# Transient & Steady State Response-1 Study Notes

By Yash Bansal|Updated : July 29th, 2021

In this article, you will find the Study Notes on Steady-State Equation & Analysis which will cover the topics such as Introduction, Response of the first order Systems, Response of the second-order Systems and Steady-State error.

In this article, you will find the Study Notes on Steady-State Equation & Analysis which will cover the topics such as Introduction, Response of the first order Systems, Response of the second order Systems and Steady State error.

### 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 the 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 if  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 K1 and K2 . 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-dash pot 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 the equilibrium point, and it is assumed that y(0) = y(0)' = 0.

• Hence,
• On taking Laplace transforms, we obtain,

• where K =1/ M , a1 = µ / M , a2 = 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, p1 and p2 are both real and unequal. The equation can be written as

critically damped system:

• In this case, the poles are equal: p1 = p2 = a1 / 2 = p , and

under-damped system:

• In this case, the poles p1 and p2 are complex conjugate having the form

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(= a1M) 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 rising 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.

• 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 that indicate the steady-state error when the system is subjected to three different standard reference inputs r(s).

step input:  r(t) = ku(t)

called the position error constant, then

Ramp input: r(t) = ktu(t)

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

Parabolic input: r(t) = 1/2 kt2 u(t)

• In this case,   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,  K1 is called the gain of the transfer function.
• Hence the steady-state error ess is summarized in Table

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