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    Hey guys. Got a beauty of a Matrix question in a recent assignment. http://personal.maths.surrey.ac.uk/s...h/em2/ass2.pdf this is a link to the assignment, it's question number 1 that is giving me grief.

    I first apply KVL around each of the three loops to gain:

    i1(1/jw + 4jw) + i2(-jw) + i3(-3jw) = 0
    i1(-jw) + i2(1/3jw + 3jw) + i3(-2jw) = 0
    i1(-3jw) + i2(-2jw) + i3(1/2jw + 5jw) = 0 (edited the i2 component)

    That seems pretty easy to me. Checked it a few times, and I'm happy with it. With some simple manipulation of the terms a 1/xjw and xjw together, you can put it in matrix form like this, TAKING OUT 1/jw COMMON TO ALL:

    (1 - 4w^2, w^2, 3w^2).......... (i1)
    (w^2, 1-9w^2, 2w^2)..........(i2)
    (3w^2, 2w^2, 1-10w^2)....... (i3)

    Now part b) asks you to solve for the non-trivial case that there are currents flowing, and the way we've been told to do this is to solve for determinant of the matrix equal to zero.

    This is a lot of algebra so I won't try to write it all here, but I ended up with this answer:

    -241w^6 + 152w^4 - 23w^2 + 1 = 0 (EDIT)

    Which is horrible looking lol. I understand this can give me three possible values of w^2 as it's a cubic expression but anyone able to check my working or offer any advice? It looks too complicated to me, but even if I can show some method of solving it I'll get some effort marks hopefully!

    This is a beast....
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    oh damn, my bad sorry I meant 241. I was hoping the errors wouldn't be before as Kirchoffs Voltage Law is pretty simple to apply, or at least it should be

    sorry, here's the link http://personal.maths.surrey.ac.uk/s...h/em2/ass2.pdf
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    (Original post by ghostwalker)
    Edit: The final line of your matrix looks wrong; you seem to have multipled the terms by different things. In fact the same looks true of the second line.
    There is one term in the third equation that is incorrect, but the correct version has been used in the matrix. In loop 3, i2 causes a voltage of -i2(jwL) where L =2 then we have i2(-2jw) as used in the matrix.

    Remember that everything in the matrix has been divided by 1/jw, so

    1/jw times 2w^2 = -2jw as required etc.
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    That's fine. i didn't want answers anyway.

    I think I'm getting to the stage where the message boards arent that useful for help anymore :\ never mind.

    are you a moderator? if so can you delete this thread. cheers.
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    (Original post by Maurice_Syzlak)
    That's fine. i didn't want answers anyway.

    I think I'm getting to the stage where the message boards arent that useful for help anymore :\ never mind.

    are you a moderator? if so can you delete this thread. cheers.
    I did this assignment last night - I'm not convinced your KVLs are correct. Your evaluation of the matrix you have is correct, but we have different matrices. Evaluating mine lead to a disguised quadratic, which is significantly easier to solve.
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    hey dude, its fraser.

    this was my first bash at it. Still get tempted to run to student room when I get stuck, but its becoming less fruitful each time!

    That matrix is definitely incorrect, I went back to it from scratch and came up with a much more agreeable determinant, that had cancelling w^6 terms. It is so "pretty" that it makes me sure it's right.
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    (Original post by EEngWillow)
    I did this assignment last night - I'm not convinced your KVLs are correct. Your evaluation of the matrix you have is correct, but we have different matrices. Evaluating mine lead to a disguised quadratic, which is significantly easier to solve.
    p.s. remember the good old days on here before year 1, brings back a tear to the old [japs] eye.
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    (Original post by Maurice_Syzlak)
    hey dude, its fraser.

    this was my first bash at it. Still get tempted to run to student room when I get stuck, but its becoming less fruitful each time!

    That matrix is definitely incorrect, I went back to it from scratch and came up with a much more agreeable determinant, that had cancelling w^6 terms. It is so "pretty" that it makes me sure it's right.
    Ah, hey. Craig here This is blatently going to become an extension of last night's FB conversation at this rate! I got a similarly "pretty" solution so guess we can both be happy and hand this in on Monday with 100%?

    Haha, yes I remember. I still regular here though
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    Ha! Yeah I knew it was you, Willow!

    I will have to find some effort to transcribe my demented scrawlings into nice work before I hand it in, so maybe next Monday!
 
 
 
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