Overdamped System MCQ Quiz - Objective Question with Answer for Overdamped System - Download Free PDF

The correct answer is option 4):(ζ > 1)

Concept:

The transfer function of the standard second-order system is:       

$TF=\frac{C\left(s\right)}{R\left(s\right)}=\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$

ζ is the damping ratio ωn is the undamped natural frequency

Characteristic equation:  

The roots of the characteristic equation are: $s^2+2\zeta {\omega }_n+{\omega }_n^2=0$

$-\zeta {\omega }_n\pm j{\omega }_n\sqrt{1-{\zeta }^2}=-\alpha \pm j{\omega }_d$

 α is the damping factor

ζ = 0, the system is undamped

ζ = 1, the system is critically damped

0 < ξ < 1, the system is underdamped

ζ > 1, the system is overdamped

Concept:

Explanation:

Given an overdamped system consisting of poles at -4 and -6a/4

For overdamped case ζ > 1

When the poles lie at (-3 ± j4), these poles will present in the s plane as shown above underdamped system. (practical system).

So, for underdamped system 0 < ζ < 1

Hence, the damping ratio decreases when the system moves from an overdamped case to an underdamped case.

Concept:

The standard second-order system is given by $\frac{{\omega }_n^2}{s^2+2\xi {\omega }_{n}s+{\omega }_n^2}$

Where ξ is the damping ratio

ωn is the natural frequency

If ξ = 1, then the system is critically damped.

If ξ < 1, then the system is underdamped.

If ξ > 1, then the system is order damped.

If ξ = 0, the system is said to be undamped

Calculation:

$TF=\frac{32}{s^2+15s+32}$

By comparing with a standard second-order transfer function,

ωn2 = 32 ⇒ ωn = √32

$2\xi {\omega }_n=15\Rightarrow \xi =\frac{15}{2\sqrt{32}}>1$

So, it is an overdamped system.

Concept:

General expression for the transfer function of a second order system

$\frac{C\left(s\right)}{R\left(s\right)}=\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$ 

When denominator is equated to zero it gives characteristic equation i.e.

s² + 2ζω_ns + ω²_n ^{= 0}

$\mathrm{\%}overshoot=e^{-\frac{\zeta \pi }{\sqrt{1-{\zeta }^2}}}\times 100$ 

Calculation:

$\frac{C\left(s\right)}{R\left(s\right)}=\frac{1}{s^2+1}$ 

Comparing the denominator ζ = 0

$\mathrm{\%}overshoot=e^{-\frac{\zeta \pi }{\sqrt{1-{\zeta }^2}}}\times 100=100$ 

$\begin{array}{l}\mathrm{G}\left(\mathrm{s}\right)=\frac{\mathrm{K}}{\begin{array}{c}s\left(\mathrm{s}+10\right)\end{array}}\\ \mathrm{C}\mathrm{L}\mathrm{T}\mathrm{F}=\frac{\mathrm{K}}{{\mathrm{s}}^2+10\mathrm{s}+\mathrm{K}}\end{array}$

Characteristics equation of given CLTF is:

s²+10s+k = 0

Compare this with the standard equation,

s2+2ζw_ns+w_n²=0wωωnωnωnωn

we get:ωnωn

$\begin{array}{l}{\omega }_{\mathrm{n}}=\sqrt{\mathrm{K}}\mathrm{\&}2\zeta {\omega }_{\mathrm{n}}=10\\ 2\times .25\times \sqrt{\mathrm{K}}=10\Rightarrow \mathrm{K}=400\end{array}$

Concept:

The standard second-order system is given by $\frac{{\omega }_n^2}{s^2+2\xi {\omega }_{n}s+{\omega }_n^2}$

Where ξ is the damping ratio

ωn is the natural frequency

If ξ = 1, then the system is critically damped.

If ξ < 1, then the system is underdamped.

If ξ > 1, then the system is order damped.

If ξ = 0, the system is said to be undamped

Calculation:

$TF=\frac{32}{s^2+15s+32}$

By comparing with a standard second-order transfer function,

ωn2 = 32 ⇒ ωn = √32

$2\xi {\omega }_n=15\Rightarrow \xi =\frac{15}{2\sqrt{32}}>1$

So, it is an overdamped system.

$\begin{array}{l}\mathrm{G}\left(\mathrm{s}\right)=\frac{\mathrm{K}}{\begin{array}{c}s\left(\mathrm{s}+10\right)\end{array}}\\ \mathrm{C}\mathrm{L}\mathrm{T}\mathrm{F}=\frac{\mathrm{K}}{{\mathrm{s}}^2+10\mathrm{s}+\mathrm{K}}\end{array}$

Characteristics equation of given CLTF is:

s²+10s+k = 0

Compare this with the standard equation,

s2+2ζw_ns+w_n²=0wωωnωnωnωn

we get:ωnωn

$\begin{array}{l}{\omega }_{\mathrm{n}}=\sqrt{\mathrm{K}}\mathrm{\&}2\zeta {\omega }_{\mathrm{n}}=10\\ 2\times .25\times \sqrt{\mathrm{K}}=10\Rightarrow \mathrm{K}=400\end{array}$

Concept:

Explanation:

Given an overdamped system consisting of poles at -4 and -6a/4

For overdamped case ζ > 1

When the poles lie at (-3 ± j4), these poles will present in the s plane as shown above underdamped system. (practical system).

