Often we study a physical system by putting in a short pulse and then seeing what the system does. The resulting behavior is often called impulse response, and understanding it tells us how the system responds to any input. Let us see what we mean by a pulse. The simplest kind of a pulse is a rectangular pulse defined by
\begin{equation}
\varphi(t) =
\begin{cases}
0 & \text{if } \; \phantom{a \leq {}} t < a , \\
M & \text{if } \; a \leq t < b , \\
0 & \text{if } \; b \leq t .
\end{cases}
\end{equation}
We want \(b\) to be very small; we wish to have the pulse be very short and very tall. By letting \(b\) go to zero, we arrive at the concept of the Dirac delta function. The limit of \(\frac{1-e^{-bs}}{bs}\) as \(b \to 0\) is \(1\text{,}\) so we are looking for a “function” whose Laplace transform is \(1\text{.}\)
is not exactly a function; it is sometimes called a generalized function. We avoid unnecessary details and simply say that it is an object that does not really make sense unless we integrate it. The motivation is that we would like a “function” \(\delta(t)\) such that for any continuous function \(f(t)\text{,}\)
The formula should hold if we integrate over any interval that contains 0, not just \((-\infty,\infty)\text{.}\) So \(\delta(t)\) is a “function” with all its “mass” at the single point \(t=0\text{.}\) For any interval 2
It is important that we consider \(c\) and \(d\) as part of the interval. One could write this integral as \(\int_{c-}^{d+}\text{.}\)
Unfortunately, there is no such function in the classical sense. You could informally think that \(\delta(t)\) is zero for \(t \neq 0\) and somehow infinite at \(t=0\text{.}\)
A good way to think about \(\delta(t)\) is as a limit of pulses of decreasing length, each having an integral of \(1\text{.}\) For example, consider a rectangular pulse \(\varphi(t)\) as above with \(a=0\) and \(M=\nicefrac{1}{b}\text{,}\) that is, \(\varphi(t) = \frac{u(t) - u(t-b)}{b}\text{.}\) Compute
If \(f(t)\) is continuous at \(t=0\text{,}\) then for very small \(b\text{,}\) the function \(f(t)\) is approximately equal to \(f(0)\) on the interval \([0,b]\text{.}\) We approximate the integral
Let us therefore accept \(\delta(t)\) as an object that is possible to integrate. We often want to shift \(\delta\) to another point, for example \(\delta(t-a)\text{.}\) In that case,
Note that \(\delta(a-t)\) is the same object as \(\delta(t-a)\text{.}\) With that, the convolution of \(\delta(t)\) with \(f(t)\) is again \(f(t)\text{,}\)
The Laplace transform of \(\delta(t-a)\) would be the Laplace transform of the derivative of the Heaviside function \(u(t-a)\text{,}\) if the Heaviside function had a derivative. First,
To obtain what the Laplace transform of the derivative would be, we multiply by \(s\) to obtain \(e^{-as}\text{,}\) which is the Laplace transform of \(\delta(t-a)\text{.}\) We see the same thing using integration,
This line of reasoning allows us to talk about derivatives of functions with jump discontinuities. We can think of the derivative of the Heaviside function \(u(t-a)\) as being somehow infinite at \(a\text{,}\) which is precisely our intuitive understanding of the delta function.
