CHAPTER 1

INTRODUCTION

1.1 Time Response

This chapter demonstrates application of the system repetition by

evaluating the transient response for the system model. Based on response

characteristic shown in figure above, the system output response is equal to

the natural response plus with the force response. The natural response is

the response from the input, while the force response is as input of the

system.

Figure 1.1: Response characteristic

After describing a valuable analysis and design tool, poles, and zeros, we

begin analyzing our model to find the step response of first and second

order system. The order refer to the denominator of the transfer function

after cancellation of common factors in the numerator or the number of

simultaneous first order equation required for the state space

representation. The definition for the order of a system is being the highest

power of derivative in the differential equation, or being the highest power

in the denominator of the transfer function. [1]A first-order system has only

one parameter in the denominator, while a second-order system has two

parameters in the denominator for switching. The typical signals such as

impulse function, step function and ramp function was used for analyzing

system characteristics may be determined by the form of the input that the

system will be subjected to most frequently under normal operation. In a

first order system, there is only one unit step response. The initial conditions

are assumed to be zero. In a second order system, the damping ratio, ζ

indicate the unit step response for each case. Whenζ is greater than 1, it is

an overdamped case. When the damping ratio, ζis equal to 1, it is a critically

damped case. When ζis between 0 and 1, it is aunderdamped case. Whenζ

is equal to zero, it is aundamped case. Many systems in daily activities are

approximately to be first order such as velocity of a car on the road, electric

systems where energy storage is essentially in one capacitor or one

inductor, incompressible fluid flow in a pipe, level control of a tank and

pressure control in a gas tank and etc. However, for the application in

second order system were car absorber, toaster, car break system, and etc.

[2]

1.2 Poles and Zero

Pole and zero are used to determine the time response of control system.

Zeroes are determined by equating the numerator of the transfer function

to zero

poles are determined by equating the denominator of the transfer function

to zero

zeroes means the output at those frequencies is zero while pole means that

output at that frequency is infinite (or very large) poles and zeroes are

related to stability of the system.

1.2.1Example for calculation

Figure 1.2: System showing input and output

For example a system with all poles on the left side of the s-plane is

considered to be stable. The pole of a transfer function are the value of

Laplace Transform variable, s, that cause the transfer function to become

infinite or any roots of the denominator of the transfer function that are

common to roots of the numerator. Given that the transfer function as

shown as figure 1.2, the pole exist at s = -7 and a zero exist at s = -6. These

values are plotted on the complex s-plane in figure 1.3, using an X for the

pole and an O for the zero.

s+6s+7

R (s )=1s C ( s )

Figure 1.3: Pole-Zero plots

The pole of the input function generates the form generates the form of

the forced response, thus the pole of the transfer function generated the

form natural response.

1.3 First Order System

1.3.1 Introduction of Firs Order System

First order systems are, by definition, systems whose input-output

relationship is a first order differential equation. A first order differential

equation contains a first order derivative but no derivative higher than first

order. The order of a differential equation is the order of the highest order

derivative present in the equation. [3] A first order system without zeros can

be described by the transfer function shown in figure below;

Figure 1.5: First order system and pole plot

Figure 4.2 shown, if the input is unit step, where R(s) = 1/s, the Laplace

transform of the step response is C(s), where

C ( s )=R (s )G(s)= as(s+a)

Taking the inverse transform, the step response is given by c(t) = cf(t)

+cn(t) = 1-e-atwhere the input pole at the origin generated the force

response cf(t) = 1, and the system pole at –a, as shown in figure 1.5,

generated the natural response cn(t) = 1-e-at. The equation c(t) = cf(t) +cn(t)

= 1-e-atis plotted in figure 1.6.

