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# Study Notes for Transient & Steady State Response-1 Electrical Engineering (EE) Notes | EduRev

## Networking Theory

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## Electrical Engineering (EE) : Study Notes for Transient & Steady State Response-1 Electrical Engineering (EE) Notes | EduRev

The document Study Notes for Transient & Steady State Response-1 Electrical Engineering (EE) Notes | EduRev is a part of the Electrical Engineering (EE) Course Networking Theory.
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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 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-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 , 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. 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 p1,2 = α + iβ where α = a1/2 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(= 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 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.

• 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)

•   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,  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,  K1 is called the gain of the transfer function. Hence the steady state error ess depends on j and r(t) as summarized in Table
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