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ESE 499 – Feedback Control Systems
SECTION 3: LAPLACE TRANSFORMS & TRANSFER FUNCTIONS
K. Webb ESE 499
This section of notes contains an introductionto Laplace transforms. This should mostly be areview of material covered in your differentialequations course.
Introduction – Transforms2
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3
Transforms
What is a transform? A mapping of a mathematical function from one domain to
another A change in perspective not a change of the function
Why use transforms? Some mathematical problems are difficult to solve in their
natural domain Transform to and solve in a new domain, where the problem is
simplified Transform back to the original domain
Trade off the extra effort of transforming/inverse-transforming for simplification of the solution procedure
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Transform Example – Slide Rules
Slide rules make use of a logarithmic transform
Multiplication/division of large numbers is difficult Transform the numbers to the logarithmic domain Add/subtract (easy) in the log domain to multiply/divide
(difficult) in the linear domain Apply the inverse transform to get back to the original
domain Extra effort is required, but the problem is simplified
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Laplace Transforms5
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Laplace Transforms
An integral transform mapping functions from the time domain to the Laplace domain or s-domain
𝑔𝑔 𝑡𝑡 ↔ℒ
𝐺𝐺 𝑠𝑠 Time-domain functions are functions of time, 𝑡𝑡
𝑔𝑔 𝑡𝑡 Laplace-domain functions are functions of 𝒔𝒔
𝐺𝐺 𝑠𝑠 𝑠𝑠 is a complex variable
𝑠𝑠 = 𝜎𝜎 + 𝑗𝑗𝑗𝑗
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Laplace Transforms – Motivation
We’ll use Laplace transforms to solve differential equations
Differential equations in the time domain difficult to solve
