Resonant Peak MCQ Quiz - Objective Question with Answer for Resonant Peak - Download Free PDF

Concept:

DC gain = 5

$M_r=\frac{10}{\sqrt{3}}=\frac{1}{2\xi \sqrt{1-{\xi }^2}}$

100 [4ε² (1 – ε²)] = 3

400 ε²– 400 ε⁴ = 3

400x² – 400 x + 3 = 0  

x² – x + 0.0075 = 0

ξ² = 0.99244 , ξ² = 0.00755

ξ = 0.99 , ξ = 0.08689

If resonant peak > 1 then

 $\zeta \le \frac{1}{\sqrt{2}}=0.707$

∴ ξ = 0.08689

$5\sqrt{2}={\omega }_n\sqrt{1-2{\xi }^2}$

$5\sqrt{2}={\omega }_n\sqrt{1-2\times 0.00755}$

ωn = 7 rad/s

$s^2+2\xi {\omega }_{n}s+{\omega }_n^2$

= s² + 1.21 s + 49

Now ${\frac{K}{s^2+1.21s+49}|}_{s=0}=5$ 

K = 49 × 5 = 245

So the required transfer function = $\frac{245}{\left(s^2+1.21s+49\right)}$

The bode magnitude plot is shown

At resonant frequency, (ωp), magnitude is maximum and hence slope is 0.

DC gain = 5

$M_r=\frac{10}{\sqrt{3}}=\frac{1}{2\xi \sqrt{1-{\xi }^2}}$ 

100 [4ε² (1 – ε²)] = 3

400 ε² – 400 ε⁴ = 3

400 x² – 400 x + 3 = 0  

x² – x + 0.0075 = 0

ξ² = 0.99244 , ξ² = 0.00755

ξ = 0.99 , ξ = 0.08689

If resonant peak > 1 then

 $\zeta \le \frac{1}{\sqrt{2}}=0.707$

∴ ξ = 0.08689

$5\sqrt{2}={\omega }_n\sqrt{1-2{\xi }^2}$ 

$5\sqrt{2}={\omega }_n\sqrt{1-2\times 0.00755}$ 

ω_n = 7 rad/s

$s^2+2\xi {\omega }_{n}s+{\omega }_n^2$ 

= s² + 1.21 s + 49

Now ${\frac{K}{s^2+1.21s+49}|}_{s=0}=5$ 

K = 49 × 5 = 245

Concept:

The closed-loop transfer function for a standard 2^nd order system is defined as

$CLTF=\frac{{\omega }_n^2}{s^2+2\xi {\omega }_{n}s+{\omega }_n^2}$

The Resonant Peak is one of the frequency domain specifications for a 2^nd order system, which is defined at the peak (maximum) value of the magnitude of the CLTF and is calculated at resonant frequency ω_r, given by

${\omega }_r={\omega }_n\sqrt{1-2{\xi }^2}$

Where, ω_n = Natural frequency and

ξ = Damping ratio

Calculation:

The given feedback is a unity feedback system,

$CLTF=\frac{G\left(s\right)}{1+G\left(s\right)}=\frac{\frac{16}{s(s+4)}}{1+\frac{16}{s(s+4)}}$

$=\frac{16}{s(s+4)+16}=\frac{16}{s^2+4s+16}$

The characteristic equation is given by;

s² + 4s + 16 = 0

Comparing this with the standard 2^nd order characteristic equation $s^2+2\xi {\omega }_{n}s+{\omega }_n^2=0$, we see that,

${\omega }_n^2=16$.

So, ω_n = 4

And, 2ξω_n = 4

So, $\xi =0.5=\frac{1}{2}$

So, the resonant frequency is given by;

${\omega }_r={\omega }_n\sqrt{1-2{\xi }^2}$

${\omega }_r=4\sqrt{1-2{\left(\frac{1}{2}\right)}^2}$

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

$\left|T\left(j\omega \right)\right|=\left|\frac{k{\omega }_n^2}{{\omega }_n^2+2\zeta {\omega }_n\left(j\omega \right)-{\omega }^2}\right|$

