Control System Simulator

Scientech 2454

Product Tutorial
Ver.1.1

Designed & Manufactured by:

An ISO 9001:2008 company
Scientech Technologies Pvt. Ltd.
94, Electronic Complex, Pardesipura, Indore - 452 010 India,
+ 91-731 4211100, : info@scientech.bz , : www.ScientechWorld.com
Scientech 2454

Control System Simulator

Scientech 2454

1. Safety Instructions 4
2. Introduction 5
3. Features 6
4. Technical Specifications 7
5. Theory 8
6. Experiments
• Experiment 1 48
To observe the First Order control system for different values of the
Damping Ratio at different values of resistance
• Experiment 2 50
To observe the Second Order control system for different values of the
Damping Ratio at different values of resistance
• Experiment 3 52
To observe the Third Order control system for different values of the
Damping Ratio at different values of resistance
• Experiment 4 54
To observe the Type0 control system Steady State Error (Ess) for Unit
Step or Square wave input
• Experiment 5 57
To observe the Type0 control system Steady state error (Ess) for
Ramp as input
• Experiment 6 60
To observe the Type0 control system Steady state error (Ess) for
Parabolic as input
• Experiment 7 63
To observe the Type1 control system Steady State Error (Ess) for Unit
Step or Square wave input
• Experiment 8 66
To observe the Type1 control system Steady state error (Ess) for Ramp
as input

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• Experiment 9 69
To observe the Type 1 control system Steady state error (Ess) for
Parabolic as input
• Experiment 10 72
To observe the Type 2 control system Steady State Error (Ess) for Unit
Step or Square wave input
• Experiment 11 75
To observe the Type 2 control system Steady state error (Ess) for
Ramp as input
• Experiment 12 78
To observe the Type 2 control system Steady state error (Ess) for
Parabolic as input
7. Additional Experiments
● Experiment 1 81
To observe the First Order control system for different values of the
Damping Ratio at different values of resistance
• Experiment 2 82
To observe the Second Order control system for different values of the
Damping Ratio at different values of resistance
• Experiment 3 83
To observe the Third Order control system for different values of the
Damping Ratio at different values of resistance
8. Warranty 84
9. List of Accessories 84

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Safety Instructions
Read the following safety instructions carefully before operating the instrument. To
avoid any personal injury or damage to the instrument or any product connected to the
instrument.
Do not operate the instrument if you suspect any damage within.
The instrument should be serviced by qualified personnel only

For your safety:

Use proper Mains cord : Use only the mains cord designed for this instrument.
Ensure that the mains cord is suitable for your
country.
Ground the Instrument : This instrument is grounded through the protective
earth conductor of the mains cord. To avoid electric
shock the grounding conductor must be connected to
the earth ground. Before making connections to the
input terminals, ensure that the instrument is properly
grounded.
Observe Terminal Ratings : To avoid fire or shock hazards, observe all ratings and
marks on the instrument.
Use only the proper Fuse : Use the fuse type and rating specified for this
instrument.
Use in proper Atmosphere : Please refer to operating conditions given in the
manual.
● Do not operate in wet / damp conditions.
● Do not operate in an explosive atmosphere.
● Keep the product dust free, clean and dry.

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Introduction
Scientech TechBooks are compact and user friendly learning platforms to provide a
modern, portable, comprehensive and practical way to learn Technology. Each
TechBook is provided with detailed Multimedia learning material which covers basic
theory, step by step procedure to conduct the experiment and other useful information.
Scientech 2454 Control System Simulator helps the users to gain invaluable
knowledge about Order and type of Control System. Square wave, Ramp wave,
Parabolic wave, Unit step signal and variable DC supply are provided on board as
standard inputs. On board Resistance, Capacitor and Inductor banks for studying
different combination for the order of a system are also available.

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Features
• Study of I Order System
• Study of II Order System
• Study of III Order System
• Study of Type 0 System
• Study of Type 1 System
• Study of Type 2 System
• Additional Resistance Bank
• Additional Capacitance Bank
• Additional Inductance Bank
• Variable Voltage Output
• Unit Step Output
• Square Wave Output
• Ramp Output
• Parabolic Output
• Buffers
• Ess Block

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Technical Specifications
Variable Voltage : -10V to +10V
Unit Step Signal
Square Wave : 100Hz
Ramp Wave : 100Hz
Parabolic Wave : 100Hz
Resistance Bank : 100R, 1K, 10K, 10K, 50K, 100K
Inductor Bank : 1μH, 680μH, 10mH, 10mH, 68mH, 68mH
Capacitor Bank : 1nF, 10nF, 10nF, 100nF, 1μF, 1μF
Dimensions (mm) : W 326 x D 252 x H 52
Power Supply : 100V - 240V AC, 50/60Hz
Weight : 1.5Kg (approximately)
Learning Material : CD (Theory, procedure, reference results etc),
Online (optional)

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Theory
Open Loop Systems:
Those systems in which the output has no effect on the control action are called open
loop control systems. In other words, in an open loop control system the output is
neither measured nor fed back for comparison with the input. Thus, to each reference
input there corresponds to a fixed operating condition; as a result, accuracy of the
system depends upon calibration. In the presence of disturbances, an open loop
control system will not perform desired task. Open loop system can be used only if
the relationship between input and output is known and if there are neither internal
nor external disturbances. Note that any system that operates on time basis is open
loop.

Figure 1
Closed-loop systems:
Feedback control systems are often referred to as closed loop control systems. In
practice, the terms feedback control and close loop control are used interchangeable.
Let us start with the concept of a closed-loop feedback system. An amplifier is
presented with signals from a summing junction. Output voltage is modified by a
factor B, subtracted from the input voltage, and the result is the signal that the
amplifier is given to amplify.

