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First order systems – Transient Response

A first order system is one that is defined by a first order differential equation of the type:

In Figure 1 the transfer function G(s) of the previous equation is represented:

Figure 1

Figure 2 shows the location of the pole of G(s), located in s=-a0:

Figure 2

For analytical convenience we are going to order G(s) as follows:

Where K and τ are The System Gain and The Time Constant respectively. K and τ are the most important parameters of a first order system as they help to quickly understand (or anticipate) the behavior of the system. This is possible due to the unit step is usually used as the test input u(t) to most systems to know how the output y(t) is going to behave.

Applying Laplace’s anti-transform to the equation Y(s)/U(s), in terms of the parameters defined above, we can easily see that the output y(t) of a first-order system subjected to a step input ue(t) of amplitude A, is:

Very interesting in the previous expression is to realize the following:

  • That if the constant a0 is positive, the exponential response of the system ends up stabilizing. This agrees with the theory that the poles of a transfer function must be located on the left side of the s-plane for a system to be stable.
  • That the time constant τ is positive if the constant a0 is positive. Which makes a lot of sense since τ is a time interval. If a0 were negative, the constant τ is negative and the system is unstable,
Example 1

The following example and its graph in Matlab (Figure 3) allows us to see the standard form of y(t) as the output of a first order system subjected to a step input ue(t) of amplitude 1, K=2, τ=1:

>> G=tf([2],[1 1]);

>> step(G)

Figure 3

In Figure 3 we can see the qualitative meaning of the constants K and τ. The constant K is the final value of the first order system when a long time has passed (steady state). In the example and thanks to the graph in Figure 2, we see that this final value is 2, as specified by G(s). Hence the usefulness of knowing the value of K if the system is of the first order …. We already know what the final value of the system is in steady state for a unit step input.

For its part, by definition, the constant τ is the time it takes a system to reach 63.2% of its final value. We can check this in our example. If the final system value at the unit step input is equal to 2, 63.2% of 2 is 1,264. In our Matlab graph we can mark the value of the output with a right click on the blue curve, and then drag this point until τ = 1, and notice that the value is as expected, Figure 4:

Figure 4

n the following graph, Figure 5, we can observe in general the meaning of the parameters defined for a first order system:

Figure 5. Response of a 1st order system to a unit step input.

Another important characteristic is the time in which the system reaches the stable state, known as ts (settlement time) for which there are two criteria, the 2% criterion or the 5% criterion. Settling time ts is the time required for transient damped oscillations to reach and remain within ± 2% or ± 5% of the final value or steady state value.

According to Figure 5, and according to the 5% criterion, the ts is reached at t = 3τ. On the other hand, according to the 2% criterion, the ts is reached at t = 4τ. We can corroborate these statements in our example and its graph in Matlab (Figure 6), knowing that 2% of 2 is 0.04 (the output has to be worth 1.96 or less after t = 4 s), while 5% of 2 is 0.1 (output must be 1.90 or less after t = 3 s)

Figure 6

The equation that allows to relate the settlement time ts with the constant τ is the following:

Where tR is the delay time if it comes to show up.

Ejemplo 2

To incorporate a 2 seconds delay in our previous example (Figure 3), we modify our G(s) as follows (Figure 7):

>> s=tf(‘s’);

>> H=exp(-2*s);

>> G2=G*H;

>> step(G2)

Figure 7

Therefore, the settling time ts in our example, according to both criteria, is:

Figure 8
Example 3. 

Considering a first-order system defined by:

Obtain the response of the system when a step input U(t)=9 is applied to it, as well as its final value y(∞):

Solution:

We know that the output of a first-order system is of the type:

Where A is the amplitude of the step input. Therefore, we only have to obtain the value of both parameters K and τ from the function G(s):

We also know that in a first order system the function G(s) has the following form

Then we can assert that:

Thus:

From where we can also deduce the final value of y(t):

Ejemplo 4

Hablemos ahora de un sistema físico muy común, un circuito RC como el de la Figura:

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Demostrar que la respuesta del sistema para una entrada escalón unitario es:

null

Respuesta:

Sabemos que la salida de un sistema de primer orden es del tipo:

Que a su vez es la solución para la ecuación diferencial del tipo:

Comparamos las siguientes ecuaciones:

De la comparación podemos asignar lo siguiente:

Entonces la expresión para Vc es del tipo:

La expresión anterior supone que la condición inicial V(0) del capacitor es cero. Caso contrario, debe incluirse de la manera siguiente:

Queda así demostrado el enunciado del problema, ya que en un circuito RC la constante τ=RC.

Fuente:

  1. Ingeniería de Control Moderno 3ra. Ed. Katsuhiro Ogata.
  2. Control Systems Engineering, Nise
  3. Automática – Tema 4 – Respuesta temporal
  4. Introducción a los sistemas de control con matlab

Literature review made by:

Prof. Larry Francis Obando – Technical Specialist – Educational Content Writer

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