Frequency Domain Specifications MCQ Quiz in मल्याळम - Objective Question with Answer for Frequency Domain Specifications - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 15, 2025
Latest Frequency Domain Specifications MCQ Objective Questions
Top Frequency Domain Specifications MCQ Objective Questions
Frequency Domain Specifications Question 1:
Match the following
| 1) Resonant frequency | P) \({\omega _n}\sqrt {1 - 2{\xi ^2} + \sqrt {2 - 4{\xi ^2} + 4{\xi ^4}} } \) |
| 2) Bandwidth | Q) \({\omega _n}\sqrt {1 - 2{\xi ^2}}\) |
| 3) Resonant peak | R) \({\omega _n}\sqrt {1 - {\xi ^2}}\) |
| 4) Damping ratio | S) \(\frac{1}{{2\xi }}\sqrt {1 - {\xi ^2}}\) |
| T) \(\frac{1}{{2Q}}\) |
Answer (Detailed Solution Below)
\(\rm 1) → Q, 2) → P, 3) →S, 4) →T\)
Frequency Domain Specifications Question 1 Detailed Solution
Resonant frequency, \(\rm \omega _r={\omega _n}\sqrt {1 - 2{\xi ^2}}\)
Damping Ratio \(\rm \xi =\frac{1}{{2Q}}\)
Resonant Peak \(\rm M_r = \frac{1}{{2\xi }}\sqrt {1 - {\xi ^2}}\)
Bandwidth \(\rm BW={\omega _n}\sqrt {1 - 2{\xi ^2} + \sqrt {2 - 4{\xi ^2} + 4{\xi ^4}} } \)
Frequency Domain Specifications Question 2:
The unit step response of a second-order system is shown below
The resonant peak of the system is:
Answer (Detailed Solution Below)
Frequency Domain Specifications Question 2 Detailed Solution
Given:
Maximum overshoot = 0.37
\(e\frac{{ - \zeta \pi }}{{\sqrt {1 - {\zeta ^2}} }} = 0.37\)
\(\zeta = 0.3\)
Resonant Peak:
\( {M_r} = \frac{1}{{2\zeta \sqrt {1 - {\zeta ^2}} }}\)
Putting the value of ζ, we get
Mr = 1.747
Frequency Domain Specifications Question 3:
Match the following
1) Resonant frequency P) \({\omega _n}\sqrt {1 - 2{\xi ^2} + \sqrt {2 - 4{\xi ^2} + 4{\xi ^4}} } \)
2) Bandwidth Q) \({\omega _n}\sqrt {1 - 2{\xi ^2}}\)
3) Resonant peak R) \({\omega _n}\sqrt {1 - {\xi ^2}}\)
4) Damping ratio S) \(\frac{1}{{2\xi }}\sqrt {1 - {\xi ^2}}\)
T) \(\frac{1}{{2Q}}\)
Answer (Detailed Solution Below)
Frequency Domain Specifications Question 3 Detailed Solution
Resonant frequency, \(\rm \omega _r={\omega _n}\sqrt {1 - 2{\xi ^2}}\)
Damping Ratio \(\rm \xi =\frac{1}{{2Q}}\)
Resonant Peak \(\rm M_r = \frac{1}{{2\xi }}\sqrt {1 - {\xi ^2}}\)
Bandwidth \(\rm BW={\omega _n}\sqrt {1 - 2{\xi ^2} + \sqrt {2 - 4{\xi ^2} + 4{\xi ^4}} } \)
Frequency Domain Specifications Question 4:
Range of ϕr with ξ variation is –
(ϕr = phase at resonant freq.)
