lecture3_time_domin.pdf

SPC318: System Modeling and Linear Systems
Lecture 3: Time Domain Analysis of 1st Order Systems
Dr. Haitham El-Hussieny
Adjunct Lecturer
Space and Communication Engineering
Zewail City of Science and Technology
Fall 2016
Dr. Haitham El-Hussieny SPC318: System Modeling and Linear Systems 1 / 31

Lecture Outline:
1 Modeling in State-Space Representation.
2 Introduction to Time Domain Analysis.
3 Time Response of 1st Order Systems.
4 Identification of the 1st Order Transfer Function.
5 First Order System with a Zero.

Table of Contents

Modeling in State-Space Representation:
Definition
State: The state of a dynamic system is the smallest set of variables (called state variables)
such that knowledge of these variables at t = t0, together with knowledge of the input for
t ≥ t0, completely determines the behavior of the system for any time t ≥ t0.
For example, the system:
ÿ = aẏ + by + cu
The variables that complete the knowledge about the system are: the input u and two state
variables: x1 = y and x2 = ẏ.
State Vector: The n state variables are considered the n components of a vector x.
x =

x1
x2

State-space: The n-dimensional space whose coordinate axes are x1, x2, . . . , xn.

State-space Equations:
In state-space analysis we are concerned with three variables: input, output and states.
In state-space the LTI system is described as:
ẋ = Ax + Bu
y = Cx + Du
A: State matrix.
B: Input matrix.
C: Output matrix.
D: Transmission matrix.
Block diagram of the LTI system represented in
state-space.

Correlation Between Transfer Functions and State-Space Equations:
G(s) =
Y (s)
U(s)
= C(sI − A)−1
B + D I : Identity Matrix
Example
Solution

Table of Contents

Introduction to Time Domain Analysis:
Time Domain Analysis:
It is a method to analyze a certain system’s performance.
It measures the response of a dynamic system to an input as a function of time.
To perform a time-domain analysis we need:
I The mathematical model for the physical system to be analyze.
I An analytical expression for the input signal in time domain.

Problem:
In practice, the input signal to a system is
not known ahead of time. So, the input
cannot be expressed analytically!

Problem:
In practice, the input signal to a system is
not known ahead of time. So, the input
cannot be expressed analytically!
Solution:
The performance of a dynamic system is
therefore judged and compared under
application of standard test signals: an
impulse, a step, a ramp, and parabolic
signals.

Standard Test Input Signals:
Impulse
Sudden shock
δ(t) =

A, t = 0
0, t 6= 0

If A = 1: Unit Impulse
L{δ(t)} = δ(s) = A

Impulse
Sudden shock
δ(t) =

A, t = 0
0, t 6= 0

L{δ(t)} = δ(s) = A
Step
Sudden change
u(t) =

A, t ≥ 0
0, t 0

If A = 1: Unit Step
L{u(t)} = U(s) =
A
s

Impulse
Sudden shock
δ(t) =

A, t = 0
0, t 6= 0

L{δ(t)} = δ(s) = A
Step
Sudden change
u(t) =

A, t ≥ 0
0, t 0

If A = 1: Unit Step
L{u(t)} = U(s) =
A
s
Ramp
Ramp change
r(t) =

At, t ≥ 0
0, t 0

If A = 1: Unit Ramp
L{r(t)} = R(s) =
A
s2

Impulse
Sudden shock
δ(t) =

A, t = 0
0, t 6= 0

L{δ(t)} = δ(s) = A
Step
Sudden change
u(t) =

A, t ≥ 0
0, t 0

If A = 1: Unit Step
L{u(t)} = U(s) =
A
s
Ramp
Ramp change
r(t) =

At, t ≥ 0
0, t 0

If A = 1: Unit Ramp
L{r(t)} = R(s) =
A
s2
Parabolic
Parabolic change
p(t) =

At2, t ≥ 0
0, t 0

If A = 1: Unit Parabolic
L{p(t)} = P(s) =
2A
s3

Time Response of Physical Systems:
It measures the response (output) of the physical system in time-domain when subjected to
one of the standard test signals as an (input).

It measures the response (output) of the physical system in time-domain when subjected to
one of the standard test signals as an (input).
To find the time-response, y(t):
1 Write the system TF:
G(s) =
Y (s)
U(s)
2 Find Y (s):
Y (s) = G(s) U(s)
3 Apply a test input U(s).
4 Find y(t):
y(t) = L−1
{G(s) U(s)}

When the response of the system is changed form rest or equilibrium it takes some time to
settle down.

settle down.
The time response of a system
consists of two components:
1 Transient response:
The response of a system
from rest or equilibrium to
settle down at steady state.
2 Steady-state response:
The response of the system
after the transient response.

settle down.
Remarks on time-response analysis:
1 The transient response depends upon the system
poles only and not on the type of input.
2 It is sufficient to analyze the transient response using a
step input.
3 The steady-state response depends on system
dynamics and the input quantity.
4 The steady-state response is examined using different
test signals by final value theorem.
The time response of a system
consists of two components:
1 Transient response:
The response of a system
from rest or equilibrium to
settle down at steady state.
2 Steady-state response:
The response of the system
after the transient response.

