Control Systems Test 1

The characteristic equation of the system is:

s⁴ + K₁s³ + s² + K₂ s + 1 = 0

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

For the system having two imaginary poles, the required condition is:

\(K_1^2 - {K_1}{K_2} + K_2^2 = 0\)  

But the above equation has no real roots, so there is no relationship between K₁ and K₂ that will yield just two jω-poles. 

For high pass filter zero is closer to pole.

\(T.F = \frac{{s + a}}{{s + b}}\)

\(\left| a \right| < \left| b \right|\)

Let a = 1

b = 2

\(\phi = {\tan ^{ - 1}}\left( \omega \right) - {\tan ^{ - 1}}\left( {\frac{\omega }{2}} \right)\)

\(M = \left| {GH} \right| = \sqrt {\frac{{1 + {\omega ^2}}}{{4 + {\omega ^2}}}}\)

At ω = 0

\(M = \frac{1}{2} = 0.5\)

At ω → ∞

m = 1

At ω = 2

\(\phi = {\tan ^{ - 1}}\left( 2 \right) - {\tan ^{ - 1}}\left( 1 \right) = + ve\)

So the correct polar plot will be:

A closed loop system is:

• More accurate: It uses feedback to correct errors, improving the precision of the system's output.
• More stable: Feedback helps maintain desired performance despite disturbances or changes in system parameters.
• Comparison to open loop: Open loop systems lack feedback, making them less capable of correcting deviations and maintaining stability

Thus, option A is correct: A closed loop system is more accurate and more stable than an open loop system.

Linear system obeys both reciprocity theorem and principle of maximum power transfer.

The standard transfer function of a PI controller is

\(T\left( s \right) = {K_p} + \frac{{{K_I}}}{s} = {K_p}\;\left( {1 + \frac{1}{{{T_I}s}}} \right)\)

Where \({T_I} = \frac{{{K_p}}}{{{k_I}}}\) is known as integral or reset time.

Proportional band \(\left( {PB} \right) = \frac{{100}}{{{K_p}}}\)

The transfer function of PI controller is

\(T\left( s \right) = 2.5\;\left( {1 + \frac{1}{{0.25s}}} \right)\)

\(= 2.5\;\left( {1 + \frac{4}{s}} \right)\)

A step function is commonly used as a input signal in a control system because:

• If we go through first order system, here we calculate steady-state error by providing input function then we got zero steady-state error.
• Actually in step function, output follows input always that means steady error is minimum.

Step Function:

Concept

The steady-state value of the output can be calculated using final value theorem

\(\mathop {{\rm{lt}}}\limits_{s \to 0} s\left[ {y\left( s \right)} \right]\)

\(x\left( t \right)\mathop \to \limits^{L.T} s\;X\left( s \right)\)

\(u\left( t \right)\mathop \to \limits^{L.T} \frac{1}{s}\)

Applications:

\({s^2}\;Y\left( s \right) + \;16\;sY\left( s \right)\; + \;15\;Y\left( s \right)\; = 1 + \frac{5}{s}\)

\(\left( {{s^2} + 16s + 15} \right)Y\left( s \right) = 1 + \frac{5}{s}\)

\(Y\left( S \right) = \frac{{s + 5}}{{s\left( {{s^2}+16s + 15} \right)}}\)

The steady-state value

\(\mathop {\lim }\limits_{S \to 0} \frac{{s\left( {s + 5} \right)}}{{s\left( {{s^2} + 16s + 15} \right)}} = \frac{1}{3}\)

• The main difference between an open-loop system and a closed-loop system is that the closed-loop system has the ability to self-correct while the open-loop system doesn't.
• Consequently, closed-loop systems are often called feedback control systems while open-loop systems are also known as non-feedback controls.

• A closed-loop system automatically corrects errors and disturbances by using feedback. so C is correct 
• It continuously compares the output with the desired value and adjusts accordingly.
• Open-loop systems do not correct errors since they lack feedback.
• While closed-loop systems reduce steady-state error, they may not always eliminate it completely (hence D is incorrect).
• They are more complex and costly than open-loop systems, making A incorrect.
• Closed-loop systems require feedback, so B is incorrect.

Ramp function occurring at t = 0 is denoted by r(t) = tu(t).

Ramp function occurring at t - a will be denoted: 

- In open loop control systems, the control action is independent of the desired output signal.
- The system operates without feedback, meaning it doesn't compare the output with the desired value.
- Control actions are predefined and do not adjust based on output changes.
- This can lead to inaccuracies if the system or external conditions change.
- Open loop systems are typically simpler but less accurate compared to closed loop systems, where feedback adjusts the control action.

Reference Input in Closed-Loop Control Systems
- In a closed-loop control system, the reference input is a standard signal that is used for comparison with the output of the system.
- It represents the desired or target value that the system is trying to achieve.
- The reference input is typically a known signal that is used to evaluate the performance of the system.
- By comparing the output of the system to the reference input, the controller can make adjustments to ensure that the system operates as desired.
- The reference input can take various forms, such as a step input, ramp input, sinusoidal input, etc.
- The controller uses the error between the reference input and the actual output of the system to adjust the control signals and minimize the error.
- The reference input plays a crucial role in closed-loop control systems by providing a benchmark for performance evaluation and adjustment of control signals.

