# Why the delta at the end of the approximation?

The equation to approximate an input signal with a unit impulse in Continuous Time, is shown below, before we take the limit $\hat{x}(t)=\frac{lim}{\Delta\rightarrow0}\sum^{\infty}_{-\infty}x(k\Delta)\delta_\Delta(t-k\Delta)\Delta$ <-- why is there a final $\Delta$ multiplying the $\delta_\Delta(t-k\Delta)$?

• try defining formally $\delta_\Delta(t)$. what is the height of that function, given a specific $\Delta$? – robert bristow-johnson May 21 '14 at 19:58
• Could you clarify what do you mean? Is it the discretation process you're after? – Royi May 21 '14 at 21:10
• This is to do with the convolution integral and dissertation sounds like it has something to do with representing a CT signal as as series of discrete signals. i think the $\delta_\Delta(t-k\Delta)\Delta$ part does this? – Sach May 22 '14 at 7:19
• the height of the function at every $\Delta$ is given by $x(k\Delta)$ for the relevant value of k. – Sach May 22 '14 at 7:20
• yes @Princ3Sach, the height of the function, $\hat{x}(t)$ when $k\Delta \le t < (k+1)\Delta$ is $x(k\Delta)$, but the height of $\delta_\Delta(t-k\Delta)$ is $\frac{1}{\Delta}$ because the width is $\Delta$ and the area has to be 1. so now do you see why you need that factor of $\Delta$ in there? – robert bristow-johnson May 23 '14 at 3:00

The actual equation is , $$x(t)=\int_{-\infty}^{\infty}{x(t_0)\delta(t-t_0)dt}$$
In your equation ${lim}_{\Delta\rightarrow0}\sum^{\infty}_{-\infty}$ stands for integration so $\Delta$ is required at the end, which stands for dt in integral.
$\delta_\Delta(t-k\Delta)$ is a spike with value 1. $\Delta$ is the width. It is like you are approximating the area under $x(t)$ with a bunch of rectangles of height $x(k\Delta)$ and width $\Delta$. So area is width by height, hence the multiplication. $x(k\Delta)\delta_\Delta(t-k\Delta)$ is a just a way of writing the discrete set of values at $x(k\Delta)$.