EEL6667: Homework #1 Solutions

Transcription

1 EEL6667: Homework #1 Solutions Problem 1: Note: homework1.nb is a Mathematica notebook that solves many of the problems in this homework. (a) See homework1.nb. (b) See homework1.nb. Problem :[raig, Exercise.14 Generalizing example.9 in raig, ' ' ' ' where frames { ' } and { ' } are defined on page 53 and illustrated on page 54 of raig (Figure.0). Each individual transform in equation (1) is given by, (1) ' I 3 P () ' ' ( θ) 0 (3) ' I 3 P (4) ombining equations (1) through (4): ( θ) [ I 3 ( θ) P (5) Problem 3:[raig, Exercise.15 We solve this problem by assuming, e κθ to be correct and showing that no contradiction results. Now, we differentiate two different ways: (1) element by element, and () by applying, (6) de κθ κe κθ (7) From homework1.nb, element-wise differentiation yields: ( k x 1) sin( θ) k z cos( θ) + k y cos( θ) + k z cos( θ) ( k y 1) sin( θ) k x cos( θ) k y cos( θ) + k x cos( θ) ( k z 1) sin( θ) (8) - 1 -

2 pplying equation (7) above, we get that, κ which simplifies to, (9) ( k y + k z ) sin( θ) k z cos( θ) + k y cos( θ) + k z cos( θ) ( k x + k z ) sin( θ) k x cos( θ) k y cos( θ) + k x cos( θ) ( k x + k y ) sin( θ) (10) (See homework1.nb.) Note that all the off-diagonal terms in equations (8) and (10) are equivalent, and that the diagonal terms can be made to be equivalent by using the identity, k x + k y + k z 1 (11) so that, for example, ( k x 1) ( k y + k z ) (1) which shows that the r 11 element is equivalent in both matrices. he same can be done for r and r 33. (See homework1.nb.) hus, our initial assumption must be correct. Problem 4:[raig, Exercise.3 he transformation is given by, POG (13) hus, we need to determine and P from the given information. y definition, POG P1 We also know that, OG. (14) Ŷ Ẑ (15) From the problem statement: Ẑ ( P P ) P P1 ( P3 P1) ( P3 P ) 1 (16) (17) Ŷ Ẑ ogether, equations (13) through (18) completely define. (18) Problem 5: (a) From Figure 1, - -

3 Z ( π) (19) POG 300 (0) POG (1) () (See homework1.nb.) (b) From Figure 1, Y ( π ) X ( π + π 3) (3) [using from eq. (19) (4) POG 30 (5) POG (6) (See homework1.nb for numeric solution.) (c) Note that Figure (Figure.6 in raig) is inconsistent. he top-surface triangle cannot at the same time have sides of length 3 and 4 respectively, while at the same time have an angle of 30. I chose to ignore the indicated length of the longer side, letting it be 3 3 (instead of 4) to conform with the 30 angle specification. hen, Y ( π) X ( π ) Z ( π 6) (7) POG 300 (8) POG (9) (See homework1.nb for numeric solution.) (d) With the same caveat as in part (c) above, Z ( π) X ( π ) (30) [using from eq. (7) (31) POG (3) - 3 -

4 P OG (33) Problem 6: (See homework1.nb for intermediate and numeric solutions.) (a) From the definition of the distance metric: dqp min[ Eqp, Eq (, p) dqp (b) For a unit quaternion q, min[ ( s q s p ) + ( x q x p ) + ( y q y p ) + ( z q z p ), ( s q + s p ) + ( x q + x p ) + ( y q + y p ) + ( z q + z p ) min[ ( s p s q ) + ( x p x q ) + ( y p y q ) + ( z p z q ), ( s p + s q ) + ( x p + x q ) + ( y p + y q ) + ( z p + z q ) dpq q [ s, ( xyz,, ) the equivalent rotation matrix is given by, (34) (35) (36) 1 ( y + z ) ( xy sz) ( xz + sy) ( xy + sz) 1 ( x + z ) ( yz sx) ( xz sy) ( yz + sx) 1 ( x + y ) (37) For a unit quaternion q [ s, ( x, y, z), note that you get exactly the same rotation matrix since the negatives in q cancel out in every product permutation of { s, x, y, z} that appears in. Hence, q and q represent the same rotation. Now, if q p, Eqp 0 while if q p (38) Eq (, p) 0. (39) herefore, dqp 0 if q p or q p; in other words, if q and p represent the same rotation. From (35) it is self-evident that in all other cases, dqp 0. (c) See homework1.nb. he results are: dqq' dq' (, q'' ) dqq'' It is easy to verify that: dqq' + dq' (, q'' ) dqq'' (40) (41) (4) (43) - 4 -

5 dq' (, q) + dqq'' dq' (, q'' ) dqq'' + dq'' (, q' ) dqq' (d) See homework1.nb. Note that, θ qq' θ q'q'' π θ qq''.0555 (44) (45) (46) (47) where θ qp denotes the angle of rotation from unit quaternion q to p. Observe that the results in (46) and (47) are consistent with the results in (40) through (43). When the angle of rotation between two unit quaternions is equal, so is the distance measure between them. Larger angles of rotation correspond to large distances. he distance measure appears to be independent of kˆ. (e) See homework1.nb. Our approach to this problem is as follows. First, we show that the distance metric dpu between an arbitrary unit quaternion, p [ cos( θ ), sin( θ ) ( k x, k y, k z ) (48) and the zero-rotation unit quaternion, u [ 1000,,, is dependent only on θ (which is the angle of rotation between p and u in this case). So, dpu min[ ( c 1) + ( k x s) + ( k y s) + ( k z s), ( c + 1) + ( k x s) + ( k y s) +( k z s) (49) (50) where c cos( θ ) and s sin( θ ). Since k x + k y + k z 1, dpu min[ ( c 1) + ( k x + k y + k z )s, ( c + 1) + ( k x + k y + k z )s dpu min[ ( c 1) + s, ( c + 1) + s dpu min[ cos( θ ), + cos( θ ) n alternative expression for dpu is given by, (51) (5) (53) dpu min[ cos( θ 4), sin( θ 4) Now, assume that both p and u are multiplied by some arbitrary unit quaternion q [ s, ( xyz,, ). Since p and u are rotated by the same rotation q, the relative distance between them should not change, and results(53) and (54) will still hold for dqpq. his is verified in homework1.nb so that for two arbitrary vectors p and q : dpq min[ cos( θ 4), sin( θ 4) which is plotted in homework1.nb. **(f) Since we know that dpq depends only on θ between p and q, we can rewrite the triangle inequality as: (54) (55) d( θ 1 ) + d( θ ) d( θ 1 + θ ) (56) Put another way, f( θ 1, θ ) d( θ 1 ) + d( θ ) d( θ 1 + θ ) 0, θ 1, θ. (57) where d( θ) denotes the distance between two unit quaternions separated by an angle θ. In homework1.nb, f( θ 1, θ ) is plotted and shown to be greater than zero for θ 1, θ [ 0π,. (58) - 5 -