Figure 2
Then:
V2= AV1
V1=Vin - BV2

Substituting,
V2= A (Vin - BV2)
Rearranging,

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V2 (1+BA) = Avin
A
V2 = Vin
BA + 1
Or
V2 A
=
V10 BA + 1
or
Op amp gains (A) are typically 50,000 to 100,000, at least at DC. Note, therefore, that
the gain of this block is quite insensitive to A. For B = .01,
V1 10000 V
A=10,000 = = 99.01 If A=100,0300, 1 = 99.90 The term A (which
Vin 100 + 1 Vin
could include other terms in addition to the gain of the amplifier as will be shown)
changed by a factor of 10 but the overall gain changed by 0.9%
This is the fundamental relationship for feedback control systems, and it's very
powerful. Note that in general, A and B are differential equations or Laplace
transforms that describe the behavior of these functions with frequency and time.
Because the transfer function above is a ratio of polynomials, (A could be in series
with some function G rather than just a gain block) and since the time and frequency
nature of networks can similarly be described as ratios of polynomials in s (Laplace
transforms), many functions can be realized with this structure. Oscillators, filters,
amplifiers, impedance changers, negative-impedance blocks comprise just a few. For
now we'll confine ourselves to a simple DC case.
Because A is large and BA is therefore large compared to 1, the transfer function can
often be simplified to
V1 1
≈
Vin B
Note that if other functions are in series with A, their transfer functions could be
lumped with A and would cancel out as they did above. This means that you can
cancel or minimize the effects of functions you cannot control simply by including
them "inside the loop" - i.e., in series with A.

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Standard Inputs
Standard inputs are usually listed in following order
An Impulse:
This is an instantaneous change in i lasting for zero length of time and returning to
the initial value. This is mostly applied to digital system where instantaneous values
are sampled by digital to analog converters. It is widely used as a standard input to a
system to compare the responses of different systems.

Figure 3
A Step Change
This is an instantaneous change in the input which then remains at the new value.

Figure 4
i = H at all values of time after t = 0.
H is the change or height of step.

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• A Ramp or Velocity Change

This is when the input changes at a constant rate. It is also called a velocity input

=c or i = ct

Figure 5
C is the rate of change (Velocity).

• A parabolic or Acceleration Change

This is when

=a or i=

Figure 6
This is also known as acceleration since the rate of rate of change is a constant ‘a’.

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First-Order Systems
Consider the first-order system shown in (1). Physically, this system may represent an
RC circuit, thermal system, or the like. A simplified block diagram is shown in
Figure 7. The input-output relationship is given by
( )
= (1)
( )

Figure 7
(A) Block diagram of a first-order system
(B) Simplified block diagram
In the following, we shall analyze the system responses to such inputs as the unit-step,
unit-ramp, and unit-impulse functions. The initial conditions are assumed to be zero.

Note that all systems having the same transfer function will exhibit the same output in
response to the same input. For any given physical system, the mathematical response
can be given a physical interpretation.

Unit-step response of first-order systems Since the Laplace transform of the unit-step
function is 1/ s, substituting R(s) = 1/s into Equation, we obtain

( ) = (1)

Expanding C(s) into partial fractions gives

( ) = − = − (2)

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inverse Laplace transform of (2), we obtain

( ) = 1 − . For t ≥ 0 (3)
Equation (3) states that initially the output c(t) is zero and finally it becomes unity.
One important characteristic of such an exponential response curve c(f) is that at t =
T the value of c(i) is 0.632, or the response c(t) has reached 63.2% of its total
change. This may be easily seen by substituting f = Tin c(t).That is,

( ) = 1 − = 0.632

Note that the smaller the time constant T, the faster the system response. Another im-
portant characteristic of the exponential response curve is that the slope of the tangent
line at f = 0 is 1/T, since

(4)

The output would reach the final value at t = T if it maintained its initial speed of re-
sponse. From Equation (4) we see that the slope of the response curve c(t) decreases
monotonically from 1/T at f = 0 to zero at f = ∞

The exponential response curve c(t) given by Equation (4-3) is shown in Figure 8. In
one time constant, the exponential response curve has gone from 0 to 63.2% of the fi-
nal value. In two time constants, the response reaches 86.5% of the final value. At f =
3T, 4T, and 5T, the response reaches 95%, 98.2%, and 99.3%, respectively, of the
final value. Thus, for t ≥ 4T, the response remains within 2% of the final value. As
seen from Equation (4-3), the steady state is reached mathematically only after an
infinite time. In practice, however, a reasonable estimate of the response time is the
length of time the response curve needs to reach the 2% line of the final value, or four
time constants.

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Exponential response curve

Figure 8
Consider the system shown in Figure 9.

Figure 9
A general system

To determine experimentally whether or not the system is of first order, plot the curve
log |c(t) - c( )|, where c(t) is the system output, as a function of t. If the curve turns
out to be a straight line, the system is of first order. The time constant T can be read
from the graph as the time T that satisfies the following equation:

C(T) – c (∞) = 0.368 [ c(0)-c (∞)]

Note that instead of plotting log |c(t) - c( )| versus t it is convenient to plot |c(t) - c(
)| / |c(0) - c( )| versus t on semi-log paper, as shown in Figure 10.

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Figure 10

Plot of | c (T) – c (∞)| / |c (o) – c(∞)| versus t on semilog paper

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First-Order System Step Response

The general form of a transfer function for a first-order system is G(s) = K / (s+a).
The time constant associated with this system is τ =1/a . The time constant tells how
quickly the system responds. For example, if a =1, then the system responds on the
order of τ =1 second; however, if a =100, then the system responds on the order of τ =
0.01 seconds. So, systems that respond quickly have large values of a, and systems
that respond slowly have small values of a. The 2% settling time for a first order
system is Ts =4τ =4a. This represents the amount of time required for the system to
reach and stay within 2% of the final value. The plot below shows the step response
for three different systems with a values of 0.5, 1, and 5. Setting K = a allows all
systems to attain a final value of 1, but note that the larger the value of a , the more
quickly the system reaches the 98% value. The 2% settling times for the three systems
are 4 5, 4, and 8 seconds, respectively.

Figure 11

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Unit-ramp response of first-order systems. Since the Laplace transform of the unit-
ramp function is 1/s2, we obtain the output of the system of figure 11 as

Expanding C(s) into partial fractions gives

(5)
Taking the inverse Laplace transform of Equation (4-5), we obtain

The error signal e(t) is then

As t approaches infinity, e-t/T approaches zero, and thus the error signal e(t)
approaches T or

The unit-ramp input and the system output are shown in figure 12. The error in fol-
lowing the unit-ramp input is equal to T for sufficiently large t. The smaller the time
constant T, the smaller the steady-state error in following the ramp input.

