Basics of Control Systems MCQ Quiz - Objective Question with Answer for Basics of Control Systems - Download Free PDF

Concept:

A linear time-invariant (LTI) system is said to be:

• Stable if all poles of the transfer function lie in the left half of the s-plane (have negative real parts).
• Unstable if any pole lies in the right half of the s-plane (has positive real part).
• Marginally stable if poles lie on the imaginary axis (pure imaginary), and none are repeated.

Given:

$\text{Transfer function:}\frac{4s+1}{4s^2+1}$

Calculation:

The denominator polynomial determines the poles of the system.

$4s^2+1=0\Rightarrow s^2=-\frac{1}{4}\Rightarrow s=\pm j\frac{1}{2}$

The poles are purely imaginary and non-repeated.

Conclusion:

Since the poles lie on the imaginary axis and are simple (not repeated), the system is marginally stable.

Concept:

The error signal in a negative feedback control system is the difference between the input and the feedback signal.

The transfer function for error is given by:

$\frac{E(s)}{R(s)}=\frac{1}{1+GH}$

Where:

R(s) = Reference Input,

E(s) = Error Signal,

G = Forward Path Gain,

H = Feedback Path Gain.

Calculation:

Given a negative feedback system, the output is fed back through H and compared with input R(s).

The error signal E(s) is: $E(s)=R(s)-H\cdot C(s)$

With unity feedback or feedback factor H, and forward gain G, the output becomes: $C(s)=\frac{G}{1+GH}R(s)$

Hence, the error signal is: $E(s)=R(s)-H\cdot C(s)=\frac{1}{1+GH}R(s)$

Answer:

$\frac{1}{1+GH}$ is the correct error transfer function for a negative feedback system.

Concept:

Poles of a transfer function are the values of s that make the denominator zero.

$H(s)=\frac{1}{(s+3{)}^2}$

Set the denominator equal to zero to find the poles.

Calculation:

$(s+3{)}^2=0\Rightarrow s+3=0\Rightarrow s=-3$

Since the power is 2, it means the pole is of multiplicity 2 (repeated pole).

Hence the correct option is 3

Explanation:

Correct Option Analysis:

The correct option is:

Option 1: Asymptotic Stability

This option correctly describes a type of stability in Linear Time Invariant (LTI) systems where, in the absence of input, the output tends towards zero irrespective of initial conditions. To understand why this is the correct option, let’s delve deeper into the concept of asymptotic stability in LTI systems.

Asymptotic Stability:

In control theory, an LTI system is said to be asymptotically stable if, when the input to the system is zero, the output not only remains bounded but also approaches zero as time tends to infinity. This means that any initial perturbations or deviations will eventually die out, and the system will settle back to the equilibrium state. The system's response to any initial condition will decay to zero over time.

Mathematically, an LTI system is asymptotically stable if all the poles of its transfer function have negative real parts. Poles with negative real parts indicate that the system's natural response components will decay exponentially with time, leading to a zero output in the absence of an input.

To illustrate this, consider the following example:

Let the transfer function of an LTI system be given by:

The pole of this system is at $s=-2$, which has a negative real part. This implies that the system is asymptotically stable. If the input to the system is zero, the output will decay to zero over time, regardless of the initial conditions.

Additional Information:

To further understand the analysis, let’s evaluate the other options:

Option 2: Selective Stability

This term is not a standard term used in control theory for describing the stability of LTI systems. Stability classifications in control theory typically include asymptotic stability, bounded-input bounded-output (BIBO) stability, marginal stability, and absolute stability, among others. "Selective stability" does not describe any recognized stability property of an LTI system.

Option 3: Notion Stability

This term also does not correspond to any well-defined concept in control theory related to the stability of LTI systems. Stability concepts are well-established and include terms like asymptotic stability, BIBO stability, and so on. "Notion stability" is not one of them and does not describe the behavior of LTI systems.