Apply the Laplace transform Transform to the s-domain
Differential equations become algebraic equations easy to solve
Transform the s-domain solution back to the time domain
Transforming back and forth requires extra effort, but the solution is greatly simplified
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Laplace Transform
Laplace Transform:
ℒ 𝑔𝑔 𝑡𝑡 = 𝐺𝐺 𝑠𝑠 = ∫0∞𝑔𝑔 𝑡𝑡 𝑒𝑒−𝑠𝑠𝑠𝑠𝑑𝑑𝑡𝑡 (1)
Unilateral or one-sided transform Lower limit of integration is 𝑡𝑡 = 0 Assumed that the time domain function is zero for all
negative time, i.e.
𝑔𝑔 𝑡𝑡 = 0, 𝑡𝑡 < 0
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In the following subsection of notes, we’ll derive a few important properties of the Laplace transform.
Laplace Transform Properties9
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Laplace Transform – Linearity
Say we have two time-domain functions:𝑔𝑔1 𝑡𝑡 and 𝑔𝑔2 𝑡𝑡
Applying the transform definition, (1)
ℒ 𝛼𝛼𝑔𝑔1 𝑡𝑡 + 𝛽𝛽𝑔𝑔2 𝑡𝑡 = �0
∞𝛼𝛼𝑔𝑔1 𝑡𝑡 + 𝛽𝛽𝑔𝑔2 𝑡𝑡 𝑒𝑒−𝑠𝑠𝑠𝑠𝑑𝑑𝑡𝑡
= �0
∞𝛼𝛼𝑔𝑔1 𝑡𝑡 𝑒𝑒−𝑠𝑠𝑠𝑠𝑑𝑑𝑡𝑡 + �
0
∞𝛽𝛽𝑔𝑔2 𝑡𝑡 𝑒𝑒−𝑠𝑠𝑠𝑠𝑑𝑑𝑡𝑡
= 𝛼𝛼�0
∞𝑔𝑔1 𝑡𝑡 𝑒𝑒−𝑠𝑠𝑠𝑠𝑑𝑑𝑡𝑡 + 𝛽𝛽�
0
∞𝑔𝑔2 𝑡𝑡 𝑒𝑒−𝑠𝑠𝑠𝑠𝑑𝑑𝑡𝑡
= 𝛼𝛼 � ℒ 𝑔𝑔1 𝑡𝑡 + 𝛽𝛽 � ℒ 𝑔𝑔2 𝑡𝑡
ℒ 𝛼𝛼𝑔𝑔1 𝑡𝑡 + 𝛽𝛽𝑔𝑔2 𝑡𝑡 = 𝛼𝛼𝐺𝐺1 𝑠𝑠 + 𝛽𝛽𝐺𝐺2 𝑠𝑠 (2)
The Laplace transform is a linear operation
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Laplace Transform of a Derivative
Of particular interest, given that we want to use Laplace transform to solve differential equations
ℒ �̇�𝑔 𝑡𝑡 = �0
∞�̇�𝑔 𝑡𝑡 𝑒𝑒−𝑠𝑠𝑠𝑠𝑑𝑑𝑡𝑡
Use integration by parts to evaluate
∫ 𝑢𝑢𝑑𝑑𝑢𝑢 = 𝑢𝑢𝑢𝑢 − ∫ 𝑢𝑢𝑑𝑑𝑢𝑢 Let
𝑢𝑢 = 𝑒𝑒−𝑠𝑠𝑠𝑠 and 𝑑𝑑𝑢𝑢 = �̇�𝑔 𝑡𝑡 𝑑𝑑𝑡𝑡then
𝑑𝑑𝑢𝑢 = −𝑠𝑠𝑒𝑒−𝑠𝑠𝑠𝑠𝑑𝑑𝑡𝑡 and 𝑢𝑢 = 𝑔𝑔 𝑡𝑡
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Laplace Transform of a Derivative
ℒ �̇�𝑔 𝑡𝑡 = 𝑒𝑒−𝑠𝑠𝑠𝑠𝑔𝑔 𝑡𝑡 �0
∞− �
0
∞𝑔𝑔 𝑡𝑡 −𝑠𝑠𝑒𝑒−𝑠𝑠𝑠𝑠 𝑑𝑑𝑡𝑡
= 0 − 𝑔𝑔 0 + 𝑠𝑠 �0
∞𝑔𝑔 𝑡𝑡 𝑒𝑒−𝑠𝑠𝑠𝑠𝑑𝑑𝑡𝑡 = −𝑔𝑔 0 + 𝑠𝑠ℒ 𝑔𝑔 𝑡𝑡
The Laplace transform of the derivative of a function is the Laplace transform of that function multiplied by 𝑠𝑠 minus the initial value of that function
