# Analogous Systems

In our previous blog we studied about the modelling of mechanical systems. Now, in this one, we will talk about what are analogous systems, their advantages, what are electrical analogies of mechanical systems, force-voltage analogy, force-current analogy and will also know about the analogous terms of mechanical and electrical systems. There is a lot to cover in this blog, so let’s start.

The systems which are physically different but can be represented by same type of differential equations are called as Analogous Systems.

We can say two systems are analogous to each other if the following conditions are satisfied:

1. These systems are physically different from each-other.
2. The differential equation modelling of these systems should be same i.e. they should be represented by same type of differential equations.

In analogous systems, a non-electrical system is expressed in terms of its equivalent electrical system.

The electrical equivalent systems of mechanical systems are called as Analogous Systems.

It is possible to draw an electrical system which will behave exactly similar to the given mechanical system. It is called as the Electrical Analogous of the given mechanical system.

While working on a non-electrical system, it is advantageous to convert them into their analogous electrical networks. Some of the advantages are:

1. It is easier to study and analyse an electrical system as compared to a non-electrical or mechanical system.

2. All the techniques of electrical circuit theory such as network theorems can be applied in the analysis of the given non-electrical system.

Hence, in this way by using the network theorems of electrical system, we can easily study and analyse the given mechanical system without even actually analysing it because in reality, we are analysing electrical system. But as both the systems are analogous to each other therefore by analysing electrical system we can also comment on the behaviour of the mechanical system.

3. Electrical components can be changed easily by connecting and disconnecting them in the circuit. This helps in model construction and testing.

Suppose if there is a need to analyse a given mechanical system by changing some of its components then this will be done by changing the components of its analogous electrical network because it is easier to change the components of an electrical system as compared to a mechanical system. This is very helpful in model construction and testing.

4. Once the circuit diagram of the analogous electrical system is determined, it is possible to visualize and even predict the behaviour of the given mechanical system such as resonance, passband, damping coefficient, time constant, etc.

5. If analogous terms are known then equation of one system can easily be converted into other system.

For example, mechanical translational system and mechanical rotational system are analogous to each other. Their analogous terms and respective differential equation models are given in the table below:

## Methods of Obtaining Electrical Analogous Networks

When Force or Torque of mechanical system is assumed to be analogous with Voltage of electrical system then we get our Force-Voltage Analogy and Torque-Voltage Analogy.

When Force or Torque of mechanical system is assumed to be analogous with Current of electrical system then we get our Force-Current Analogy and Torque-Current Analogy.

Now to write these analogies we must find out the analogous terms of these mechanical and electrical systems. And for that we must first write the differential equation models for these systems and after that by simply comparing those differential equations, we can easily establish analogous terms of these systems and write all the analogies.

## Mechanical Translational System

Suppose, there is a body of mass M which is placed on a surface. The friction present between the body and the surface is shown by a damper element having a damping or friction constant B and the stiffness of the body is shown by a spring element having a spring constant K as shown in the figure given below.

Now, when we apply an external force F on the body then it will produce acceleration a. There will be three resisting forces due to the mass of the body, friction and stiffness of the body which are F1, F2 and F3 respectively. These resisting forces will act in the opposite direction of the applied force.

External Force = F

Now as we know that,

Resisting Forces:

(i) Force due to Mass Element

$$F_1=Ma=M\frac{dv}{dt}=M\frac{d^2X}{dt^2}$$

(ii) Force due to Damper Element (Friction)

$$F_2=Bv=B\frac{dX}{dt}$$

(iii) Force due to Spring Element (Stiffness)

$$F_3=K\int v\cdot dt=KX$$

Now, according to d’Alembert’s principle

$$F_{ext}+F_{res}=0$$

$$F+\left[-F_1-F_2-F_3\right]=0$$

As the resisting forces acts in the opposite direction of the externally applied force, therefore (-)ive sign is taken.

$$F=F_1+F_2+F_3$$

$$\Rightarrow F=M\frac{dv}{dt}+Bv+K\int v\cdot dt$$

$$\Rightarrow F=M\frac{d^2X}{dt^2}+B\frac{dX}{dt}+KX\;\;\;\_\_\_\_\_\left(1\right)$$

## Mechanical Rotational System

Suppose, there is a body having moment of inertia J which is placed on a surface and there is some friction between the body and the surface which is represented by torsional damping constant B and the stiffness of the body is represented by the torsional spring constant K as shown in the figure given below.

Now, when an external torque T is applied on the body then it will produce angular acceleration α. There will be three resisting torques due to the inertia of the body, friction and stiffness of the body which are T1, T2 and T3 respectively. These resisting torques will act in the opposite direction of the applied torque.

