Question

The Laplace transform of the step response of a system is given by
\[ Y(s)=\dfrac{100}{s(s+100)} \]
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 (\(T_r\)), settling time (\(T_s\)), and time constant (\(T_c\)), all expressed in seconds, are

• A. \(T_r=0.022,\ T_s=0.04,\ T_c=0.01\)
• B. \(T_r=0.22,\ T_s=0.404,\ T_c=0.01\)
• C. \(T_r=2.2,\ T_s=4.04,\ T_c=1.01\)
• D. \(T_r=22,\ T_s=40.4,\ T_c=10.1\)

Hint:

Find the time constant from the pole location, then use the standard 2.2 times Tc rise time and about 4 times Tc settling time formulas for a first-order step response.

Answer 1

The Correct Option is A

Instead of deriving the time response from scratch, we can start from the standard benchmark formulas for a first order system and just verify them.

For a first order system with time constant $\tau$, two widely used approximations are: rise time (10% to 90%) $T_r\approx2.2\tau$, and settling time (within 2% of the final value) $T_s\approx4\tau$ (the more precise value is $3.91\tau$).

Identify $\tau$ from the pole of the given transfer function. $Y(s)=\dfrac{100}{s(s+100)}$ has its finite pole at $s=-100$. For a first order form $\dfrac{a}{s+a}$, the time constant is $\tau=1/a$, so here:

\[ \tau=T_c=\frac{1}{100}=0.01\ \text{s} \]

Plugging into the two formulas:

\[ T_r=2.2\times0.01=0.022\ \text{s}, \qquad T_s=4\times0.01=0.04\ \text{s} \]

Now verify these numbers against the exact response $y(t)=1-e^{-100t}$. At $t=0.022$: $y=1-e^{-2.2}=1-0.111=0.889$, close to the 90% mark, confirming the rise time estimate. At $t=0.04$: $y=1-e^{-4}=1-0.0183=0.982$, matching the 98% settling criterion closely.

Both checks confirm the same triple of values.

\[ \boxed{T_r=0.022\ \text{s},\ T_s=0.04\ \text{s},\ T_c=0.01\ \text{s}} \]