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Solution to time offset vector equation?

  1. Feb 15, 2010 #1
    Hello math goers,

    My education in linear algebra is limited to an Intro course I took a year ago. So I am posting this to see if such a solution exists in the first place, at least so I can start learning about it.

    The problem is: solve A for equation
    [tex] u(t) = A \cdot v(t+t_o) [/tex]

    [tex] \left[ \begin{array}{c} u(t_1)\\ u(t_2)\\ \vdots\\ u(t_n) \end{array} \right]=\left[ \begin{array}{cccc} a_1 & a_2 & \cdots & a_n \end{array} \right] \cdot \left[ \begin{array}{c} v(t_1+t_o) \\ v(t_2+t_o) \\ \vdots \\ v(t_n+t_o) \end{array} \right] [/tex]

    Where [tex] t_o [/tex] is some known offset with respect to [tex] t [/tex] .

    Essentially what these represent are two data signals [tex] v(t_v) [/tex] and [tex] u(t_u) [/tex], these two signals have some small time offset that I can calculate using a reference peak that both signals contain. However, getting A is not as simple as solving for A because the data is not continuous.

    The "easy" way around this is to manually align every single value [tex] v(t_v) [/tex] with [tex] u(t_u)[/tex], but this is computationally expensive.

    Any thoughts?
     
    Last edited: Feb 15, 2010
  2. jcsd
  3. Feb 16, 2010 #2
    Your notation looks a bit strange. The right hand side would be a scalar since it is the inner product of two vectors. On the other hand, the left side is a vector. Shouldn't A be a matrix?

    Torquil
     
  4. Feb 16, 2010 #3
    Yes, sorry about that, it should look like:

    [tex] \left[ \begin{array}{c} u(t_1)\\ u(t_2)\\ \vdots\\ u(t_n) \end{array} \right]=\left[ \begin{array}{c} a_1 \\ a_2 \\ \cdots \\ a_n \end{array} \right] \cdot \left[ \begin{array}{c} v(t_1+t_o) \\ v(t_2+t_o) \\ \vdots \\ v(t_n+t_o) \end{array} \right] [/tex]

    Where, [tex] a_n [/tex] is some scaling constant. For simplicity, I am just going to go ahead and let [tex] a_n=1; [/tex] for all n. So that the equation now is:

    [tex] u(t) = v(t+t_o) [/tex]

    Which I just realized isn't really an equation at all... my problem is a sampling problem. If I have some time vector
    [tex] t=\left( \begin{array}{c}1\\ 2\\ 3\\ 4\\ 5\\ 6 \end{array}\right) [/tex]

    and signal vectors, representing step function:
    [tex] u=\left( \begin{array}{c} 0\\ 0\\ 0\\ 1\\ 1\\ 1\end{array}\right); \; v=\left( \begin{array}{c} 0\\ 2\\ 2\\ 2\\ 0\\ 0 \end{array}\right) [/tex]

    Then [tex] t_o= -2[/tex], and [tex] a_n= \frac{u(t_n)}{v(t_n-2)} [/tex]
    which happens to be equal to 2 or 0 in this case.

    Because this is a discrete signal, whenever [tex]t_o[/tex] is larger than the length of the signal, this will not work. I think I need to insert "fake" points in the data vectors for this case.

    Perhaps this thread belongs in the "Computing & Technology" forum?
     
  5. Apr 12, 2010 #4
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