${\left|T\left(j\omega \right)\right|}^2=\frac{k^2{\omega }_n^4}{{\left({\omega }_n^2-{\omega }^2\right)}^2+4{\zeta }^2{\omega }_n^2{\omega }^2}$

From fig. |T(j0)|= 1

${\left|T\left(j0\right)\right|}^2=\frac{k^2{\omega }_n^4}{{\omega }_n^4}\Rightarrow 1=k^2$

k = 1

$M_r=\frac{1}{2\zeta \sqrt{1-{\zeta }^2}}=2.5$

Solving for ζ gives

= 0.2

We know that the characteristic equation for a unity feedback system is written as:

1 + G(s) = 0

$\Rightarrow 1+\frac{\mathrm{K}}{\mathrm{s}\left(\mathrm{s}+2\right)}=0$ 

⇒ s² + 2s + k = 0

Comparing this with the standard 2^nd order characteristic equation, we get:

ω_n² = k

ω_n = √k

Also 2ξω_n = 2

$\xi =\frac{1}{\sqrt{\mathrm{k}}}$

${\mathrm{M}}_{\mathrm{r}}=\frac{1}{2\xi \sqrt{1-{\xi }^2}}=2$

$\xi \sqrt{1-{\xi }^2}=\frac{1}{4}$

${\xi }^2=.933,.067$

For resonance peak ξ < 0.707.

So, we can neglect 0.933

${\xi }^2=.067=\frac{1}{\mathrm{k}}$ 

Concept:

DC gain = 5

$M_r=\frac{10}{\sqrt{3}}=\frac{1}{2\xi \sqrt{1-{\xi }^2}}$

100 [4ε² (1 – ε²)] = 3

400 ε²– 400 ε⁴ = 3

400x² – 400 x + 3 = 0  

x² – x + 0.0075 = 0

ξ² = 0.99244 , ξ² = 0.00755

ξ = 0.99 , ξ = 0.08689

If resonant peak > 1 then

 $\zeta \le \frac{1}{\sqrt{2}}=0.707$

∴ ξ = 0.08689

$5\sqrt{2}={\omega }_n\sqrt{1-2{\xi }^2}$

$5\sqrt{2}={\omega }_n\sqrt{1-2\times 0.00755}$

ωn = 7 rad/s

$s^2+2\xi {\omega }_{n}s+{\omega }_n^2$

= s² + 1.21 s + 49

Now ${\frac{K}{s^2+1.21s+49}|}_{s=0}=5$ 

K = 49 × 5 = 245

So the required transfer function = $\frac{245}{\left(s^2+1.21s+49\right)}$

DC gain = 5

$M_r=\frac{10}{\sqrt{3}}=\frac{1}{2\xi \sqrt{1-{\xi }^2}}$ 

100 [4ε² (1 – ε²)] = 3

400 ε² – 400 ε⁴ = 3

400 x² – 400 x + 3 = 0  

x² – x + 0.0075 = 0

ξ² = 0.99244 , ξ² = 0.00755

ξ = 0.99 , ξ = 0.08689

If resonant peak > 1 then

 $\zeta \le \frac{1}{\sqrt{2}}=0.707$

∴ ξ = 0.08689

$5\sqrt{2}={\omega }_n\sqrt{1-2{\xi }^2}$ 

$5\sqrt{2}={\omega }_n\sqrt{1-2\times 0.00755}$ 

ω_n = 7 rad/s

$s^2+2\xi {\omega }_{n}s+{\omega }_n^2$ 

= s² + 1.21 s + 49

Now ${\frac{K}{s^2+1.21s+49}|}_{s=0}=5$ 

K = 49 × 5 = 245

Concept:

The closed-loop transfer function for a standard 2^nd order system is defined as

$CLTF=\frac{{\omega }_n^2}{s^2+2\xi {\omega }_{n}s+{\omega }_n^2}$

The Resonant Peak is one of the frequency domain specifications for a 2^nd order system, which is defined at the peak (maximum) value of the magnitude of the CLTF and is calculated at resonant frequency ω_r, given by

${\omega }_r={\omega }_n\sqrt{1-2{\xi }^2}$

Where, ω_n = Natural frequency and

ξ = Damping ratio

Calculation:

The given feedback is a unity feedback system,

$CLTF=\frac{G\left(s\right)}{1+G\left(s\right)}=\frac{\frac{16}{s(s+4)}}{1+\frac{16}{s(s+4)}}$

$=\frac{16}{s(s+4)+16}=\frac{16}{s^2+4s+16}$

The characteristic equation is given by;

s² + 4s + 16 = 0

Comparing this with the standard 2^nd order characteristic equation $s^2+2\xi {\omega }_{n}s+{\omega }_n^2=0$, we see that,

${\omega }_n^2=16$.