Option 4: Relative Stability

Relative stability refers to a measure of how stable a system is, not whether it is stable or not. It involves comparing the degrees of stability of different systems or the stability of a system under different conditions. While it is a useful concept in control theory, it does not directly describe the behavior of the output of an LTI system in the absence of input.

Conclusion:

Understanding asymptotic stability is crucial for analyzing the behavior of LTI systems. Asymptotic stability ensures that any initial disturbances will diminish over time, leading the system output to approach zero in the absence of input. This characteristic is essential for the reliable operation of control systems, ensuring that they return to equilibrium after disturbances. While other terms like selective stability, notion stability, and relative stability might appear relevant, they do not accurately describe the fundamental stability property of LTI systems that asymptotic stability does.

Concept

The gain of a system with a given transfer function can be determined by evaluating the transfer function at specific frequencies. In this case, we are asked to find the gain for a.d.c. (zero frequency) input.

The transfer function given is:

$G(s)=\frac{s+2}{(s+1)(s+3)(s+4)}$

For a.d.c. input, the frequency is zero. Therefore, we substitute s=0" id="MathJax-Element-8-Frame" role="presentation" style="position: relative;" tabindex="0">s=0s=0" id="MathJax-Element-15-Frame" role="presentation" style="position: relative;" tabindex="0">s=0$s=0$ s=0 $s = 0$ into the transfer function to find the gain.

Calculation

Substitute s = 0 into the transfer function:s=0" id="MathJax-Element-9-Frame" role="presentation" style="position: relative;" tabindex="0">s=0s=0" id="MathJax-Element-16-Frame" role="presentation" style="position: relative;" tabindex="0">s=0$s=0$

$G(0)=\frac{0+2}{(0+1)(0+3)(0+4)}=\frac{2}{1\cdot 3\cdot 4}=\frac{2}{12}=\frac{1}{6}$

The gain of the system for a.d.c. input is 16" id="MathJax-Element-10-Frame" role="presentation" style="position: relative;" tabindex="0">1616" id="MathJax-Element-17-Frame" role="presentation" style="position: relative;" tabindex="0">16$\frac{1}{6}$ 16 $\frac{1}{6}$ .

Concept:

A transfer function is defined as the ratio of Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output]/L[input]

$TF=\frac{C\left(s\right)}{R\left(s\right)}$

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, the transfer function is also known as the impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

Calculation:

Given the differential equation is,

$\frac{d^{2}y\left(t\right)}{d{t}^2}+4y\left(t\right)=6r\left(t\right)$

By applying the Laplace transform,

s² Y(s) + 4 Y(s) = 6 R(s)

$\Rightarrow \frac{Y\left(s\right)}{R\left(s\right)}=\frac{6}{s^2+4}$

Poles are the roots of the denominator in the transfer function.

⇒ s² + 4 = 0

Concept:

Closed-loop transfer function = $\frac{G(s)}{1+G(s)H(s)}$

For unity negative feedback system Open-loop transfer function (G(s) H(s)) can be found by subtracting the numerator term from the denominator term 

Application:

Open-loop transfer Function

$=\frac{s+4}{s^2+7s+13-s-4}=\frac{s+4}{s^2+6s+9}$

For DC gain s = 0

∴ open-loop gain $=\frac{4}{9}$

The transfer function of a control system is defined as the ratio of the Laplace transform of the output variable to Laplace transform of the input variable assuming all initial conditions to be zero.

It is also defined as the Laplace transform of the impulse response.

If the input is represented by R(s) and the output is represented by C(s), then the transfer function will be:

$\frac{T}{F}=\frac{C\left(s\right)}{R\left(s\right)}$

Closed-loop control system:

These are the systems in which the control action depends on the output. These systems have a tendency to oscillate.

Ex: Temperature controllers, speed control of the motor, systems having sensors, Human eye, etc.

The closed-loop control system can be described by a block diagram as shown in the figure below.

