2 1 -3 2 0 2
A=
1 -3 0 2
-1 0 0 2
-1 0 0 -3
-2 -3 2 2
1 0 2 2
1 0-2 -3
-2 1 2 0
1 -1 2 0
1 -1-2 1
2 -2 -22 2 -23 3 -1
=B = cofactor =
2 2 3-2 2 3-2 -2 -1
C=BT = adjoint=
2. Matrices
-
- -
-
5. Math 22
Determinant 1 -1 0-2 1 -3 2 0 2
determinant of A =
The determinant of A = dot product of any row in A times the corresponding column in the adjoint matrix =
dot product of any row (or column) in A timesthe corresponding row (or column) in the cofactor matrix
The determinant of A = dot product of any row in A times the corresponding column in the adjoint matrix =
dot product of any row (or column) in A timesthe corresponding row (or column) in the cofactor matrix
1 -1 0
=4
2-2-2
= 4
2. Matrices
5. Math 23
Inverse matrix (1 of 3)
B = A-1 =adjoint(A)/determinant(A) = 0.5 0.5 0.75-0.5 0.5 0.75-0.5 -0.5 -0.25
1 -1 0-2 1 -3 2 0 2
0.5 0.5 0.75-0.5 0.5 0.75-0.5 -0.5 -0.25
1 0 00 1 00 0 1
=
2. Matrices
InverseInverse
5. Math 24
Inverse matrix (2 of 3)
Matrix inverse using ExcelMatrix inverse using Excel
2. Matrices
A inv(A)1 -1 0 0.5 0.5 0.75-2 1 -3 -0.5 0.5 0.752 0 2 -0.5 -0.5 -0.25
1. Highlight area for answer2. Type "= MINVERSE(", or use INSERT FUNCTION3. Highlight area of matrix4. Type CTL+SHIFT+ENTER
5. Math 25
Inverse matrix (3 of 3)(AB)-1 = B-1 A-1
1 -1 0-2 1 -3 2 0 2
1 -1 -1 0 4 2-1 0 1
A= B= 0.25 0.75 1.625 0 0 -0.5 -0.25 0.35 1.375
(AB)-1 =
0.5 0.5 0.75 -0.5 0.5 0.75-0.5 -0.5 -0.25
A-1 =
B-1 =
B-1A-1 = 0.25 0.75 1.625 0 0 -0.5 -0.25 0.35 1.375 2 0.5 1
-1 0 -1 2 0.5 2
Inverse of a productInverse of a product
2. Matrices
5. Math 26
Orthogonal matrix
An orthogonal matrix is a matrix whose inverse is equal to its transpose.
1 0 00 cos sin 0 -sin cos
1 0 00 cos -sin 0 sin cos
1 0 00 1 00 0 1
=
2. Matrices
5. Math 27
Hermetian matrix (1 of 3)
A Hermetian matrix is a matrix that is equal to its own Hermetian transpose• A = AH
The Hermetian transpose of A is the complex conjugate transpose of A• AH = AT
Hermetian matrixHermetian matrix
2. Matrices
5. Math 28
Hermetian matrix (2 of 3)
1 1-I 21+I 3 i2 -i 0
A =
1 1+I 21-I 3 - i2 +i 0
AT =
AT = 1 1-I 21+I 3 i2 -i 0
= A
ExampleExample
2. Matrices
5. Math 29
Hermetian matrix (3 of 3)
Hermetian matrix using ExcelHermetian matrix using Excel
2. Matrices
H H' conj(H')1 1-i 2 1 1+i 2 1 1-i 2
1+i 3 -i 1-i 3 i 1+i 3 -i2 i 0 2 -i 0 2 i 0
1. Use COMPLEX to enter complex numbers 2. Use IMCONJUGATE to convert cell-by-cell Note: Cell operations; not matrix
5. Math 30
Unitary matrix
A matrix is unitary if its inverse equals its Hermetian transpose• U-1 = UH
DFT and inverse DFT are unitary matrices
2. Matrices
5. Math 32
Example 1 (1 of 9)
x + 2y + 3z = 14-2x + + z = 1 2x + y = 4
1 2 3-2 0 1 2 1 0
A = -1 3 2 2 -6 -7 -2 3 4
A-1 = -1/3 b =14 1 4
xyz
= A-1 b = 1 2 3
Solve 3 equations and 3 unknownsSolve 3 equations and 3 unknowns3. Least squares
5. Math 33
Example 1 (2 of 9)
x + 2y + 3z = 14-2x + + z = 1 2x + y = 4 3x + y - z = 2
xyz
= 1 2 3
x + 2y + 3z = 13-2x + + z = 1 2x + y = 4 3x + y - z = 3
xyz
= ?
