acp693 102 Report post Posted February 23, 2010 Hi, I'm stuck with a problem for a few days, it's for structural engineering project. Say there is a point on an xy plane at zero. if the point oscillates in the x direction with a magnitude of 6 and a phase of 25 degrees, and also oscillates in the y direction with a magnitude of 8 and phase of 90 degrees, how do I find the maximum extension from zero that the point will make? and at what phase will that occur? I first thought it would be simple addition of complex numbers, but that doesn't seem to work. Any ideas? Best regards Albert 0 Share this post Link to post Share on other sites
haegarr 7373 Report post Posted February 23, 2010 Quote:Original post by acp693... if the point oscillates in the x direction with a magnitude of 6 and a phase of 25 degrees, and also oscillates in the y direction with a magnitude of 8 and phase of 90 degrees...Does this mean that the point is given as a vector of 2 sine functions with the same frequency but different phases (and amplitudes, of course)? 0 Share this post Link to post Share on other sites
acp693 102 Report post Posted February 23, 2010 Hi, Yes that's right. 0 Share this post Link to post Share on other sites
Emergent 982 Report post Posted February 23, 2010 Quote:how do I find the maximum extension from zero that the point will make? and at what phase will that occur? I first thought it would be simple addition of complex numbers, but that doesn't seem to work.It should work. Maybe you're getting the sign on the phase wrong? Think about it with phasors, or equivalently remember the trig identity,M cos(omega t - phi) = A cos(omega t) + B sin(omega t)whereA = M cos(phi)B = M sin(phi) .Notice that the relationship between (M, phi) and (A,B) is precisely the relationship between polar and cartesian coordinates. 0 Share this post Link to post Share on other sites
acp693 102 Report post Posted February 24, 2010 Thanks, That's what I first thought, however consider the simpler case of x direction vector= Magnitude 6, phase 0 degrees, (6+0i) y direction vector= magnitude 6, phase 90 degrees,(0+6i)the point should then describe a circle of radius 6? However addition of the complex numbers 6+0i + 0+6i = 6+6i.magnitude = |6+6i| = 8.4853 ? That would be the correct answer if x and y had the same phase.I don't know where I'm going wrong with this.Best regardsAlbert 0 Share this post Link to post Share on other sites
haegarr 7373 Report post Posted February 24, 2010 Quote:Original post by acp693Thanks, That's what I first thought, however consider the simpler case of x direction vector= Magnitude 6, phase 0 degrees, (6+0i) y direction vector= magnitude 6, phase 90 degrees,(0+6i)the point should then describe a circle of radius 6? However addition of the complex numbers 6+0i + 0+6i = 6+6i.magnitude = |6+6i| = 8.4853 ? AFAIS, you're dropping the time dependency here with the result that the vector you construct for the addition has x and y components at different times.The correct vector would bev(t) := [ 6 * sin( wt ), 6 * sin( wt + pi/2 ) ]^{t} = [ 6 * sin( wt ), 6 * cos( wt ) ]^{t}with an expected magnitude of|v(t)| = sqrt( 36 * sin^{2}( wt ) + 36 * cos^{2}( wt ) ) = 6Now, the complex number you have used are results e.g. atv( t=0 ) = [ 0, 6 ]^{t}andv( t=pi/(2w) ) = [ 6, 0 ]^{t}where obviously 2 different times are used. Hence is is not correct to add them up.EDIT: Uhh, I recently saw that I've exchanged the phases in comparison to your original example. But the reasoning is still the same, though. 0 Share this post Link to post Share on other sites
acp693 102 Report post Posted February 24, 2010 Thank you Haegarr for your help. I'm still stuck with this, although your example seems to work for the simple example, I can't get it to work with my original example? I have drawn on graph paper the expected path of the point of the original example using 5 degrees as increments. The maximum magnitude would seem to be 8.64How do I determine what t would be at maximum magnitude?Best regardsAlbert 0 Share this post Link to post Share on other sites
