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

The correct answer is 1

Explanation:

In an open-loop transfer function, when written as a standard polynomial in 's', the highest power of 's' in the denominator represents the Order of the system.

• Order of the system: This is defined as the highest power of 's' in the denominator polynomial of the transfer function, after cancelling any common factors in the numerator and denominator. It indicates the number of energy storage elements (like capacitors or inductors in electrical systems, or masses and springs in mechanical systems) in the system and is crucial for determining the system's dynamic behavior, such as stability and transient response.
• Type of the system: This refers to the number of poles at the origin ($s=0$) in the open-loop transfer function. It's related to the steady-state error characteristics of the system for various types of inputs.
• Number of differentiators/integrators: These relate to specific components, but the highest power of 's' in the denominator describes the overall system order, not just the number of differentiators or integrators, although integrators contribute to the order by adding poles at the origin.

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

Transfer Function Analysis of the Given Circuit:

The transfer function H(s) of an electrical circuit represents the relationship between the input and output signals in the Laplace domain. It is expressed as a ratio of the Laplace transform of the output signal to the Laplace transform of the input signal. In this problem, we aim to determine the correct transfer function of the given circuit.

Correct Option:

The correct transfer function is:

Option 2: H(s)=Rs2LRC+sL+R" id="MathJax-Element-32-Frame" role="presentation" style="position: relative;" tabindex="0">H(s)=Rs2LRC+sL+R$\mathrm{H}(\mathrm{s})=\frac{\mathrm{R}}{{\mathrm{s}}^2\mathrm{L}\mathrm{R}\mathrm{C}+\mathrm{s}\mathrm{L}+\mathrm{R}}$ $\rm H(s) = \frac{R}{s^2 L R C + s L + R}$

Derivation:

To derive the transfer function, we analyze the circuit using fundamental principles such as Kirchhoff's Voltage Law (KVL) and Laplace transform techniques.

Step 1: Circuit Configuration and Parameters

The circuit under consideration is likely an RLC circuit, comprising a resistor (R), inductor (L), and capacitor (C). The configuration of the circuit determines how these components interact, influencing the transfer function. Based on the mathematical representation in the options, we assume it is a series RLC circuit.

Step 2: Applying Kirchhoff's Voltage Law (KVL)

For a series RLC circuit, KVL states that the sum of voltage drops across the resistor, inductor, and capacitor equals the applied input voltage:

Vin(t)=VR(t)+VL(t)+VC(t)" id="MathJax-Element-33-Frame" role="presentation" style="position: relative;" tabindex="0">Vin(t)=VR(t)+VL(t)+VC(t)$V_{in}(t)=V_R(t)+V_L(t)+V_C(t)$ $V_{in}(t) = V_R(t) + V_L(t) + V_C(t)$

Using the Laplace transform, the voltage drops can be expressed as:

• VR(s)=RI(s)" id="MathJax-Element-34-Frame" role="presentation" style="position: relative;" tabindex="0">VR(s)=RI(s)$V_R(s)=RI(s)$ $V_R(s) = R I(s)$
• VL(s)=sLI(s)" id="MathJax-Element-35-Frame" role="presentation" style="position: relative;" tabindex="0">VL(s)=sLI(s)$V_L(s)=sLI(s)$ $V_L(s) = s L I(s)$
• VC(s)=I(s)sC" id="MathJax-Element-36-Frame" role="presentation" style="position: relative;" tabindex="0">VC(s)=I(s)sC$V_C(s)=\frac{I(s)}{sC}$ $V_C(s) = \frac{I(s)}{s C}$

Substituting these into the KVL equation:

Vin(s)=RI(s)+sLI(s)+I(s)sC" id="MathJax-Element-37-Frame" role="presentation" style="position: relative;" tabindex="0">Vin(s)=RI(s)+sLI(s)+I(s)sC$V_{in}(s)=RI(s)+sLI(s)+\frac{I(s)}{sC}$ $V_{in}(s) = R I(s) + s L I(s) + \frac{I(s)}{s C}$

Factorizing I(s)" id="MathJax-Element-38-Frame" role="presentation" style="position: relative;" tabindex="0">I(s)$I(s)$ $I(s)$ :

Vin(s)=I(s)(R+sL+1sC)" id="MathJax-Element-39-Frame" role="presentation" style="position: relative;" tabindex="0">Vin(s)=I(s)(R+sL+1sC)$V_{in}(s)=I(s)\left(R+sL+\frac{1}{sC}\right)$ $V_{in}(s) = I(s) \left( R + s L + \frac{1}{s C} \right)$

I(s)=Vin(s)R+sL+1sC" id="MathJax-Element-40-Frame" role="presentation" style="position: relative;" tabindex="0">I(s)=Vin(s)R+sL+1sC$I(s)=\frac{V_{in}(s)}{R+sL+\frac{1}{sC}}$ $I(s) = \frac{V_{in}(s)}{R + s L + \frac{1}{s C}}$

Step 3: Output Voltage Relation

The output voltage is typically taken across one of the components. Based on the options provided, the output voltage is likely across the resistor. Therefore:

Vout(s)=VR(s)=RI(s)" id="MathJax-Element-41-Frame" role="presentation" style="position: relative;" tabindex="0">Vout(s)=VR(s)=RI(s)$V_{out}(s)=V_R(s)=RI(s)$ $V_{out}(s) = V_R(s) = R I(s)$

