The impulse response

So you wonder how impulse responses are born? They are born out of linearity.

Before we get to deal with impulses, recall the following strategy for calculating the response of a linear system, say $G$, to a certain input which can be written as a linear combination of “simpler” inputs $u_k(t)$’s

$$u(t) = a_1 u_1(t) + a_2 u_2(t) + \cdots = \sum_{k} a_k u_k(t)$$

A typical scenario is one in which the $u_k(t)$’s are some simpler functions, such as polynomials, as in a Taylor series, or exponentials, as in a Fourier series.

Suppose one has calculated the response of the system to each of the individual inputs $u_k(t)$’s, that is, one has a collection of signals

$$y_k(t) = G(u_k(t), t), \quad k = 1, 2, \cdots$$

produced by a dynamic linear system $G$ in response to each of the $u_k(t)$’s. Because $G$ is linear the response to the original signal $u(t)$ can be obtained by combining these individual responses using superposition, that is

$$y(t) = a_1 y_1(t) + a_2 y_2(t) + \cdots = \sum_{k} a_k y_k(t)$$

That brings us back to impulses. Recall the sifting property of the impulse:

$$u(t) = \int_{-\infty}^{\infty} u(\tau) \, \delta(t – \tau) \, d\tau$$

which can be though of as writing the original input signal $u(t)$ as a combination of delayed impulses. The key here is to realize that $u(\tau)$ is not a function of $t$ but a “coefficient” of the delayed impulse $\delta(t – \tau)$. An integral replaces the summation above because it is necessary to use a continuum of impulse functions, one for each possible delay.

If a system is linear then all that one needs to do is to evaluate the response to the delayed impulses and later assemble the response to the original input by superposition. That is, calculate the delayed impulse response:

$$y_\tau(t) = g(t, \tau) = G(\delta(t – \tau), t)$$

and assemble the response

$$y(t) = \int_{-\infty}^{\infty} u(\tau) \, g(t, \tau) \, d\tau$$

in which we have used the same “coefficients” $u(\tau)$ associated with each delayed impulse $\delta(t – \tau)$.

With linearity alone that is as far as one can go. Enter time-invariance. If the linear system is time-invariant, then the response to a delayed impulse is nothing but the delayed response to an impulse. That is the impulse response to a delayed impulse

$$y_\tau(t) = g(t – \tau) = G(\delta(t – \tau), t)$$

is now a function only of the difference $t – \tau$, and not the absolute $t$ and $\tau$. Using time-invariance one obtains the well know convolution formula

$$y(t) = \int_{-\infty}^{\infty} u(\tau) \, g(t – \tau) \, d\tau = \int_{-\infty}^{\infty} u(t – \tau) \, g(\tau) \, d\tau $$

A final simplification is possible if $G$ is also causal, in which case $g(t) = 0$ for all $t < 0$, and $u(t) = 0$, $t < 0$. In that case, the extremes of integration can be tightened

$$y(t) = \int_{0}^{t} u(\tau) \, g(t – \tau) \, d\tau = \int_{0}^{t} u(t – \tau) \, g(\tau) \, d\tau $$

The above discussion can be found at Section 3.2.