Mathematical Modeling and Representation of Systems MCQ Quiz - Objective Question with Answer for Mathematical Modeling and Representation of Systems - Download Free PDF

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.

The correct option is 1

Concept:

Output = e(t)

Input =  (t)

\({\rm{e}}\left( {\rm{t}} \right) \propto \frac{{{\rm{d\theta }}\left( {\rm{t}} \right)}}{{{\rm{dt}}}}\)

Taking LT,

E(s) = Kts θ(s)

\(\Rightarrow {\rm{TF}} = \frac{{E\left( S \right)}}{{\theta\left( S \right)}} = {{\rm{k}}_t}s\)

Explanation:

• Boolean operation is an important way in geometry modeling.
• It is the main way to build a complex model from simple models, and it is widely used in computer-aided geometry design and computer graphics.
• Traditional Boolean operation is mainly used in solid modeling to build a complex solid from a primary solid e.g. cube, column, cone, sphere, etc.
• With the development of computer applications, there are many ways to represent digital models, such as parametric surfaces, meshes, point models, etc.
• Models become more and more complex, and features on models such as on statutory artworks are more detailed.

Concept:

Mechanical - Electrical analogies

Steps for circuit construction in Force - Voltage analogy:

• Force is represented by the Voltage source
• Parallel connected elements of the mechanical translation system are connected in series by replacing them with respective analogies mentioned in the table.
• Series connected elements of the mechanical translation system are connected in parallel by replacing them with respective analogies mentioned in the table.

Application:

In the given figure B₁ and K are replaced with the R₁ and C in the electrical circuit.

B₂ is replaced by R₂ connected in parallel to the R₁ and C

Force is replaced with the voltage source as shown in the below figure

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:

Given the differential equation is:

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

Taking the Laplace transform on both the sides, we get:

s² Y(s) + 4 Y(s) = 6 R(s)

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

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

s² + 4 = 0

Concept:

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

\(\frac{{{d^2}y}}{{d{t^2}}} + 7\frac{{dy}}{{dt}} + 10y\left( t \right) = 4x\left( t \right) + 5\frac{{dx}}{{dt}}\)

Apply Laplace transform on both sides,

s²Y(s) + 7sY(s) + 10Y(s) = 4 X(s) + 5s X(s)

⇒ (s² + 7s + 10) Y(s) = (4 + 5s) X(s)

\(\Rightarrow \frac{{Y\left( s \right)}}{{X\left( s \right)}} = \frac{{\left( {4 + 5s} \right)}}{{{s^2} + 7s + 10}}\)

\(\Rightarrow \frac{{Y\left( s \right)}}{{X\left( s \right)}} = \frac{{4 + 5s}}{{\left( {s + 5} \right)\left( {s + 2} \right)}} = \frac{7}{{\left( {s + 5} \right)}} - \frac{2}{{\left( {s + 2} \right)}}\)

Apply inverse laplace transform 

Explanation:

• Boolean operation is an important way in geometry modeling.
• It is the main way to build a complex model from simple models, and it is widely used in computer-aided geometry design and computer graphics.
• Traditional Boolean operation is mainly used in solid modeling to build a complex solid from primary solid e.g. cube, column, cone, sphere, etc.
• With the development of computer applications, there are many ways to represent digital models, such as parametric surface, meshes, point model, etc.
• Models become more and more complex, and features on models such as on statutary artworks are more detailed.

Output = e(t)

Input =  (t)

\({\rm{e}}\left( {\rm{t}} \right) \propto \frac{{{\rm{d\theta }}\left( {\rm{t}} \right)}}{{{\rm{dt}}}}\)

Taking LT,

E(s) = K_ts θ(s)

\(\Rightarrow {\rm{TF}} = \frac{{E\left( S \right)}}{{\theta\left( S \right)}} = {{\rm{k}}_t}s\)

By applying KCL at node A,

\(\frac{{{V_A} - {e_i}}}{R} + \frac{{{V_A} - {V_B}}}{R} + \frac{{{V_R}}}{{{X_C}}} = 0\)

\(\Rightarrow {V_A}\left[ {\frac{2}{R} + \frac{1}{{{X_C}}}} \right] - \frac{{{e_i}}}{R} - \frac{{{V_B}}}{R} = 0\)      …1)

By applying KCL at node B,

\(\frac{{{V_B} - {V_A}}}{R} + \frac{{{V_B}}}{{{X_c}}} = 0\)

\({V_B}\left[ {\frac{1}{R} + \frac{1}{{{X_c}}}} \right] = \frac{{{V_A}}}{R} \Rightarrow {V_A} = {V_B}\left[ {1 + \frac{R}{{{X_c}}}} \right]\)     …2)

From equation 1) and 2)-

\( \Rightarrow {V_B}\left[ {1 + \frac{R}{{{X_C}}}} \right]\left[ {\frac{2}{R} + \frac{1}{{{X_c}}}} \right] - \frac{{{V_B}}}{R} = \frac{{{e_i}}}{R}\)