Figure 1.7: First order unit step response

Let us examined the significant of parameter a, the only parameter need

to describe the transient response. When t = 1/a

e-at |t=1/a = e-1 = 0.37 or

c(t)|t=1/a = 1 – e-at|t=1/a = 1-0.37 = 0.63

1.3..1.1 Time Constant

We now use both of the equation to define the transient response

performances specification. The time constant was called 1/a of the

response. The time constant is the time it takes for the system to rise 63%

of its final value and the time constant can be described as the time for e -1to

decay to 37% of the initial value for an impulse response. More generally, it

represents the time scale for which the dynamics of the system are

significant. [4]

1.3.1.2 Rise Time

Rise timed is defined as the time for the waveform to go from 0.1 to 0.9 of

its final value. Rise time is found by solving the equation below.

1.3.2Pole and Zero for a First Order system: An Example

To show the properties of the pole and zeros, let us find the unit step

response of the system. Multiplying the transfer function of figure above by

a step function yields.

C ( s )= s+6s (s+7)

= AS

+ BS+7

=¿

Where,

A=S+6S+7 S→0

=67

B=S+6S S→−7

=17

Thus,

C ( s )= s+6s (s+7)

=AS

+BS+7

=¿

67S

+

17

S+7

C ( t )=67+ 17e−7 t

1.3.3First Order system: An Example system

1.3.3.1 Mass DamperSystem

First order systems are an extremely important class of systems.

Many practical systems are first order; for example, the mass-damper

system and the mass heating system. Understanding first order systems

and their responses is an important aspect to design and analysis of

systems in general. Followed the entire step below to for the unitstep

response of first order systems;[5]

i. R(s) = 1/s, and therefore the unit-step response is:

Y(s) = 1

s (Ts+1)

ii. Expanding Y(s) into partial fraction:

Y(s) = 1s−T

Ts+1 = 1s−1

s+1/T

iii. Take the inverse Laplace Transform:

y(t) = 1-e-t/T , t ≥ 0

iv. The solution has two parts: a steady state response: yss(t) = 1, and a

transient response y1(t) = e-t/T, which decays to zero as t ∞

Figure 1.8: Unit step response

v. The slope of the tangent line at t = 0 is 1/T while the pole location in

the s plane: s = -1/T.

vi. At t = T, y(T) = 1 – e-1 = 0.632. T is called the time constant, and it is

the time it takes for the step response to rise to 63% of its final value.

vii. For t ≥ 4T, the response y(t) remains within 2% of the final value, this

time is known as the settling time, Ts. It can be see when y(2T) =

0.865; y(3T) = 0.95; y(4T) = 0.982; y(5T) = 0.993.

viii. The rise time, Tris defined as the time for the waveform to go from

10% to 90% of its final value.

ix. The steady-state error is the error after the transient response has

decayed leaving only the continuous response. The error signal:

e(t) = r(t) - y(t) = 1 - 1 + e-t/T = e-t/T

x. As t approaches infinity, e-t/T approaches zero and the steady-state

error is:

ess = e(∞) = limt →∞

[r (t )− y (t ) ] = 0

xi. The larger the time constant T is, the slower the system response is. It

is noted that the transient response dominates the response of the

system at time immediately after the input is applied and can make

significant contribution to the system response when the time

constant is large.

1.4 Second Order System

1.4.1 Introduction of Second Order

Second order systems can be definition, the systems whose inputoutput

relationship is a second order differential equation. A second order

differential equation contains a second order derivative but no derivative

higher than second order.Compared to the simplicity of a first order system,

a second ordersystem exhibits a wide range of responses that must be

analysed and described.Whereas varying a first order system parameter

simply changes the speed of theresponse, changes in the parameters of a

second-order system can change the form ofthe response. Thus, the

second-order system can display characteristics much like a first order

system or depending on component values, display damped or pure

oscillations for its transient response. There are four type of second order

system which is underdamped response, critically damped response,

undamped response, and overdamped response. WHAT IS DAMPING????

N masukn formula?????!!~

ξ= exponential decay frequencynatural frequency

= σωn

=a/2ωn

For example of electrical system shown in figure 1.9, the parameter in

determined the type of the system is the damping coefficient ξ. The

parameter was set by the value of resistance, capacitance and inductance.

