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

 

[blue_raver22]

I would suggest exploring the use of Fourier analysis to solve this time offset vector equation. This approach involves transforming the data signals into the frequency domain, where the time offset can be easily identified and corrected for. This can be more efficient and accurate compared to manually aligning each value. Additionally, you can look into techniques such as cross-correlation and time shifting to further refine your solution. I would also recommend consulting with experts in signal processing for further guidance on this problem.