ℒ �̇�𝑔 𝑡𝑡 = 𝑠𝑠𝐺𝐺 𝑠𝑠 − 𝑔𝑔(0) (3)
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Higher-Order Derivatives
The Laplace transform of a second derivative is
ℒ �̈�𝑔 𝑡𝑡 = 𝑠𝑠2𝐺𝐺 𝑠𝑠 − 𝑠𝑠𝑔𝑔 0 − �̇�𝑔 0 (4)
In general, the Laplace transform of the 𝒏𝒏𝒕𝒕𝒕𝒕 derivativeof a function is given by
ℒ 𝑔𝑔 𝑛𝑛 = 𝑠𝑠𝑛𝑛𝐺𝐺 𝑠𝑠 − 𝑠𝑠𝑛𝑛−1𝑔𝑔 0 − 𝑠𝑠𝑛𝑛−2�̇�𝑔 0 −⋯− 𝑔𝑔 𝑛𝑛−1 0 (5)
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Laplace Transform of an Integral
The Laplace Transform of a definite integral of a function is given by
ℒ ∫0𝑠𝑠 𝑔𝑔 𝜏𝜏 𝑑𝑑𝜏𝜏 = 1
𝑠𝑠𝐺𝐺 𝑠𝑠 (6)
Differentiation in the time domain corresponds to multiplication by 𝒔𝒔 in the Laplace domain
Integration in the time domain corresponds to division by 𝒔𝒔 in the Laplace domain
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Next, we’ll derive the Laplace transform of some common mathematical functions
Laplace Transforms of Common Functions
15
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Unit Step Function
A useful and common way of characterizing a linear system is with its step response The system’s response (output) to a unit step input
The unit step function or Heaviside step function:
1 𝑡𝑡 = �0, 𝑡𝑡 < 01, 𝑡𝑡 ≥ 0
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Unit Step Function – Laplace Transform
Using the definition of the Laplace transform
ℒ 1 𝑡𝑡 = �0
∞1 𝑡𝑡 𝑒𝑒−𝑠𝑠𝑠𝑠𝑑𝑑𝑡𝑡 = �
0
∞𝑒𝑒−𝑠𝑠𝑠𝑠𝑑𝑑𝑡𝑡
= −1𝑠𝑠𝑒𝑒−𝑠𝑠𝑠𝑠 �
0
∞= 0 − −
1𝑠𝑠
=1𝑠𝑠
The Laplace transform of the unit step
ℒ 1 𝑡𝑡 = 1𝑠𝑠
(7)
Note that the unilateral Laplace transform assumes that the signal being transformed is zero for 𝑡𝑡 < 0 Equivalent to multiplying any signal by a unit step
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Unit Ramp Function
The unit ramp function is a useful input signal for evaluating how well a system tracks a constantly-increasing input
The unit ramp function:
𝑔𝑔 𝑡𝑡 = �0, 𝑡𝑡 < 0𝑡𝑡, 𝑡𝑡 ≥ 0
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Unit Ramp Function – Laplace Transform