Figure 12
Unit-ramp response of the system shown in figure

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Unit-impulse response of first-order systems- For the unit-impulse input. R(s) = 1 and
the output of the system of figure 12 can be obtained as

( )=
+
Or

( )= (6)
The response curve given by Equation (6) is shown in Figure 13

Figure 13
Unit-impulse response of the system shown in figure
An important property of linear time-invariant systems. In the analysis above, it has
been shown that for the unit-ramp input the output c(t) is

( )= − + for t ≥ 0
For the unit-step input, which is the derivative of unit-ramp input, the output c(t) is

( )= 1− − for t ≥ 0
Finally, for the unit-impulse input, which is the derivative of unit-step input, the out-
put c (t) is

( )= for t ≥ 0

Comparison of the system responses to these three inputs clearly indicates that the re-
sponse to the derivative of an input signal can be obtained by differentiating the re-
sponse of the system to the original signal. It can also be seen that the response to the
integral of the original signal can be obtained by integrating the response of the
system to the original signal and by determining the integration constants from the
zero output initial condition. This is a property of linear time-invariant systems.
Linear time-varying systems and nonlinear systems do not possess this property.

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Second-Order Systems
In this section, we shall obtain the response of a typical second-order control system
to a step input, ramp input, and impulse input. Here we consider a dc servomotor as
an example of a second-order system. Conventional dc motors use mechanical
brushes and commutators that require regular maintenance. Due to improvements that
have been made in the brushes and commutators, however, many dc motors used in
servo systems can be operated almost maintenance free. Some dc motors use
electronic commutation. They are called brushless dc motors.
The standard form of closed loop transfer function of second order system is given by
C(s)/R(s) = ωn2/[s2 + 2δωns + ωn2] (7)
Where ωn = undamped natural frequency, rad/sec.
δ = damping ratio.

Case 1: Undamped system, δ = 0,

Case 2: Under damped system, 0 < δ < 1,
Case 3: Critically damped system, δ = 1,
Case 4: Over damped system, δ > 1.

The characteristics equation of second order system is

s2 + 2δωns + ωn2 = 0 (8)
It is a quadratic equation and the roots are given by
δω ±√( δ ω ω )
s1, s2 = (9)

δω ±√( ω (δ ))
= (10)

= -δωn ± ωn2√ (δ2-1) (11)

When δ = 0, s1, s2 = ± jωn; roots are purely imaginary and the system is undamped.
When δ = 1, s1, s2 = - ωn; roots are real and equal and the system is critically damped.
When δ > 1, s1, s2 = - δωn ± ωn2√ (δ2-1);
Roots are real and unequal and the system is over damped.
When 0 < δ < 1, s1, s2 = - δωn ± ωn√ (δ2-1)
= - δωn ± ωn√ (-1) (1-δ2)
= - δωn ± jωn√ (1-δ2)
= - δωn± ωd;

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Roots are complex conjugate and the system is under damped.

Where ωd = ωn√ (1-δ2)
Here ωd is called damped frequency of oscillation of the system and its unit is rad/sec.
Unit step response of Second Order systems
r(t) = A u(t).
Where, u(t) = 1; t > 0.
U(t) = 0; t < 0.
Which has Laplace transform as given by
R(s) = 1/s.
From equation (7)
C(s) = {ωn2/ [s2+2δωns+ωn2] } R(s) (12)
C(s) = {ωn2/[s2+2δωns+ωn2 ]}[1/s]
Referring to the table of inverse Laplace transforms, we have the response of the
system given by
√
( )= sin + tan (13)
√

Where δ = damping ratio.

Ωn = undamped natural frequency of the system
ωd = damped frequency of oscillations of the system
and ωd = ωn√(1-δ2)
From equation (10), it can be seen that the frequency of transient oscillations is the
damped natural frequency wd and thus varies with the damping ratio.
● Typical response of undamped second order system with step input [δ=0]
C(s) /R(s) = ωn2/ [s2+2δωns+ωn2]………………………………………..(14)
For step input
R(s) = 1/s.
The time domain response is
C(t) = 1- cos ωnt (15)
The response of undamped second order system for unit step input is completely
oscillatory.

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● Typical response of under-damped second order system with step input [0 <
δ <1]
The time domain response is

( ) =1− (sin cos + cos sin )

√
( )
=1− (16)
√
√1 −
= tan

The response of under damped second order system oscillates before settling to
a final value. The oscillation depends on value of the damping ratio.
● Typical response of critically damped second order system with step Input
[δ=1].
The time domain response is
( ) =1− (1 + ) (17)
The response of critically damped second order system has no oscillations.
● Typical response of over damped second order system with step input
[δ>1].
The time domain response is

( ) =1− − (18)
√
Where

= − − 1
= + − 1
The response of over damped second order system has no oscillations but it
takes longer time for response to reach the final steady value.
Time Response:
The time response of the system is the output of the closed loop system as a function
of time. It is denoted by C(t). The time response can be obtained by solving the
differential equation governing the system. Alternatively, the response C(t) can be
obtained from the transfer function of the system and the input to the system. The
output in s-domain, C(s) is given by the product of the transfer function and the input
R(s). On taking inverse Laplace transform of this product the time domain response,
C(t) can be obtained.

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Closed loop system

Figure 14
The closed loop transfer function,
( ) ( )
= ( ) ( )
(19)
( )
Response in s-domain,
( ) ( )
( )= (20)
( ) ( )
Response in time domain,
( ) ( )
( )= [ ( )] = (21)
( ) ( )

The time response of a control system consists of two parts: the transient and steady
state response. The transient response shows the response of the system when the
input changes from one state to another. The steady state response shows the response
as time, t approaches infinity.
The standard test signals is
Step Signal: The step signal is a signal whose value changes from zero to A at t=0
and remains constant at A for t >0. A special case of step signal is unit step in which
A is unity.
r(t) = A u(t).
Where, u(t) = 1; t > 0.
u(t) = 0; t < 0.

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Time Domain Specification:

The desired performance characteristics of control systems are specified in terms of
time domain specifications. Systems with energy storage elements cannot respond
instantaneously and will exhibit transient responses, wherever they are subjected to
input or disturbances. The desired performance characteristics of a system of any
order may be specified in terms of the transient response to a unit step input signal.
The response of a second order system for unit-step input with various values of
damping ratio is shown in figure15.

Figure 15
The time response of II order system for various values of δ
The transient response of a practical control system often exhibits damped oscillations
before reaching steady state. A typical oscillatory response of the second order system
is shown figure16.

