format = charfcn or piecewise

method = function or matrix

The ImpulseResponse command computes the impulse response of sys, a System object.

The ImpulseResponse command returns an algebraic Matrix, except in the case when the option format=piecewise is selected, as described under Options, and the output may be a piecewise expression of Matrices.

Two methods are provided for computing the impulse response of a system.

The function method generates a transfer-function model of sys, then computes the inverse Laplace transformation or the inverse z-transformation, for continuous and discrete system, respectively.

The matrix method generates a state-space model of sys, then computes the impulse response from the state-transition matrix.

$\mathrm{with}\left(\mathrm{DynamicSystems}\right)\:$

$\mathrm{sys}≔\mathrm{NewSystem}\left(⟨⟨\frac{s}{s^2+{\mathrm{\omega }}^2}\,\frac{s+1}{s^2+3}⟩⟩\right)\:$

$\mathrm{imp}≔\mathrm{ImpulseResponse}\left(\mathrm{sys}\right)$

$\mathrm{imp}≔\left[\begin{array}{c}\cos \left(\mathrm{\omega }t\right)\\ \cos \left(\sqrt{3}t\right)+\frac{\sqrt{3}\sin \left(\sqrt{3}t\right)}{3}\end{array}\right]$

$\mathrm{with}\left(\mathrm{inttrans}\right)\:$

$\mathrm{map}\left(\mathrm{laplace}\,\mathrm{imp}\,t\,s\right)$

$\left[\begin{array}{c}\frac{s}{s^2+{\mathrm{\omega }}^2}\\ \frac{s+1}{s^2+3}\end{array}\right]$

$\mathrm{sys}≔\mathrm{NewSystem}\left(⟨⟨\frac{1}{z+1}\,\frac{z-a}{z+b}⟩⟩\,\mathrm{discrete}\right)\:$

$\mathrm{imp}≔\mathrm{ImpulseResponse}\left(\mathrm{sys}\right)$

$\mathrm{imp}≔\left[\begin{array}{c}-{\left(\mathrm{-1}\right)}^q+{\mathrm{charfcn}}_0\left(q\right)\\ \frac{{\left(-b\right)}^{q}a+{\left(-b\right)}^{q}b-a{\mathrm{charfcn}}_0\left(q\right)}{b}\end{array}\right]$

$\mathrm{simplify}\left(\mathrm{map}\left(\mathrm{ztrans}\,\mathrm{imp}\,q\,z\right)\right)$

$\left[\begin{array}{c}\frac{1}{z+1}\\ \frac{z-a}{z+b}\end{array}\right]$

$\mathrm{simplify}\left(\mathrm{ImpulseResponse}\left(\mathrm{sys}\,\mathrm{method}=\mathrm{matrix}\,\mathrm{format}=\mathrm{piecewise}\right)\right)$

$\left\{\begin{array}{cc}\left[\begin{array}{c}0\\ 1\end{array}\right]& q\le 0\\ \left[\begin{array}{c}\frac{b\left(-{\left(-b\right)}^{q-1}-{\left(\mathrm{-1}\right)}^q\right)-{\left(-b\right)}^q+{\left(\mathrm{-1}\right)}^q}{-1+b}\\ -{\left(-b\right)}^{q-1}\left(a+b\right)\end{array}\right]& 0<q\end{array}\right.$