Control system MCQ Questions and Answers with explanation -2021

Answer.4. None of the above

Explanation

The characteristic equation of the system is:

s⁴ + K₁s³ + s² + K₂ s + 1 = 0

For the characteristic equation, we form the Routh’s array method as:

For the system having two imaginary poles, the required condition will be:

[katex]{K_1}^2 – {K_1}{K_2} + {K_2}^2[/katex]

Since the above equation has no real roots, therefore there is no relationship between K₁ and K₂ that will have just two jω-poles.

Answer.2. 3

Explanation

The characteristic equation of the system is given by the signal flow graph (SFG) , i.e.

Δ = 1 – (L₁ + L₂ + L₃) + L₁₂

Where L₁, L_2, and L₃ are the individual loop gains and L₁₂ is the gain of a pair of two-nontouching loops. For the given SFG of the system, we have

L₁ = 1/s

L₂ = −3/s

L₂ = −k/s²

L1₂ = −3/s²

The determinant for the SFG can be calculated as:

[katex]\begin{array}{l} \Delta = 1 – \left( {\dfrac{1}{s} – \dfrac{3}{s} – \dfrac{K}{{{s^2}}}} \right) + \left( {\dfrac{{ – 3}}{{{s^2}}}} \right)\\ \\ = 1 + \dfrac{2}{s} + \dfrac{{K – 3}}{{{s^2}}}\\ \\ q\left( s \right) = 1 + \dfrac{2}{s} + \dfrac{{K – 3}}{{{s^2}}} = 0 \end{array}[/katex]

According to the Routh array characteristics Methods

Thus, the required condition for the stability of the system is: A block diagram is shown below. The transfer function for this system is

K – 3 > 0

Or K > 3

Answer.4. [katex]\dfrac{{2s\left( {2s + 1} \right)}}{{4{s^2} + 13s + 5}}[/katex]

Explanation:-

The given block diagram is shown below:

Here, the two branches having gains 1/s and 2 are in parallel hence they will be added.

Now, we change the take-off position of the block of 2s

Again, both feedback branches are parallel, therefore adding the gain we get:

Thus, using feedback formula, we obtain the overall transfer function as:

[katex]\begin{array}{l} T\left( s \right) = \dfrac{{G\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\\ \\ = \dfrac{{\dfrac{{\left( {2s + 1} \right)2s}}{s}}}{{1 + \dfrac{{\left( {2s + 1} \right)2s}}{s} \times \dfrac{{\left( {5 + 2s} \right)}}{{2s}}}}\\ \\ = \dfrac{{2s\left( {2s + 1} \right)}}{{s + \left( {2s + 1} \right)\left( {5 + 2s} \right)}}\\ \\ = \dfrac{{2s\left( {2s + 1} \right)}}{{4{s^2} + 13s + 5}} \end{array}[/katex]

Answer.2. 4th order and unstable

Explanation:-

From the block diagram,

[katex]\begin{array}{l} G\left( s \right) = \frac{1}{{{s^2}\left( {s + 1} \right)}}\\ \\ H\left( s \right) = \frac{{20}}{{\left( {s + 20} \right)}} \end{array}[/katex]

As the given feedback is negative, the transfer function of the closed-loop system is

[katex]\begin{array}{l} \frac{{Y\left( s \right)}}{{R\left( s \right)}} = \frac{{G\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\\ \\ = \frac{{\frac{1}{{{s^2}\left( {s + 1} \right)}}}}{{1 + \frac{1}{{{s^2}\left( {s + 1} \right)}}\frac{{20}}{{\left( {s + 20} \right)}}}}\\ \\ = \frac{{s + 20}}{{{s^2}\left( {s + 1} \right)\left( {s + 20} \right) + 20}}\\ \\ = \frac{{s + 20}}{{{s^4} + 21{s^3} + 20{s^2} + 20}} \end{array}[/katex]

In the above transfer function denominatior has highest value of 4. Therefore, the order of the system is 4.

The coefficient of ‘s’ term is zero in the characteristic equation (denominator of above transfer function). Therefore, the system is unstable.