Figure 16
The time response of II order system for under damped case

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In specifying the transient response characteristics of a control system to a unit-step

input, we usually specify the following.
● Delay time (td): It is time taken for response to reach 50% of the final value, for
the very first time.
td = [1+ 0.7δ] /ωn seconds.
● Rise time (tr): It is the time taken for response to raise from 0 to 100% for the
very first time. For under damped system, the rise time is calculated from 0 to
100%. But for over damped system, it is the time taken by the response to raise
from 10% to 90%. For critically damped system it is the time taken for response
to raise from 5% to 95%.
∏−
=

● Peak time (tp): It is the time taken for the response to reach the peak value for
the very first time, or it is the time taken for the response to reach the peak
overshoot, MP.

= /

● Peak overshoot (Mp): It is defined as he ratio of the maximum peak value

measured from final value to the Mp.
% peak overshoot Mp = [e-δΠ/(√(1-δ))] x100.

● Settling time (ts): It is the time taken by the response to reach & stay within a
specified error. i.e., either 2% or 5% of the final value.
ts = 4/ δωn for ± 2% tolerance.
ts = 3/ δωn for ± 5% tolerance.

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Analysis of Second Order System

The characteristics equation of second order system is
+ 2 + =0 (22)

Where =
√

Figure 17
For series RLC circuit,
1
+ + =

+ + =0

+ + =0 (23)

on comparing equation (22) & (23), we get,

=2

= (24)

Third Order System:

= ( )

= ( )( )

= + +
( ) ( )

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Take inverse Laplace transform the time response can be found. The time constant of
first order term has a great impact on the time response of the third order system. The
relative location of the first order pole with respect to the quadratic pole decides the
nature of time response of the third order system. It can be shown that the effect of
real pole is to reduce the maximum overshoot and increase the settling time.

1−
( )
− 2 cos 1− + sin ( 1 − )}–
( ) ( )
( )

(t≥ 0)
Where, =

The unit step response of this third-order system with = 0.5 for different values of α is
shown in figure

Figure 18
Response of a third –order system to unit step input for different values of α
Most of the control systems, use time as its independent variable, so it is important to
analyze the response given by the system for the applied excitation which is function
of time. Analysis of response means to see the variation of output with respect to
time. The evaluation of system is based on the analysis of such response. This output
behavior with respect to time should be within specified limits to have satisfactory
performance of the system. The complete base of stability analysis lies in the time
response analysis. The system stability, system accuracy and complete evaluation is
always based on the time response analysis and corresponding results.

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Definition and Classification of Time Response :

Time response of a control system means, how output behaves with respect to time.
So it can be defined as below.
Definition: Time Response: The response given by the system which is function of
the time, to the applied excitation is called time response of a control system.
In any practical system, output of the system takes some finite time to reach to its
final value. This time varies from system to system and is dependent on different
factors. Similarly final value achieved by the output also depends on the different
factors like friction, mass or inertia of moving elements, some non linearity present
etc.
For example consider a simple ammeter as a system. It is connected in a system so as
to measure current of magnitude 5 A. Ammeter pointer hence must deflect to show us
5 A reading on it. So 5 A is its ideal value that it must show. Now pointer will take
some finite time to stabilize to indicate some reading and after stabilizing also, it
depends on various factors like friction, pointer inertia etc. whether it will show us
accurate 5 A or not.
Based on this example, we can classify the total output response into two parts. First
is the part of output during the time, it takes to reach to its final value. And second is
the final value attained by the output which will be near to its desired value if system
is stable and accurate.
This can be further explained by considering another practical example. Suppose we
want to travel from city A to city B. So our final desired position is city B. But it will
take some finite time to reach to city B. Now this time depends on whether we travel
by a bus or a train or a plane. Similarly whether we will reach to city B or not depends
on number of factors like vehicle condition, road condition, weather condition etc. So
in short we can classify the output as,
● Where to reach ?
● How to reach ?
Successfulness and accuracy of system depends on the final value reached by the
system output which should be very close to what is desired from that system. While
reaching to its final value, in the mean time, output should behave smoothly.
Thus final state achieved by the output is called steady state while output variations
within the time it takes to achieve the steady state is called transient response of the
system.

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Definition: Transient Response

The output variation during the time, it takes to achieve its final value is called as
transient response. The time required to achieve the final value is called transient
period.
This can also be defined as that part of the time response which decays to zero after
some time as system output reaches to its final value.
Key Point: The transient response may be exponential or oscillatory in nature.
Symbolically it is denoted as ct (t)

To get the desired output, system must pass satisfactorily through transient period.
Transient response must vanish after some time to get the final value closer to the
desired value. Such systems in which transient response dies out after some time are
called Stable Systems.
Mathematically for stable operating

lim ( )=0
→

From transient response we can get following information about the system,
● When the system has started showing its response to the applied excitation?
● What is the rate of rise of output? From this, parameters of system can be
designed which can withstand such rate of rise. It also gives indication about
speed of the system.
● Whether output is increasing exponentially or it is oscillating.
● If output is oscillating, whether it is over shooting its final value.
● When it is settling down to its final value?
All this information matters much at the time of designing the systems.
Definition : Steady State Response
It is that part of the time response which remains after complete transient response
vanishes from the system output. This also can be defined as response of the system
as time approaches infinity from the time at which transient response completely dies
out. The steady state response is generally the final value achieved by the system
output. Its significance is that it tells us how far away the actual output is from its
desired value.

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Key Point : The steady state response indicates the accuracy of the system. The
symbol for steady state output is C0.
From steady state response we can get following information about the system :
● How much away the system output is from its desired value which indicates
error?
● Whether this error is constant or varying with time? So the entire information
about system performance can be obtained from transient and steady state
response.
Hence total rime response c (t) we can write as,

C (t) = Css + Ct (t)

The difference between the desired output and the actual output of the system is called
steady state error which is denoted as ess. This error indicates the accuracy and plays
an important role in designing the system.
The above definitions can be shown in the waveform as in the Figure 19 where input
applied to system is step type of input

(A) Ct (t) is exponential (B) Ct (t) is oscillatory

Figure 19
Standard Test Inputs
In practice, many signals are available which are the functions of time and can be
used as reference inputs for the various control systems. These signals are step, ramp,
saw tooth type, square wave, triangular etc. But while analyzing the systems it is
highly impossible to consider each one of it as an input and study the response. Hence
from the analysis point of view, those signals which are most commonly used as
reference inputs are defined as Standard Test Inputs. The evaluation of the system can
be done on the basis of the response given by the system to the standard test inputs.
Once system behaves satisfactorily to a test input, its time response to actual input is
assumed to be up to the mark.

