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

\(2\xi {{\rm{\omega }}_n} = 15 \Rightarrow \xi = \frac{15}{{2\sqrt {32} }} > 1\)

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}

\(\% \;overshoot = {e^{ - \frac{{\zeta \pi }}{{\sqrt {1 - {\zeta ^2}} }}}} \times 100\) 

Calculation:

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

Comparing the denominator ζ = 0

\(\% \;overshoot = {e^{ - \frac{{\zeta \pi }}{{\sqrt {1 - {\zeta ^2}} }}}} \times 100 = 100\) 

\(\begin{array}{l} {\rm{G}}\left( {\rm{s}} \right) = \frac{{\rm{K}}}{{\begin{array}{*{20}{c}} {s\left( {{\rm{s}} + 10} \right)} \end{array}}}\\ {\rm{CLTF}} = \frac{{\rm{K}}}{{{\rm{s^2}} + 10{\rm{s}} + {\rm{K}}}} \end{array}\)

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

\(\begin{array}{l} {{\rm{\omega }}_{\rm{n}}} = \sqrt {\rm{K}} {\rm{\;}}\& {\rm{\;\;}}2{\rm{\zeta }}{{\rm{\omega }}_{\rm{n}}}{\rm{\;}} = {\rm{\;}}10\\ 2 \times .25 \times \sqrt {\rm{K}} = 10 \Rightarrow {\rm{K}} = 400 \end{array}\)

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

\(2\xi {{\rm{\omega }}_n} = 15 \Rightarrow \xi = \frac{15}{{2\sqrt {32} }} > 1\)

So, it is an overdamped system.

\(\begin{array}{l} {\rm{G}}\left( {\rm{s}} \right) = \frac{{\rm{K}}}{{\begin{array}{*{20}{c}} {s\left( {{\rm{s}} + 10} \right)} \end{array}}}\\ {\rm{CLTF}} = \frac{{\rm{K}}}{{{\rm{s^2}} + 10{\rm{s}} + {\rm{K}}}} \end{array}\)

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

\(\begin{array}{l} {{\rm{\omega }}_{\rm{n}}} = \sqrt {\rm{K}} {\rm{\;}}\& {\rm{\;\;}}2{\rm{\zeta }}{{\rm{\omega }}_{\rm{n}}}{\rm{\;}} = {\rm{\;}}10\\ 2 \times .25 \times \sqrt {\rm{K}} = 10 \Rightarrow {\rm{K}} = 400 \end{array}\)

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}

\(\% \;overshoot = {e^{ - \frac{{\zeta \pi }}{{\sqrt {1 - {\zeta ^2}} }}}} \times 100\) 

Calculation:

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

Comparing the denominator ζ = 0

\(\% \;overshoot = {e^{ - \frac{{\zeta \pi }}{{\sqrt {1 - {\zeta ^2}} }}}} \times 100 = 100\) 

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

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

So, the system is overdamped.

2) s² + 24s + 225 = 0

ω²_n = 225 ⇒ ω_n = 15

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

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
\({\rm{\zeta }} = \frac{R}{2}\sqrt {\frac{C}{L}} \)
For no oscillations:

\({\rm{\zeta }} \ge 1 \to {\rm{\zeta }} = \frac{R}{2}\sqrt {\frac{C}{L}} \ge 1\)
\(R \ge 2\sqrt {\frac{L}{C}} \)

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

\(2\xi {{\rm{\omega }}_n} = 15 \Rightarrow \xi = \frac{15}{{2\sqrt {32} }} > 1\)

So, it is an overdamped system.

\(\begin{array}{l} {\rm{G}}\left( {\rm{s}} \right) = \frac{{\rm{K}}}{{\begin{array}{*{20}{c}} {s\left( {{\rm{s}} + 10} \right)} \end{array}}}\\ {\rm{CLTF}} = \frac{{\rm{K}}}{{{\rm{s^2}} + 10{\rm{s}} + {\rm{K}}}} \end{array}\)

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

\(\begin{array}{l} {{\rm{\omega }}_{\rm{n}}} = \sqrt {\rm{K}} {\rm{\;}}\& {\rm{\;\;}}2{\rm{\zeta }}{{\rm{\omega }}_{\rm{n}}}{\rm{\;}} = {\rm{\;}}10\\ 2 \times .25 \times \sqrt {\rm{K}} = 10 \Rightarrow {\rm{K}} = 400 \end{array}\)

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.