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**1. Introduction**

Many applications of control theory are to servomechanisms which are systems using the feedback principle designed so that the output will follow the input. Hence there is a need for studying the time response of the system. The time response of a system may be considered in two parts:

- Transient response: this part reduces to zero as t → ∞
- Steady-state response: response of the system as t → ∞

**2. Response of the first order systems**

- Consider the output of a linear system in the form Y(s) = G(s)U(s) where Y(s) : Laplace transform of the output, G(s) : transfer function of the system and U(s) : Laplace transform of the input.
- Consider the first order system of the form ay + y = u , its transfer function is
- For a transient response analysis it is customary to use a reference unit step function u(t) for which

- It then follows that
- On taking the inverse Laplace of equation, we obtain
- The response has an exponential form. The constant 'a' is called the time constant of the system.
- Notice that when t = a, then y(t) = y(a) = 1-e
^{-1}= 0.63. The response is in two-parts, the transient part e^{-t/a}, which approaches zero as t →∞ and the steady-state part 1, which is the output when t → ∞. - If the derivative of the input are involved in the differential equation of the system, that is then its transfer function is

- where

K = b / a

z =1/ b : the zero of the system

p =1/ a : the pole of the system - When U(s) =1/s , Equation can be written as
- Hence,
- With the assumption that z>p>0 , this response is shown in
- We note that the responses to the systems have the same form, except for the constant terms K
_{1}and K_{2}. It appears that the role of the numerator of the transfer function is to determine these constants, that is, the size of y(t), but its form is determined by the denominator.

**3. Response of second order systems **

- An example of a second order system is a spring-dashpot arrangement, Applying Newton’s law, we find
- where k is spring constant, µ is damping coefficient, y is the distance of the system from its position of equilibrium point, and it is assumed that y(0) = y(0)' = 0.
- Hence,
- On taking Laplace transforms, we obtain,

- where K = 1/ M , a
_{1}= µ/M , a_{2}= k/M. Applying a unit step input, we obtain - where are the poles of the transfer function that is, the zeros of the denominator of G(s).
- There are there cases to be considered:

**over-damped system:**

- In this case p
_{1}and p_{2}are both real and unequal. Equation can be written as

**critically damped system:**

- In this case, the poles are equal: p
_{1}= p_{2}= a_{1}/ 2 = p , and

**under-damped system:**

In this case, the poles p_{1} and p_{2} are complex conjugate having the form p_{1,2} = α __+__ iβ where α = a_{1}/2 and

The three cases discussed above are plotted as:

There are two important constants associated with each second order system:

- The undamped natural frequency ωn of the system is the frequency of the response shown in Fig.
- The damping ratio ξ of the system is the ratio of the actual damping µ(= a
_{1}M) to the value of the damping µc , which results in the system being critically damped. - also,

**Some definitions:**

**Overshoot:**defined as- Time delay τd: the time required for a system response to reach 50% of its final value
- Rise time: the time required for the system response to rise from 10% to 90% of its final value
- Settling time: the time required for the eventual settling down of the system response to be within (normally) 5% of its final value
- Steady-state error ess: the difference between the steady state response and the input.

**4. Steady state error **

- Consider a unity feedback system
- where

r(t) : reference input

c(t) : system output

e(t) : error - We define the error function as
- e(t) = r(t) − c(t)
- hence, Since E(s) = R(s) − A(s)E(s) , it follows that and by the final value theorem
- We now define three error coefficients which indicate the steady state error when the system is subjected to three different standard reference inputs r(s).

**step input: r(t) = ku(t)**

**Ramp input:** r(t) = ktu(t)

- In this case, is called the velocity error constant.

**Parabolic input:** r(t) = 1/2 kt^{2} u(t)

- In this case, where is called the acceleration error constant.
- From the definition of the error coefficients, it is seen that ess depends on the number of poles at s = 0 of the transfer function. This leads to the following classification. A transfer function is said to be of type N if it has N poles at the origin. Thus if
- At s = 0, K
_{1}is called the gain of the transfer function. Hence the steady state error e_{ss}depends on j and r(t) as summarized in Table

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