Could easily evaluate the transform integral Requires integration by parts
Alternatively, recognize the relationship between the unit ramp and the unit step Unit ramp is the integral of the unit step
Apply the integration property, (6)
ℒ 𝑡𝑡 = ℒ �0
𝑠𝑠1 𝜏𝜏 𝑑𝑑𝜏𝜏 =
1𝑠𝑠�
1𝑠𝑠
ℒ 𝑡𝑡 = 1𝑠𝑠2
(8)
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Exponential – Laplace Transform
𝑔𝑔 𝑡𝑡 = 𝑒𝑒−𝑎𝑎𝑠𝑠
Exponentials are common components of the responses of dynamic systems
ℒ 𝑒𝑒−𝑎𝑎𝑠𝑠 = �0
∞𝑒𝑒−𝑎𝑎𝑠𝑠𝑒𝑒−𝑠𝑠𝑠𝑠𝑑𝑑𝑡𝑡 = �
0
∞𝑒𝑒−(𝑠𝑠+𝑎𝑎)𝑠𝑠𝑑𝑑𝑡𝑡
= −𝑒𝑒− 𝑠𝑠+𝑎𝑎 𝑠𝑠
𝑠𝑠 + 𝑎𝑎�0
∞= 0 − −
1𝑠𝑠 + 𝑎𝑎
ℒ 𝑒𝑒−𝑎𝑎𝑠𝑠 = 1𝑠𝑠+𝑎𝑎
(9)
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Sinusoidal functions
Another class of commonly occurring signals, when dealing with dynamic systems, is sinusoidal signals –both sin 𝑗𝑗𝑡𝑡 and cos 𝑗𝑗𝑡𝑡
𝑔𝑔 𝑡𝑡 = sin 𝑗𝑗𝑡𝑡
Recall Euler’s formula
𝑒𝑒𝑗𝑗𝑗𝑗𝑠𝑠 = cos 𝑗𝑗𝑡𝑡 + 𝑗𝑗 sin 𝑗𝑗𝑡𝑡
From which it follows that
sin 𝑗𝑗𝑡𝑡 =𝑒𝑒𝑗𝑗𝑗𝑗𝑠𝑠 − 𝑒𝑒−𝑗𝑗𝑗𝑗𝑠𝑠
2𝑗𝑗
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Sinusoidal functions
ℒ sin 𝑗𝑗𝑡𝑡 =12𝑗𝑗�0
∞𝑒𝑒𝑗𝑗𝑗𝑗𝑠𝑠 − 𝑒𝑒−𝑗𝑗𝑗𝑗𝑠𝑠 𝑒𝑒−𝑠𝑠𝑠𝑠𝑑𝑑𝑡𝑡
=12𝑗𝑗�0
∞𝑒𝑒− 𝑠𝑠−𝑗𝑗𝑗𝑗 𝑠𝑠 − 𝑒𝑒− 𝑠𝑠+𝑗𝑗𝑗𝑗 𝑠𝑠 𝑑𝑑𝑡𝑡
=12𝑗𝑗�0
∞𝑒𝑒− 𝑠𝑠−𝑗𝑗𝑗𝑗 𝑠𝑠𝑑𝑑𝑡𝑡 −
12𝑗𝑗�0
∞𝑒𝑒− 𝑠𝑠+𝑗𝑗𝑗𝑗 𝑠𝑠𝑑𝑑𝑡𝑡
=12𝑗𝑗
𝑒𝑒− 𝑠𝑠−𝑗𝑗𝑗𝑗 𝑠𝑠
− 𝑠𝑠 − 𝑗𝑗𝑗𝑗 �0
∞−
12𝑗𝑗
𝑒𝑒− 𝑠𝑠+𝑗𝑗𝑗𝑗 𝑠𝑠
− 𝑠𝑠 + 𝑗𝑗𝑗𝑗 �0
∞
=12𝑗𝑗
0 +1
𝑠𝑠 − 𝑗𝑗𝑗𝑗−
12𝑗𝑗
0 +1
𝑠𝑠 + 𝑗𝑗𝑗𝑗=
12𝑗𝑗
2𝑗𝑗𝑗𝑗𝑠𝑠2 + 𝑗𝑗2
ℒ sin 𝑗𝑗𝑡𝑡 = 𝑗𝑗𝑠𝑠2+𝑗𝑗2 (10)
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Sinusoidal functions
It can similarly be shown that
ℒ cos 𝑗𝑗𝑡𝑡 = 𝑠𝑠𝑠𝑠2+𝑗𝑗2 (11)
Note that for neither sin(𝑗𝑗𝑡𝑡) nor cos 𝑗𝑗𝑡𝑡 is the function equal to zero for 𝑡𝑡 < 0 as the Laplace transform assumes
Really, what we’ve derived is
ℒ 1 𝑡𝑡 � sin 𝑗𝑗𝑡𝑡 and ℒ 1 𝑡𝑡 � cos 𝑗𝑗𝑡𝑡
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More Properties and Theorems24
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Multiplication by an Exponential, 𝑒𝑒−𝑎𝑎𝑠𝑠
We’ve seen that ℒ 𝑒𝑒−𝑎𝑎𝑠𝑠 = 1𝑠𝑠+𝑎𝑎
What if another function is multiplied by the decaying exponential term?
