*7.4. Impulse Response to Fractional Oscillators in Class III*

We present the impulse response to fractional oscillators in Class III with Theorem 15.

**Theorem 15** (Impulse response III)**.** *Let h*3(*t*) *be the impulse response to a fractional oscillator in Class III. For t* ≥ *0, 1 < α* ≤ *2, 0 < β* ≤ *1, it is in the form*

$$h\_3(t) = \frac{e^{-\frac{m\omega^{\alpha-1}\sin\frac{4\pi}{2} + c\omega^{\beta-1}\sin\frac{\beta\pi}{2}}{2\sqrt{-(m\omega^{\alpha-2}\cos\frac{4\pi}{2} + c\omega^{\beta-2}\cos\frac{\beta\pi}{2})^k}}\omega\_{\text{eqn},3}t}{-\left(m\omega^{\alpha-2}\cos\frac{\alpha\pi}{2} + c\omega^{\beta-2}\cos\frac{\beta\pi}{2}\right)\omega\_{\text{eq},3}},\tag{203}$$

,

*where*

$$
\omega\_{\text{eqn}3} = \frac{\omega\_n}{\sqrt{-\left(\omega^{a-2}\cos\frac{a\pi}{2} + \frac{c}{m}\omega^{\beta-2}\cos\frac{\beta\pi}{2}\right)}},
$$

*and*

$$\omega\_{eqd,3} = \frac{\omega\_n \sqrt{1 - \frac{\left(m\omega^{a-1}\sin\frac{a\pi}{2} + c\omega^{\beta-1}\sin\frac{\beta\pi}{2}\right)^2}{4\left[-\left(m\omega^{a-2}\cos\frac{a\pi}{2} + c\omega^{\beta-2}\cos\frac{\beta\pi}{2}\right)k\right]}}}{\sqrt{-\left(\omega^{a-2}\cos\frac{a\pi}{2} + \frac{c}{m}\omega^{\beta-2}\cos\frac{\beta\pi}{2}\right)}}$$

**Proof.** With (196), we ge<sup>t</sup>

$$h\_3(t) = e^{-\zeta\_{\rm eq}\wp\omega\_{\rm eqn}\mathfrak{z}t} \frac{1}{m\_{\rm eq}\wp\omega\_{\rm eqd,3}} \sin\omega\_{\rm eqd,3}t, t \ge 0. \tag{204}$$

.

In the above expression, substitute *meq*3 with the one in Section 4, *ςeq*3, *<sup>ω</sup>eqd*,3, *<sup>ω</sup>eqn*,<sup>3</sup> by those in Section 5, respectively, we have, for *t* ≥ 0,

$$h\_3(t) = \frac{c^{-\frac{m\omega^{\alpha-1}\sin\frac{\alpha\pi}{2} + c\omega^{\beta-1}\sin\frac{\beta\pi}{2}}{2}}\omega\_{\text{eqn},3}t}{-\left(m\omega^{\alpha-2}\cos\frac{\alpha\pi}{2} + c\omega^{\beta-2}\cos\frac{\beta\pi}{2}\right)^{k}}\sin\omega\_{\text{eq},3}t}{-\left(m\omega^{\alpha-2}\cos\frac{\alpha\pi}{2} + c\omega^{\beta-2}\cos\frac{\beta\pi}{2}\right)\omega\_{\text{eq},3}}t$$

The right side on the above is (203). Thus, the proof completes. 

The plots of *h*3(*t*) with fixed *ω* are shown in Figure 39, with variable *ω* in Figure 40, and in t-ω plane by Figure 41.

**Figure 39.** Impulse response *h*3(*t*) for *m* = *c* = 1, *k* = 25 (*<sup>ω</sup>n* = 5). (**a**) (*<sup>α</sup>*, *β*) = (1.8, 0.3), solid line: *ω* = 2 (*ςeq*<sup>3</sup> = 0.03); dot line: *ω* =1(*ςeq*<sup>3</sup> = 0.02). (**b**) (*<sup>α</sup>*, *β*) = (1.5, 0.8), solid line: *ω* =2(*ςeq*<sup>3</sup> = 0.20); dot line: *ω* = 1 (*ςeq*<sup>3</sup> = 0.10). (**c**) (*<sup>α</sup>*, *β*) = (1.8, 0.5), solid line: *ω* =2(*ςeq*<sup>3</sup> = 0.05); dot line: *ω* =1(*ςeq*<sup>3</sup> = 0.02). (**d**) (*<sup>α</sup>*, *β*) = (2, 1), solid line: *ω* =2(*ςeq*<sup>3</sup> = 0); dot line: *ω* =1(*ςeq*<sup>3</sup> = 0).

**Figure 40.** Impulse response *h*3(*t*) to a fractional oscillator in Class III for *m* = *c* = 1, *k* = 25 (*<sup>ω</sup>n* = 5). (**a**) (*<sup>α</sup>*, *β*) = (1.8, 0.8), *ω* = 1, 2, ..., 5 (0.09 ≤ *ςeq*<sup>3</sup> ≤ 0.45). (**b**) (*<sup>α</sup>*, *β*) = (1.5, 0.8), *ω* = 1, 2, ..., 5 (0.09 ≤ *ςeq*<sup>3</sup> ≤ 0.20). (**c**) (*<sup>α</sup>*, *β*) = (1.3, 0.8), *ω* = 1, 2, ..., 5 (0.48 ≤ *ςeq*<sup>3</sup> ≤ 0.67).

**Figure 41.** Impulse response to a fractional oscillator in Class III in t-ω plane for *m* = *c* = 1, *k* = 25 (*<sup>ω</sup>n*<sup>=</sup> 5) with *t* = 0, 1, ..., 30; *ω* = 1, 2, ..., 5. (**a**) *α* = 1.8, *β* = 0.8 (0.09 ≤ *ςeq*<sup>3</sup> ≤ 0.45). (**b**) *α* = 1.8, *β* = 0.4 (0.07 ≤ *ςeq*<sup>3</sup> ≤ 0.15). (**c**) *α* = 1.5, *β* = 0.8 (0.33 ≤ *ςeq*<sup>3</sup> ≤ 0.91). (**d**) *α* = 2, *β* =1(*ςeq*<sup>3</sup> = 0).

**Note 7.3:** The impulse response *h*3(*t*) degenerates to the conventional one when *α* = 2 and *β* = 1. Indeed,

$$\left.h\_{3}(t)\right|\_{a=2,\beta=1} = \left[\frac{e^{\frac{ma^{a-1}\sin\frac{\omega\pi}{2}+c\omega\beta-\sin\frac{\beta\pi}{2}}\sin\frac{\beta\pi}{2}}}{-\left(m\omega^{a-2}\cos\frac{\omega\pi}{2}+c\omega\beta^{-2}\cos\frac{\beta\pi}{2}\right)}\sin\omega\_{\text{eq},3}t}\right]\_{a=2,\beta=1} = \frac{e^{-\zeta\omega\_{\text{n}}t}}{m\omega\_{d}}\sin\omega\_{d}t.\tag{205}$$
