Convolution and Leibniz's Law

**ragnar** · June 2nd 2011, 10:24 PM

I'm reviewing this document:

http://ocw.mit.edu/courses/mathemati...18_03S10_i.pdf

The short story is that $w(t)$ is the weight function (solution to the associated homogeneous equation) of some generic equation $my'' + cy' + ky = f(t)$ . We want to break the interval between $0 < t < x$ into smaller intervals, $[t_{i}, t_{i+1}], i = 0, 1, 2, 3, ..., n$ and define $f_{i}(t) \begin{cases}f(t_{i}) \text{ if } t_{i} < t < t_{i+1}\\0 \text{ elsewhere}\end{cases}$ so that over the interval it's a constant function. This means that $f(t) \approx f_{0}(t) + f_{1}(t) + ... f_{n-1}(t)$ .

I get everything until page 3 equation (8). If $w(t)$ is the weight function for $f(t)$ , then I would think that for $f_{i}(t)$ , the weight function would just be $w(t-t_{i})$ (restricted to the interval). I don't understand why we need to amplify this by $f(t_{i})\Delta t$ .

(I forgot to include a part about Leibniz's Formula but I'm too tired.)

**Aryth** · June 4th 2011, 07:57 AM

The response of the system to an impulse of area A delayed by time $t_i$ is $Aw(t - t_{i})$ .

The Area of an impulse at time $t_i$ is $F(t_{i})\Delta t$