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Unformatted text preview: Rotations Rotations (in 2 dimensions)
Consider rotating the vector u to get the vector v Hull = llvll = a This can be expressed as: v= [:2] a(cos 0 cos ¢ — sin 0 sin qb)
a(sin0 cos 925 + cos 0 sin qb) a(cos 0 cos <15 — sin 0 sin (1))
a(si110 cos ([5 + cos 0 sin d9) cos0 —sin0 acosqﬁ
sin9 cos0 a sin 45
_ cos 0 — sin 0 U1
_ sin 0 cos 0 uz
W
P Roy Smith, ECE 210a, 10: 1 Rotations Rotations (in 2 dimensions) P = [cose —s1n6] sin 6 cos 0 Note that PT is a rotation in the opposite direction so PT 2 P’1 Therefore PT P = I and so P is unitary Roy Smith, ECE 210a, 10:2 Rotations Rotations (in 3 dimensions) Rotation about a coordinate axis (2) Projection onto the x,y plane: V
1 0 0 PM = 0 1 0 y
0 0 0 i) nyv gives the 2D version of P from before, cos6 —sin6
P _ [sin6 cos6] Now the 2 component in unchanged and we get, cos 6 — sin 6 0
P2 (6) = sin 6 cos 6 0
0 0 1 Roy Smith, ECE 210a, 10:3 Rotations Rotations (in 3 dimensions) The complete set of coordinate rotations in 3D are: 1 0 0 Pg, (6) = 0 cos 6 — Sin 6 Rotation about the x axis
0 sin 6 cos 6 cos 6 0 sin 6 P7, (6) = 0 1 0 Rotation about the y axis
— sin 6 0 cos 6 cos 6 — sin 6 0
PZ (6) = sin 6 COS 0 0 Rotation about the z axis
0 0 1 Roy Smith, ECE 210a, 10:4 Rotations Rotations (in 3 dimensions) We can describe any rotation in 3D by a rotation about each of the axes: P = Py(0)PZ(¢) Rotations are not commutative. In general, they can be described by Euler angles: (¢, 9,1/1) To specify a rotation we must specify the axes, the order,
and whether the axis also moves. Roy Smith, ECE 210a, 10:5 Rotations
Rotations Z
/' Euler angle rotations:
\ P = PzW) PM) Pz(¢>)
y
The second and third rotations
X are about the new x and z axes.
There are 12 possible conventions for these angles: 6 of the zxz type 6 of the xyz type Properties of rotation matrices (in 3dimensions) eig(P) = {Laijﬂ}, trace(P) 2 14—20050, det(P) = 1. Reference: wikipedia: Euler angles
Roy Smith, ECE 210a, 10: 6 Rotations Rotations
We can also consider the case where the rotation
axes remain ﬁxed.
There is an equivalent ﬁxed axis set of
rotations.
If (and only if) the z and z’ axes start aligned,
these rotations are:
P : P2(¢’)Px(0,)Pz(¢l)
with,
W — (b,
0’ = 0,
¢' = 1&
Roy Smith, ECE 210a, 10: 7
Rotations
Rotations Euler’s theorem: A113D rotations can be described
by a single rotation about a given vector, u. Three parameters are required to uniquely
specify a rotation (u can be normalized). Roy Smith, ECE 210a, 10:8 Rotations Rotations: Aerospace conventions roll This is useful in aerospace applications for describing the
effects of bodyaxis rotations. "Euler angles:” successive rotations to move the vehicle
carried local vertical to the body (or wind) frame. P = P’s/(0) PZ(¢) qb Azimuth angle
0 Elevation angle 1/} Bank angle The body and wind frames are related by the rotations: Py (a) (angleof—attack)
and P2 (—ﬂ) (sideslip). Reference: Etkin, “Dynamics antmosphBris Flight,” Wiley, 1972 Roy Smith, ECE 210a, 10:9 Rotations Plane Rotations I 0
c s <— z'th row
Deﬁne P as: Pij = I
—s c <— jth row
0 I
with c2 + 82 = l (for example: cos 0, sin 0)
Then for a given x,
331
371 0:12, + sac]
:L' 2 PM :1) =
2:71 —sa:,, + cxj
$n Roy Smith, ECE 210a, 10:10 Rotations Plane Rotations 11:1
0127, + sac]
Pij $ =
—s:ri + 0an
$7L
. 302‘ 957'
If xi aé 0 and acj aé 0, then the ch01ce, c = — 3 = —7 gives,
as? + as? + $1 + P. t I = _ Note that we can use this to selectively put zeros
m 1 into a vector.
