The Laplace transform of the step response of a system is given by Y(s) = 100 /s(s+100) . The rise time is defined as the time taken for the response to go from 0.1 to 0.9 of its final value. The settling time is defined as the time taken for the response to reach 0.98 of its final value. For this system, the rise time (Tr), settling time (Ts), and time constant (Tc), all expressed in seconds, are

The correct option is T_r = 0.022, T_s = 0.04, T_c = 0.01.

Here is the step-by-step educational explanation of how to arrive at this solution:

Step 1: Finding the time-domain response y(t)
The Laplace transform of the output response is given as:
$$Y(s)=\frac{100}{s(s+100)}$$
We can expand this expression using partial fractions:
$$\frac{100}{s(s+100)}=\frac{1}{s}-\frac{1}{s+100}$$
Taking the inverse Laplace transform, we get the time-domain response:
$$y(t)=1-e^{-100t}$$
for t ≥ 0.

Step 2: Identifying the Time Constant (T_c)
A standard first-order system response has the form:
$$y(t)=A(1-e^{-t/T_c})$$
Comparing this with our system's response $y\left(t\right)=1-e^{-100t}$, we have:
$$\frac{1}{T_c}=100\Rightarrow T_c=\frac{1}{100}=0.01\text{seconds}$$
Also, note that the final steady-state value of the response as t approaches infinity is:
$$y_{\text{final}}=\underset{t\to \infty }{lim}(1-e^{-100t})=1$$

Step 3: Calculating the Rise Time (T_r)
The rise time is the time taken to go from 0.1 to 0.9 of its final value (which is 1). Let t₁ and t₂ be the times when the response reaches 0.1 and 0.9 respectively.
For t₁ (0.1 of final value):
$$1-e^{-100t_1}=0.1\Rightarrow e^{-100t_1}=0.9\Rightarrow t_1=\frac{-ln(0.9)}{100}\approx 0.00105\text{seconds}$$
For t₂ (0.9 of final value):
$$1-e^{-100t_2}=0.9\Rightarrow e^{-100t_2}=0.1\Rightarrow t_2=\frac{-ln(0.1)}{100}\approx 0.02303\text{seconds}$$
Therefore, the rise time is:
$$T_r=t_2-t_1=\frac{ln(0.9)-ln(0.1)}{100}=\frac{ln(9)}{100}\approx \frac{2.197}{100}\approx 0.022\text{seconds}$$

Step 4: Calculating the Settling Time (T_s)
The settling time is the time taken to reach 0.98 of its final value:
$$1-e^{-100T_s}=0.98\Rightarrow e^{-100T_s}=0.02$$
Taking the natural logarithm of both sides:
$$-100T_s=ln(0.02)\approx -3.912$$
$$T_s=\frac{3.912}{100}\approx 0.03912\approx 0.04\text{seconds}$$

Conclusion:
The calculated parameters are T_r = 0.022 seconds, T_s = 0.04 seconds, and T_c = 0.01 seconds, which matches the option Tr = 0.022, Ts = 0.04, Tc = 0.01.