Where, R(s) is Reference Input,

Actuating Signal E(s) = R(s) - C(s)

G(s) is unity feedback open loop gain.

H(s) is Feedback gain.

C(s) is the Output and it is given as a feedback signal to input through the feedback gain function H(s)

From the above figure, we can find out closed-loop transfer function.

CLTF = C(s) / R(s)

From this open-loop transfer function is calculated as,

$OPLT={\displaystyle \frac{CLTF}{1-CLTF}}={\displaystyle \frac{\frac{C(s)}{R(s)}}{1-\frac{C(s)}{R(s)}}}$

$OLTF={\displaystyle \frac{C(s)}{R(s)-C(s)}}={\displaystyle \frac{C(s)}{E(s)}}$

OLTF = Output / actuating signal

The various relation in force voltage and force current analogy is given in the table below:

Concept:

The transfer function is defined as the ratio of Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output]/L[input]

$TF=\frac{C\left(s\right)}{R\left(s\right)}$

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, transfer function is also known as impulse response of the system.

Transfer function = L[IR]

IR = L^-1 [TF]

Calculation:

c(t) = 1 – e^-3t

Applying Laplace transform, i.e. time domain into s- domain

$\Rightarrow C\left(s\right)=\frac{1}{s}-\frac{1}{s+3}=\frac{3}{s\left(s+3\right)}$

Input is unit step, i.e. r(t) = u(t)

$\Rightarrow R\left(s\right)=\frac{1}{s}$

Transfer Function

The transfer function is defined as the ratio of the Laplace transform of the output to the Laplace transform of the input keeping initial conditions zero.

The transfer function of the closed-loop system is:

$\frac{C(s)}{R(s)}=\frac{G}{1+GH}$

where, G = Open loop gain

H = Feedback gain

Sensitivity of the transfer function

Case 1: Sensitivity with respect to the open loop gain (G)

$S=\frac{1}{1+GH}$

Case 2: Sensitivity with respect to the feedback gain (H)

$S=\frac{-GH}{1+GH}$

In both cases, if the value of (1 + GH) is less than 1, then sensitivity increases.

Concept:

A transfer function is defined as the ratio of the Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output]/L[input]

$TF=\frac{C\left(s\right)}{R\left(s\right)}$

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, the transfer function is also known as the impulse response of the system.

Transfer function = L[IR]

IR = L^-1 [TF]

Calculation:

The given differential equation is,

$\frac{dc\left(t\right)}{dt}+2c\left(t\right)=r\left(t\right)$

By applying the Laplace transform, we get

⇒ s C(s) + 2 C(s) = R(s)

$\Rightarrow G\left(s\right)=\frac{C\left(s\right)}{R\left(s\right)}=\frac{1}{s+2}$

From the block diagram,

$G\left(s\right)=\frac{1}{s^2\left(s+1\right)}$

$H\left(s\right)=\frac{20}{\left(s+20\right)}$

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

$\frac{Y\left(s\right)}{R\left(s\right)}=\frac{G\left(s\right)}{1+G\left(s\right)H\left(s\right)}$

$=\frac{\frac{1}{s^2\left(s+1\right)}}{1+\frac{1}{s^2\left(s+1\right)}\frac{20}{\left(s+20\right)}}$

$=\frac{s+20}{s^2\left(s+1\right)\left(s+20\right)+20}$

$=\frac{s+20}{s^4+21s^3+20s^2+20}$

The denominator of the above transfer function has the highest degree of 4. Therefore, the order of the system is 4.

Concept:

$\begin{array}{l}r\left(t\right)=\delta \left(t\right),R\left(s\right)=1\\ TF=\frac{L\left[C\left(s\right)\right]}{L\left[R\left(s\right)\right]}\end{array}$

Transfer function $=L\left[C\left(s\right)\right]$

Hence, the Laplace of the response is the same as the system function for unit impulse input.