Let us compute \({\mathcal{L}}^{-1} \left\{ \frac{s+1}{s} \right\}\text{.}\) So far, we only computed the inverse transform of proper rational functions in the \(s\) variable. That is, the numerator was of lower degree than the denominator. Not so with \(\frac{s+1}{s}\text{.}\) We can use the delta function to compute,
As we said before, in the differential equation \(L x = f(t)\text{,}\) we think of \(f(t)\) as the input, and \(x(t)\) as the output. We think of the delta function as an impulse, and so to find the response to an impulse, we use the delta function in place of \(f(t)\text{.}\) The solution to
Perhaps an astute reader will notice that it does not seem like \(x'(0)=0\text{.}\) However, we really want to think that \(x(t)=0\) for \(t \leq 0\text{,}\) so \(x(t)\) has a “corner” at \(t=0\text{,}\) that is, \(x'\) has a jump discontinuity there, which is what produces the \(\delta(t)\) when we take \(x''\text{.}\) See Remark 6.4.1. The initial condition really is \(x'(0-) = \lim_{t \uparrow 0} x'(0) = 0\text{.}\)
Let us notice something about the example above. In Example 6.3.4, we found that when the input is \(f(t)\text{,}\) the solution to \(Lx = f(t)\) is given by
That is, the solution for an arbitrary input is given as convolution with the impulse response. Let us see why. For functions \(f(t)\) and \(h(t)\) that are zero for negative \(t\text{,}\)
You may be worried about the fact that our \(h\) is not really twice differentiable, in fact, \(h'(0)\) does not exist as a normal derivative, and \(h''\) involves the \(\delta\) function. You can pretend that \(h\) is twice differentiable for this computation and either apply the standard Leibniz rule (that \(h(t)=0\) for \(t < 0\) is important) or note that for such functions, the integral is actually over \((-\infty,\infty)\) rather than over \([0,t]\text{.}\)
. Suppose that \(h(t)\) is the impulse response (solution to \(h''+\omega_0^2 h = \delta(t)\)). If we convolve the entire equation (6.3), the left-hand side becomes
The procedure works for any constant-coefficient linear equation \(Lx = f(t)\text{.}\) If you find the impulse response \(h\) (solution to \(Lh = \delta(t)\)), you also know how to obtain the output \(x(t)\) for any input \(f(t)\text{.}\) We simply convolve, \(x(t) = (f * h)(t)\text{.}\) As you may have noticed in the example, the impulse response is in fact just the inverse Laplace transform of the transfer function, that is, \(h(t) = {\mathcal{L}}^{-1} \bigl\{ H(s) \bigr\}\text{.}\)
A quite different example application where the delta function appears is in representing point loads on a steel beam. Consider a beam of length \(L\text{,}\) resting on two simple supports at the ends. Let \(x\) denote the position on the beam, and let \(y(x)\) denote the deflection of the beam in the vertical direction. The deflection \(y(x)\) satisfies the Euler–Bernoulli equation 4
\begin{equation}
EI \frac{d^4 y}{dx^4} = F(x) ,
\end{equation}
where \(E\) and \(I\) are constants 5
\(E\) is the elastic modulus and \(I\) is the second moment of area. Let us not worry about the details and simply think of these as some given constants.
and \(F(x)\) is the force applied per unit length at position \(x\text{.}\) The situation we are interested in is when the force is applied at a single point as in Figure 6.4.
\begin{equation}
EI \frac{d^4 y}{dx^4} = -F \delta(x-a) ,
\end{equation}
where \(x=a\) is the point where the mass is applied. The constant \(F\) is the force applied and the minus sign indicates that the force is downward, that is, in the negative \(y\) direction. The end points of the beam satisfy the conditions,
Suppose that the length of the beam is 2, and \(EI=1\) for simplicity. Further suppose that a force \(F=1\) is applied at \(x=1\text{.}\) That is, we have the equation
We could integrate, but using the Laplace transform is even easier. We apply the transform in the \(x\) variable rather than the \(t\) variable. We again denote the transform of \(y(x)\) by \(Y(s)\text{.}\)
We still need to apply two of the endpoint conditions. As the conditions are at \(x=2\) we can simply replace \(u(x-1) = 1\) when taking the derivatives. Therefore,
Hence, \(C_2 = \frac{1}{2}\text{.}\) Solving for \(C_1\) using the first equation, we obtain \(C_1 = \frac{-1}{4}\text{.}\) Our solution for the beam deflection is
Suppose that \(f(t)\) and \(g(t)\) are differentiable functions (and the derivatives are continuous) and suppose that \(f(t) = g(t) = 0\) for all \(t \leq 0\text{.}\) Show that
Suppose that \(L x = \delta(t)\text{,}\)\(x(0) = 0\text{,}\)\(x'(0) = 0\text{,}\) has the solution \(x(t) = t e^{-t}\) for \(t > 0\text{.}\) Find the solution to \(Lx = t^2\text{,}\)\(x(0) = 0\text{,}\)\(x'(0) = 0\) for \(t > 0\text{.}\)
Suppose we have a beam of length \(1\) simply supported at the ends and suppose that a force \(F=1\) is applied at \(x=\nicefrac{3}{4}\) in the downward direction. Suppose that \(EI=1\) for simplicity. Find the beam deflection \(y(x)\text{.}\)
Suppose that \(L x = \delta(t)\text{,}\)\(x(0) = 0\text{,}\)\(x'(0) = 0\text{,}\) has the solution \(x(t) = e^t \sin(t)\) for \(t > 0\text{.}\) Find (in closed form) the solution to \(Lx = e^t\text{,}\)\(x(0) = 0\text{,}\)\(x'(0) = 0\) for \(t > 0\text{.}\)
