Understanding Convolution Take:


  $$
(u*v)(k) = \sum_{i=-\infty}^\infty u(i)v(k-i).
$$
  
  The $k$ is there, it's because you want to define
  $$
\ldots\ldots, (u*v)(-3), (u*v)(-2), (u*v)(-1), (u*v)(0), (u*v)(1), (u*v)(2), (u*v)(3), \ldots\ldots
&gt;$$
  etc.  The number in the parentheses is $k$.  Thus, for example, when $k=4$, we have
  \begin{align}
(u*v)(4) = \sum_{i=-\infty}^\infty u(i)v(4-i)
\end{align}
  $$
= \cdots\cdots+u(-3)v(7)+u(-2)v(6)+u(-1)v(5)+u(0)v(4)+u(1)v(3)+u(2)v(2)
&gt;$$
  $$
\phantom{={}} {}+u(3)v(1)+ u(4)v(0) + u(5) v(-1) + u(6)v(-2)+u(7)v(-3)+u(8)v(-4)+\cdots&gt;\cdots.
&gt;$$


Could you give examples of what the 'k' value could stand for?
Can it only be time offset?

Could you give an intuitive explaination why you need to multiply u(-3) [a point left from the y-axes] with v(7) [a point right from the y-axes]! And not u(-3) with v(-3) for example? With a diagram for example.
I think the reason you are getting confused is that you are only considering the convolution of two signals, whereas convolution is not usually between signals but between a signal and an impulse response.

The best way to visualise it is probably with a filter.  For example, in 


we can see that the input comes in, and is multiplied by $b_0$ to give its contribution to the output $y$.  Then in the next time step, that input value is multiplied by $b_1$ to give its contribution to the output, and so forth.  In this case your $k$ variable specifies which time in the output sequence you want.  The output signal is given by $y[n] = (x * b)[n]$.

For example, consider the case of $n=4$ (that's the same as $k=4$ in your example).  That means we're looking at the 4th time step (since time = 0).  So the value at the input is $x[4]$, the fourth input value; the value to the right of the first $z^{-1}$ block is the previous input value $x[3]$; the value to the right of the next $z^{-1}$ block is $x[2]$; and so forth.  The output $y[4]$ will be given by 
$$y[4] = x[4]b_0 + x[3]b_1 + x[2]b_2 + x[1]b_3 + \cdots + x[-3]b_7 + \cdots$$
 which you should be able to see is the discrete convolution sum, and includes the product of $x[-3]$ with $b[7]$.

Any linear filter or transfer function can be implemented as a convolution, where the output is given by the convolution of the input with the system impulse response.  Try reading up on convolution in a text book, and look particularly at the sliding-strip method of calculating it, which might help with your visualisation.