What happens if we have 4 equations and 3 unknownsWhat happens if we have 4 equations and 3 unknowns
3. Least squares
5. Math 34
Example 1 (3 of 9)
e1 = x + 2y + 3z - 13e2 = -2x + + z - 1e3 = 2x + y - 4e4 = 3x + y - z - 3
Minimize J = (e12 + e2
2 + e32 + e4
2)
Minimize the sum of squaresMinimize the sum of squares
3. Least squares
5. Math 35
Example 1 (4 of 9)
Solve using Solver in ExcelSolve using Solver in Excel3. Least squares
x y x b e e^20.46 3.37 1.91
coefficients1.0 2.0 3.0 13.0 -0.1 0.0-2.0 0.0 1.0 1.0 0.0 0.02.0 1.0 0.0 4.0 0.3 0.13.0 1.0 -1.0 3.0 -0.2 0.0
sum of squares 0.11
1. Set up matrix as shown2. Select Solver3. Select the cell containing the sum of squares4. Select "minimize"5. Set "by changing cells" to the unknowns -- x, y, z6. Select solve
5. Math 36
Example 1 (5 of 9)
e1 = x + 2y + 3z - 13e2 = -2x + + z - 1e3 = 2x + y - 4e4 = 3x + y - z - 3
A = 1 2 3-2 0 1 2 1 0 3 1 1
b = 13 1 4 3
ATA s = AT b s = [ATA]-1 AT b = xyz
= 0.46 3.37 1.91
Solve using matricesSolve using matrices3. Least squares
5. Math 37
Example 1 (6 of 9)
A = a1x a1y a1z
a2x a2y a2z
a3x a3y a3z
a4x a4y a4z
b = b1
b2
b3
b4
a1x a2x a3x a4x
a1y a2y a3y a4y
a1z a2z a3z a4z
AT =
akx akx aky akx akz akx
akx aky aky aky akz aky
akx akz aky akz akz akz
a1x a2x a3x a4x
a1y a2y a3y a4y
a1z a2z a3z a4z
a1x a1y a1z
a2x a2y a2z
a3x a3y a3z
a4x a4y a4z
AT A =
=
Express matrix solution in more general termsExpress matrix solution in more general terms3. Least squares
5. Math 38
Example 1 (7 of 9)
AT b = akxbk
akxbk
akzbk
Express matrix solution in more general terms (cont)Express matrix solution in more general terms (cont)
3. Least squares
5. Math 39
Example 1 (8 of 9)
J = [a1xx + a1yy + a1zz - b1]2 + [a2xx + a2yy + a2zz - b2]2 + [a3xx + a3yy + a3zz - b3]2 + [a4xx + a4yy + a4zz - b4]2
J/ x = 2[a1xa1xx + a1ya1xy + a1za1xz - a1xb1] + [a2xa2xx + a2ya2xy + a2za2xz - a2xb2] + [a3xa3xx + a3ya3xy + a3za3xz - a3xb3] + [a4xa4xx + a4ya4xy + a4za4xz - a4xb4]
2[ akx akx x aky akx y akz akxz - akxbk ]
= 0Minimize by calculusMinimize by calculus
3. Least squares
5. Math 40
Example 1 (9 of 9)
akx akx x aky akx y akz akxz - akxbk = 0 akx aky x aky aky y akz akyz - akybk = 0 akx akz x aky akz y akz akzz - akzbz = 0
akx akx aky akx akz akx
akx aky aky aky akz aky
akx akz aky akz akz akz
xyz
- = 0 akxbk
akybk
akzbk
Minimize by calculus (continued)Minimize by calculus (continued)
3. Least squares
5. Math 41
Example 2 (1 of 3) 1.1000 1.9000 2.9000 4.0000 5.0000 6.0000
x = 2.2000 3.0000 4.1000 5.0000 6.1000 6.9000
y =
Fit a curve to the following dataFit a curve to the following data
3. Least squares
5. Math 42
Example 2 (2 of 3)
Fit z = a + b xi + c xi2
A = [[1;1;1;1;1;1], x, x.*x] =
abc
= (ATA)-1 AT b
b = y
= 1.0126 1.0949 -0.0184
1.0000 1.1000 1.2100 1.0000 1.9000 3.6100 1.0000 2.9000 8.4100 1.0000 4.0000 16.0000 1.0000 5.0000 25.0000 1.0000 6.0000 36.0000
Fit curve z to dataFit curve z to data
3. Least squares
5. Math 43
Example 2 (3 of 3)error = a + b x + c x2 - y =
1 2 3 4 5 62
2.5
3
3.5
4
4.5
5
5.5
6
6.5
7
-0.0052 0.0266 -0.0668 0.0980 -0.0726 0.0200
Error in curve fitError in curve fit
3. Least squares
5. Math 44
4. Propagation of varianceCombining varianceMultiple dimensionsExample -- propagation of positionExample -- angular rotation
4. Propagation of variables
5. Math 45
Combining variances
Variances from multiple error sources can be combined by adding variances
Example
