Control System

Concise Mathematical Modelling of Controls and Systems

On June 2, 2021
By edifythefacts
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Hello everyone, I hope that you all are doing good in your lives. The topic for today is the fundamental mathematical modelling of a control system in which we are going to study types of model, mechanical systems, and their analogies to electrical systems. Before starting with the topics we should first have our basics clear. but if you haven’t and want to go through and have a quick brief then do refer to the link below.

Control System(Features, Types etc)

Table of Contents hide

1 Mathematical Modelling

1.1 Types of Models

2 Mathematical Modelling of Mechanical Systems

2.2.4 Equilibrium Equation of a Mechanical System

3 Analogous Electrical System

3.1 Force Voltage Analogy

3.2 Force Current Analogy

4 Example on Mathematical modelling

4.1 Q.1 Solve the problem on Mechanical Modelling problem for the system below

5 Conclusion

6 Read More

Mathematical Modelling

Certainly, the representation of the control system with the help of equations is mathematical modelling. These models are efficient for the analysis (Analysis refers to obtaining the output when then the model and the inputs are available) and design of the control systems (however Designing refers to the formation of the model when the inputs and the outputs are available).

Types of Models

Firstly Differential Modeling
Transfer Modeling
and Lastly State Space analysis

Mathematical Modelling of Mechanical Systems

In Mechanical systems, we use differential equation models. However here we are going to only consider translational systems for analysis.

Translational systems are those systems that moves along a straight path. In this system we have to mainly focus on three main elements that are mass, spring and dampers. As we all that whenever a force is put on, on any kind of system then there is an opposite and equal force present that try’s to oppose the external force. This opposition is force is however applied by one of the basic elements. Now lets study each element in detail;

Mass

Mass is a fundamental quantity of all matter. It is the measure of both, a physical quantity and resistance to acceleration. Whenever a force is put onto an object of mass M then it experiences a resistive force by the mass. According to newton’s law of motion acceleration on the object is directly proportional to the force by mass. Assume that elasticity and friction being minor. Mathematical modeling of mechanical system

$F \alpha a \end.$

$F = Ma = M \frac {d^2x(t)} {dt^2} \end.$

$F = F_m = Ma = M \frac {d^2x(t)} {dt^2} \end.$

Where; F = external force and F_m = Force by mass

However by applying Laplace transform;

$F(s) = Ms^2 x(s) \end.$

Spring

Spring is an elastic material that stores potential energy. Whenever a spring K is placed under an external force an internal opposive force occurs by the elasticity of the spring. These opposing force tries to bring the spring back to its original position thus it is directly proportional to the displacement x. Moreover, we assume that mass and friction being minor. Mathematical modeling of mechanical system spring

One End is fix

$F \alpha x \end.$

$F(t) = Kx(t) \end.$

$F_k = F(t) = Kx(t) \end.$

Where; F = external force and F_k = Force by spring

However by applying laplace transform

$f(s) = Kx(s) \end.$

Both Ends Free

$F_k = F(t) = K[x_1(t) - x_2(t) ]\end.$

Since both the ends are free changes at one end will also have an effect at the other end.

However by applying Laplace transform

$F_k = F(s) = K[x_1(s) - x_2(s)] \end.$

Damper

It is also known as a dashpot. Whenever a force acts on a damper B an opposition force occurs due to the friction of the object. The force acting on the damper is directly proportional to the velocity of the damper. Moreover, we assume that elasticity and friction being minor. Mathematical modeling of mechanical system of damper

$F \alpha v \end.$

$F(t) = Bv(t) = B \frac {dx} {dt} \end.$

$F_B = F(t) = Bv(t) = B \frac {dx} {dt} \end.$

where F = external force and F_B = force due to friction

However by applying Laplace transform

$F_B = F(s) = Bsx(s) \end.$

Equilibrium Equation of a Mechanical System

However equilibrium equation is also known as force balance equation

$F = F_m + F_B + F_K \end.$

$F(t) = M \frac {d^2x(t)} {dt} + B \frac {dx(t)} {dt} + Kx(t) \end.$

However by applying Laplace transform;

$F(t) = M s^2 x(s) + B s x(s) + Kx(s) \end.$

Analogous Electrical System

A system is said to be analogous if they are having a different physical type but the same differential equation of modelling. There are mainly two analogous electrical systems i.e ;

Firstly Force Voltage Analogy
and Secondly Force Current Analogy

Force Voltage Analogy

In force voltage analogy the mathematical equation of equilibrium is compare with the Mesh equation of the electrical system.

