By Pei J., Zhang X.
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Additional resources for 1-Generator Quasi-Cyclic Codes
Q| q · q∗ For unit quaternions with |q| = 1, we have q · q∗ = 1 so that q−1 = q∗ ; hence the inverse of a unit quaternion is its conjugate. 43) to represent a rotation around the z-axis. The inverse of qφ can then be written as ∗ ˆ sin φ)∗ = (cos φ, −ˆv sin φ). q−1 φ = qφ = (cos φ, v Next, we show how to represent vectors as quaternions. A ‘pure’ vector is a quaternion for which the scalar part is absent, and can be written in various forms as r = 0 + xI + yJ + zK = [x y z 0] = (0, v) = [v, 0].
That all is not well with Euler angles can be seen when the angles ψ, θ, φ are again computed from a given rotation matrix ⎞ ⎛ A1,1 A1,2 A1,3 A = ⎝A2,1 A2,2 A2,3 ⎠. A3,1 A3,2 A3,3 One obtains the angles ψ, θ, φ from the Ai,j through the following relations: cos θ = A3,3 , cos ψ = − cos θ = A3,2 , sin θ A2,3 , sin θ sin θ = ± 1 − cos2 θ , A3,1 , sin θ A1,3 sin φ = − . sin θ sin ψ = − These equations show that a position close to θ = π/2 will lead to a divergence in the computation of the Euler angles.
86) where rb is fixed, so its time derivative vanishes. e. 86), gives d r= dt d q dt rb q∗ + qrb d ∗ q dt = (˙q) q∗ rqq∗ + qq∗ rq q˙ ∗ = q˙ q∗ r − r˙q∗ q. 87) We use that q˙ ∗ and q commute. e. the product is anti-commutative: (˙qq∗ )vec r = −r(˙qq∗ )vec . 87) into a vector equation d r = (˙qq∗ )vec × r − r × (˙q∗ q)vec dt = (˙qq∗ )vec × r + (˙q∗ q)vec × r = 2(˙qq∗ )vec × r. 85), we see that 2(˙qq∗ )vec = ω. 90) so the right-hand side is the body-fixed angular velocity ωb . 90) from the left by q gives qq∗ 2˙q = qωb , q˙ = 1 b qω .