The "delta function" is a very useful abstraction whenever you have a distribution containing "atoms." If you have a mass distribution which contains some point-like masses, you can model this by using a bump function like
d(t) = (u_a - u_{a+epsilon})/epsilon
which is concentrated in the neighborhood of a but still has mass 1. But it is possible to introduce a new symbol, delta_a(t), which idealizes this better; it is a kind of limit as epsilon goes to 0. Warning: for any t, lim_{epsilon tends to 0} d(t) = 0; this is NOT delta. The "delta function" is not a function in the usual sense, but you should think of it as a function like d for very small epsilon. We will just accept it as a notation. I am a little more honest in the optional handout. We write delta for delta_0.
Here are some key (and essentially equivalent) properties of delta_a(t):
1. delta_a = u'_a.
2. For any continuous function g(t),
integral_0^infinity g(t) delta_a(t) dt = g(a).
3. L(delta_a(t);s) = e^{-as}. In particular, L(delta(t);s) = 1, the constant function.
4. f*delta_a = f_a , defined as before by f_a(t) = f(t-a) if t is at least a, f_a(t) = 0 if t is less than a. In particular, delta is the unit for convolution. Also, delta_a * delta_b = delta_{a+b}.
Sample applications:
1. We have a bank account in which the interest is k . We save at a steady rate b dollars per year, but after one year we deposit a lump sum of c. This is modelled by:
x' - kx = b + c delta_a.
We can solve this for example using LT. Assume x(0) = 0; then
(s-k)X(s) = b/s + ce^{-as};
X(s) = b/s(s-k) + ce^{-as}/(s-k);
use partial fractions for the first term, and the t-shift and s-shift rules for the second.
Suppose we modify this by starting at x(0) = p, not necessarily zero. This would lead to
(s-k)X(s) = p + b/s + ce^{-as}.
The new term p could just as well have come from an extra term on the right hand side of the equation:
x' - kx = p delta_0 + b + c delta_a, rest initial conditions.
This makes sense: as soon as the clock starts, the value of x is kicked up by a one-time deposit of p at time 0. You get the same effect as if you had simply had an initial condition. Using this trick we can always assume rest initial conditions.
2. We have a damped pendulum, starting at neutral position but with velocity v, and we give it a kick at time a. The kick delivers a large force over a very short period of time. What matters to the system is the "impulse," which is the integral of force over time. If this integral has value k, then the ODE governing the pendulum is
x'' + px' + qx = (k/m)delta_a , x(0) = 0, x'(0) = v.
From this application, delta_a takes the name "unit impulse."
We can solve the equation, using L(x''(t);s) = s^2 L(x(t);s) - sx(0) - v:
(s^2 + ps + q)X(s) = v + (k/m)e^{-as}.
Again, this s-domain equation would have equally well come from
x'' + px' + qx = v delta_0 + (k/m)delta_a , x(0) = 0, x'(0) = 0.
The v delta_0 boosts the initial speed up to v.
(Second order initial value problems with x(0) nonzero can be brought under this method too, but you have to use yet another "generalized function," the derivative of the delta function, delta'_a(t).)
Let's alter the problem again: drop the kick at time a and take the desired initial speed to be 1:
x'' + px' + qx = delta_0, rest initial conditions.
In the s-domain,
X(s) = 1/p(s) = W(s), the transfer function,
so in the t-domain,
x(t) = w(t), the weight function.
For this reason the weight function is often called the "unit impulse response."
Let's interpret the formula x = f*w (which is called Duhamel's principle) using this observation. The convolution can be approximated as a sum of terms
w(t-tau) f(tau) Delta tau.
This approximation is equivalent to replacing f by a sum of multiples of delta functions spaced out along [0,t]. As far as integrals are concerned this is a good approximation. Each delta function gives rise to its own impulse response, and these are added together to give the full response to f. It's a beautiful picture.