CE PPT

1. Kaushik Bhuva 150050119010
Deep Chaudhary 150050119011
ME-1 , A batch
Subject: Control Engineering(2151908)
Topic: Time response analysis

2. Time Response
• Time response of system is defined as the output of a system
when subjected when to an input which is a functions of
time.
• In the block diagram representation and signal flow graphs
we studied how to obtain the transfer function of a physical
system. We have also seen how to combine individual
transfer functions to get a single transfer function using
block diagram reduction and signal flow graphs. We shall
now study how this block responds to different inputs, i.e. we
will now take closer look at the response characteristics of
the control system.

3. Inputs Supplied to a System
1. Impulse input :
Impulse represents a sudden change in input. An Impulse is infinite at t
= 0 and zero
everywhere else. The area under the curve is 1. A unit impulse has
magnitude 1 at t = 0.
r(t) = δ(t) = 1 t = 0
= 0 t ≠ 0
In the Laplace domain we have
L[r(t)] = L[δ(t)] = 1
Impulse inputs are used to derived a mathematical model of the system.

4. 2. Step input :
A step input represents a constant command such as position. The input given to an
elevator
is step input. Another example of a step input is setting the temperature of an air
conditioner.
A step signal is given by the formula,
r(t) = u(t) = A t ≥ 0
= 0 otherwise , If A = 1, it is called step.
In Laplace domain, we have
L[r(t)] = R(s) =
𝐴
𝑠
In case of a unit step, we get L [ r(t)] = R(s) =
1
𝑠

5. 3. Ramp input :
The ramp input represents a linearly increasing input command. It is given by
formula,
r(t) = At t ≥ 0 ; Here A is the slope.
= 0 t < 0
If A = 1, it is called a unit ramp.
In the Laplace domain we have,
L [r(t)] = R(s) =
𝐴
𝑠2
In case of unit ramp, we have R(s) =
1
𝑠2
System are subjected to Ramp inputs when we need to study the system
for linear increasing functions like velocity.

6. 4. Parabolic input :
Rate of change of velocity is acceleration. Acceleration is a parabolic function. It
given by the formula,
r(t) =
𝐴𝑡2
2
t ≥ 0
= 0 t < 0
If A = 1, it is called a unit parabola. In the Laplace domain we have
L [r(t)] = R(s) =
𝐴
𝑠3
In case of unit parabola, we have R(s) =
1
𝑠3

7. Steady State Response
• “Steady state response is that part of the output response where the output signal
remains constant.”
• The parameter that is important in the steady state response is the steady state error
(𝑒𝑠𝑠).
• Error in general is the difference between the input and desirable output. Steady
state error is the error at t → ∞.
∴ 𝑒𝑠𝑠 = lim
𝑡 →∞
𝐸𝑟𝑟𝑜𝑟
• By derivation of formula for steady state error,
𝑒𝑠𝑠 = lim
𝑠 →0
𝑠 ∙
𝑅(𝑠)
1+𝐺 𝑠 𝐻(𝑠)

8. Effect of Input R(s) on Steady State Error
1. Step input :
The system subjected to a step input, the steady state error is controlled by
position error coefficient 𝑘 𝑝. Refer the figure Ignore the transient part. The input is
shown by dotted line while response is shown by a firm line. The equation below is
describe steady state error for step input.
𝑒𝑠𝑠 =
𝐴
1+ 𝑘 𝑝

9. 2. Ramp input :
For ramp input; the velocity error coefficient 𝑘 𝑣 will control the steady state
error. Refer figure ignore the transient part. The input is shown by a dotted line while
the response is shown by a firm line. The equation below describe the steady state
transient error for the ramp input.
𝑒𝑠𝑠 =
𝐴
𝑘 𝑣

10. 3. Parabolic input :
When parabolic input signal is applied, the acceleration error coefficient
the steady state error of the system. Refer figure ignore transient time. The parabolic
input is shown by dotted line and response by a firm line. The equation below is
the steady state error for the parabolic time response.
𝑒𝑠𝑠 =
𝐴
𝑘 𝑎