So, ω_n = 4

And, 2ξω_n = 4

So, $\xi =0.5=\frac{1}{2}$

So, the resonant frequency is given by;

${\omega }_r={\omega }_n\sqrt{1-2{\xi }^2}$

${\omega }_r=4\sqrt{1-2{\left(\frac{1}{2}\right)}^2}$

We know that the characteristic equation for a unity feedback system is written as:

1 + G(s) = 0

$\Rightarrow 1+\frac{\mathrm{K}}{\mathrm{s}\left(\mathrm{s}+2\right)}=0$ 

⇒ s² + 2s + k = 0

Comparing this with the standard 2^nd order characteristic equation, we get:

ω_n² = k

ω_n = √k

Also 2ξω_n = 2

$\xi =\frac{1}{\sqrt{\mathrm{k}}}$

${\mathrm{M}}_{\mathrm{r}}=\frac{1}{2\xi \sqrt{1-{\xi }^2}}=2$

$\xi \sqrt{1-{\xi }^2}=\frac{1}{4}$

${\xi }^2=.933,.067$

For resonance peak ξ < 0.707.

So, we can neglect 0.933

${\xi }^2=.067=\frac{1}{\mathrm{k}}$ 

For $\zeta >\frac{1}{\sqrt{2}}(=0.707)$, there is no resonant peak in the frequency response.

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

$\left|T\left(j\omega \right)\right|=\left|\frac{k{\omega }_n^2}{{\omega }_n^2+2\zeta {\omega }_n\left(j\omega \right)-{\omega }^2}\right|$

${\left|T\left(j\omega \right)\right|}^2=\frac{k^2{\omega }_n^4}{{\left({\omega }_n^2-{\omega }^2\right)}^2+4{\zeta }^2{\omega }_n^2{\omega }^2}$

From fig. |T(j0)|= 1

${\left|T\left(j0\right)\right|}^2=\frac{k^2{\omega }_n^4}{{\omega }_n^4}\Rightarrow 1=k^2$

k = 1

$M_r=\frac{1}{2\zeta \sqrt{1-{\zeta }^2}}=2.5$

Solving for ζ gives

= 0.2

Concept:

DC gain = 5

$M_r=\frac{10}{\sqrt{3}}=\frac{1}{2\xi \sqrt{1-{\xi }^2}}$

100 [4ε² (1 – ε²)] = 3

400 ε²– 400 ε⁴ = 3

400x² – 400 x + 3 = 0  

x² – x + 0.0075 = 0

ξ² = 0.99244 , ξ² = 0.00755

ξ = 0.99 , ξ = 0.08689

If resonant peak > 1 then

 $\zeta \le \frac{1}{\sqrt{2}}=0.707$

∴ ξ = 0.08689

$5\sqrt{2}={\omega }_n\sqrt{1-2{\xi }^2}$

$5\sqrt{2}={\omega }_n\sqrt{1-2\times 0.00755}$

ωn = 7 rad/s

$s^2+2\xi {\omega }_{n}s+{\omega }_n^2$

= s² + 1.21 s + 49

Now ${\frac{K}{s^2+1.21s+49}|}_{s=0}=5$ 

K = 49 × 5 = 245

So the required transfer function = $\frac{245}{\left(s^2+1.21s+49\right)}$

The bode magnitude plot is shown

At resonant frequency, (ωp), magnitude is maximum and hence slope is 0.

Maximum overshoot is M_p = 0.25

$e^{\frac{-\xi \pi }{\sqrt{1-{\xi }^2}}}=0.25$

$\xi =0.4037$

Resonant peak $Mr=\frac{1}{2\xi \sqrt{1-{\xi }^2}}=1.3537$