2.6: An Example of Finding the Impulse Response

Last updated
Save as PDF

Page ID: 47956

Franz S. Hover & Michael S. Triantafyllou
Massachusetts Institute of Technology via MIT OpenCourseWare

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

Let's consider the differential equation \(mx''(t) + bx'(t) + cx(t) = \delta(t)\), with the initial conditions of \(x(0) = x'(0) = 0\). We have

\[ \int\limits_{-\epsilon/2}^{\epsilon/2} [mx'' + bx' + cx]\, dt \, = \int\limits_{-\epsilon/2}^{\epsilon/2} \delta(t)\, dt \, = \, 1 \]

so that \(m(x'(0^+) - x'(0^-)) = 1\).

The \(+\) superscript indicates the instant just after zero time, and the \(-\) superscript indicates the instant just before zero time. The given relation follows because at time zero the velocity and position are zero, so the acceleration must be very large. Now since \(x' (0^−) = 0\), we have \(x' (0^+) = 1/m\). This is very useful - the initial velocity after the mass is hit with a \(\delta(t)\) input. In fact, this replaces our previous initial condition \(x' (0) = 0\), and we can treat the differential equation as homogeneous from here on. With \(x(t) = c_1 e^{s_1 t} + c_2 e^{s_2 t} \), the governing equation becomes \(ms_i^2 + bs_i + k = 0\) so that

\[ s = - \dfrac{b}{2m} \pm \dfrac{\sqrt{b^2 - 4km}} {2m} .\]

Let \(\sigma = b/2m\) and

\[ \omega_d = \sqrt{ \dfrac{k}{m} - \dfrac{b^2}{4m^2}} , \]

and assuming that \(b^2 < 4km\), we find

\[ h(t) = \dfrac{1}{m \omega_d} e^{-\sigma t} sin(\omega_d t) , \, t \geq 0. \]

As noted above, once the impulse response is known for an LTI system, responses to all inputs can be found:

\[x(t) = \int\limits_{0}^{t} u(\tau) h(t - \tau) \, d\tau. \]

In the case of LTI systems, the impulse response is a complete definition of the system, in the same way that a differential equation is, with zero initial conditions.