Impulse and ramp response from first order step response

For a linear time-invariant system, the impulse response is the derivative of the step response, and the ramp response is the integral of the step response.

Transfer function

$$C(s) = \mathcal{L}\{5(1 - e^{-t/5})\} = 5\left[\frac{1}{s} - \frac{1}{s + 0.2}\right] = \frac{1}{s(s+0.2)}$$

Since $R(s) = 1/s$:

$$G(s) = \frac{1}{s + 0.2} = \frac{5}{5s + 1}$$

This is a first-order system with gain 5 and time constant $T = 5$ s.

Impulse response

$$c_\delta(t) = \frac{d}{dt}\left[5(1 - e^{-t/5})\right] = 5\cdot\frac{1}{5}e^{-t/5} = e^{-t/5},\quad t \ge 0$$

Check: $\mathcal{L}^{-1}\{1/(s+0.2)\} = e^{-0.2t}$.

Ramp response (unit ramp)

$$\begin{aligned} c_r(t) &= \int_0^t 5(1 - e^{-\tau/5})\,d\tau \\ &= 5\left[\tau + 5e^{-\tau/5}\right]_0^t \\ &= 5t + 25e^{-t/5} - 25 \\ &= 5t - 25 + 25e^{-t/5},\quad t \ge 0 \end{aligned}$$

Check: $\dfrac{1}{s^2(s+0.2)} = \dfrac{5}{s^2} - \dfrac{25}{s} + \dfrac{25}{s+0.2}$, which gives the same result.

Answer: impulse response $= e^{-t/5}$; ramp response $= 5t - 25 + 25e^{-t/5}$ for $t \ge 0$.