Actuating Signal
- The actuating signal is the difference between the output response and the reference signal in a control system.
- It is used to drive the system towards the desired setpoint by adjusting the input signals.
- The actuating signal is crucial for maintaining stability and accuracy in control systems.
- By continuously comparing the output response with the reference signal, the actuating signal ensures that the system stays on track and responds appropriately to any changes or disturbances.
- Overall, the actuating signal plays a key role in ensuring that the control system operates effectively and achieves its desired objectives.

The laplace transform of unity function i.e.

Explanation of a robust control system:

- Low sensitivities: A robust control system should have low sensitivities to uncertainties and disturbances. This means that even if there are slight variations or disturbances in the system, the control system should be able to maintain stability and performance without significant changes. Low sensitivities ensure that the system can effectively handle variations in parameters without compromising its operation.

- Stable over a wide range of parameter variation: Another important aspect of a robust control system is its ability to remain stable over a wide range of parameter variations. This means that the system should be able to maintain its stability and performance even when there are significant changes in the parameters or operating conditions. A robust control system should be able to adapt to different conditions and still perform optimally.

- Both (a) and (b): Therefore, a robust control system is one that not only has low sensitivities to uncertainties and disturbances but also remains stable over a wide range of parameter variations. By combining these two characteristics, a robust control system can effectively handle uncertainties and variations in the system, ensuring reliable and consistent performance.

The state-space representation of the system is

\(\left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}}\\ {{{\dot x}_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - \frac{1}{{{T_1}}}}&0\\ 0&{ - \frac{1}{{{T_2}}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {\frac{1}{{{T_1}}}}\\ {\frac{1}{{{T_2}}}} \end{array}} \right]\)

Comparing it to the standard state-space model, we have the matrices:

\(A = \left[ {\begin{array}{*{20}{c}} { - \frac{1}{{{T_1}}}}&0\\ 0&{ - \frac{1}{{{T_2}}}} \end{array}} \right]\;and\;B = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{{T_1}}}}\\ {\frac{1}{{{T_2}}}} \end{array}} \right]\)

\(AB = \left[ {\begin{array}{*{20}{c}} { - \frac{1}{{{T_1}}}}&0\\ 0&{ - \frac{1}{{{T_2}}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{1}{{{T_1}}}}\\ {\frac{1}{{{T_2}}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - \frac{1}{{T_1^2}}}\\ { - \frac{1}{{T_2^2}}} \end{array}} \right]\)

So, the controllability matrix is:

\({C_M} = \left[ {B\;AB} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{{T_1}}}}&{ - \frac{1}{{T_1^2}}}\\ {\frac{1}{{{T_2}}}}&{ - \frac{1}{{T_2^2}}} \end{array}} \right]\)

For complete state controllability, the matrix C_M must be of rank r = 2, i.e.

|C_M| ≠ 0

\(\left| {\begin{array}{*{20}{c}} {\frac{1}{{{T_1}}}}&{ - \frac{1}{{T_1^2}}}\\ {\frac{1}{{{T_2}}}}&{ - \frac{1}{{T_2^2}}} \end{array}} \right| \ne 0\)

\(- \frac{1}{{{T_1}T_2^2}} + \frac{1}{{{T_2}T_1^2}} \ne 0\)

\(\frac{1}{{{T_1}{T_2}}}\left( {\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}}} \right) \ne 0\)

\(\frac{1}{{{T_1}{T_2}}}\left( {\frac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}} \right) \ne 0\)

\(\frac{{\left( {{T_2} - {T_1}} \right)}}{{{{\left( {{T_1}{T_2}} \right)}^2}}} \ne 0\)

(T₂ – T₁) ≠ 0 or T₁ ≠ T₂

Therefore, the system is completely state controllable for T₁ ≠ T₂.

Radar tracking systems are generally closed-loop systems because they use feedback to continuously adjust and track an object. However, if this refers to the initial detection phase, it might not use feedback, making it an open-loop system. Without additional context, this might be incorrect.

The effect of parameter variation and internal noise is more in an open loop system.

Open-loop systems do not use feedback to compare the output with the input, so they cannot correct errors caused by disturbances or changes in system parameters; therefore they are less accurate.

Stability depends on the pole locations of the system's characteristic equation. A closed-loop system uses feedback that can modify the characteristic equation and is commonly used to improve or ensure stability. Because open-loop systems lack this corrective feedback, they are generally less stable than a suitably designed closed-loop system.

Hence, an open-loop system is less stable and less accurate, so option B is the correct choice.

The given system has characteristic equation given by: |sI - A| = 0

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {s - 1}&{ - 2}\\ { - 2}&{s - 1} \end{array}} \right| = 0\)

⇒ (s - 1)² = 4

s = +3, -1

The system has poles at s = 3 and s = -1

For option (d)

The position of poles from characteristic equation:

\(\left| {\begin{array}{*{20}{c}} {s - 27}&{42}\\ { - 16}&{s + 25} \end{array}} \right| = 0\)

(s – 27) (s + 25) + (16) (42) = 0

s² – 2s – 675 + 672 = 0

s² – 2s – 3 = 0

The characteristic equation is 1 + G(s)H(s) = 0

⇒ s(s - 0.2)(s² + s + 0.6)+K(s + 0.1) = 0
s⁴ +0.8 s³ +0.4s² +(K - 0.12)s +0.1K = 0
Routh table is as shown in fig. S.62.29

K > 0, 055 -125K > 0 ⇒ K < 0.44 -125K² +0.63K -0066 >0

(K - 0.149)(K - 0355) < 0, 0.149 < K < 0.355