# Mathematical Modelling of Physical Systems 1.2

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In this tutorial, we shall understand how we can develop mathematical models of physical systems. As we discussed earlier, a physical system is a collection of objects connected together to achieve an objective. Although physical systems include a wide variety of systems, we shall restrict ourselves to electrical and mechanical systems in this tutorial. A mathematical model is the mathematical representation of the physical system which is made using the appropriate governing laws of that system. These governing laws are Ohm’s law and Kirchhoff's laws for electrical systems and when it comes to mechanical systems, we use Newton’s laws of motion. These laws help us in obtaining the relation between the input and output of the system.

For most physical systems, a mathematical model is a set of differential equations. A system is said to be linear if its mathematical model obeys the principle of superposition and homogeneity. This means that if a system outputs *y*_{1} and *y*_{2} to *x*_{1} and *x*_{2} respectively, then the system output to the linear combination of these inputs *a*_{1}*x*_{1 }*+ **a*_{2}*x*_{2} should be the linear combination of their individual outputs *a*_{1}*y*_{1 }*+ **a*_{2}*y*_{2} for the system to be linear. Here *a*_{1} and *a*_{2} are constants.

Although most systems are non linear, they are made linear through certain assumptions in order to apply techniques developed for linear systems that we shall discuss in the upcoming tutorials. This helps make the analysis of system behaviour easier. In this entire tutorial series we shall be restricting ourselves to the analysis of** linear time invariant (LTI) systems** i.e., systems that are linear and the variables in the differential equations are independent of time. The mathematical models of LTI systems can be reshaped into a transfer function which we shall learn about in the next tutorial.

We shall now discuss electrical and mechanical systems individually and show how to develop their mathematical models.

## Modelling Electrical Systems

Electrical circuits have three basic elements, the resistor, the capacitor and the inductor. These are analyzed based on Kirchhoff’s laws as stated before.

The voltage-current relationships of these elements are as shown below in both time domain and in laplace domain.

With this, now we can proceed to develop mathematical models for a series RLC network and then for a parallel RLC network.

## Series RLC Network

A mathematical model usually defines the relation between an input and an output. Hence, first we must decide the input and the output. In this network let’s choose *v _{c}*(

*t*) as the output and

*v*(

*t*) as input as it’s the only one in this network.

Then, to find the relationship between the input and output, we start by applying Kirchhoff’s voltage law around the loop, assuming the initial conditions to be zero, obtaining,

But the goal is to find the relationship between the input and output *voltages* - which the left hand side of the equation is in terms of current. But the current is the rate of flow of charges and by replacing *i*(*t*) by *dq*(*t*)/*dt* in the above equation, we get:

Take a moment and make sure this replacement makes sense. Then notice that this is still not in terms of voltages but is one step closer. However, we know that, for a capacitor,

So we can substitute *q*(*t*) in equation (2) using equation (3), we obtain,

This equation is the mathematical model for the series RLC network. Notice that on the left hand of the equation, everything is in terms of the voltage across the capacitor, *v _{c}*(

*t*), with the input of

*v*(

*t*). We successfully created a mathematical equation that relates the input and the output. With this equation, we should be able to calculate what will happen in a real circuit given an input and component parameters. And, later, we’ll learn about how to convert this to the Laplace domain and how it’ll help us. But for now, let’s be content with what we’ve accomplished thus far.

## Parallel RLC Network

In this network, *i*(*t*) is the input and output is taken as *v*(*t*).

Applying Kirchhoff’s current law for the above network, assuming initial conditions to be zero,

The above equation represents the mathematical model of the parallel RLC network.

For now, let’s move aside and recall Faraday’s law. It tells that the induced voltage is equal to the rate of change of the magnetic flux linking the coil, which in case is referring to the magnetic flux in the inductor (L) . Read more about Faraday’s law here:

If the magnetic flux linking the inductor is φ(*t*), then the voltage across the inductor can be written as *v*(*t*) = *d*φ/*dt*. Hence the modified mathematical model in terms of this flux linkage becomes,

Any electrical network can be mathematically modelled in a similar manner using Kirchhoff's laws and other basic relations. Truly, the only difference between your typical circuit analysis and mathematical modeling of electric circuits is explicitly creating equations and relationships between an input and output, rather than solving for particular values.

## Mechanical Systems

Mechanical systems are usually classified into translational systems and rotational systems. These are modelled by three translatory and three rotatory elements as shown below.

**Translational Systems**

In translational motion, the motion is along a straight line. There are three types of elements that oppose this motion.

1) The mass element or the inertia element.

The mass element represents the system’s inertia. As per Newton’s second law of motion, the inertia force shall be equal to the rate of change of linear momentum and hence,

where,

*M* = Mass*F*(*t*) = Force*v*(*t*) = velocity*x*(*t*) = displacement

In other words, force is equal to mass times the change in velocity, or mass times acceleration.

2) The spring element.

The spring element represents the potential energy storage elements in the mechanical systems. The restoring force of a spring is directly proportional to the net displacement [ *x*_{1}(*t*) - *x*_{2}(*t*) ] of the spring.

where,

*K* = spring constant or the stiffness

In other words, the force the spring exerts is equal to the stiffness of the spring multiplied by the amount the spring is stretched.

3) The damper element.

The damper element represents the viscous friction present in the system. The damping force is directly proportional to the net velocity of the damper element.

where,

*B *= Damping coefficient

In other words, the amount of force that the damper exerts is the damping coefficient multiplied by the difference in velocity of the two sides of the damper. Why do we need to specify that it’s the difference in velocity between the two sides? Because in many situations, the entire damper is moving and it only dampens the velocity difference between its two sides. If the entire damper is moving and the two sides are maintaining the same distance from each other, the damper effectively does nothing.

