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 

       byjusexamprep

  • For a transient response analysis, it is customary to use a reference unit step function u(t) for which

        byjusexamprep

  • It then follows that 

     byjusexamprep

  • On taking the inverse Laplace of the equation, we obtain

      byjusexamprep

  • The response has an exponential form. The constant 'a' is called the time constant of the system.

byjusexamprep

  • 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 byjusexamprep then its transfer function is 

        byjusexamprep

  • 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

        byjusexamprep

  • Hence,

       byjusexamprep

  • With the assumption that z>p>0, this response is shown in

byjusexamprep

  • 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

        byjusexamprep

  • 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.

byjusexamprep

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

       byjusexamprep

  • where K =1/ M , a1 = µ / M , a2 = k / M . Applying a unit step input, we obtain

        byjusexamprep

  • where byjusexamprep are the poles of the transfer function byjusexamprep 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

      byjusexamprep

critically damped system:

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

       byjusexamprep

       byjusexamprep

under-damped system:

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

   byjusexamprep

byjusexamprep

The three cases discussed above are plotted as: 

byjusexamprep

 

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.byjusexamprep
  • 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. byjusexamprep
  • also,

       byjusexamprep 

Some definitions:

byjusexamprep

  • Overshoot: defined as 

       byjusexamprep

  • 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.

4. Steady state error 

  • Consider a unity feedback system

byjusexamprep

  • 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, byjusexamprepSince E(s) = R(s) − A(s)E(s) , it follows that byjusexamprepand by the final value theorem 

        byjusexamprep

  • 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)

     byjusexamprep

      byjusexamprep  called the position error constant, then

     byjusexamprep

 

Ramp input: r(t) = ktu(t)

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

 

Parabolic input: r(t) = 1/2 kt2 u(t)

  • In this case,  byjusexamprep 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

       byjusexamprep

  • At s = 0, byjusexamprep K1 is called the gain of the transfer function.
  • Hence the steady-state error ess is summarized in Table

byjusexamprep

 

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