Impulse response

So far, we have represented lienar time-invariant (LTI) systems in two different ways. First, as ordinary differential equations (ODEs), and second, as transfer functions (TFs). These representations are related via the Laplace transform and summarized in the following diagram, which we derived in the section on Laplace transforms.

In this section, we will derive a third representation for LTI systems: the impulse response. This representation lives entirely in the time domain, and it allows us to compute the output of an LTI system for any input signal without ever leaving the time domain. The impulse response is based on the pulse decomposition from the section on Linear systems, where we argued that any input signal can be approximated a sum of scaled and shifted pulses. We will now take this idea to its logical extreme by considering inputs that are infinitely narrow pulses, known as impulses, and obtain an exact representation for the output of an LTI system in terms of its impulse response.

Impulse function¶

Let’s define $\delta_h(t)$ to be a pulse signal of width $h$ and height $1/h$:

We use a height of $\frac{1}{h}$ so the area under the pulse always equals 1, regardless of $h$:

Now let’s find the Laplace transform of $\delta_h(t)$:

Now, let’s consider what happens as we let $h$ approach zero. This is a limit of the form $\frac{0}{0}$, so we can use L’Hôpital’s rule to evaluate it.

This motivates us to define the impulse function $\delta(t)$ as a limiting version of $\delta_h(t)$ as $h$ approaches zero. It does not make sense to think of $\delta(t)$ as an ordinary function, since we would have to write a definition that looks like

which doesn’t make much sense. Instead, we will define $\delta(t)$ based on how it behaves when we integrate it. Although we call it a “function,” $\delta(t)$ is not a traditional function. Mathematically, it is known as a distribution.

Impulse response¶

We saw earlier that if we use the unit step (Heaviside) function $H(t)$ as an input to an LTI system, the output we obtain is called the step response of the system.

Similarly, if we apply an impulse input $\delta(t)$ to an LTI system, the output we obtain is called the impulse response of the system.

Secondly, the impulse function and the unit step function are related by differentiation and integration, which means the impulse response and step response of an LTI system are also related in the same way.

Example¶

Let’s verify for the LTI system defined by the ODE below that the impulse response is the derivative of the step response.

Solution: Remember that for input $u(t)$ and output $y(t)$, we have $Y(s) = G(s) U(s)$, where $G(s)$ is the transfer function of the system. To find the output, we compute the inverse Laplace transform: $y(t) = \Lap^{-1}\{G(s) U(s)\}$. Let’s start by finding the transfer function. Taking Laplace transforms of both sides of the ODE (with zero initial conditions),

The impuse response uses $u(t) = \delta(t)$, so $U(s) = 1$. Therefore,

The step response uses $u(t) = H(t)$, so $U(s) = \frac{1}{s}$. Therefore,

We used Table 1 for the inverse Laplace transforms. Finally, let’s verify that the impulse response is the derivative of the step response:

Everything checks out!

The convolution integral¶

Let $\delta_h(t)$ be the pulse function defined earlier, and let $g_h(t)$ be response of our system $G$ corresponding to this input. We can represent this relationship as the following diagram.

Suppose our input signal is $u(t)$. We will approximate it as a sum of shifted and scaled pulses by assuming $u(t)$ is piecewise-constant over intervals of length $h$. This allows us to write:

The reason for the extra factor of $h$ in the sum is that each pulse has height $\frac{1}{h}$, so multiplying by $h$ gives the correct amplitude for the piecewise-constant approximation of $u(t)$. By superposition, we can approximate the output $y(t)$ as:

In the limit $h\to 0$, the pulse function $\delta_h(t)$ approaches the impulse function $\delta(t)$, and the response $g_h(t)$ approaches the impulse response $g(t)$. Therefore, taking the limit as $h \to 0$ in equations (17) and (18) gives:

This follows from the definition of the integral as a limit of Riemann sums. The reason the integral only goes up to $t$ rather than $\infty$ is that $g(t) = 0$ whenever $t<0$ due to causality.

This sort of integral is known as a convolution integral, and we denote it using the convolution operator $*$. Thus, we can write the output of an LTI system in response to an arbitrary input $u(t)$ as:

The convolution operation satisfies a remarkable property: it is commutative. In other words, we can swap the order of the functions being convolved without changing the result:

Since convolution produces the output of a system given its impulse response, but the Laplace transform of the output is the product of the Laplace transforms of the input and the transfer function, we can relate convolution in the time domain to multiplication in the Laplace domain. For any two functions $f(t)$ and $g(t)$, we have:

We can now update Figure 1 to include the convolution integral as yet another way to compute the output of an LTI system.

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