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
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.
\[ Y(s)=\dfrac{100}{s(s+100)} \]
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
Show 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.
Updated On: Jul 20, 2026
- \(T_r=0.022,\ T_s=0.04,\ T_c=0.01\)
- \(T_r=0.22,\ T_s=0.404,\ T_c=0.01\)
- \(T_r=2.2,\ T_s=4.04,\ T_c=1.01\)
- \(T_r=22,\ T_s=40.4,\ T_c=10.1\)
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The Correct Option is A
Solution and Explanation
Step 1: Identify the time constant of the system.
\[ Y(s)=\frac{100}{s(s+100)}=\frac{1}{s}\cdot\frac{100}{s+100} \]
so the plant transfer function is \(G(s)=100/(s+100)\). Writing this in the standard first order form \(1/(T_cs+1)\):
\[ \frac{100}{s+100}=\frac{1}{\frac{1}{100}s+1} \]
so the time constant is
\[ T_c=\frac{1}{100}=0.01\ \text{s} \]
Step 2: Find the time-domain step response.
Using partial fractions,
\[ Y(s)=\frac{1}{s}-\frac{1}{s+100} \]
so
\[ y(t)=1-e^{-100t}=1-e^{-t/T_c} \]
This rises from \(0\) toward a final value of \(1\), with time constant \(T_c=0.01\) s.
Step 3: Compute the rise time.
Solve \(1-e^{-t_1/T_c}=0.1\), giving \(e^{-t_1/T_c}=0.9\), so \(t_1=T_c\ln(1/0.9)=0.1054\,T_c\). Solve \(1-e^{-t_2/T_c}=0.9\), giving \(e^{-t_2/T_c}=0.1\), so \(t_2=T_c\ln(10)=2.3026\,T_c\). Then
\[ T_r=t_2-t_1=(2.3026-0.1054)T_c\approx2.2\,T_c \]
Substituting \(T_c=0.01\):
\[ T_r\approx2.2\times0.01=0.022\ \text{s} \]
Step 4: Compute the settling time.
Solve \(1-e^{-T_s/T_c}=0.98\), giving \(e^{-T_s/T_c}=0.02\), so \(T_s=T_c\ln(50)\approx3.91\,T_c\), commonly rounded to \(4\,T_c\). Substituting \(T_c=0.01\):
\[ T_s\approx0.0391\approx0.04\ \text{s} \]
Step 5: Final Answer.
So \(T_c=0.01\) s, \(T_r\approx0.022\) s, and \(T_s\approx0.04\) s, matching option (A). Options (B), (C), and (D) scale these values by factors of 10, 100, and 1000, which is what you would get by mistakenly reading the pole location as \(s=10\), \(s=1\), or \(s=0.1\) instead of the correct \(s=100\).
\[ \boxed{T_r=0.022\ \text{s},\ T_s=0.04\ \text{s},\ T_c=0.01\ \text{s}} \]
\[ Y(s)=\frac{100}{s(s+100)}=\frac{1}{s}\cdot\frac{100}{s+100} \]
so the plant transfer function is \(G(s)=100/(s+100)\). Writing this in the standard first order form \(1/(T_cs+1)\):
\[ \frac{100}{s+100}=\frac{1}{\frac{1}{100}s+1} \]
so the time constant is
\[ T_c=\frac{1}{100}=0.01\ \text{s} \]
Step 2: Find the time-domain step response.
Using partial fractions,
\[ Y(s)=\frac{1}{s}-\frac{1}{s+100} \]
so
\[ y(t)=1-e^{-100t}=1-e^{-t/T_c} \]
This rises from \(0\) toward a final value of \(1\), with time constant \(T_c=0.01\) s.
Step 3: Compute the rise time.
Solve \(1-e^{-t_1/T_c}=0.1\), giving \(e^{-t_1/T_c}=0.9\), so \(t_1=T_c\ln(1/0.9)=0.1054\,T_c\). Solve \(1-e^{-t_2/T_c}=0.9\), giving \(e^{-t_2/T_c}=0.1\), so \(t_2=T_c\ln(10)=2.3026\,T_c\). Then
\[ T_r=t_2-t_1=(2.3026-0.1054)T_c\approx2.2\,T_c \]
Substituting \(T_c=0.01\):
\[ T_r\approx2.2\times0.01=0.022\ \text{s} \]
Step 4: Compute the settling time.
Solve \(1-e^{-T_s/T_c}=0.98\), giving \(e^{-T_s/T_c}=0.02\), so \(T_s=T_c\ln(50)\approx3.91\,T_c\), commonly rounded to \(4\,T_c\). Substituting \(T_c=0.01\):
\[ T_s\approx0.0391\approx0.04\ \text{s} \]
Step 5: Final Answer.
So \(T_c=0.01\) s, \(T_r\approx0.022\) s, and \(T_s\approx0.04\) s, matching option (A). Options (B), (C), and (D) scale these values by factors of 10, 100, and 1000, which is what you would get by mistakenly reading the pole location as \(s=10\), \(s=1\), or \(s=0.1\) instead of the correct \(s=100\).
\[ \boxed{T_r=0.022\ \text{s},\ T_s=0.04\ \text{s},\ T_c=0.01\ \text{s}} \]
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