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These standard test signals are

i) Step Input (Position function) :
It is the sudden application of the input at a specified time as shown in the
Figure 20. Mathematically it can be described as

Figure 20
R(t) = A for t ≥ 0
=o for t < 0
If A = 1, then it is called unit step function and denoted by u(t).
Laplace transform of such input is

ii) Ramp Input (Velocity function) :

It is constant rate of change in input i.e. gradual application of input as shown in
the Figure21

Ramp
Figure 21

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Magnitude of Ramp input is nothing but its slope. Mathematically it is defined

as,
R(t) = At for t ≥ 0
=0 for t < 0
If A = 1, it is called Unit Ramp input. It is denoted as r(t). Its Laplace transform
is .

iii) Parabolic Input (Acceleration function) :

This is the input which is one degree faster than a ramp type of input as shown
in the figure 22

Figure 22
Parabolic

Magnitude of Ramp input is nothing but its slope. Mathematically it is defined

as,

R(t) = At for t ≥ 0
=0 for t < 0

( ) = for t ≤ 0
= 0 for t < 0

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where A is called magnitude of the parabolic input

Key Point: Parabolic function is expressed as t2 so that in Laplace transforms
of different standard inputs, similarity will get maintained.

If A = 1, i.e. r(t) = it is called unit parabolic input. Its Laplace transform is

iv) Impulse Input:

It is the input applied instantaneously (for short duration of time) of very high
amplitude as shown in the Figure 23

Figure 23
Impulse
It is the pulse whose magnitude is infinite while its width tends to zero i.e. t 0,
applied momentarily.
Area of the impulse is nothing but its magnitude. If its area is unity it is called Unit
Impulse Input, denoted as (t).
Mathematically it can be expressed as,

r(t) = A, for t = 0
= 0 for t = 0

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The Laplace transform of unit impulse input is always 1. The unit impulse is denoted
as δ(t).
R(t) Symbol R (S)
Unit step U (t) 1/s
Unit ramp r (t) 1/s2
Unit parabolic - 1/s3
Unit impulse δ (t) 1
Table

Steady State Analysis

Steady state is that part of the output which remains after transients completely vanish
from the output. Mainly the steady state response has following two specifications,
● How much time system takes to reach its steady state which is called settling
time which is discussed later in connection with transient response. It is related
to transient response also because same time will be required by the transients to
die out completely from the system output.
● How far away actual output is reached from its desired value which is called
steady state error (ess).
Out of the two specifications, the steady state error is the most important specification
which is related only to the steady state. So let us see on which factors it depends,
how to calculate it and how to reduce it.

Definition: Steady State Error: It is the difference between the actual output and the
desired output.
Now reference input tells us the level of desired output and actual output is fed back
through feedback element to compare it with the reference input. Hence to be precise
it can be defined as the difference between reference input and the feedback signal
(actual output).
Matthematically it is defined in Laplace domain as,
L{e(t)} = E (s) = R (S) –C (s)H(s), for non unity feedback systems
L{e(t)} = E (s) = R (S) –C(s), for unity feedback systems

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Derivation of Steady State Error

Consider a simple closed loop system using negative feedback as shown in the Figure
24

Figure 24
Where E(s) = Error signal
B(s) = Feedback signal
Now E(s) = R(s) – B(s)
But B(s) = C(s)H(s)
E(s) = R(s) – C(s)H(s)
And C(s) = E(s)G(s)
E(s) = R(s) – E(s)G(s)H(s)

E(s) + E(s)G(s)H(s) = R(s)

( )
E(s) = for non unity feedback
( ) ( )
( )
E(s) = for unity feedback
( )

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This E(s) is the error in Laplace domain and is expression in ‘s’. We want to calculate
the error value. In time domain, corresponding error will be e(t). Now steady state of
the system is that state which remains as t →∞.

Now we can relate this in Laplace domain by using final value theorem which states
that.

Therefore

Substituting E(s) from the expression derived, we can write

For negative feedback systems use positive sign in denominator while use negative
sign in denominator if system uses positive feedback.
From the above expression it can be concluded that steady state error depends on
● R(s) i.e. reference input, its type and magnitude.
● G(s)H(s) i.e. open loop transfer function.
● Dominant nonlinearities present if any.
Now we will study the effect of change in input and product G(s)H(s) on the value of
steady state error. As transfer function approach is applicable to only linear systems,
the effect of nonlinearities is not discussed.

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Effect of Input (Type and Magnitude) on Steady State Error (Static Error
Coefficient Method)
Consider a system having open loop T.F. G(s)H(s) and excited by,
● Reference Input is step of magnitude A:

( )=

( )
= lim
→ 1+ ( ) ( )
.
= lim
→ 1+ ( ) ( )

= lim
→ 1+ ( ) ( )

=
1 + lim ( ) ( )
→

Figure 25

For a system selected lim → ( ) ( ) is constant and called Positional Error

Coefficient of the system denoted as Kp

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And corresponding error is.

So whenever step input is selected as a reference input, positional error

coefficient Kp will control the error in the system along with the magnitude of
the input applied.
● Reference Input is ramp of magnitude 'A'

Figure 26

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For a selected system is constant and called Velocity Error

Coefficient as Kv.

And corresponding error is.

So whenever ramp input is selected as a reference input, velocity error

coefficient Kv will control the error in the system along with the magnitude of
input applied.
● Reference input is parabolic of magnitude 'A':

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Figure 27

So for a selected system is constant and called Acceleration Error

Coefficient as Ka

And corresponding error is,

So whenever parabolic input is selected as a reference input, acceleration error

coefficient will control the error in the system along with magnitude of input applied.
So static error coefficients are given in Table

Note: ‘A’ denotes the magnitude of the input applied

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Effect of Change in G(s)H(s) on Steady State Error (type of a System)