External Torque = T

Resisting Torques:

(i) Torque due to Inertia Element

$$T_1=J\alpha=J\frac{d\omega}{dt}=\frac{d^2\theta}{dt^2}$$

(ii) Torque due to Torsional Damper Element (Friction)

$$T_2=B\omega=B\frac{d\theta}{dt}$$

(iii) Torque due to Torsional Spring Element (Stiffness)

$$T_3=K\theta=K\int\omega\cdot dt$$

Now, according to d’Alembert’s principle

$$T_{ext}+T_{res}=0$$

$$T+\left[-T_1-T_2-T_3\right]=0$$

As the resisting torques acts in the opposite direction of the externally applied torque, therefore (-)ive sign is taken.

$$T=T_1+T_2+T_3$$

$$\Rightarrow T=J\frac{d\omega}{dt}+B\omega+K\int\omega\cdot dt$$

$$\Rightarrow T=J\frac{d^2\theta}{dt^2}+B\frac{d\theta}{dt}+K\theta\;\;\;\_\_\_\_\_\left(2\right)$$

## Electrical System with Voltage Source

Suppose, we have a series RLC circuit with a resistor R, an inductor L and a capacitor C connected in series with a voltage source which is supplying voltage V(t) to the circuit as shown in the figure given below. The current flowing through the network is i(t). The voltage across R, L and C is V1, V2 and V3 respectively.

Note: Here, parallel RLC circuit with voltage source is not taken because in the parallel RLC circuit, the voltage across all three components will be same which will be V. In this case we can’t write equation V = V1 + V2 + V3  which is very important to establish analogous terms between mechanical and electrical systems.

Applying KVL, we can write

$$V=V_1+V_2+V_3$$

$$V=Ri\left(t\right)+L\frac{di\left(t\right)}{dt}+\frac1C\int i\left(t\right)\cdot dt$$

$$∵i\left(t\right)=\frac{dq}{dt}=rate\;of\;change\;of\;charge\;wrt\;time$$

Putting the value of i(t) in above equation, we get

$$V=R\frac{dq}{dt}+L\frac d{dt}\left(\frac{dq}{dt}\right)+\frac1C\int\frac{dq}{dt}\cdot dt$$

$$\Rightarrow V=L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\left(\frac1C\right)q\;\;\;\_\_\_\_\_\left(3\right)$$

## Electrical System with Current Source

Suppose we have a parallel RLC network in which a resistor R, an inductor L and a capacitor C is connected in parallel with a current source I as shown in the figure given below. The current flowing through R, L and C is I1, I2 and I3 respectively and the voltage across these elements is V.

Note: Here, we didn’t take series RLC circuit with current source because in the series RLC circuit, the current flowing through all the components will be same which will be I. In this case we can’t write equation I = I1 + I2 + I3  which is very important to establish analogous terms between mechanical and electrical systems.

Applying KCL, we can write

$$I=I_1+I_2+I_3$$

$$I=\frac VR+\frac1L\int V\cdot dt+C\frac{dV}{dt}$$

$$∵V=\frac{d\phi}{dt}=rate\;of\;change\;of\;flux\;linkage\;wrt\;time$$

Putting the value of V in above equation, we get

$$I=\frac1R\frac{d\phi}{dt}+\frac1L\int\frac{d\phi}{dt}\cdot dt+C\frac d{dt}\left(\frac{d\phi}{dt}\right)$$

$$\Rightarrow I=C\frac{d^2\phi}{dt^2}+\left(\frac1R\right)\frac{d\phi}{dt}+\left(\frac1L\right)\phi\;\;\;\_\_\_\_\_\left(4\right)$$

## Force-Voltage Analogy / Torque-Voltage Analogy

In Force-Voltage Analogy and Torque-Voltage Analogy, Force or Torque of mechanical system is assumed to be analogous with Voltage of electrical system.

By comparing eqn(1), eqn(2) and eqn(3), we can write Force-Voltage Analogy and Torque-Voltage Analogy as shown in the table below.

$$F=M\frac{d^2X}{dt^2}+B\frac{dX}{dt}+KX\;\;\;\_\_\_\_\_\left(1\right)$$

$$T=J\frac{d^2\theta}{dt^2}+B\frac{d\theta}{dt}+K\theta\;\;\;\_\_\_\_\_\left(2\right)$$

$$V=L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\left(\frac1C\right)q\;\;\;\_\_\_\_\_\left(3\right)$$

## Force-Current Analogy / Torque-Current Analogy

In Force-Current Analogy and Torque-Current Analogy, Force or Torque of mechanical system is assumed to be analogous with Current of electrical system.

By comparing eqn(1), eqn(2) and eqn(4), we can write Force-Current Analogy and Torque-Current Analogy as shown in the table below.

$$F=M\frac{d^2X}{dt^2}+B\frac{dX}{dt}+KX\;\;\;\_\_\_\_\_\left(1\right)$$

$$T=J\frac{d^2\theta}{dt^2}+B\frac{d\theta}{dt}+K\theta\;\;\;\_\_\_\_\_\left(2\right)$$

$$I=C\frac{d^2\phi}{dt^2}+\left(\frac1R\right)\frac{d\phi}{dt}+\left(\frac1L\right)\phi\;\;\;\_\_\_\_\_\left(4\right)$$

More related to Control Systems:

Introduction to control system

Types of control systems

Sensitivity of Control System

Mathematical Models of Control Systems

What is Transfer Function of Control System