Figure 1.9: Electrical System

To find the transfer function of the electric system shown in figure above

solved the circuit by used Cramer rules method. Thus, followed the step

above;

a. STEP 1

Find the mesh equation ;

- Equation of mesh one (I1) : 2R1I1 (s) - R1I2 (s) = VI (t) ………………

1

- Equation of mesh two (I2): [R1+ L + 1C

]I2(s) - R1I1 (s) = 0

…………… 2

b. STEP 2

By used Cramer Rule find I2/Vi

- Find the determinant (∆ ¿

∆ = [ 2R1 −R1

−R1 R1+L (s )+ 1C (s) ]= [V i

0 ] ∆=R1[2L (s )C (s )]+R1C (s )+1

C (s )

Find the I2/Vi

I 2 = [ 2R1 V i

−R1 0 ] I 2=R1 (V i )

I 2V i

=C (s )

[2 L (s )C (s )]+R1C (s )+1

c. STEP 3

Find the V0/Vi

V out=V c=1

C (s )X I 2Or C ( s )V out=I 2

V out

V ¿= 1

[2LC ( s )2]+R1C ( s )+1

d. STEP 4

Find the transfer function of the parameter

1.4.2Underdamped Response: An example

When ξ <1, the system is underdamped and has oscillatory response as

shown in figure 1.10. An underdamped system overshoots the final value

and the degree of overshoot is dependent to the value of ξ. The smaller of

the values, the greater of the overshoot.Assumed R1= 1Ω ,L1= 1H, and C1=

1F.

V o

V i

= 1

[2LC (s )2]+R1C (s )+1:

V o

V i

= 1

[2(1)(1) (s )2]+(1)(1)( s )+1

V o

V i

= 12 s2+s+1

Figure 1.10: Underdaped response

By referred to the transfer function shown above, Find the damping

coefficient ratio ξ, by used this damping ratio formula;

ξ= exponential decay frequencynatural frequency

= σωn

=a/2ωn

G(s)= b

s2+a s+b= 1

2 s2+s+1

Where a =1 ,ωn= √b = √1

ξ=a /2ωn

=1/2√1

=0.5

Figure 1.11: pole Location for underdamped

By factorizes the transfer function, should be get s = - 0.25 + j 0.66

and s = - 0.25 -j 0.66.Based on the figure 1.11, for the underdamped

response system, there are two complexes at – σ1± jωd at the poles.

Damped sinusoidal with an exponential envelope whose time constant is

equal to the reciprocal of the pole real part. The radian frequencies of the

sinusoidal, the damped frequency of oscillation, is equal to the imaginary

part of the pole.

1.4.2Overdamped Response: An example

If ξ >1 the system called overdamped response and no has oscillatory

as shown in figure 1.11. In this response never overshoot the final value.

However, the approach of an overdamped system is much slower and varies

depending upon the value of ξ. If ξ =1, the system is critically damped

response and no has oscillatory. Assumed R1= 5Ω, L1= 1H, C1= 2F

V o

V i

= 1

[2LC (s )2]+R1C (s )+1

V o

V i

= 1

[2(1)(2) (s )2]+(5)(2) (s )+1

V o

V i

= 14 s2+10 s+1

Figure 1.12: Overdamped response

Referred to the transfer function above, find the damping coefficient

ratio ξ,by used this damping ratio formula;

G(s)= b

s2+a s+b= 1

4 s2+10 s+1

- Where a =10 , ωn= √b = √1

ξ=a /2ωn

=10/2√1

=5

Figure 1.13: pole location for overdamed

On figure 11.3, for the overdamped responses, there are two real – σ1

and – σ2 at the poles. Two exponential with time constant equal to the

reciprocal of the pole location. By factorizes the transfer function, should be

get s = - 0.104 and s = - 2.39.

1.4.3Critically damped Response: An example

As shown in figure 1.14, a critically damped response system is similar

to the overdamped response system. The damping ratio ξ =1. A critically

damped response system provided the fasters approach to the final value

without overshoot that is found in an underdamped response system.