This can be studied by focusing on to the dominant elements of G(s) H(s) from error
point of view. Such elements are constant of system ‘K’ and poles of G(s)H(s) at
origin if G(s)H(s) is expressed in a particular form called time constant form. This is
as shown below,

where K = Resultant system gain

j = Type of the system

‘TYPE' of the system means number of poles at origin of open loop TF. G(s)H(s) of
the system.
So j = 0 , Type zero system
= 1 , Type one system
= 2 , Type two system
= n , Type ‘n’ system
Key Point : Thus 'TYPE' is the property J open loop T.F. G(s)H(s) while ‘Order' is

the property of closed loop T.F

This is because, as defined earlier, order is the highest power of s present in the
denominator polynomial of closed loop T.F. of the system
Analysis of Type 0, 1 and 2 Systems
Note: A popular method to assess steady state performance of servomechanisms or
unit)* feedback systems is to find their error coefficients Kp, Kv and Ka.
where, Kp= Position error constant,
Kv= Velocity error constant and
Ka = Acceleration error constant.
Obviously in order to find these error constants the system must be stable, because for
an unstable system there is no steady state and Kp, Kv and Ka are undefined.
Hence before we proceed to find Kp, Kv and Ka, we must ensure (either by pole
location or by Routh table of the dosed loop system) that it is stable.
Thus the concept of Kp, Kv and Ka, is applicable only if,
● System is represented in its simple form.
● Only if the system is stable.
Consider the input selected as step of magnitude "A'

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● Let us assume that the system is of Type

‘0’

i.e. Type ‘0’ systems follow the step type of input with finite error which
can be reduced by change in ‘A’ or ‘K’ or both as per requirement.
Now ‘K’ can be increased by introducing a variable gain amplifier in the
forward path and error can be reduced. But there is limitation on the increase in
value of ‘K’ form stability point of view which is discussed later. But increase in
‘K’ is one way to reduce the error. Corresponding response will be as shown in
the Figure 28

Figure 28

● If for the same input now Type is increased to one' by adding pole at origin
in G(s)H(s).

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● Similarly if now Type Is further Increased to 'two' i.e. G(s)H(s) with 2 poles
at origin

In general, for any Type of system more than zero, Kp will be infinite (∞)and error
will be zero. Though mathematically answer for error is zero, practically small error
will be present but it will be negligibly small. Such type of responses may take one of
the forms shown in the Fig 29

A B
Figure 29
Thus Type 1 and above systems follow a step type of reference input of any
magnitude, successfully, with negligibly small error.
Let us now change the selected input from step to ramp of magnitude ‘A’ so Kv will
control the error.
● Let the system be of Type 0

i.e. Type 0 systems will not follow ramp input of any magnitude and will give large
error in the output which may damage the parameters of system or may cause the
saturation in parameters. Hence ramp input should not be applied to Type '0' systems.
The output may take the form as shown in the Fig 30

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A B
Figure 30
● If Type 1 System is subjected to Ramp input then

i.e. Type 1 systems follow the ramp type of input of magnitude ‘A’ with finite
error A/K which can be reduced as discussed earlier. The output may take the
form as shown in the Figure31

A B
Figure 31
If Type 2 system is excited by Ramp input then,

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This is true for any system of Type more than one. Hence all systems of TYPE 2
and more than two follow ramp type of input with negligible small error and
may take the form as shown in the Figure32

A B
Figure 32
Let us now change the selected input from ramp to parabolic input of magnitude
A hence coefficient K. will control the error.

● Consider Type 0 system :

● Consider Type 1 system :

For both Type ‘0’ and ‘l’ systems, error will be very large and uncontrollable if
parabolic input is used. Hence parabolic input should not be used as a reference

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to excite Type ‘0’ and Type ‘1’ systems. The output may take the form as shown
in the Figure 33 if excited by such input.

A B
Figure 33
● If Type 2 system is used i.e.

Hence Type 2 systems will follow Parabolic input with finite error A/K which
can be controlled by change in A or K or both and output may take form as
shown in the Figure 34

A B
Figure 34
And for any system of Type 3 or more if parabolic input is used, error will be
negligibly small. The results can be summarized as shown in the Table below.

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Type of Error Coefficients Error ess for

System
Kp Kv Ka Step input Ramp input Parabolic
input
0 K 0 0 ∞ ∞

1 ∞ K 0 0 ∞

2 ∞ ∞ K 0 0

Table
Disadvantages of Static Error Coefficient Method
The disadvantages of Static Error Coefficient Method are:
● Method cannot give the error if inputs are other than the three standard test
inputs.
● Most of the times, method gives mathematical answer of the error as '0' or
'infinite' and hence does not provide precise value of the error.
● Method does not provide variation of error with respect to time, which will be
otherwise very useful from design point of view.
● Error constants are defined for the loop transfer function G(s)H(s), strictly,
hence the method is applicable to only the system configuration shown in the
Figure 36
● As final value theorem is used to calculate steady state error so before applying
it is necessary to check if sE(s) has any poles on the jo axis or in the right half of
the s-plane. This means before applying this method, the system must be
checked for its stability. The method cannot be applied to unstable systems.
● When the system configuration is different than as shown above then it is
necessary to establish the expression for the error signal and apply the final
value theorem directly, without the use of error coefficients.
● The method is applicable only to stable systems.

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Figure 35
Because of these disadvantages Dynamic error coefficient method (Error series
method) is developed which eliminates above said disadvantages.

Disadvantages of Static Error Coefficient Method

● Method cannot give the error if inputs are other than the three standard test
inputs
● Most of the times, method gives mathematical answer of the error as ‘0’ or
‘infinite’ and hence does not provide precise values of the error
● Method does not provide variation of error with respect to time, which will be
otherwise very useful from design point of view

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Experiment 1
Objective: To observe the First Order control system for different values of the
Damping Ratio at different values of resistance
Equipments Needed:
● Scientech 2454 Control System Simulator
● Patch Cords
● Scientech 803/831, or equivalent
● BNC to Test Probe
● Multimeter
Procedure:
● Connect Square Wave Output to Input of I Order system. The output of the
square wave is shown in figure below.

Figure 36
● Make sure that Jumpers are connected in the I Order block.
● Connect the Test Probe to the output of I Order Block.
● Change the value of the Potentiometer given in I Order Block. The output of the
Ist order system at both extreme ends of Potentiometer is as shown in the figure
below.

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Figure 37
Extreme end 1

Extreme end 2
Figure 38
● Observe the change in Damping Ratio of the waveform in output.
● At every change at output disconnect the Jumpers by removing the cap and read
the value of Potentiometer with the help of multimeter set at Ohmmeter.