Certainly consider an electrical system consisting of a resistor, an inductor, and a capacitor all three connected in series. Besides a voltage of V(t) volts is applied at the input and current i(t) flows through the circuit. Force voltage analogy

Thus the mathematical derivation is;

$V(t) = Ri(t) + L \frac {di(t)} {dt} +\frac {1}{C}\int i(t) d(t) \end.$

So by putting i = dq(t)/dt and ∫i(t)dt = q(t);

$V(t) = R \frac {dq(t)} {dt} + L \frac {d^2q(t)} {dt^2} + \frac {1}{C} q(t) \end.$

However by applying Laplace transform;

$V(s) = R sq(s) + L s^2q(s) + \frac {1}{C} q(s) \end.$

$i.e \end.$

$V(s) = L s^2q(s) + R sq(s) + \frac {1}{C} q(s) \end.$

So by comparing the above equation with the equation of mechanical system equilibrium we get the analogous relation as;

Firstly Force (F) = Voltage (V)
Secondly Mass (M) = Inductance (L)
Damper (B) = Resistance (R)
Spring Constant (K) = Capacitance (1/C)
Displacement (x) = Charge (Q)
and Lastly Velocity (v) = Current (i)

Force Current Analogy

In the force current analogy, the mathematical equation of equilibrium is compared with the Nodal equation of the electrical system.

Consider an electrical system consisting of a resistor, an inductor and, a capacitor all three connected in parallel to each other. With a current source of i amperes Thus by applying Kirchhoff’s current law we get; Force current analogy

$I = I_R + I_L + I_C \end.$

$I(t) = \frac {1} {L} \int v(t) dt + \frac {v(t)} {R} + C \frac {dv(t)} {dt}\end.$

So by putting v = dΦ/dt

$I(t) = \frac {1} {R} \frac {d\phi} {dt} + \frac {1} {L} \phi (t) + C \frac {d^2 \phi (t)} {dt^2}\end.$

However by taking Laplace transform;

$I(s) = \frac {1} {R} s\phi (s) + \frac {1} {L} \phi (s) + C s^2 \phi (s)\end.$

$I(s) = C s^2 \phi (s) + s\frac {1} {R} \phi (s) + \frac {1} {L} \phi (s) \end.$

Thus by comparing the above equation with the equilibrium equation of mechanical system we get the analogy as;

Firstly Force (F) = Current (i)
Secondly Mass (M) = Capacitance (C)
Damper (B) = Resistance (1/R)
Spring Constant (K) = Inductance (1/L)
Displacement (x) = Flux (Φ)
and Lastly Velocity (v) = Voltage (V)

Besides there are other analogous system as well such as Torque voltage and Torque current analogy.

Example on Mathematical modelling

Q.1 Solve the problem on Mechanical Modelling problem for the system below

example on Mathematical modeling

Solution:- Firstly Calculating the total number of masses;

No. of masses = No. of displacement = No. of nodes = 2 nodal analysis

Secondly Obtain the characteristic equation; At node one

$F(t) = M_1 \frac {d^2x_1(t)} {dt^2} + B_1 \frac {dx_1(t)} {dt} + K_1x_1(t) + K_2(x_1- x_2) \end.$

However by taking Laplace transform;

$F = M_1s^2 x_1(s) + B_1s x_1(s) + K_1 x_1(s) + k_2 [x_1(s) - x_2(s)]\end.$

naming the above equation as 1. However at node 2;

$0 = M_2 \frac {d^2x_2(t)} {dt^2} + K_2(x_2- x_1) \end.$

$-M_2 \frac {d^2x_2(t)} {dt^2} = -K_2(x_1- x_2) \end.$

Thus by applying Laplace transform

$M_2 s^2 x_2 (s) = K_2(x_1 (s)- x_2 (s))\end.$

Naming the equation above as 2. However by force current analogy

$v = \frac {d\phi} {dt} \end.$

$Taking ~L.T \end.$

$v(s) = s \phi (s) \end.$

$\phi (s) = \frac {v(s)} {s} \end.$

Certainly, by putting the values in equation 1 in 2

$i(s) = s^2 C \frac {v(s)} {s} + \frac {1} {R} s \frac {v(s)} {s} + \frac {v(s)} {LS} + \frac {v_1} {L_2s} + \frac {v_2(s)} {L_2s}\end.$

$~= s C v_1(s) + \frac {1} {R} v_1(s) + \frac {v_1(s)} {LS} + \frac {v_1} {L_2s}+ \frac {v_2(s)} {L_2s}\end.$

$C_2 s^2 \frac {v_2 (s)} {s} = \frac {v_1(s)} {L_2s} - \frac {v_2(s)} {L_2s}\end.$

Thus from both the equations above we get the force current analogy as; Force current of a mathematical modelanalogy

Conclusion

Here we are at the end of the blog, I hope that all of your doubts are clear. However if there’s any doubt regarding the topic then please free to ask down below. Besides if you do like the blog then please do make sure to share it others and also would love have your suggestion on the topic which you like to read next on.

Have a nice day 🙂

Regards.

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