xorig = standard deviation in original position = 1 mvorig = standard deviation in original velocity = 0.5 m/sT = time between samples = 2 secxcurrent = error in current position
= square root of [(xorig)2 + (vorig * T)2] = sqrt(2)
4. Propagation of variables
5. Math 46
Multiple dimensions When multiple dimensions are included, covariance matrices can be added
When an error source goes through a linear transformation, resulting covariance is expressed as follows
P1 = covariance of error source 1P2 = covariance of error source 2P = resulting covariance = P1 + P2
T = linear transformationTT = transform of linear transformationPorig = covariance of original error sourceP = T * P * TT
4. Propagation of variables
5. Math 47
Example -- propagation of position
xorig = standard deviation in original position = 2 mvorig = standard deviation in original velocity = 0.5 m/sT = time between samples = 4 secxcurrent = error in current position
xcurrent = xorig + T * vorig
vcurrent = vorig
1 4 0 1
T = Porig =22
00
0.52
Pcurrent = T * P orig * TT = 1 4 0 1
1 0 4 1
40
00.25
= 164
40.25
4. Propagation of variables
5. Math 48
Example -- angular rotation
5. Statistics
Xoriginal = original coordinates
Xcurrent = current coordinates
T = transformation corresponding to angular rotation
cos -sin sin cos
T = where = atan(0.75)
Porig =1.64 -0.48-0.48 1.36
Pcurrent = T * P orig * TT = 0.8 -0.60.6 0.8
= 20
01
1.64 -0.48-0.48 1.36
0.8 0.6-0.6 0.8
x’
y’
x
y
5. Math 49
5. Geometry
Unit vectorsAngle between two linesPerpendicular to a planePointing
5. Geometry
5. Math 50
Unit vectors
A unit vector is a vector of length 1.Unit vectors are frequently used to denote
vectors that have the same direction, such as those parallel to a chosen axis of a coordinate system
5. Geometry
5. Math 51
Angle between two lines (1 of 10)
The dot product is the result of multiplying the length of a vector A times the length of the component of vector B that is parallel to A
A • B = |A| |B| cos , where is the angle between the vectors
Dot productDot product
5. Geometry
5. Math 52
Angle between two lines (2 of 10)
To find the angle between two lines,• Establish a vector A and a vector B
along each line• Solve for = arccos[A • B /( |A| |B| )]• 0
Solving for using dot productSolving for using dot product5. Geometry
5. Math 53
Angle between two lines (3 of 10)A = [1 2], B = [2 1]|A| = SQRT(12 + 22) = SQRT(5)|B| = SQRT(22 + 12) = SQRT(5)A • B = [1 2] • [2 1]T = [2 • 1 + 2 • 1] = 44 = SQRT(5) • SQRT(5) cos cos = 4/5
A
Bx
y
Example using dot productExample using dot product5. Geometry
5. Math 54
Angle between two lines (4 of 10)
Using Excel to compute valuesUsing Excel to compute values5. Geometry
A B A' B' A'*A B'*B A'*B1 2 1 2 2 1 5 5 42 1 2.24 2.24
1. Use dot product to compute square of hypotenuse
angle(radians) angle(degrees) = angle (radians)*180/pi0.64 36.9
1. Use ACOS to compute angle in radians2. Use 180/pi to convert angle to degrees 3. Use PI function to compute pi Note: PI must be followed by "(" if typed
5. Math 55
Angle between two lines (5 of 10)
The cross product is the result of multiplying the length of a vector A times the length of the component of vector B that is perpendicular to A
A x B = |A| |B|sin , where is the angle between the vectors
The vector A x B is perpendicular to the plane containing A and B
Cross productCross product5. Geometry
5. Math 56
Angle between two lines (6 of 10)
To find the angle between two lines,• Establish a vector A and a vector B
along each line• Solve for = arcsin[A x B /( |A| |B| )]• - /2 /2
