KEC-602 Control System Unit-3 gandgfdghhg

Control System (KEC-602)
Unit-3

Objectives and Outcomes of Unit-3
• Analyze the output of a control system and present it in time domain
• Analysis of system is done in s-domain using Laplace transform
• Inverse Laplace transform is applied to describe the system
• Able to describe different types of systems in time domain
• Calculate steady state error of a system
• Obtain the transient and steady state response of the given system
• Able to calculate Time domain parameters

Time Domain Analysis
• Time domain analysis refers to the analysis of system performance in time i.e., the study of
evolution of system variables (specifically output) with time
• There are two common ways of analysing the response of systems:
1. Natural response and forced response
2. Transient response and steady state response
• In both cases, the complete response of the system is given by the combination of both responses
i.e., natural and forced responses or transient and steady state responses

Natural and Forced Responses
•Natural response (Zero input response) :
–System's response to initial conditions with all external forces set to zero
–E.g. In RLC circuits, this would be the response of the circuit to initial conditions (inductor
currents or capacitor voltages) with all the independent voltage and current sources set to zero
•Forced response (Zero state response) :
–System's response to external forces with zero initial conditions
–E.g. In RLC circuits, this would be the response of the circuit to only external voltage and
current source, and zero initial conditions

Transient and Steady State Responses
•Transient response 𝒚tr(𝒕)
–Part of the time response that goes to zero as time tends to be large
–Transient response can be tied to any event that affects the equilibrium of a system viz.
switching, disturbance, change in input, etc. lim(𝑡→∞) 𝑦tr(𝑡) = 0
•Steady state response 𝒚𝒔𝒔(𝒕):
–Steady state response is the time response of a system after transient practically vanishes and
as time goes to infinity
𝑦(𝑡)=𝑦𝑡𝑟(𝑡) + 𝑦𝑠𝑠(𝑡)

Standard Test Inputs
• In most cases, the input signals to a control system are not known prior to design of
control system
• Hence to analyse the performance of a control system, it is excited with standard test
signals
• In general, control system design specifications are also based on the response of the
system to such test signals
• Standard test signals include:
– Unit impulse, unit step (sudden change), ramp (constant velocity), parabolic
(constant acceleration) and sinusoidal
– These inputs are chosen because they capture many of the possible variations
that can occur in an arbitrary input signal

Standard Test Inputs
• Unit impulse signal:
– A signal which is non-zero only at 𝑡=0 and
integrates to one
ℒ[𝛿𝑡]=1
• Unit step signal:
– A signal that switches to one at a time instant
and stays there indefinitely
U(t) = 1 ∀ 𝑡 > 0
0 ∀ 𝑡 < 0
ℒ[u(t)] =
1
𝑠

• Ramp signal:
– A signal which increases linearly with time
𝑥(r)𝑡 = 𝐴𝑡 ∀ 𝑡≥0
0 ∀ 𝑡<0
ℒ[𝑥(𝑡)] =
𝐴
𝑠2
• Parabolic signal:
𝑥(𝑡) = (𝐴/2)𝑡2 ∀ 𝑡 ≥ 0
0 ∀ 𝑡<0
ℒ[𝑥(𝑡)] =
𝐴
𝑠3

Standard Inputs in time and s-domain
1. Unit Impulse Signal
𝛿 𝑡 = 1 𝑓𝑜𝑟 𝑡 = 0 𝑎𝑛𝑑 𝐿 𝛿 𝑡 = 1
2. Unit Step Signal
u 𝑡 = 1 𝑓𝑜𝑟 𝑡 ≥ 0 𝑎𝑛𝑑 𝐿 𝑢(𝑡) =
1
𝑠
3. Unit Ramp Signal
r 𝑡 = 𝑡 𝑓𝑜𝑟 𝑡 ≥ 0 𝑎𝑛𝑑 𝐿 𝑟(𝑡) =
1
𝑠2
4. Unit Parabolic Signal
𝑝 𝑡 =
𝑡2
2
𝑓𝑜𝑟 𝑡 ≥ 0 𝑎𝑛𝑑𝐿 𝑝 𝑡 =
1
𝑠3

Some terminologies
• Define the following
1. Type and Order of a System
2. Poles and Zeroes of a System
3. Open loop gain
4. Closed loop gain
5. Loop Gain
6. Unity feedback system
7. Characteristic Polynomial/Equation

1st Order Systems
• Systems with only one pole are called 1st order systems
𝝉: System time constant
– It characterizes the speed of response of a system to an input
– Higher the time constant, slower the response and vice-versa
Block Diagram of a 1st order system

Time Response of First Order Systems
Q1. Derive the expression for the output response of a 1st order system
when different standard test signals are applied at the input.
Q2. Identify the transient and steady state components of the time
response.

Impulse Response of 1st Order Systems

KEC-602 Control System Unit-3 gandgfdghhg

Step Response of 1st Order Systems

Ramp Response of 1st Order Systems

Parabolic Response of 1st Order Systems

Parameters of 2nd Order Systems

Impulse Response of 2nd Order Systems

Step Response of 2nd Order Systems

Steady State Error for Standard Inputs

Features of Steady State Error

Steady State Error for Different Types

Static Error Coefficients
There are 3 types of static error coefficients. These are
1. Position error coefficient (Kp)
𝐾𝑝 = lim
𝑠→0
𝐺 𝑠 𝐻(𝑠)
2. Velocity error coefficient (Kv)
𝐾𝑣 = lim
𝑠→0
𝑠. 𝐺 𝑠 𝐻(𝑠)
3. Acceleration error coefficient (Ka)
𝐾𝑎 = lim
𝑠→0
𝑠2. 𝐺 𝑠 𝐻(𝑠)

Input
Type
Unit Step Unit Ramp Unit Parabolic
Type-0 1
1 + 𝐾𝑝
∞ ∞
Type-1 0
1
𝐾𝑣
∞
Type-2 0 0
1
𝐾𝑎
Relation between steady state error and static error coefficients

Dynamic Error Coefficients
• Static error coefficients are limited to standard inputs like step, ramp and parabolic signals
• Static error coefficients do not describe the variation in error w.r.t. time
• To solve these issues we define dynamic error and dynamic error coefficients
Dynamic error is written as:
𝑒𝑑 𝑡 = 𝐶0𝑟 𝑡 + 𝐶1
𝑑𝑟(𝑡)
𝑑𝑡
+ 𝐶2
𝑑2𝑟(𝑡)
𝑑𝑡2 + 𝐶3
𝑑3𝑟(𝑡)
𝑑𝑡3 + _ _ _ _ _ _
where, C0, C1, C2, C3, and so on are the dynamic error coefficients and given as:
𝐶0 = lim
𝑠→0
𝑇𝑒(𝑠)
𝐶1 = lim
𝑠→0
𝑑𝑇𝑒(𝑠)
𝑑𝑠
𝐶2 = lim
𝑠→0
𝑑2𝑇𝑒(𝑠)
𝑑𝑠2
𝐶3 = lim
𝑠→0
𝑑2
𝑇𝑒(𝑠)
𝑑𝑠2
where, 𝑇𝑒(𝑠) =
𝐸(𝑠)
𝑅(𝑠)
is called the error transfer function.

KEC-602 Control System Unit-3 gandgfdghhg

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