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Experiment 2
Objective: To observe the Second Order control system for different values of the
Damping Ratio at different values of resistance
Equipments Needed:
● Scientech 2454 Control System Simulator
● Patch Cords
● Scientech 803/831, or equivalent
● BNC to Test Probe
● Multimeter
Procedure:
● Connect Square Wave Output to Input of II Order system. The output of the
square wave is shown in figure below.

Figure 39
● Make sure that Jumpers are connected in the II Order block.
● Connect the Test Probe to the output of II Order Block.
● Change the value of the Potentiometer given in II Order Block. The output of
the IInd order system at both extreme ends of Potentiometer is as shown in the
figure below.

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Figure 40
Extreme end 1

Figure 41
Extreme end 2

● Observe the change in Damping Ratio of the waveform in output.

● At every change at output disconnect the Jumpers by removing the cap and read
the value of Potentiometer with the help of multimeter set at Ohmmeter.

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Experiment 3
Objective: To observe the Third Order control system for different values of the
Damping Ratio at different values of resistance
Equipments Needed:
● Scientech 2454 Control System Simulator
● Patch Cords
● Scientech 803/831, or equivalent
● BNC to Test Probe
● Multimeter
Procedure:
● Connect Square Wave Output to Input of III Order system. The output of the
square wave is shown in figure below.

Figure 42
● Make sure that Jumpers are connected in the III Order block.
● Connect the Test Probe to the output of III Order Block.
● Change the value of the Potentiometer given in III Order Block. The output of
the IIIrd order system at both extreme ends of Potentiometer is as shown in the
figure below.

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Figure 43
Extreme end 1

Figure 44
Extreme end 2

● Observe the change in Damping Ratio of the waveform in output.

● At every change at output disconnect the Jumpers by removing the cap and read
the value of Potentiometer with the help of multimeter set at Ohmmeter.

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Experiment 4
Objective: To observe the Type0 control system Steady State Error (Ess) for Unit
Step or Square wave input.
Equipments Needed:
● Scientech 2454 Control System Simulator
● Patch Cords
● Scientech 803/831, or equivalent
● BNC to Test Probe
Procedure:
● Connect Unit Step output to Type0 input. The output of square wave is shown in
figure below.

\
Figure 45
● Connect the Test Probe to the output of Type 0 Block.
● Connect the Input of Unit Step or square wave to the buffer input. The output of
the buffer should be connected to Vi and also connect it to the input of type 0
system. Connect Output of Type0 block to Vo. The connection is as shown
below

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Figure 46
● Switch the toggle switch from Low to High in case of Unit step input.
● Observe the Output of Type0 block. The output of the Type 0 with Square wave
input is as shown in the figure below.

Figure 47

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● Connect the Test Probe to Ess output.

● Observe the output at Ess. The output will be shown in figure below.

Figure 48

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Experiment 5
Objective: To observe the Type0 control system Steady state error (Ess) for Ramp as
input.
Equipments Needed:
● Scientech 2454 Control System Simulator
● Patch Cords
● Scientech 803/831, or equivalent
● BNC to Test Probe
Procedure:
● Connect Ramp output to Type0 input.
● Connect the Test Probe to the output of Type 0 Block.
● Select the Ramp Function with the toggle switch. The output of ramp function is
as shown in figure below.

Figure 49
● Connect the Input of Ramp wave to the buffer input. The output of the buffer
should be connected to Vi and also connect it to the input of type 0 system.
Connect Output of Type0 block to Vo. The connection is as shown below

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Figure 50
● Observe the Output of Type0 block. The output will be shown in figure below.

Figure 51

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● Connect the Test Probe to Ess.

● Observe the output at Ess. The output will be as shown in figure below.

Figure 52

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Experiment 6
Objective: To observe the Type0 control system Steady state error (Ess) for Parabolic
as input.
Equipments Needed:
● Scientech 2454 Control System Simulator
● Patch Cords
● Scientech 803/831, or equivalent
● BNC to Test Probe
Procedure:
● Connect Parabolic output to Type0 input.
● Connect the Test Probe to the output of Type 0 Block.
● Select the Parabolic Function with the toggle switch. The output will be as
shown in the figure below.

Figure 53
● Connect the Input of Ramp wave to the buffer input. The output of the buffer
should be connected to Vi and also connect it to the input of type 0 system.
Connect Output of Type0 block to Vo. The connection is as shown below

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Figure 54
● Observe the Output of Type0 block. The output will be as shown in figure
below.

Figure 55

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● Connect the Test Probe to Ess.

● Observe the output at Ess. The output will be as shown in the figure below.

Figure 56

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Experiment 7
Objective: To observe the Type1 control system Steady State Error (Ess) for Unit
Step or Square wave input.
Equipments Needed:
● Scientech 2454 Control System Simulator
● Patch Cords
● Scientech 803/831, or equivalent
● BNC to Test Probe
Procedure:
● Connect Unit Step output to Type1 input. The output of square wave is shown in
figure below.

\
Figure 57
● Connect the Test Probe to the output of Type 1 Block.
● Connect the Input of Unit Step or square wave to the buffer input. The output of
the buffer should be connected to Vi and also connect it to the input of type 1
system. Connect Output of Type1 block to Vo. The connection is as shown
below

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Figure 58
● Switch the toggle switch from Low to High in case of Unit step input.
● Observe the Output of Type1 block. The output of the Type 1 with Square wave
input is as shown in the figure below.

Figure 59

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● Connect the Test Probe to Ess output.

● Observe the output at Ess. The output will be shown in figure below.

Figure 60

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Experiment 8
Objective: To observe the Type1 control system Steady state error (Ess) for Ramp as
input.
Equipments Needed:
● Scientech 2454 Control System Simulator
● Patch Cords
● Scientech 803/831, or equivalent
● BNC to Test Probe
Procedure:
● Connect Ramp output to Type1 input.
● Connect the Test Probe to the output of Type 1 Block.
● Select the Ramp Function with the toggle switch. The output of ramp function is
as shown in figure below.

Figure 61
● Connect the Input of Ramp wave to the buffer input. The output of the buffer
should be connected to Vi and also connect it to the input of type 1 system.
Connect Output of Type1 block to Vo. The connection is as shown below

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Figure 62
● Observe the Output of Type1 block. The output will be shown in figure below.

Figure 63

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● Connect the Test Probe to Ess.

● Observe the output at Ess. The output will be as shown in figure below.