\( \Rightarrow {V_B}\left[ {\frac{2}{R} + \frac{1}{{{X_c}}} + \frac{2}{{{X_c}}} + \frac{R}{{X_c^2}} - \frac{1}{R}} \right] = \frac{{{e_i}}}{R}\)      …3)

As we can see from the circuit diagram, V_B = e₀

\( \Rightarrow {e_0}\left[ {\frac{1}{R} + \frac{3}{{{X_c}}} + \frac{R}{{X_c^2}}} \right] = \frac{{{e_i}}}{R}\)

\(\Rightarrow \frac{{{e_0}}}{{{e_i}}} = \frac{1}{{R\left[ {\frac{1}{R} + \frac{3}{{{X_c}}} + \frac{R}{{X_c^2}}} \right]}}\)

\(= \frac{1}{{{{\left( {CsR} \right)}^2} + 3RCs + 1}}\)

Time constant, T = RC

\(h\left( T \right) = \frac{1}{{{T^2}{s^2} + 3Ts + 1}}\)

• A dynamic system is a kind of system whose behavior is a function of time.
• Static system analysis does not give an accurate analysis while Dynamic system analysis does give an accurate analysis.
• Example: An aircraft is subjected to time-varying stress during the flight through turbulent air hard landing is an example of a Dynamic system.
• Dynamic system analysis is more complex than static system analysis since the conclusion based on static system analysis is not correct.
• In the dynamic system, the transfer function does not change by a linear transformation of the state.
• The dynamic modeling starts with the physical component description of the system's understanding of component behavior to create the mathematical model.
• A given state description in the mathematical model can be transformed to a controllable canonical form if the controllability matrix is non-singular. And zero cannot be computed from this matrix.

​The dynamic system state variable is used for:

• Analysis, Identification, and Synthesis of Dynamic System.
• Predict the future behavior (Y) of the system when subjected to future input variables (U) and present (X).

Output = e(t)

Input =  (t)

\({\rm{e}}\left( {\rm{t}} \right) \propto \frac{{{\rm{d\theta }}\left( {\rm{t}} \right)}}{{{\rm{dt}}}}\)

Taking LT,

E(s) = Kts θ(s)

\(\Rightarrow {\rm{TF}} = \frac{{E\left( S \right)}}{{\theta\left( S \right)}} = {{\rm{k}}_t}s\)

The correct option is 1

Concept:

Output = e(t)

Input =  (t)

\({\rm{e}}\left( {\rm{t}} \right) \propto \frac{{{\rm{d\theta }}\left( {\rm{t}} \right)}}{{{\rm{dt}}}}\)

Taking LT,

E(s) = Kts θ(s)

\(\Rightarrow {\rm{TF}} = \frac{{E\left( S \right)}}{{\theta\left( S \right)}} = {{\rm{k}}_t}s\)

Concept:

Damping force:

\(F = f\frac{{d\left( {{x_1} - {x_2}} \right)}}{{dt}} = f\left( {{v_1} - {v_2}} \right)\)

F: Damper force

f: Damper constant

x₁, x₂: Displacement at side 1 and side 2 of the damper

v₁, v₂: Velocity at side 1 and side 2 

Spring force

\(F = k\left( {{x_1} - {x_2}} \right) = k\mathop \smallint \nolimits \left( {{v_1} - {v_2}} \right)dt\)

k: Spring constant

Calculation:

Method 1:

Given damper constant is D and the Spring constant is k

Assuming that velocities at side 1 and 2

K ∫(y - x) dt + D (y – 0) = 0

Dy +k ∫ y dt – k ∫ x dt = 0

Applying the Laplace Transform

\(DY\left( s \right) + \frac{k}{s}Y\left( s \right) - \frac{k}{s}X\left( s \right) = 0\)

\(Y\left( s \right)\left[ {\frac{{Ds + k}}{s}} \right] = \frac{k}{s}X\left( s \right)\)

\(\frac{{Y\left( s \right)}}{{X\left( s \right)}} = \frac{k}{{Ds + k}}\)

Pole is present at s = - k / D

Method 2:

Given damper constant is D and the Spring constant is k

Assuming displacement at side 1 and 2

\(k\left( {y - x} \right) + D\frac{{d\left( {y - 0} \right)}}{{dt}} = 0\)

Applying the Laplace Transform

k Y(s) – k X(s) + D sY(s) = 0

Y(s) (k + Ds) – k X(s) = 0

\(\frac{{Y\left( s \right)}}{{X\left( s \right)}} = \frac{k}{{Ds + k}}\)

Pole is present at s = - k / D

Explanation:

• Boolean operation is an important way in geometry modeling.
• It is the main way to build a complex model from simple models, and it is widely used in computer-aided geometry design and computer graphics.
• Traditional Boolean operation is mainly used in solid modeling to build a complex solid from a primary solid e.g. cube, column, cone, sphere, etc.
• With the development of computer applications, there are many ways to represent digital models, such as parametric surfaces, meshes, point models, etc.
• Models become more and more complex, and features on models such as on statutory artworks are more detailed.