Solution to time offset vector equation?

[Xuser]

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
u(t) = A \cdot v(t+t_o)

\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]

Where t_o is some known offset with respect to t .

Essentially what these represent are two data signals v(t_v) and u(t_u), 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 v(t_v) with u(t_u), but this is computationally expensive.

Any thoughts?

 

[torquil]

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

 

[Xuser]

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

Yes, sorry about that, it should look like:

\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]

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

u(t) = v(t+t_o)

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

and signal vectors, representing step function:
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)

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

Because this is a discrete signal, whenever t_o 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?

 

[Xuser]

[solved] Re: Solution to time offset vector equation?

This is a mathematical problem, the solution to it is called cross-correlation. I do not understand it completely, but if you are interested check out: http://en.wikipedia.org/wiki/Cross-correlation .

Also here is a link to the website I used to figure out how to implement this in Matlab:
http://www.mathworks.com/support/solutions/en/data/1-3M6NCT/index.html?product=SG&solution=1-3M6NCT