Figure 64

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Experiment 9
Objective: To observe the Type1 control system Steady state error (Ess) for Parabolic
as input.
Equipments Needed:
● Scientech 2454 Control System Simulator
● Patch Cords
● Scientech 803/831, or equivalent
● BNC to Test Probe
Procedure:
● Connect Parabolic output to Type1 input.
● Connect the Test Probe to the output of Type 1 Block.
● Select the Parabolic Function with the toggle switch. The output will be as
shown in the figure below.

Figure 65

● Connect the Input of Ramp wave to the buffer input. The output of the buffer
should be connected to Vi and also connect it to the input of type 1 system.
Connect Output of Type1 block to Vo. The connection is as shown below

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Figure 66
● Observe the Output of Type1 block. The output will be as shown in figure
below.

Figure 67

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● Connect the Test Probe to Ess.

● Observe the output at Ess. The output will be as shown in the figure below.

Figure 68

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Experiment 10
Objective: To observe the Type2 control system Steady State Error (Ess) for Unit
Step or Square wave input.
Equipments Needed:
● Scientech 2454 Control System Simulator
● Patch Cords
● Scientech 803/831, or equivalent
● BNC to Test Probe
Procedure:
● Connect Unit Step output to Type2 input. The output of square wave is shown in
figure below.

Figure 69

● Connect the Test Probe to the output of Type 2 Block.

● Connect the Input of Unit Step or square wave to the buffer input. The output of
the buffer should be connected to Vi and also connect it to the input of type 2
system. Connect Output of Type2 block to Vo. The connection is as shown
below

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Figure 70
● Switch the toggle switch from Low to High in case of Unit step input.
● Observe the Output of Type2 block. The output of the Type 2 with Square wave
input is as shown in the figure below.

Figure 71

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● Connect the Test Probe to Ess output.

● Observe the output at Ess. The output will be shown in figure below.

Figure 72

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Experiment 11
Objective: To observe the Type2 control system Steady state error (Ess) for Ramp as
input.
Equipments Needed:
● Scientech 2454 Control System Simulator
● Patch Cords
● Scientech 803/831, or equivalent
● BNC to Test Probe
Procedure:
● Connect Ramp output to Type2 input.
● Connect the Test Probe to the output of Type 2 Block.
● Select the Ramp Function with the toggle switch. The output of ramp function is
as shown in figure below.

Figure 73

● Connect the Input of Ramp wave to the buffer input. The output of the buffer
should be connected to Vi and also connect it to the input of type 2 system.
Connect Output of Type2 block to Vo. The connection is as shown below

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Figure 74
● Observe the Output of Type2 block. The output will be shown in figure below.

Figure 75

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● Connect the Test Probe to Ess.

● Observe the output at Ess. The output will be as shown in figure below.

Figure 76

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Experiment 12
Objective: To observe the Type2 control system Steady state error (Ess) for Parabolic
as input.
Equipments Needed:
● Scientech 2454 Control System Simulator
● Patch Cords
● Scientech 803/831, or equivalent
● BNC to Test Probe
Procedure:
● Connect Parabolic output to Type1 input.
● Connect the Test Probe to the output of Type 2 Block.
● Select the Parabolic Function with the toggle switch. The output will be as
shown in the figure below.

Figure 77

● Connect the Input of Ramp wave to the buffer input. The output of the buffer
should be connected to Vi and also connect it to the input of type 2 system.
Connect Output of Type2 block to Vo. The connection is as shown below

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Figure 78
● Observe the Output of Type2 block. The output will be as shown in figure
below.

Figure 79

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● Connect the Test Probe to Ess.

● Observe the output at Ess. The output will be as shown in the figure below.

Figure 80

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Scientech 2454

Additional Experiments
Experiment 1
Objective: To observe the First Order control system for different values of the
Damping Ratio at different values of resistance
Equipments Needed:
● Scientech 2454 Control System Simulator
● Patch Cords
● Scientech 803/831, or equivalent
● BNC to Test Probe
Procedure:
● Connect Square Wave to any of the resistances given in resistance bank.
● Now connect the other and of the resistance to the capacitor given in capacitor
bank.
● Connect the capacitor’s other end to ground.
● Connect the test probe to the junction of resistor and capacitor.

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Scientech 2454

Experiment 2
Objective: To observe the Second Order control system for different values of the
Damping Ratio at different values of resistance
Equipments Needed:
● Scientech 2454 Control System Simulator
● Patch Cords
● Scientech 803/831, or equivalent
● BNC to Test Probe
Procedure:
● Connect Square Wave to any of the resistances given in resistance bank.
● Now connect the other and of the resistance to the inductor given in inductor
bank.
● Connect the inductor’s other end to the capacitor given in capacitor bank.
● Connect capacitor’s other end to ground.
● Connect the test probe to the junction of inductor and capacitor.

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Scientech 2454

Experiment 3
Objective: To observe the Third Order control system for different values of the
Damping Ratio at different values of resistance
Equipments Needed:
● Scientech 2454 Control System Simulator
● Patch Cords
● Scientech 803/831, or equivalent
● BNC to Test Probe
Procedure:
● Connect output of Square Wave to any of the resistance R1 given in resistance
bank.
● Now connect the other and of the resistance to the capacitor C1 given in
capacitor bank.
● Connect the junction of resistor R1 and capacitor C1 to the inductor given in
inductor bank.
● Connect Resistor R2 given in resistor bank to the other end of the inductor.
● Capacitor C2 from the other end of the resistor R2
● Connect the capacitor C1 and C2’s other end to ground.
Connect the test probe to the junction of resistor R2 and capacitor C2.

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Scientech 2454

Warranty
● We guarantee the product against all manufacturing defects for 24 months from
the date of sale by us or through our dealers.
● The guarantee will become void, if
• The product is not operated as per the instruction given in the operating
manual.
• The agreed payment terms and other conditions of sale are not followed.
• The customer resells the instrument to another party.
• Any attempt is made to service and modify the instrument.
● The non-working of the product is to be communicated to us immediately giving
full details of the complaints and defects noticed specifically mentioning the
type, serial number of the product and date of purchase etc.
● The repair work will be carried out, provided the product is dispatched securely
packed and insured. The transportation charges shall be borne by the customer.

List of Accessories
● Patch Cord 8" ......................................................................................... 16 Nos.
● Mains Cord................................................................................................. 1 No.
● Learning Material (CD) .............................................................................. 1 No.
● TechBook Power Supply ............................................................................ 1 No.

----------------------------------- The End--------------------------------------------

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