Categories: Mathematics, Physics.

# Impulse response

Given a system whose behaviour is described by a linear operator $\hat{L}$,
its **impulse response** $u_\delta(t)$ is defined as the systemâ€™s response
when forced by the Dirac delta function $\delta(t)$:

This can be used to find the response $u(t)$ of $\hat{L}$
to *any* forcing function $f(t)$,
by simply taking the convolution with $u_\delta(t)$:

Starting from the definition of $u_\delta(t)$, we shift the argument by some constant $\tau$, and multiply both sides by $f(\tau)$:

$\begin{aligned} \hat{L} \{ u_\delta(t - \tau) \} &= \delta(t - \tau) \\ \hat{L} \{ f(\tau) \: u_\delta(t - \tau) \} &= f(\tau) \: \delta(t - \tau) \end{aligned}$Where $f(\tau)$ was moved inside thanks to the linearity of $\hat{L}$. Integrating over $\tau$ gives us:

$\begin{aligned} \int_0^\infty \hat{L} \{ f(\tau) \: u_\delta(t - \tau) \} \dd{\tau} &= \int_0^\infty f(\tau) \: \delta(t - \tau) \dd{\tau} = f(t) \end{aligned}$The integral and $\hat{L}$ are operators of different variables, so we reorder them, and recognize that the resulting integral is a convolution:

$\begin{aligned} f(t) &= \hat{L} \int_0^\infty f(\tau) \: u_\delta(t - \tau) \dd{\tau} = \hat{L} \Big\{ (f * u_\delta)(t) \Big\} \end{aligned}$Because $\hat{L} \{ u(t) \} = f(t)$ by definition, we then see that $(f * u_\delta)(t) = u(t)$.

This is useful for solving initial value problems, because any initial condition can be satisfied thanks to linearity, by choosing the initial values of the homogeneous solution $\hat{L}\{ u_0(t) \} = 0$ such that the total solution $(f * u_\delta)(t) + u_0(t)$ has the desired values.

For boundary value problems, there is the related concept of a fundamental solution.

## References

- O. Bang,
*Applied mathematics for physicists: lecture notes*, 2019, unpublished.