Answer (Detailed Solution Below)
Frequency Domain Specifications Question 4 Detailed Solution
Range ξ is \(0<ξ<1/√2\)
\({\left. {{\phi _r}} \right|_{\xi = 0}} = - {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 - 2{\xi ^2}} }}{\xi }} \right] = - 90^\circ \)
\({\left. {{\phi _r}} \right|_{\xi = \frac{1}{{\sqrt 2 }}}} = - {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 - 2{\xi ^2}} }}{\xi }} \right] = 0^\circ \)
Frequency Domain Specifications Question 5:
Frequency response of a second order system is given below. The value of K is ______
\(G\left( s \right) = \frac{{K\left( {36} \right)}}{{{s^2} + 6s + 36}}\)
Answer (Detailed Solution Below) 10
Frequency Domain Specifications Question 5 Detailed Solution
\(G\left( s \right) = \frac{{K\left( {36} \right)}}{{{s^2} + 6s + 36}}\)
characteristic equation:
s2 + 6s + 36 = 0
ω2n = 36 ⇒ ωn = 6 rad/sec
2ξωn = 6 ⇒ ξ = 0.5
From the given response,
\({M_r} = \frac{{20}}{{\sqrt 3 }}\)
\({M_r} = \frac{1}{{2\xi \sqrt {1 - {\xi ^2}} }}\;for\;0 \le \xi \le \frac{1}{{\sqrt 2 }}\)
\(= 1\;for\;\xi \ge \frac{1}{{\sqrt 2 }}\)
\(\Rightarrow \frac{{20}}{{\sqrt 3 }} = \frac{K}{{2\left( {0.5} \right)\sqrt {1 - \left( {\frac{1}{4}} \right)} }}\)
\( \Rightarrow \frac{K}{{\sqrt {\frac{3}{4}} }} = \frac{{20}}{{\sqrt 3 }} \Rightarrow K = 10\)Frequency Domain Specifications Question 6:
Range of ϕr with ξ variation is –
(ϕr = phase at resonant freq.)
Answer (Detailed Solution Below)
Frequency Domain Specifications Question 6 Detailed Solution
Range ξ is \(0<ξ<1/√2\)
\({\left. {{\phi _r}} \right|_{\xi = 0}} = - {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 - 2{\xi ^2}} }}{\xi }} \right] = - 90^\circ \)
\({\left. {{\phi _r}} \right|_{\xi = \frac{1}{{\sqrt 2 }}}} = - {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 - 2{\xi ^2}} }}{\xi }} \right] = 0^\circ \)
Frequency Domain Specifications Question 7:
Phase at resonant frequency if ξ = 0.4 is
Answer (Detailed Solution Below)
Frequency Domain Specifications Question 7 Detailed Solution
\(\phi = {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 - 2{\xi ^2}} }}{\xi }} \right]\)
\(=64.12°\)
since control systems are Low pass filters that give more negative phase. So Ø = -64.12 degrees
Note :
Cos-1ξ cannot be used here because this is resonance condition
Frequency Domain Specifications Question 8:
If damping ratio ξ = 0.8 then resonant peak is
Answer (Detailed Solution Below)
Frequency Domain Specifications Question 8 Detailed Solution
For \(\zeta\ > \frac{1}{\sqrt 2}(=0.707)\), there is no resonant peak in the frequency response.
Frequency Domain Specifications Question 9:
If \(\omega_n\ = \omega_b\)then damping ratio is – (\(\omega_b\) = bandwidth, \(\omega_n\) = natural freq.)
Answer (Detailed Solution Below)
Frequency Domain Specifications Question 9 Detailed Solution
\({\omega _b} = {\omega _n}\sqrt {\left( {1 - 2{\xi ^2}} \right) + \sqrt {4{\xi ^4} - 4{\xi ^2} + 2} } \)
For \(\xi = \frac{1}{{\sqrt 2 }}\) we will get \({\omega _n} = {\omega _b}\)
Frequency Domain Specifications Question 10:
Consider the following systems:
System 1: \(G\left( s \right) = \frac{1}{{2s + 1}}\)
System 2: \(G\left( s \right) = \frac{1}{{5s + 1}}\)
The true statement regarding the system isAnswer (Detailed Solution Below)
Frequency Domain Specifications Question 10 Detailed Solution
System 1: \(G\left( s \right) = \frac{1}{{2s + 1}}\)
Time constant (τ1) = 2 sec.
System 2: \(G\left( s \right) = \frac{1}{{5s + 1}}\)
Time constant (τ2) = 5 sec
Bandwidth is inversely proportional to time constant.
So, bandwidth of system 1 is greater than the bandwidth of system 2.