Table of Contents

Time Response of 1st
Order Systems:
The first order system has only one pole.
G(s) =
Y (s)
U(s)
=
K
Ts + 1
K is the D.C gain and T is the time constant of
the system.
Time constant T is a measure of how quickly a
1st order system responds to a unit step input.
D.C Gain K of the system is ratio between the
input signal and the output signal at the steady
state. (i.e. s = 0 or t = ∞)

Order Systems:
G(s) =
Y (s)
U(s)
=
K
Ts + 1
the system.
state. (i.e. s = 0 or t = ∞)
Examples:
[1] G(s) =
10
3s + 1
K = 10 and T = 3

Order Systems:
G(s) =
Y (s)
U(s)
=
K
Ts + 1
the system.
state. (i.e. s = 0 or t = ∞)
Examples:
[1] G(s) =
10
3s + 1
K = 10 and T = 3
[2] G(s) =
3
s + 5
in standard form:
G(s) =
3/5
(1/5)s + 1
K = 3/5 and T = 1/5

Impulse response of 1st
Order Systems:
Y (s) = G(s)δ(s)
since δ(s) = 1 (unit impulse),
Y (s) =
K
Ts + 1
rearrange,
Y (s) =
K/T
s + 1/T
Find response y(t) = L−1{Y (s)}:
y(t) = L−1
{K
1/T
s + 1/T
}
y(t) =
K
T
e−t/T

Order Systems:
Example
Find the unit impulse
response for a first
order system with
T = 2sec. and K = 3.
y(t) =
K
T
e−t/T
y(t) =
3
2
e−t/2

Order Systems:
Example
Find the unit impulse
order system with
y(t) =
K
T
e−t/T
y(t) =
3
2
e−t/2
Or

Step response of 1st
Order Systems:
Y (s) = G(s)U(s)
since U(s) =
1
s
(unit step),
Y (s) =
K
s(Ts + 1)
by partial fraction,
Y (s) =
K
s
−
KT
Ts + 1
Find response y(t) = L−1{Y (s)}:
y(t) = L−1
{K(
1
s
−
T
Ts + 1
)}
y(t) = K(1 − e−t/T
)

Order Systems:
Example
Find the unit step
order system with
y(t) = K(1 − e−t/T
)
y(t) = 3(1 − e−t/2
)
Or
When t = T:
y(t) = K(1−e−1
) = 0.632K

Order Systems:
System takes five time constants (i.e. t = 5T) to
reach its final value K.
The smaller the time constant T, the faster the
system response.
The slope of the tangent line at t = 0 is 1/T,

Order Systems:
Example
Find the unit step response of a first order system:
1 When T = 2 and K = 1, 3 and 10.
2 When K = 10 and T = 2, 4 and 10.
(1) Changing D.C. gain:

Order Systems:
Example
Find the unit step response of a first order system:
1 When T = 2 and K = 1, 3 and 10.
2 When K = 10 and T = 2, 4 and 10.
(2) Changing Time constant:

Relation Between Step and Impulse response:
The step response of the first order system is:
y(t) = K(1 − e−t/T
)
dy(t)
dt
=
d
dt
{K(1 − e−t/T
)}
dy(t)
dt
=
K
T
e−t/T
Impulse response
The Impulse response of the first order
system is:
y(t) =
K
T
e−t/T
Z
y(t)dt =
Z
K
T
e−t/T
dt
Z
y(t)dt = −Ke−t/T
+ C Step response
C is constant and could be found by the initial
condition y(t = 0) = 0.

Example
The Impulse response of a first order
system is given by:
y(t) = 4e−0.4t
Find:
1 Time constant T. (
1
T
= 0.4 T = 2.5)
2 D.C Gain K. (
K
T
= 4 K = 10)
3 Transfer function. G(s) =
10
2.5s + 1
4 Step Response of the system.

Example
The Impulse response of a first order
system is given by:
y(t) = 4e−0.4t
Find:
1 Time constant T. (
1
T
= 0.4 T = 2.5)
2 D.C Gain K. (
K
T
= 4 K = 10)
3 Transfer function. G(s) =
10
2.5s + 1
4 Step Response of the system.
To find Step response ys(t), we should
integrate the impulse response:
ys(t) =
Z
4e−0.4t
dt = 10e−0.4t
+ C
at initial condition ys(0) = 0, C = −10:
ys(t) = 10(1 − e−0.4t
)

Ramp response of 1st
Order Systems:
Y (s) = G(s)R(s)
since R(s) =
1
s2
(unit ramp),
Y (s) =
K
s2(Ts + 1)
Find response y(t) = L−1
{Y (s)}:
y(t) = K(t − T + Te−t/T
)
Example: K = 1 and T = 1:
y(t) = t − 1 + e−t
If T increases, Error will increase. (Try it!)

Parabolic response of 1st
Order Systems:
Y (s) = G(s)P(s)
P(s) =
2
s3
(unit parabolic)
Do it yourself!
Find y(t) and,
Use MATLAB to draw the parabolic
response.
To find the system
response in s-domain to
a different input signal,
use the lsim()
MATLAB command.

Table of Contents

Identification of the 1st
Order Transfer Function:
Often the system transfer function could not be available analytically.
With the step response of the system, we can identify the D.C. gain and the time
constant.
Finally, the first order transfer function can be written as:
G(s) =
K
Ts + 1
K = 0.72 Steady state value
T = 0.13 sec time constant
G(s) =
0.72
0.13s + 1

Table of Contents

First Order System with a Zero:
Consider the first order system with a zero in the numerator:
G(s) =
K(1 + αs)
Ts + 1
This system has a zero at −
1
α
and a pole at −
1
T
.
Studying the Step response
Y (s) =
K(1 + αs)
s(Ts + 1)
Y (s) =
K
s
+
K(α − T)
Ts + 1
y(t) = K +
K
T
(α − T)e−t/T

Lecture Assignment (LA3):
1 A thermometer requires 1 min to indicate 98% of the response to a step input. Assuming
the thermometer to be a first-order system, find the time constant.
2 If the thermometer is placed in a bath, the temperature of which is changing linearly at a
rate of 10 deg /min, how much error does the thermometer show?
3 With the help of MATLAB, draw the ramp response for the thermometer system.

End of Lecture
Best Wishes
haitham.elhussieny@gmail.com

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