Question 3.2: Find the unit impulse response h(t) of the single-DOF system......

In both cases, the response to a unit impulse at t=0 is equal to the free response of the system when the initial conditions are z = 0, \dot{z} = (1/m), at t = 0. The method is then the same as that used in Section 2.2 to find the free response

The equation of motion is \ddot{z} + \omega ^{2}_{n}z = 0. The displacement response is, from Eq. (2.28), in Chapter 2, noting that \gamma = 0 and \omega_{d} = \omega_{n} in this case:

z = e^{-\gamma \omega _{n}t}(A \cos \omega _{d}t + B \sin \omega _{d}t )            (2.28)

z = A \cos \omega _{n}t + B \sin \omega _{n}t             (A)

where A and B are given by substituting the initial conditions. Differentiating Eq. (A):

\dot{z} = -A  ω_{n}\sin \omega _{n}t + B  ω_{n}\cos \omega _{n}t           (B)

Substituting z = 0 and t = 0 in Eq. (A), and \dot{z} = (1/m) and t = 0 in Eq. (B), we find that A = 0 and B = (1/m ω_{n}), so from Eq. (A), the unit impulse response of the system is

z = \frac{1}{m \omega _{n}}\sin \omega _{n}t = h(t)             (C)

Case  (b):  with  non-zero  damping,  but  less  than  critical:
The equation of motion is

\ddot{z} + 2\gamma \omega _{n}\dot{z} + \omega ^{2}_{n}z = 0        (D)

From Eqs (A) and (B) of Example 2.2, the free response is given by:

z = e^{-\gamma \omega _{n}t}(A \cos \omega _{d}t + B \sin \omega _{d}t )            (E)

and

\dot{z} = e^{-\gamma \omega _{n}t}[(B\omega _{d} – A\gamma \omega _{n})\cos \omega _{d}t  –  (A\omega _{d} + B\gamma \omega _{n}) \sin \omega _{d}t]         (F)

Substituting the same initial conditions as for Case (a) into Eqs (E) and (F), we find that A = 0, and B = (1/m ω_{d}), so from Eq. (E), the required unit impulse response is

z = \frac{1}{m \omega _{d}}\left(e^{-\gamma \omega _{n}t}\sin \omega _{d}t \right) = h(t)            (G)

where the damped natural frequency, \omega _{d}, is given by \omega _{d} = \omega _{n}\sqrt{1-\gamma ^{2}}, and \omega _{n} is the undamped natural frequency.