11. Effect of Open Loop Transfer Function on
Steady State Error
• The steady state error(𝑒𝑠𝑠) also depends on G(s)H(s). Actually 𝑒𝑠𝑠 depends on the
Type of system G(s)H(s).
• Type of system is defined by the number of open loop Poles i.e., poles of G(s)H(s)
that are present at the origin.
• The open loop transfer function written in time constant form is,
𝐺 𝑠 𝐻 𝑠 =
𝑘1 1+𝑇𝑧1 𝑠 1+𝑇𝑧2 𝑠 …
𝑠 𝑛 1+𝑇𝑝1 𝑠 1+𝑇𝑝2 𝑠 …
• The open loop transfer function written in pole-zero form is written as
𝐺 𝑠 𝐻 𝑠 =
𝑘 𝑠+𝑧1 𝑠+𝑧2 …
𝑠 𝑛 𝑠+𝑝1 𝑠+𝑝2 …
• Here n is number of poles at the origin. It is very easy to obtain one from the other.

12. Subjecting a Type 0,1 & 2 Systems to a
Step, Ramp & Parabolic Input
Sr. No. Type Step Input Ramp Input Parabolic Input
𝒌 𝒑 𝒆 𝒔𝒔 𝒌 𝒗
𝒆 𝒔𝒔 𝒌 𝒂
𝒆 𝒔𝒔
(1) Type
Zero
K 𝐴
1 + 𝑘
0 ∞ 0 ∞
(2) Type
One
∞ 0 k 𝐴
𝑘
0 ∞
(3) Type ∞ 0 ∞ 0 K 𝐴
𝑘

13. Transient Response
• In the earlier section, we discussed the steady state response in detail and found a
method of calculating the steady state error.
• We realized that 𝑒𝑠𝑠 was dependent on the Type of the system i.e. the number of
poles, the system had at the origin.
• The transient response of the system depends on the order of system. Order of a
system is the highest power of s in the denominator of closed loop transfer
function.
• Hence for transient response, we need to work with the closed loop transfer
function,
𝐺(𝑠)
1+𝐺 𝑠 𝐻(𝑠)

14. Analysis of First Order Systems
• General form:
• Problem: Derive the transfer function for the following circuit
1)(
)(
)(


s
K
sR
sC
sG

1
1
)(


RCs
sG

15. • Transient Response: Gradual change of output from initial to the desired condition.
• Block diagram representation:
• By definition itself, the input to the system should be a step function which is given
by the following:
C(s)R(s)
1s
K

Where,
K : Gain
 : Time constant
s
sR
1
)( 

16. • General form:
• Output response:
1)(
)(
)(


s
K
sR
sC
sG

)()()( sRsGsC 
1
1
1
)(
















s
B
s
A
s
K
s
sC




t
e
B
Atc 
)(

17. Analysis of Second Order Systems
• General form:
• Roots of denominator:
  22
2
2 nn
n
ss
K
sG




Where,
K : Gain
ς : Damping ratio
n : Undamped natural frequency
02 22
 nnss 
12
2,1   nns

18. • Natural frequency, n
Frequency of oscillation of the system without damping.
• Damping ratio, ς
Quantity that compares the exponential decay frequency of the
envelope to the natural frequency.
(rad/s)frequencyNatural
frequencydecaylExponentia


19.  Step responses for second-order system damping cases

20. • When 0 < ς < 1, the transfer function is given by the following.
• Pole position:
 
  dndn
n
jsjs
K
sG




2 Where,
2
1   nd

21. Transient Response Specifications
1. Delay time (Td) :
It is the time required for the response to reach 50% of the final value in the first
attempt. It is given by the formula,
1+0.7𝜉
𝜔 𝑛
sec
2. Rise time (Tr) :
It is the time required by response to rise from 10% to 90% of the final value for
a overdamped system. For a underdamped system (our case) the rise time is the taken
for the response to rise from 100% of the final value in the attempt. It is given by the
formula,
𝜋−Θ
𝜔 𝑑
sec

22. 3. Peak time (Tp) :
It is the time required by the response to reach its first peak. The first
peak is always the maximum peak,
𝜋
𝜔 𝑑
sec
4. Setting time (Ts) :
It is defined as the time required for the transient damped oscillations
to reach and stay within a specified tolerance band (usually 2% of the input
value).
4
𝜉𝜔 𝑛
sec

23. 5. Peak overshoot (Mp) :
It is the maximum peak value of the response measured from the input signal
value. It is also maximum error between input and output. It is generally written in
terms of percentage,
%𝑀 𝑝 = 𝑒
−𝜉𝜋
1−𝜉2
∗ 100

24. Thank You