0
11.77, Roy Smith, ECE 210a, 10:11 Rotations Plane Rotations Applying this once gives, P12 :3 : Applying it to the next component gives, P13P12 x llivll
0 And after nl rotations, PM . . . P13P12 a: = Roy Smith, ECE 210a, 10:12 Plane Rotations Notethat,
T
IIijII=Hm1 x§+x§ 0 95,]
andso, Pij =1 and,infact,P,jisunitary 13512:] = I, P77 = Pfl 1.] U We can use plane rotations to orthogonalize a matrix and get a QR decomposition. llwll
0
P1nP13P12T Rotations Roy Smith, ECE 210a, 10:13 Rotations Givens Rotations Recall the ﬁrst step of the Householder reﬂection method: For the Givens case use H1 2 PM . . . P13P12. The next step in the Householder procedure used, 1 0
HP [0 ml For the Givens case use7 H2 2 [PgnP2n_1 . . . P23]
_v_/
n — 2 plane rotations Roy Smith, ECE 210a, 10:14 Rotations QR decompositions Consider the computation effort (in terms of the number of multiplications) Given, AERW" A = QR Gaussian elimination (scaled, partial pivoting): n3/ 3 GramSchmidt (classical & modiﬁed): n3
Householder: 2713 / 3
Givens: 4713 / 3 Givens is very useful if A is sparse. Roy Smith, ECE 210a, 10:15 Rotations QR decompositions Q1 R1 = Q2 R2 Is the QR decomposition unique? A Deﬁne U = ngl ( =Q'2TQ2R2RI1)
= Rng1 Note that R2 R1— 1 is also uppertriangular and has positive elements on the diagonal. Consider the ﬁrst column of U U11 0
U>k1 : . But = 1 2} U11 2 2:1. As R2 Rfl also has positive diagonal elements, um = 1 Roy Smith, ECE 210a, 10:16 QR decomposition QR decompositions Now consider the second column, U12
U22 U*2 = 0 But UEUQ = 0 (orthogonal) => U12 = 0
0 Again U22 = ::1 and the positivity of the diagonal of R2R1—1 implies that 1422 = 1. Repeat this argument for the other columns, U=I andso U=Q2TQ1=L This implies that Q; = Qfl = Q? SO Q1 : Q2 and R1 = = : R2_ Roy Smith, ECE 210a, 10:17 Rotations Computational stability Forward stability: Do small errors in the problem imply a Close answer? Very small perturbation matrix: F << F Consider a series of unitary operations, P = Pk . . . P1 “(Pk    P1)(A + E)p = IIPA + PEllp llerrorllp = llPEllp = llEllp << llAllp = llAllp Backward stability: Does the answer correspond to a “Close” problem? For the QR decomposition A = Q R Q+E = Q, R+F = 1%. This is a QR decomposition for some Q R = fl Is A a: A ? Roy Smith, ECE 210a, 10:18 Rotations Computational stability Backward stability: Is A w A ? A = QR = (Q+E)(R+F)
QR + QF + ER + EF EF is negligible Now bound each of the error terms:
“QFHF = “FIIF “AHF = IIQRIIF = “RHF “ERHF S llEllpllRllp = llEllpllAllp This gives,
llAfillp S “FHF + HEIIFHAIIF + “EHFHFIIF SoAzA Roy Smith, ECE 210a, 10:19 ...
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 Spring '08
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