2.4: The Impulse Response and Convolution

A fundamental property of LTI systems is that they obey the convolution operator. This operator is defined by

\[y(t) = \int\limits_{-\infty}^{\infty} u(t_1) h(t-t_1)\, dt_1 = \int\limits_{-\infty}^{\infty} u(t-t_1) h(t_1)\, dt_1.\]

The function \(h(t)\) above is a particular characterization of the LTI system known as the impulse response (see below). The equality between the two integrals should be clear since the limits of integration are infinite. The presence of the \(t_1\) and the \(−t_1\) term inside the integrations tells you that we have integrals of products - but that one of the signals is turned around. We will describe the meaning of the convolution more fully below. To understand the impulse response, first we need the concept of the impulse itself, also known as the delta function \(\delta(t)\). Think of a rectangular box centered at time zero, of width (time duration) \(\epsilon\), and height (magnitude) \(1/ \epsilon\); the limit as \( \epsilon \longrightarrow 0\) is the \(\delta\) function. The area clearly equals 1 in any case.

Figure \(\PageIndex{1}\): graph of rectangle with area = 1 centered at \(t=0\), showing how the limit changes as the rectangle width \(\epsilon\) approaches 0.

The inner product of the delta function with any function is the value of the function at zero time:

\[ \int\limits_{-\infty}^{\infty} f(t) \delta(t)\,dt \, = \int\limits_{-\epsilon/2}^{\epsilon/2} f(t) \delta(t)\,dt \, = \, f(t=0) \int\limits_{-\epsilon/2}^{\epsilon/2} \delta(t)\, dt = f(0). \]

More usefully, the delta function can pick out the function value at a given, nonzero time \(\xi\):

\[ \int\limits_{-\infty}^{\infty} f(t) \delta(t-\xi)\, dt = f(\xi). \]

Returning now to the impulse response function \(h(t)\), it is, quite simply, the output of the LTI system, when driven by the delta function as input, that is \(u(t) = \delta(t)\), or \(h(t) = F[\delta(t)]\). In practical terms, we can liken \(h(t)\) to the response of a mechanical system when it is struck very hard by a hammer!

Next we put the delta function and the convolution definition together, to show explicitly that the response of a system to arbitrary input \(u(t)\) is the convolution of the input and the impulse response \(h(t)\). This is what is stated in the definition given at the beginning of this section. First we note that

\begin{align} u(t) &= \int\limits_{-\infty}^{\infty} u(\xi) \delta(\xi - t) \, d\xi \\[4pt] &= \int\limits_{-\infty}^{\infty} u(\xi) \delta(t - \xi) \, d\xi && \text{(because the input is symmetric about zero time).} \nonumber \end{align}

Now set the system response \(y(t) = F[u(t)]\), where \(F\) is an LTI system - we will use its two properties below.

\begin{align*} y(t) &= F \left[ \int\limits_{-\infty}^{\infty} u(\xi) \delta(t - \xi)\, d\xi \right] \\[4pt] &= \int\limits_{-\infty}^{\infty} u(\xi) F[\delta(t - \xi)] \, d\xi && \text{(using linearity)} \\[4pt] &= \int\limits_{-\infty}^{\infty} u(\xi) h(t - \xi) \, d\xi && \text{(using time-invariance),} \end{align*}

and this indeed is the definition of convolution, often written as \(y(t) = h(t) \times u(t)\).

An intuitive understanding of convolution can be gained by thinking of the input as an infinite number of scaled delta functions, placed very closely together on the time axis. Explaining the case with the integrand \(u(t − \xi)h(\xi) \), we see the convolution integral will call up all these virtual impulses, referenced to time \(t\), and multiply them by the properly shifted impulse responses. Consider one impulse only that occurs at time \(t = 2\), and we are interested in the response at \(t = 5\). Then \(u(t) = \delta(t − 2)\) or \(u(t − \xi) = \delta(t − 2 − \xi) \). The integrand will thus be nonzero only when \(t − 2 − \xi\) is zero, or \( \xi = t − 2\). Now \(h(\xi) = h(t − 2)\) will be \(h(3)\) when \(t = 5\), and hence it provides the impulse response three time units after the impulse occurs, which is just what we wanted.