haegarr 7373 Report post Posted February 24, 2010 OMG, you need a bag full of ugly math! The following is a correct mathematical way AFAIK, but I'm not a mathematician, so be warned. Hence I invite people to check that solution, of course.The vector is defined asv(t) := [ A_{x} sin( ωt + φ_{x} ), A_{y} sin( ωt + φ_{y} ) ]^{T}and its length is|v(t)| = sqrt( A_{x}^{2} sin^{2}( ωt + φ_{x} ) + A_{y}^{2} sin^{2}( ωt + φ_{y} ) )There is a correspondence in trigonometry that sayssin^{2}( α ) = ( 1 - cos( 2α ) ) / 2So we get|v(t)| = sqrt( A_{x}^{2} + A_{y}^{2} - A_{x}^{2} cos( 2ωt + 2φ_{x} ) - A_{y}^{2} cos( 2ωt + 2φ_{y} ) ) / sqrt( 2 )=: sqrt( p(t) ) / sqrt( 2 )To get the t where |v(t)| is maximum, the usual way is to compute the derivative, set it to 0, resolve for t, and check whether the function has the correct sign reversal. Now, if |v(t)| is maximal, then also p(t) must be maximal (AFAIK), so we actually computed p(t) / dt = 0what makes things easier.So we getd p(t) / dt = A_{x}^{2} 2ω sin( 2ωt + 2φ_{x} ) + A_{y}^{2} 2ω sin( 2ωt + 2φ_{y} )Setting this to 0 and resolving for t ... is not in my power! But there is the superposition principle of harmonics to our rescue! It says thatA_{1} sin( ωt + φ_{1} ) + A_{2} sin( ωt + φ_{2} ) == A sin( ωt + φ )whereA = ... (not of particular interest here)tan( φ ) = ( A_{1} sin( φ_{1} ) + A_{2} sin( φ_{2} ) ) / ( A_{1} cos( φ_{1} ) + A_{2} cos( φ_{2} ) )With the substitutionsA_{1} := A_{x}^{2} 2ωA_{2} := A_{y}^{2} 2ωφ_{1} := 2φ_{x}φ_{2} := 2φ_{y}ω := 2ωwe can compute the φ.Now, the formula (using the original ω)A sin( 2ωt + φ ) == 0means that2ωt + φ == n π (( err, that writing is "en by pi"))where n is a natural number. Hence it is possible to compute a t, verify it, set it into the original formula of |v(t)| and yield in the maximum magnitude.BTW: I will damn everybody who provides an easier solution after this post ;) There was so much time from the beginning of this thread up to now... 0 Share this post Link to post Share on other sites
acp693 102 Report post Posted February 24, 2010 Wow, thank you very much Haegarr, It will take me a while to look through this,Actually, my problem is really in 3 dimensions and not 2 as in my example, I just thought it would be easier to try and find out how to do it in 2 dimensions first.Thanks againAlbert 0 Share this post Link to post Share on other sites
haegarr 7373 Report post Posted February 24, 2010 Quote:Original post by acp693Actually, my problem is really in 3 dimensions and not 2 as in my example, I just thought it would be easier to try and find out how to do it in 2 dimensions first.If the way shown above works well, then its extension to 3 dimensions is not a problem: The derivation works the same for 3 dimensions, and the superposition principle can be extended to any dimensions (according to my Bronstein, just for the case you know it ;) 0 Share this post Link to post Share on other sites
Emergent 982 Report post Posted February 24, 2010 Quote:Original post by acp693Thanks, That's what I first thought, however consider the simpler case of x direction vector= Magnitude 6, phase 0 degrees, (6+0i) y direction vector= magnitude 6, phase 90 degrees,(0+6i)Oh, your problem is not the problem I thought you had...Anyway, here's how I'd think about it. From my last post you see how you can represent a phase-shifted sinusoid either in "polar" (magitude/phase), or "cartesian" (cosine term, sine term) form. Well that means that you can write your point (x,y)(t) as,[ x(t) ] [ A1 B1 ] [ cos(t) ][ ] = [ ] [ ][ y(t) ] [ A2 B2 ] [ sin(t) ]or more simply, justp(t) = M u(t)where u(t) is a unit vector for all time, and p=(x,y).What unit vector u maximizes the norm of the expression "M u" ? Simply the one corresponding to the largest singular value of M! Equivalently: The eigenvector corresponding to the largest eigenvalue of M^{T} M. 0 Share this post Link to post Share on other sites
alvaro 21272 Report post Posted February 24, 2010 I was about to post the exact same thing Emergent posted, only less clearly exposed. :) 0 Share this post Link to post Share on other sites