A system can be defined by a model, real (scale model) or virtual (mathematic) model. A dynamic systems, for example, is described by differential equations. The solution of the differential equation shows how the variables of the system depend on the time.

Let’s take as example the translational mass with spring-damper.

The equation of motion of the body mass is (for a complete explanation of how to derive the equation, read the article Mechanical systems modeling using Newton’s and D’Alembert equations):

\[{m \cdot \frac{d^2x}{dt^2} + c \cdot \frac{dx}{dt} + k \cdot x = F} \tag{1}\]where:

m [kg] – mass

k [N/m] – spring constant (stiffness)

c [N⋅s/m] – damping coefficient

F [N] – external force acting on the body

x [m] – displacement of the body

### Free response of a system

**The free response of a system** is the solution of the describing differential equation of the system, when the input is zero.

In our case the input into the system is the force F [N]. Therefore, in order to verify the free response of the system we have to solve the differential equation:

\[{m \cdot \frac{d^2x}{dt^2} + c \cdot \frac{dx}{dt} + k \cdot x = 0} \tag{2}\]The easiest way to to this is to integrate the differential equation in Xcos. For a complete explanation how handle it, read the article How to solve (integrate) a differential equation in Xcos.

The equation (2) is for the system in equilibrium. This means that, if we solve it in Xcos, the output of the system (displacement x [m] of the body) will be zero. To be able to see the free response, we need an **initial condition**. For our example we are going to set the initial conditions:

x(0) &= 0.4\\

v(0) &= 0

\end{split} \]

which translates into: at time t = 0 (when simulation starts), the position of the mass is 0.4 m to the right and the speed is 0 m/s. The **free response** of the mass spring-damper system will be the variation in time of the displacement x [m].

To integrate equation (2), we have to write it in the following form:

\[\frac{d^2x}{dt^2} = \frac{1}{m} \cdot \left ( – c \cdot \frac{dx}{dt} – k \cdot x \right ) \tag{3}\]Equation (3) is modelled and integrated in Xcos (see block diagram below).

The parameters of the model are defined directly in the blocks of the Xcos diagram. The simulation is run for `50 s`

and the Clock block samples data at `0.01 s`

. Notice that the initial condition of the first integrator block (speed) is set at `0 m/s`

and the second integrator block (position/displacement) is set at `0.4 m`

. This meas that the initial value of the speed will be `0 m/s`

and the position will be `0.4 m`

.

Running the Xcos block diagram will output the following plot:

At time `t = 0`

, the position x [m] of the mass is `0.4 m`

. After being released it begins to oscillate around the equilibrium value, `0 m`

, with smaller and smaller amplitudes. Due to the damping coefficient, after a while, it will stabilise at `0 m`

(the equilibrium position).

The **general equation of a free response system** has the differential equation in the form:

The solution *x(t)* of the equation (4) depends only on the *n* initial conditions.

### Forced response of a system

**The forced response of a system** is the solution of the differential equation describing the system, taking into account the impact of the input. In our case the input is force F [N].

Therefore, in order to verify the forced response of the system, we have to solve the differential equation:

\[{m \cdot \frac{d^2x}{dt^2} + c \cdot \frac{dx}{dt} + k \cdot x = F(t)} \tag{3}\]To visualise the forced system response of a system, all the initial conditions must be zero:

\[ \begin{split}x(0) &= 0\\

v(0) &= 0

\end{split} \]

which translates into: at time t = 0 (when simulation starts), the position of the mass is 0 m and the speed is 0 m/s. At time t = 10 s, the input force will become 0.5 N and it will pull the mass to the right. The **forced respons**e of the mass spring-damper system will be the variation in time of the displacement x [m].

To integrate equation (3), we have to write it in the following form:

\[\frac{d^2x}{dt^2} = \frac{1}{m} \cdot \left (F – c \cdot \frac{dx}{dt} – k \cdot x \right ) \tag{4}\]We can reuse the Xcos model from the free system response, the only differences being that there is an input step force added and the initial condition of the position is 0 m.

The input step force will have value 0.5 N when the simulation time is 10 s.

Running the Xcos block diagram will output the following plot:

In this case, with an input force of 0.5 N, the equilibrium position of the system is around 0.25 m. We can notice how the position oscillates around the equilibrium point, settling down in time due to the damping coefficient.

### Total response of a system

When combining the free and forced response of a system we get the total response of the system. **The total response of a system** is the solution of the differential equation with an input and initial conditions different than zero.

Therefore, in order to verify the total response of the system we have to solve the differential equation (same as forced response):

\[{m \cdot \frac{d^2x}{dt^2} + c \cdot \frac{dx}{dt} + k \cdot x = F} \tag{5}\]with the initial conditions as (same as free response):

\[ \begin{split}x(0) &= 0.4\\

v(0) &= 0

\end{split} \]

which translates into: at time t = 0 (when simulation starts), the position of the mass is 0.4 m and the speed is 0 m/s. At time t = 10 s, the input force will become 0.5 N and it will pull the mass to the right. The **total respons**e of the mass spring-damper system will be the variation in time of the displacement x [m].

The Xcos block diagram is the same as in the forced response case, the only difference being the initial condition setting of the position (0.4 m).

The input step force will have value 0.5 N when the simulation time is 10 s.

Running the Xcos block diagram will output the following plot:

Notice how the mass is released from the initial position (0.4 m) and tries to settle around the equilibrium position 0 m. When the simulation time is 10 s, the input force becomes 0.5 N and pulls the mass towards a new equilibrium point 0.25 m.

The free and forced responses of a system are useful to understand the dynamic behaviour of a system (plant). Further, the parameters and the time responses of the plant will be useful when designing a control system, for a particular application.