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Time Response of First Order System

The type of system having ‘1’ as the maximum power of ‘s’ in the denominator of the transfer function of the control system is known as the first-order system. Thus, we can say, that the order of the system is specified by the highest power of s.

Basically the order of the system shows information regarding the closed-loop poles of the system. Thus for the first-order system, we can say that there will be one closed-loop pole.

Here we can see the block diagram of unity negative feedback first order control system:

block diagram of first order control system

Here in this article, we are about to discuss the time response of the first-order control system. Now the question arises-

What is time response?

The time response of the system is defined as the output of the system obtained by providing specific input to the system, where both input and output must be the function of time.

This means that the time response of the system provides an idea about the variation in output when a certain input is provided with respect to time.

Basically the time response of the system is composed of steady-state response and transient response and is given as:

eq1

The transient response represents the fluctuation in the output of the system on applying input before achieving the final value.

While steady-state response represents the finally achieved output of the system.

Sometimes the finally achieved value shows variation from that of the desired value. Thus the difference of desired value and achieved value shows the steady-state error of the system, represented by ess.

The closed-loop transfer function of the system is given as:

eqn''

For unity negative feedback system, the characteristic equation gives the poles of the system

eq2

So, in general form, the first order equation will be:

eq3

Thus the closed-loop pole will be:

eq4

Time Response of First-Order System

We know that to determine the response of the system, some input must be provided to it. So, here we will consider different inputs and will see the response of each input on the first-order control system.

For unit step signal as input

Since the open-loop gain of the first-order system is given as:

eq5

We know that for a unit step input

Eq6'

Thus

eqn

Taking the Laplace transform of r(t). So,

eq7

We know that the closed-loop transfer function is given as:

eq8

: C(s) is the controlled output and R(s) is the reference input

Sinceeq9

On substituting the open-loop gain, the above equation:eq10

On simplifying,

eq11

Therefore,

eq12

As we have already determined, R(s), for unit step unit. Hence substituting the value of R(s), we will have

eq13

On solving the partial fraction of the above equation,

eq14

On simplifying,

eq15

On equating constant terms

eq16

Further substituting A = 1 in the general equation

eq17

Now, equating the coefficient of s, we will have

eq18

Therefore,

eq19

On substituting the values of A and B, we will have

eq20

Further,

eq21 Now, taking inverse Laplace transform of the above equation,

eq22

Since,eq23

Whileeq24

So,eq25

Thus,eq26

Here u(t) represents the steady-state response while the other term is the transient response of the first-order control system.

The response of the unit step signal is given as:

unit step response

For unit ramp signal as input

We know the unit ramp signal is mathematically expressed as:

eq27

Therefore

eq28

The open-loop gain of the first-order control system is

eq29

Therefore,

eq30

Further on solving

eq31

On simplifying

eq32

On equating the constant term

eq33

On equating the coefficient of s after substituting the value of A

eq34

Further equating the coefficient of s2

eq35

So, on substituting the values of all the values in the general equation for unit ramp signal

eq36

Therefore,

eq37

Applying the inverse Laplace transform in the above equation

eq39

So, we will have

eq40

Here

eq41

represents the steady-state response while

eq42

shows the transient response of the first-order system with unit ramp unit.

The unit ramp response is:

unit ramp response

For unit impulse signal as input

The unit impulse input in the time domain is given as:

eq43

Taking Laplace transform

eq44

Since the closed-loop transfer function is

eq45

Substituting the value of R(s)

eq46

Thus

eq47

Taking inverse Laplace transformeq48

This is the time response of the first-order system with unit impulse input.

The unit impulse response is given as:

unit impulse response of first order system

For unit parabolic signal as input

The reference input in the time domain is given as

eq49

As the open-loop gain is

eq5

The transfer function is given as

eq51

So,

eq52

On solving the partial fraction

eq53

So, on simplifying

eq54

On comparing the constant

eq55

On comparing the coefficient of s

eq56

On comparing the coefficient of s2

eq57

On comparing the coefficient of s3

eq58

On substituting the values, we will get

eq59

On applying the inverse Laplace transform, we will have

eq60

Therefore,

eq61

Here,

eq62

represents the steady-state response and

eq63

is the transient response.

Generally, for the time domain analysis of first-order control systems, step signals are majorly used as input. This is so because the response of ramp and parabolic input signals in the case of the first-order system increases even with infinite time.

However, as step and impulse functions provide stable responses thus are majorly used. But due to the absence of a steady-state term in the impulse response of the first-order system, step signals are mostly preferred.

Related Terms:

  1. Time Domain Analysis of Control System
  2. Proportional Derivative (PD) Controller
  3. Time Response of Second Order System
  4. Closed-Loop Control System
  5. Stability of Control System

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