Disgusting Matrix question!!

Discussion in 'Math' started by Fraser_Integration, Mar 24, 2011.

  1. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
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    Hey guys. Got a beauty of a Matrix question in a recent assignment. http://personal.maths.surrey.ac.uk/st/J.Deane/Teach/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

    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^2m, 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:

    -261w^6 + 152w^4 - 23w^2 + 1 = 0

    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?

    This is a beast....
     
    Last edited: Mar 24, 2011
  2. steveb

    Senior Member

    Jul 3, 2008
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    There are a couple of typos in what you wrote, but more importantly your final matrix is not correct.
     
  3. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
    142
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    hey steve, thanks for taking the time to look through it, appreciate it.

    could you tell me which elements need looking at again? hopefully it's not all of them! please remember that each element in the matrix is divided by 1/jw, I just couldnt find a better way to show that in the post.
     
  4. steveb

    Senior Member

    Jul 3, 2008
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    The approach you use seems reasonable, but look at the second row, second column. Also the third row, second and third columns.
     
  5. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
    142
    5
    Hey.

    Second row, second column: in the KVL equation, for loop two, i2 has voltage drops coming from the 3C capacitor, a 2L inductor and a 1L inductor. This combines to give a voltage of i2(1/3jw + 3jw) which equals i2( [1 -9w^2]/jw)

    and from there you can see how I got it in the form that ends up in the matrix.

    Same goes for third row, third column.

    Third row, second column. This comes from the contribution of i2 in loop 3. Which from the circuit equations is i2(-2jw) as i2 passes backwards through a 2L inductor. To take 1/jw out of this expression, we need to put in the matrix 2w^2, as 2w^2 / jw = 2w/j = 2jw

    Therefore I cannot see the error in those steps you pointed out.

    Can you advise further?

    Thanks,
    Fraser
     
  6. steveb

    Senior Member

    Jul 3, 2008
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    Sure.

    The original matrix looks OK. However, as I mentioned in the first post, the errors are in your final matrix.

    You are free to divide by jw, but you have to divide the entire row by the same exact value. The elements I mentioned didn't look correct to my in my cursory look at it. (EDIT: looking again, I think one of the ones I mentioned is actually ok, so I think the 2,2 and 3,3 elements are wrong)

    I think your method is correct.

    By the way, solving cubic equations algebraically is doable, as I think you know, but it is a little tedious, as I think you know also. You may want to solve these using the root solve on your calculator. Note that although there are 3 solutions for omega squared, there are 6 solutions for a sixth order equation. But, as you can tell, the negative frequencies are not relevant. So, I think you should end up with 3 positive frequencies that are the answers.
     
  7. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
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    Yes! I see what you mean now, thank you so much. can be so hard to spot those silly mistakes in your own working. If I get something pretty, ill post the outcome. My lecturer tends to give questions that don't need calculators
     
  8. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
    142
    5
    Hey mate!

    Just completed it. The w^6 terms cancel, and get left with the term 1 -23w^2 + 121w^4 = 0 which factorises to w^2 = 1/11 and 1/12

    Very nice, easy numbers that makes me sure it is correct!

    This was the final matrix I used then, again all divided by jw:

    (1-4w^2), (w^2), (3w^2)
    (w^2), (1/3 - 3w^2), (2w^2)
    (3w^2), (2w^2), (1/2 - 5w^2)

    after some simple but tiresome algebra you end up with the simple quadratic above.

    Thank you very much for your help steveb, it's appreciated here in "Old" England. :)
     
  9. steveb

    Senior Member

    Jul 3, 2008
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    You are very welcome, but please double check your work. I haven't worked this out fully, but I'm surprised that the sixth power terms cancel out.
     
  10. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
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    Okay, I will.

    But this assignment was from a straigh Mathematics module, not Electronics, so is not meant to be too realistic. Plus, as I said, the lecturer is all for questions that can be worked out in the head alone.
     
    steveb likes this.
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I actually obtain ω=0.3503734 rad/sec as one of the solutions which isn't a nice number, but I have some confidence in the answer.

    Another solution seems to be ω=0.259465 rad/sec.

    A simulation check of the circuit supports these ω values. Also tested using matrix math in Scilab - see my comments below on negative signs in the matrix.

    These values both satisfy the condition that detA=0, suggesting a resonance condition exists for the given ω solutions.

    I would have thought the matrix would have some negative terms given the assumed current directions in the original circuit, but I'm not sure if that's important. I would have all the terms excepting those bracketed in your penultimate post as negative.
     
  12. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
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    hmmm. maybe I'll have another looksie at it. thanks for taking the time to plough through it.
     
  13. steveb

    Senior Member

    Jul 3, 2008
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    I did it out in full. Tnk is right, and your quadratic , or quartic actually, is correct. The w^6 terms do cancel, and you get two positive real solutions. I think you just made a mistake in getting the numerical answer.

    Note that you won't get a nice round number or rational because of the square roots that show up in the quadratic solutions and also because you first solve for w^2.
     
  14. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Fraser_Integration - Your matrix is indeed correct - my comments about the negative signs arose because I missed the point that you had probably done a consistent sign change throughout the entire matrix. Apologies for not looking carefully enough.

    Hopefully we eventually agree on the final solution! Good luck.
     
  15. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
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    Hi guys.

    Thanks for the input.

    I realised I made yet another silly error, working too fast yesterday I assumed 1 -23w^2 + 121w^4 = 0 factorised to (1-11w^2)(1-12w^2) = 0

    It almost does, but not quite!

    Therefore I must solve for w^2 using the quadractic equation, and then take roots to find w. I will do all of this when I have my calculator, I dont fancy typing it all in to the Windows Calculator!
     
  16. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    If I type this matrix into my HP50 calculator (no need to divide through by 1/jω) and then execute "DET", I get:

    \frac{j(121 \omega^4 - 23 \omega^2 + 1)}{6 \omega^3}

    I then edit away everything but:

    121 \omega^4 - 23 \omega^2 + 1=0

    and execute "SOLVE" and get the two positive roots:

    \omega = \frac{3 \sqrt{5} + 1}{22}

    \omega = \frac{3 \sqrt{5} - 1}{22}
     
  17. t_n_k

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    That's real nice Electrician. I only have a HP35s.

    This is what I get using Scilab ....
     
  18. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
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    I'm glad your fancy-pants calculators agree with my pen and paper effort.
     
  19. The Electrician

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    Oct 9, 2007
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    I should certainly hope so!

    You didn't say that those who would help shouldn't use a "fancy-pants calculator". Did you think we were going to do any more work than we had to? ;)
     
  20. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I'm quite fond of my "fancy pants calculator". It even has a special drawer in my desk. It also likes to go on vacation with me. It's the Reverse Polish Notation thing that makes it so attractive.:rolleyes:

    Anyway - All power to the old fashioned neck top calculator! Mind you Fraser_Integration (may I call you F_I for short?) - would you have picked up your error if we hadn't pulled our HP's (or matrix programs) from their holster and done some fancy pants keystrokes?

    Actually my calculator isn't as fancy as Electrician's. Mine can solve equations but it doesn't have the same bells & whistles .... so I used Scilab. BTW F_I, Scilab is free and quite useful.
     
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