Two linear time-invariant systems with transfer functions

${\mathrm{G}}_1(\mathrm{s})=\frac{10}{{\mathrm{s}}^2+\mathrm{s}+1}$ and ${\mathrm{G}}_2(\mathrm{s})=\frac{10}{{\mathrm{s}}^2+\mathrm{s}\sqrt{10}+10}$

have unit step responses y₁(t) and y₂(t), respectively. Which of the following statements is/are true?

System Analysis: Comparing Linear Time-Invariant Systems

To determine the true statements regarding the unit step responses ${\mathrm{y}}_1(\mathrm{t})$ and ${\mathrm{y}}_2(\mathrm{t})$ of the given linear time-invariant systems, ${\mathrm{G}}_1(\mathrm{s})$ and ${\mathrm{G}}_2(\mathrm{s})$, we need to analyze their characteristics. Both are second-order systems, and their behavior can be described by comparing them to the standard second-order transfer function form:

$$\mathrm{G}(\mathrm{s})=\frac{\mathrm{K}{\omega }_{\mathrm{n}}^2}{{\mathrm{s}}^2+2\zeta {\omega }_{\mathrm{n}}\mathrm{s}+{\omega }_{\mathrm{n}}^2}$$

where:

- ${\omega }_n$ is the natural frequency
- $\zeta$ is the damping ratio
- $K$ is the DC gain, which is also the steady-state value for a unit step input if the numerator is $K{\omega }_n^2$, or more generally, the steady-state value is $\mathrm{G}(0)$.

Let's find the parameters for each system.

Parameters for G₁(s)

The first transfer function is given as: ${\mathrm{G}}_1(\mathrm{s})=\frac{10}{{\mathrm{s}}^2+\mathrm{s}+1}$

Comparing this to the standard form:

- Natural frequency squared: ${\omega }_{n1}^2=1⟹{\omega }_{n1}=1\text{ rad/s}$
- Damping ratio: $2{\zeta }_1{\omega }_{n1}=1⟹2{\zeta }_1(1)=1⟹{\zeta }_1=0.5$
- Damped frequency of oscillation: ${\omega }_{d1}={\omega }_{n1}\sqrt{1-{\zeta }_1^2}=1\sqrt{1-(0.5{)}^2}=\sqrt{1-0.25}=\sqrt{0.75}=\frac{\sqrt{3}}{2}\text{ rad/s}$
- Steady-state value for unit step response (${\mathrm{y}}_{ss1}$): ${\mathrm{y}}_{ss1}={\lim }_{\mathrm{s}\to 0}\mathrm{s}\cdot \frac{1}{\mathrm{s}}\cdot {\mathrm{G}}_1(\mathrm{s})={\mathrm{G}}_1(0)=\frac{10}{0^2+0+1}=10$
- 2% Settling time ($T_{s1}$): $T_{s1}=\frac{4}{{\zeta }_1{\omega }_{n1}}=\frac{4}{0.5\times 1}=\frac{4}{0.5}=8\text{ seconds}$

Parameters for G₂(s)

The second transfer function is given as: ${\mathrm{G}}_2(\mathrm{s})=\frac{10}{{\mathrm{s}}^2+\mathrm{s}\sqrt{10}+10}$

Comparing this to the standard form:

- Natural frequency squared: ${\omega }_{n2}^2=10⟹{\omega }_{n2}=\sqrt{10}\text{ rad/s}$
- Damping ratio: $2{\zeta }_2{\omega }_{n2}=\sqrt{10}⟹2{\zeta }_2(\sqrt{10})=\sqrt{10}⟹{\zeta }_2=0.5$
- Damped frequency of oscillation: ${\omega }_{d2}={\omega }_{n2}\sqrt{1-{\zeta }_2^2}=\sqrt{10}\sqrt{1-(0.5{)}^2}=\sqrt{10}\sqrt{0.75}=\frac{\sqrt{30}}{2}\text{ rad/s}$
- Steady-state value for unit step response (${\mathrm{y}}_{ss2}$): ${\mathrm{y}}_{ss2}={\lim }_{\mathrm{s}\to 0}\mathrm{s}\cdot \frac{1}{\mathrm{s}}\cdot {\mathrm{G}}_2(\mathrm{s})={\mathrm{G}}_2(0)=\frac{10}{0^2+0+10}=1$
- 2% Settling time ($T_{s2}$): $T_{s2}=\frac{4}{{\zeta }_2{\omega }_{n2}}=\frac{4}{0.5\times \sqrt{10}}=\frac{8}{\sqrt{10}}\approx 2.53\text{ seconds}$

Comparison of System Characteristics

Let's summarize the calculated parameters in a table for easier comparison:Evaluating the Statements

Now, let's evaluate each statement based on our calculated values:

- **Statement 1: ${\mathrm{y}}_1(\mathrm{t})$ and ${\mathrm{y}}_2(\mathrm{t})$ have the same percentage peak overshoot.** The percentage peak overshoot ($\%M_p$) for a second-order system is given by the formula: $$\%M_p=e^{-\frac{\zeta \pi }{\sqrt{1-{\zeta }^2}}}\times 100\%$$ Since both systems have the same damping ratio (${\zeta }_1=0.5$ and ${\zeta }_2=0.5$), their percentage peak overshoots will be identical. This statement is TRUE.
- **Statement 2: ${\mathrm{y}}_1(\mathrm{t})$ and ${\mathrm{y}}_2(\mathrm{t})$ have the same steady-state value.** From our calculations, ${\mathrm{y}}_{ss1}=10$ and ${\mathrm{y}}_{ss2}=1$. These values are different. This statement is FALSE.
- **Statement 3: ${\mathrm{y}}_1(\mathrm{t})$ and ${\mathrm{y}}_2(\mathrm{t})$ have the same damped frequency of oscillation.** From our calculations, ${\omega }_{d1}=\frac{\sqrt{3}}{2}$ and ${\omega }_{d2}=\frac{\sqrt{30}}{2}$. These values are different. This statement is FALSE.
- **Statement 4: ${\mathrm{y}}_1(\mathrm{t})$ and ${\mathrm{y}}_2(\mathrm{t})$ have the same 2% settling time.** From our calculations, $T_{s1}=8\text{ s}$ and $T_{s2}=\frac{8}{\sqrt{10}}\text{ s}$. These values are different. This statement is FALSE.

Based on the analysis, only the first statement is true.