thomas49th said:
Now I picture this as a rectangle with no width and infinite height. In fact I think of the width (along the x axis) as (1/inf = 0) and the height being inf. So the area (integral) is 1/inf * inf = 1
This is not a good way to think about it, since ∞ is not a number, and 1/∞ ≠ 0. It is undefined.
However, you are correct that the Dirac delta can thought of as the limiting case of a series of rectangular pulses each of which is getting progressively narrower and taller, but in such a way that the products of their widths and heights (their areas) are always constant and equal to unity.
thomas49th said:
However if the width is 0, then why does the integral have limits between -inf and +inf?
You're right that any integral of δ(t-T) over the interval [a,b] will be equal to 1 provided that the point T is included within the bounds of integration (i.e. it lies somewhere between a and b). But the point is that if you extend the limits of integration extend all the way to infinity in either direction, it's guaranteed to be true that the integral is equal to 1 (for any T).
thomas49th said:
Am I right in the thinking when a signal g(t) is multiplied by the Dirac function it just turns a signal "on" for an infintately small amount of time?
Sort of. A good way to think about it is that doing this "samples" the value of the function at t = T, (since the Dirac delta is zero everywhere else except at this point of interest).
Imagine if you took one of our finite rectangular pulses of height h and width w such that homework = 1. "Sweep" this pulse over your signal of interest (with it centred at t = T) and take the integral over their product, and you have something like:
h\int_{T-w/2}^{T+w/2} g(t)\, dt
Now, if we make w very small, we can sort of "pretend" (as an approximation to the integral) that g(t) is constant and equal to g(T), the value of the function in the middle of the integration range. In this case, the integral reduces to:
h g(T)\int_{T-w/2}^{T+w/2} dt = hwg(T) = g(T)
since homework = 1. The smaller w is, the better this approximation is. It is in this way that integrating over a narrow pulse of unit area can "sample" the value of the signal at the point of interest, excluding all other points.
thomas49th said:
Can someone please explain how
\int_{-\infty}^{\infty} \delta(t-T)g(t)dt = g(T)
The proof is simple enough. Since δ(t-T) is 0 everywhere except at t = T, we can replace g(t)δ(t-T) with g(T)δ(t-T) (again, we have "sampled" that particular value). Since g(T) is constant, it comes outside the integral, and we have:
g(T)\int_{-\infty}^{\infty} \delta(t-T)\,dt = g(T) (1)
I hope that this result is easier to understand conceptually in light of my example with the finite pulse above. The properties of the Dirac delta function are such that using it is equivalent to taking the pulse example to the limiting case in which the approximation becomes exact.
thomas49th said:
I am having problems relating this to the real world
Ha, yeah, well in the real world we don't deal with signals that are continuous with time. Well, we do, but we can't store them for later manipulation in such a way that we can capture the value of that signal at any (and every) time value. We can really only record the value of the signal at a discrete set of (usually evenly-spaced) time steps. So what we end up is a set of numbers that are samples of the signal at regular intervals in time. In order to select out one of these samples it is sufficient to multiply the discretized signal by a pulse of height 1 that is narrow enough that it spans only one time step. Doing this is the equivalent for discrete-time signals of what convolving with the Dirac delta function does for continuous-time signals.