Solving for using cross productSolving for using cross product
5. Geometry
5. Math 57
Angle between two lines (7 of 10)
A = i j kAx Ay AzBx By Bz
=
i j k1 2 02 1 0
= -3k
Example using cross productExample using cross product5. Geometry
5. Math 58
Angle between two lines (8 of 10)
A = [1 2], B = [2 1]|A| = SQRT(12 + 22) = SQRT(5)|B| = SQRT(22 + 12) = SQRT(5)A x B = -3 k-3 = SQRT(5) • SQRT(5) sin sin = -3/5
Example using cross product (continued)Example using cross product (continued)
5. Geometry
5. Math 59
Angle between two lines (9 of 10)
= atan2(sin , cos )
Combining dot product and cross productCombining dot product and cross product
5. Geometry
5. Math 60
Angle between two lines (10 of 10)
Using Excel to compute arctangentsUsing Excel to compute arctangents
5. Geometry
x y ATAN ATAN2-0.6 -0.8 53.13 -127
1. Use ATAN2 for four quadrant arctangentNote: First argument is X and not Y as in FORTRAN
5. Math 61
Perpendicular to a plane The cross product defines the direction
perpendicular to the plane defined by the two vectors A and B
5. Geometry
5. Math 62
Pointing (1 of 14)
A (3,1,1)
B (2,3,2)
camera x0
y0
Change pointing of camera so that points A and B are on the same level
Point camera as directedPoint camera as directed5. Geometry
5. Math 63
Pointing (2 of 14)
A (3,1,1)
B (2,3,2)
x0
y0
camera
x1
y1
z0 and z1 are positive out of page
Pan camera to point at A in the x0-y0 planePan camera to point at A in the x0-y0 plane5. Geometry
5. Math 64
Pointing (3 of 14)
= atan2(3,1) = 18.4o
cos sin 0-sin cos 0 0 0 1
311
=3.160.001.00
cos sin 0-sin cos 0 0 0 1
232
=2.852.222.00
T01
Determine T01 as followsDetermine T01 as follows
5. Geometry
5. Math 65
Pointing (4 of 14)
A (3.16,0,1)
B (2.85,2.22,2)
x1
y1
camera
z1 is positive out of page
Redraw problem in x1-y1Redraw problem in x1-y1
5. Geometry
5. Math 66
Pointing (5 of 14)
A (3.16,0,1)
B (2.85,2.22,2)
x1
z1
camera
y1 is positive into page
View x1-z1 planeView x1-z1 plane
5. Geometry
5. Math 67
Pointing (6 of 14)
A (3.16,0,1)
B (2.85,2.22,2)
x1
z1
camera
x2
z2
y1 and y2 are positive into page
Elevate camera to point at A in x1-z1 planeElevate camera to point at A in x1-z1 plane5. Geometry
5. Math 68
Pointing (7 of 14)
= atan2(1,3.16) = 17.5o
cos 0 sin 0 1 0 -sin 0 cos
=3.160.000.00
=2.852.222.00
3.160.001.00
cos 0 sin 0 1 0 -sin 0 cos
3.322.211.05
T12
Determine T12 as followsDetermine T12 as follows5. Geometry
5. Math 69
Pointing (8 of 14)
A (3.16,0,0)
B (3.32,2.21,1.05)
z2x2 is positive into page
y2
View y2-z2 planeView y2-z2 plane
5. Geometry
5. Math 70
Pointing (9 of 14)
A (3.16,0,0)
B (3.32,2.21,1.05)
z2
x2 and x3 are positive into page
y2
z3y3
Roll camera so that A and B are on horizontal lineRoll camera so that A and B are on horizontal line5. Geometry
5. Math 71
Pointing (10 of 14)
= atan2(1.05.2.21) = 25.4o
= 1 0 0 0 cos sin 0 -sin cos
3.322.450.00
3.322.211.05
T23
Determine T23 as followsDetermine T23 as follows
5. Geometry
5. Math 72
Pointing (11 of 14)
A (3.16,0,0)
B (3.32,2.45,0)
z3x3 is positive into page
y3
View y3-z3 planeView y3-z3 plane5. Geometry
5. Math 73
Pointing (12 of 14)
T01T T12T T23T 001
-0.12-0.49 0.86
=
Express unit vector perpendicular to AB in x0-y0-z0 planeExpress unit vector perpendicular to AB in x0-y0-z0 plane5. Geometry
5. Math 74
Pointing (13 of 14)
A = i j kAx Ay AzBx By Bz
=
i j k3 1 12 3 2
= (- i - 4j +7k)/sqrt(66)
-0.12-0.49 0.86
=
Compare perpendicular unit vector to cross productCompare perpendicular unit vector to cross product5. Geometry