Answer.4. a = 1, b = 2

Explanation

General form of lead compensator is given as

Gc(s) = (s + z)/(s + p)

Here

z = 1/T, P = 1/αT

For lead compensator

α = Z/P < 1

The pole-zero location of lead compensator in s-plane is given as

For phase lead compensator, a < b

Option (4) is correct i.e a = 1, b = 2

Answer.3. ω = 3 rad/sec

sinωt → G(s) → Y(s)=|G(s)| ⋅ sin(ωt+∠G(s))

The output will be zero when

|G(s)| = 0

Put s = jω

[katex]\left| {\frac{{\left( { – {\omega ^2} + 9} \right)\left( {j\omega + 2} \right)}}{{\left( {j\omega + 1} \right)\left( {j\omega + 3} \right)\left( {j\omega + 4} \right)}}} \right| = 0[/katex]

For the value ω = 3   , |G(jω)| = 0

Answer.1. K > 1

Explanation

Given that characteristic equation is,

s³ + Ks² + (K + 2)s + 3 = 0

For stable system,

K > 0, K (K + 2) – 3 > 0

⇒ K > 0, K² + 2K – 3 > 0

⇒ K > 0, (K + 3) (K – 1) > 0

⇒ K > 0, K > -3, K > 1

⇒ K > 1

Answer.1. A ⇒ 1, B ⇒ 2, C ⇒ 3

Explanation:-

The second-order system is given by the equation

[katex]\frac{{\omega _n^2}}{{{s^2} + 2\xi {\omega _n}s + \omega _n^2}}[/katex]

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.

[katex]A = \frac{{15}}{{{s^2} + 5s + 15}}[/katex]

By comparing the second order transfer function,

ω_n² = 15 ⇒ ω_n = √15

[katex]2\xi {\omega _n} = 5 \Rightarrow \xi = \frac{5}{{2\sqrt {15} }} < 1[/katex]

Hence it is an underdamped equation

[katex]B=\frac{{25}}{{{s^2} + 10s + 25}}[/katex]

ω_n² = 25 ⇒ ω_n = 5

2 ξ ω_n = 10 ⇒ ξ = 1

So, it si critically damped system.

[katex]C = \frac{{35}}{{{s^2} + 18s + 45}}[/katex]

ω_n² = 35 ⇒ ω_n = √45

[katex]2\xi {\omega _n} = 18 \Rightarrow \xi = \frac{9}{{\sqrt {35} }} > 1[/katex]

So, it is overdamped system.

Answer.4. 2, 0.75

Explanation:-

Characteristic equation:

s³ + as² + (2 + k) s + (k + 1) = 0

[katex]\frac{{a\left( {k + 2} \right) – \left( {k + 1} \right)}}{a} = 0[/katex]

a = (K + 1)/(K + 2)—— (1)

For oscillator

as² + k + 1 = 0

s² = ω² = -4

-4a + k + 1 = 0     – – – – (2)

Solving equation (1) & (2) we get

a = 0.75 & k = 2

Answer. [katex]\frac{{ – 1.58}}{{{s^2} – 2.59s + 1.12}}[/katex]

Explanation-

[katex]T\left( s \right) = \frac{{k\:\omega _n^2}}{{{s^2} + 2\xi {\omega _n}s + \omega _n^2}}[/katex]

Under damped system ξ < 1

[katex]\xi = \frac{{1.91}}{{2\sqrt {1.91} }} = 0.69[/katex]

Only option 2 is correct

Answer.4. All of the above

Explanation:-

Lead compensator increases relative stability by increasing phase margin.

For a LEAD COMPENSATOR, bandwidth INCREASES as zero is added to the transfer function, ie. s in the numerator, as an effect, the damping factor increases thereby increasing the bandwidth. [bandwidth = 2x(damping factor)]

Due to phase-lead controller (compensator) amplification at higher frequencies, it increases the bandwidth of the compensated system. The phase-lead controllers are used to improve the gain, resonant frequency, and phase stability margins and to increase the system bandwidth (decrease the system response rise time).