Assumed R1= 4Ω, L1= 125mH, C1= 500mF :

V o

V i

= 1

[2LC (s )2]+R1C (s )+1

V o

V i

= 1

[2(125m)(500m)( s )2]+(4 )(500m) (s )+1

V o

V i

= 10.125 s2+2 s+1

Figure 1.14: Critically damped Response

Referred to the transfer function above, find the damping coefficient

ratio ξ,by used this formula;

ξ= exponential decay frequencynatural frequency

= σωn

=a/2ωn

G(s)= b

s2+a s+b= 1

0.125 s2+2 s+1

- Where a =10 , ωn= √b = √1

ξ=a /2ωn

=2/2√1

=1

σ

Figure 1.15: Pole location for critically damped

Figure 1.15 are shown the pole location of critically damped response

system, there are two real at – σ1 pole locations. By used factorization we

get s1 dan s2 =4. One term is an exponential whose time constant is equal

to the reciprocal of the pole location. Another term is the product of time, t,

and an exponential with time constant equal to the reciprocal of the pole

location.

1.4.4UndampedResponse: An example

If ξ = 0 this is undamped response system has sinewave oscillation as

shown in figure1.16. An undamped response system commonly model for

physical problem, display unique behavior that must be itemized. An

undamped response system, a detailed description of the underdamped

response is necessary for the both analysis and design. Assumed R1=

0Ω,L1= 1H, C1= 1F :

V o

V i

= 1

[2LC (s )2]+R1C (s )+1

V o

V i

= 1

[2(1)(1) (s )2]+(0)(1) (s )+1

V o

V i

= 12 s2+1

Figure 1.15: Undamped Response

Referred to the transfer function above, find the damping coefficient

ratio ξ, by used this formula;

ξ= exponential decay frequencynatural frequency

= σωn

=a/2ωn

G(s)= b

s2+a s+b= 1

2 s2+2

- Where a =10 , ωn= √b = √1

ξ=a /2ωn

=0 /2√1

=0

Figure 1.16: Pole location for undamped response

Figure 1.16, shown the undamped response system pole location

there are two imaginary at ± jω1. Bu used factorization s = j and = -

j.Undamped sinusoidal with radian frequency equal to the imaginary part of

the poles.

CHAPTER 2

COMPARISON FIRST ORDER AND SECOND ORDER

BUAT DALAM TABLE:

First Order System Second Order System

CHAHPER 3

DAMPING RESPONSE IN OUR DAILY LIFE

Figure 3.1: Example system

The car break system is one of second order system example that

apply in our daily activities. By inspection, the overdamped response

indicate that the car stop before the white line after break apply on a red

signal. We assumed that the white line is the steady state of a system.

When a traffic light turns red, we apply break but the car stop before the

white line. So, the car will take times to stop at the white line. In a step

response for overdamped, it takes longer time to achieve constant

compared to others similarly with the car that stop before the white line.

Besides, the critically damped response indicate that the car stop at the

white line after break apply on a red signal. When a traffic light turns red,

we apply break but the car stop exactly on the white line. This mean it

requires less time. In a step response for critically damped, it takes shorter

time to achieve constant compared to others similarly with the car that stop

on the white line. For instant, the underdamped response indicate that the

car stop beyond the white line after break apply on a red signal. When a

traffic light turns red, we apply break but the car stop beyond the white line.

This mean the system is unstable. In a step response for underdamped, it

will fluctuate and finally comes to constant similarly with the car that stop

beyond the white line. However, the undampedresponse indicates that the

car could not stop after break applies on a red signal. When a traffic light

turns red, we apply break but the car could not stop due to the failure of car

break system. This mean the system is very unstable. In a step response for

undamped, it will oscillate non-stop similarly with the car that could not

stop.

Reference

[1] buku control

[2]http://people.exeter.ac.uk/mmaziz/ecm2105/ecm2105_n4.pdf

[3] http://people.exeter.ac.uk/mmaziz/ecm2105/ecm2105_n4.pdf

[4] bukucontrol