So, for underdamped system 0 < ζ < 1

Hence, the damping ratio decreases when the system moves from an overdamped case to an underdamped case.

The correct answer is option 4):(ζ > 1)

Concept:

The transfer function of the standard second-order system is:       

$TF=\frac{C\left(s\right)}{R\left(s\right)}=\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$

ζ is the damping ratio ωn is the undamped natural frequency

Characteristic equation:  

The roots of the characteristic equation are: $s^2+2\zeta {\omega }_n+{\omega }_n^2=0$

$-\zeta {\omega }_n\pm j{\omega }_n\sqrt{1-{\zeta }^2}=-\alpha \pm j{\omega }_d$

 α is the damping factor

ζ = 0, the system is undamped

ζ = 1, the system is critically damped

0 < ξ < 1, the system is underdamped

ζ > 1, the system is overdamped

Concept:

General expression for the transfer function of a second order system

$\frac{C\left(s\right)}{R\left(s\right)}=\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$ 

When denominator is equated to zero it gives characteristic equation i.e.

s² + 2ζω_ns + ω²_n ^{= 0}

$\mathrm{\%}overshoot=e^{-\frac{\zeta \pi }{\sqrt{1-{\zeta }^2}}}\times 100$ 

Calculation:

$\frac{C\left(s\right)}{R\left(s\right)}=\frac{1}{s^2+1}$ 

Comparing the denominator ζ = 0

$\mathrm{\%}overshoot=e^{-\frac{\zeta \pi }{\sqrt{1-{\zeta }^2}}}\times 100=100$ 

The characteristic equation of standard second order system is given by

s² + 2ξ ω_n s + ω²_n = 0

The system is said to be

a) undamped if ξ = 0

b) critically damped if ξ = 1

c) underdamped if ξ < 1

d) overdamped if ξ > 1

1) s² + 8s + 15 = 0

ω²_n = 15 ⇒ ω_n = √15

$2\xi {\omega }_n=8\Rightarrow 2\xi \left(\sqrt{15}\right)=8\Rightarrow \xi =\frac{4}{\sqrt{15}}>1$

So, the system is overdamped.

2) s² + 24s + 225 = 0

ω²_n = 225 ⇒ ω_n = 15

$2\xi {\omega }_n=24\Rightarrow 2\xi \left(15\right)=24\Rightarrow \xi =\frac{12}{15}<1$

So, the system is underdamped.

3) s² + 20.25 = 0

${\omega }_n^2=20.25\Rightarrow {\omega }_n=\sqrt{20.25}$

2ξω_n = 0 ⇒ ξ = 0

So, the system is undamped.

4) s² + 20s + 100 = 0

ω²_n = 100 ⇒ ω_n = 10

2ξω_n = 20 ⇒ 2ξ (10) = 20 ⇒ ξ = 1

Transfer function will be:

$T(s)=\frac{\frac{1}{sC}}{R+sL+\frac{1}{sC}}$

$T(s)=\frac{1}{s^{2}LC+sRC+1}$

$T(s)=\frac{\frac{1}{LC}}{s^2+s\frac{R}{L}+\frac{1}{LC}}$
${\omega }_n=\frac{1}{\sqrt{LC}}$
2 ζω_n = R/L
$\zeta =\frac{R}{2}\sqrt{\frac{C}{L}}$
For no oscillations:

$\zeta \ge 1\to \zeta =\frac{R}{2}\sqrt{\frac{C}{L}}\ge 1$
$R\ge 2\sqrt{\frac{L}{C}}$

Concept:

The standard second-order system is given by $\frac{{\omega }_n^2}{s^2+2\xi {\omega }_{n}s+{\omega }_n^2}$

Where ξ is the damping ratio

ωn is the natural frequency

If ξ = 1, then the system is critically damped.

If ξ < 1, then the system is underdamped.

If ξ > 1, then the system is order damped.

If ξ = 0, the system is said to be undamped

Calculation:

$TF=\frac{32}{s^2+15s+32}$

By comparing with a standard second-order transfer function,

ωn2 = 32 ⇒ ωn = √32

$2\xi {\omega }_n=15\Rightarrow \xi =\frac{15}{2\sqrt{32}}>1$

So, it is an overdamped system.

$\begin{array}{l}\mathrm{G}\left(\mathrm{s}\right)=\frac{\mathrm{K}}{\begin{array}{c}s\left(\mathrm{s}+10\right)\end{array}}\\ \mathrm{C}\mathrm{L}\mathrm{T}\mathrm{F}=\frac{\mathrm{K}}{{\mathrm{s}}^2+10\mathrm{s}+\mathrm{K}}\end{array}$

Characteristics equation of given CLTF is:

s²+10s+k = 0

Compare this with the standard equation,

s2+2ζw_ns+w_n²=0wωωnωnωnωn

we get:ωnωn

$\begin{array}{l}{\omega }_{\mathrm{n}}=\sqrt{\mathrm{K}}\mathrm{\&}2\zeta {\omega }_{\mathrm{n}}=10\\ 2\times .25\times \sqrt{\mathrm{K}}=10\Rightarrow \mathrm{K}=400\end{array}$

Concept:

Explanation:

Given an overdamped system consisting of poles at -4 and -6a/4

For overdamped case ζ > 1

When the poles lie at (-3 ± j4), these poles will present in the s plane as shown above underdamped system. (practical system).

So, for underdamped system 0 < ζ < 1

Hence, the damping ratio decreases when the system moves from an overdamped case to an underdamped case.