**Rotational Systems**

In rotational motion, the motion is about a fixed axis. There are three types of elements that oppose this motion. Conceptually, they can be related to their linear counterpart in how they work but mathematically, they must be treated very differently.

1) The inertia element.

This element represents the moment of inertia of the system. This represents the stored kinetic energy of the rotational motion. The inertia torque is the rate of change of angular momentum.

where,

*J* = Moment of inertia in Kg.m^{2}*T*(

*t*) = Torque in N.m

ω(

*t*) = angular velocity in rad/s

θ(

*t*) = angular displacement radians

2) The torsional spring element.

This element represents the stored potential energy of the system. The spring torque is directly proportional to net torsional displacement [ θ_{1}(t) - θ_{2}(t) ] of the element.

where,

*K* = spring constant or the torsional stiffness

3) The damper element.

The damper element is responsible for the viscous friction in the system. The damping torque is directly proportional to the net angular velocity of the element.

where,

*B *= Damping coefficient

Now, let’s model a simple translational and a rotational system.

**Translational System**

Consider the below translational system with the usual notations as discussed earlier.

Here, *F*(*t*) is the force that is serving as the input to the system and we consider the displacement *x*(*t*) as the output.

When we equate this applied force to the sum of all the opposing forces on the system, we obtain

The above equation represents the mathematical model of the given translational system. It’s easier than the circuit models we went over earlier simply because we didn’t need to change the units we were working in.

**Rotational System**

Consider the below rotational system with the usual notations as discussed earlier. We consider a shaft which is connected to a wheel of moment of inertia *J* at one end and rigidly mounted on the other side. External torque *T*(*t*) is applied on the wheel which causes an angular displacement θ(*t*) and *B* represents the coefficient of friction betweens the wheel and the ground.

Here, *T*(*t*) is the external torque that serves as the input to the system and we consider the angular displacement θ(*t*) as the output.

When we equate this applied torque to the sum of all the opposing torques on the system, we obtain

The above equation represents the mathematical model of the given rotational system.

In a similar manner, all mechanical systems can be modelled.

## The Analogy

Until now, we have understood how we can model electrical and mechanical systems. Now let’s take a look at some interesting relations.

Let’s now compare the mathematical models of a series RLC network (equation (2)) and the discussed translational system [equation (8)] and rotational system [equation (9)].

Do you find any similarities in these three equations? That’s right, they have the same structure. Here, the force *F*(*t*) and torque *T*(*t*) are analogous to the voltage *v*(*t*) and hence, this analogy is called the force-voltage analogy. Now let’s tabulate the other analogous quantities in this analogy.

Now again, let’s now compare the mathematical models of a parallel RLC network [equation (7)] and the discussed translational system [equation (8)] and rotational system [equation (9)].

Do you see any similarities? Again, it’s the same structure but here the force *F* and torque *T* are analogous to the current *i*. Hence this type of analogy is called the force-current analogy.

In a similar fashion, we shall tabulate the other analogous quantities in this analogy.

We can convert a mechanical system into an electrical network by the above analogies and apply network analysis techniques to analyse the mechanical system.

## Practical Example

Now we shall use the methods we’ve learned to model a DC motor. We shall then, in the next tutorial, use this model to obtain the transfer function of the DC motor and simulate it with SciLab XCOS.

A DC motor is represented as shown below.

This motor is excited by the magnetic field produced by an external electromagnet or a permanent magnet. This means that the flux linkage remains constant here.

Now, the motor torque is directly proportional to current flowing in its armature. Hence,

where,

*K _{T} *= motor torque constant

The back emf of the motor *e _{b}* is proportional to the speed of the motor and hence,

where,

*K _{b} *= back emf constant

Let’s now consider the electrical part of the DC motor.

Applying Kirchhoff’s voltage law to the motor circuit, we get,

Now, consider the mechanical section of the DC motor. Here, J represents the moment of inertia of the load and the motor and B represents the friction involved with the rotation of the motor.

The torque can be written as,

In the above equation, the torsional spring element is ignored. This is because the motor shaft is assumed rigid and there is no difference in the angular displacement between the ends of the shaft.

The above obtained 4 equations represent the mathematical model of a DC motor.

Now, let’s simplify it further.

From equations (10) and (13), we obtain,

Substituting for *i* *and* *e _{b} * from equation (14) and (11) in equation (13), we obtain

Upon further simplification,

Suppose we consider the speed of the motor as the output, then we substitute ω =*d*θ/*d**t* in the above equation.

This may look complicated but we stepped through the process of arriving at this conclusion and each step made sense, so if you need to go over it again to get the picture, that’s fine. Don’t get lost in the details, remember what we’re working with, our goals that we’re trying to achieve with this, and what the math represents. This entire equation boils down to - if we know the different parameters of the motor, we can calculate the output speed with a given voltage.

In this tutorial, we learned how we can develop mathematical models for electrical and mechanical systems using simple examples. The same procedure can be followed to develop models of complex systems. Then we looked at the similarities between the electrical and the mechanical systems and finally we developed a practical mathematical model of a DC motor which is an electromechanical system. In the next tutorial, we shall learn about transfer functions and how we use these mathematical models to obtain transfer functions. Till then, take a few electrical networks and try to obtain their mathematical model for practice.

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