Substituting I(s)" id="MathJax-Element-42-Frame" role="presentation" style="position: relative;" tabindex="0">I(s)$I(s)$ $I(s)$ into Vout(s)" id="MathJax-Element-43-Frame" role="presentation" style="position: relative;" tabindex="0">Vout(s)$V_{out}(s)$ $V_{out}(s)$ :

Vout(s)=R⋅Vin(s)R+sL+1sC" id="MathJax-Element-44-Frame" role="presentation" style="position: relative;" tabindex="0">Vout(s)=R⋅Vin(s)R+sL+1sC$V_{out}(s)=R\cdot \frac{V_{in}(s)}{R+sL+\frac{1}{sC}}$ $V_{out}(s) = R \cdot \frac{V_{in}(s)}{R + s L + \frac{1}{s C}}$

Thus, the transfer function H(s)" id="MathJax-Element-45-Frame" role="presentation" style="position: relative;" tabindex="0">H(s)$H(s)$ $H(s)$ becomes:

H(s)=Vout(s)Vin(s)=RR+sL+1sC" id="MathJax-Element-46-Frame" role="presentation" style="position: relative;" tabindex="0">H(s)=Vout(s)Vin(s)=RR+sL+1sC$H(s)=\frac{V_{out}(s)}{V_{in}(s)}=\frac{R}{R+sL+\frac{1}{sC}}$ $H(s) = \frac{V_{out}(s)}{V_{in}(s)} = \frac{R}{R + s L + \frac{1}{s C}}$

Simplify the denominator:

H(s)=Rs2LRC+sL+R" id="MathJax-Element-47-Frame" role="presentation" style="position: relative;" tabindex="0">H(s)=Rs2LRC+sL+R$H(s)=\frac{R}{s^{2}LRC+sL+R}$ $H(s) = \frac{R}{s^2 L R C + s L + R}$

This matches Option 2, confirming its correctness.

Important Information:

To further analyze the other options, let’s evaluate their mathematical expressions:

Option 1: H(s)=Cs2L+sLRC+R" id="MathJax-Element-48-Frame" role="presentation" style="position: relative;" tabindex="0">H(s)=Cs2L+sLRC+R$\mathrm{H}(\mathrm{s})=\frac{\mathrm{C}}{{\mathrm{s}}^2\mathrm{L}+\mathrm{s}\mathrm{L}\mathrm{R}\mathrm{C}+\mathrm{R}}$ $\rm H(s) = \frac{C}{s^2 L + s L R C + R}$

This option incorrectly places C" id="MathJax-Element-49-Frame" role="presentation" style="position: relative;" tabindex="0">C$C$ $C$ in the numerator instead of R" id="MathJax-Element-50-Frame" role="presentation" style="position: relative;" tabindex="0">R$R$ $R$ . Moreover, the denominator does not match the standard form derived from the circuit analysis. Therefore, it is incorrect.

Option 3: H(s)=Rs2L+sLRC+R" id="MathJax-Element-51-Frame" role="presentation" style="position: relative;" tabindex="0">H(s)=Rs2L+sLRC+R$\mathrm{H}(\mathrm{s})=\frac{\mathrm{R}}{{\mathrm{s}}^2\mathrm{L}+\mathrm{s}\mathrm{L}\mathrm{R}\mathrm{C}+\mathrm{R}}$ $\rm H(s) = \frac{R}{s^2 L + s L R C + R}$

Although this option has R" id="MathJax-Element-52-Frame" role="presentation" style="position: relative;" tabindex="0">R$R$ $R$ in the numerator, the denominator differs from the correct one derived earlier. Specifically, it omits the term s2LRC" id="MathJax-Element-53-Frame" role="presentation" style="position: relative;" tabindex="0">s2LRC$s^{2}LRC$ $s^2 L R C$ , making it inconsistent with the circuit's transfer function.

Option 4: H(s)=Cs2LRC+sL+R" id="MathJax-Element-54-Frame" role="presentation" style="position: relative;" tabindex="0">H(s)=Cs2LRC+sL+R$\mathrm{H}(\mathrm{s})=\frac{\mathrm{C}}{{\mathrm{s}}^2\mathrm{L}\mathrm{R}\mathrm{C}+\mathrm{s}\mathrm{L}+\mathrm{R}}$ $\rm H(s) = \frac{C}{s^2 L R C + s L + R}$

This option swaps R" id="MathJax-Element-55-Frame" role="presentation" style="position: relative;" tabindex="0">R$R$ $R$ for C" id="MathJax-Element-56-Frame" role="presentation" style="position: relative;" tabindex="0">C$C$ $C$ in the numerator. However, the denominator matches the correct transfer function. Since the numerator should be R" id="MathJax-Element-57-Frame" role="presentation" style="position: relative;" tabindex="0">R$R$ $R$ , this option is incorrect.

Conclusion:

The correct transfer function for the given circuit is H(s)=Rs2LRC+sL+R" id="MathJax-Element-58-Frame" role="presentation" style="position: relative;" tabindex="0">H(s)=Rs2LRC+sL+R$\mathrm{H}(\mathrm{s})=\frac{\mathrm{R}}{{\mathrm{s}}^2\mathrm{L}\mathrm{R}\mathrm{C}+\mathrm{s}\mathrm{L}+\mathrm{R}}$ $\rm H(s) = \frac{R}{s^2 L R C + s L + R}$ , corresponding to Option 2. This result is obtained through a systematic application of circuit analysis principles, including KVL and Laplace transform techniques.

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.