ℒ 𝑔𝑔 𝑡𝑡 𝑒𝑒−𝑎𝑎𝑠𝑠 = �0
∞𝑔𝑔 𝑡𝑡 𝑒𝑒−𝑎𝑎𝑠𝑠𝑒𝑒−𝑠𝑠𝑠𝑠𝑑𝑑𝑡𝑡 =�
0
∞𝑔𝑔 𝑡𝑡 𝑒𝑒− 𝑠𝑠+𝑎𝑎 𝑠𝑠𝑑𝑑𝑡𝑡
This is just the Laplace transform of 𝑔𝑔 𝑡𝑡 with 𝑠𝑠replaced by 𝑠𝑠 + 𝑎𝑎
ℒ 𝑔𝑔 𝑡𝑡 𝑒𝑒−𝑎𝑎𝑠𝑠 = 𝐺𝐺 𝑠𝑠 + 𝑎𝑎 (12)
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Decaying Sinusoids
The Laplace transform of a sinusoid is
ℒ sin 𝑗𝑗𝑡𝑡 =𝑗𝑗
𝑠𝑠2 + 𝑗𝑗2
And, multiplication by an decaying exponential, 𝑒𝑒−𝑎𝑎𝑠𝑠, results in a substitution of 𝑠𝑠 + 𝑎𝑎 for 𝑠𝑠, so
ℒ 𝑒𝑒−𝑎𝑎𝑠𝑠 sin 𝑗𝑗𝑡𝑡 =𝑗𝑗
𝑠𝑠 + 𝑎𝑎 2 + 𝑗𝑗2
and
ℒ 𝑒𝑒−𝑎𝑎𝑠𝑠 cos 𝑗𝑗𝑡𝑡 =𝑠𝑠 + 𝑎𝑎
𝑠𝑠 + 𝑎𝑎 2 + 𝑗𝑗2
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Time Shifting
Consider a time-domain function, 𝑔𝑔 𝑡𝑡
To Laplace transform 𝑔𝑔 𝑡𝑡we’ve assumed 𝑔𝑔 𝑡𝑡 = 0 for 𝑡𝑡 < 0, or equivalently multiplied by 1(𝑡𝑡)
To shift 𝑔𝑔 𝑡𝑡 by an amount, 𝑎𝑎, in time, we must also multiply by a shifted step function, 1 𝑡𝑡 − 𝑎𝑎
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Time Shifting – Laplace Transform
The transform of the shifted function is given by
ℒ 𝑔𝑔 𝑡𝑡 − 𝑎𝑎 � 1 𝑡𝑡 − 𝑎𝑎 = �𝑎𝑎
∞𝑔𝑔 𝑡𝑡 − 𝑎𝑎 𝑒𝑒−𝑠𝑠𝑠𝑠𝑑𝑑𝑡𝑡
Performing a change of variables, let
𝜏𝜏 = 𝑡𝑡 − 𝑎𝑎 and 𝑑𝑑𝜏𝜏 = 𝑑𝑑𝑡𝑡
The transform becomes
ℒ 𝑔𝑔 𝜏𝜏 � 1 𝜏𝜏 = �0
∞𝑔𝑔 𝜏𝜏 𝑒𝑒−𝑠𝑠 𝜏𝜏+𝑎𝑎 𝑑𝑑𝜏𝜏 = �
0
∞𝑔𝑔 𝜏𝜏 𝑒𝑒−𝑎𝑎𝑠𝑠𝑒𝑒−𝑠𝑠𝜏𝜏𝑑𝑑𝜏𝜏 = 𝑒𝑒−𝑎𝑎𝑠𝑠 �
0
∞𝑔𝑔 𝜏𝜏 𝑒𝑒−𝑠𝑠𝜏𝜏𝑑𝑑𝜏𝜏
A shift by 𝑎𝑎 in the time domain corresponds to multiplication by 𝑒𝑒−𝑎𝑎𝑠𝑠 in the Laplace domain
ℒ 𝑔𝑔 𝑡𝑡 − 𝑎𝑎 � 1 𝑡𝑡 − 𝑎𝑎 = 𝑒𝑒−𝑎𝑎𝑠𝑠𝐺𝐺 𝑠𝑠 (13)
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Multiplication by time, 𝑡𝑡
The Laplace transform of a function multiplied by time:
ℒ 𝑡𝑡 ⋅ 𝑓𝑓 𝑡𝑡 = − 𝑑𝑑𝑑𝑑𝑠𝑠𝐹𝐹(𝑠𝑠) (14)
Consider a unit ramp function:
ℒ 𝑡𝑡 = ℒ 𝑡𝑡 ⋅ 1 𝑡𝑡 = −𝑑𝑑𝑑𝑑𝑠𝑠
1𝑠𝑠 =
1𝑠𝑠2
Or a parabola:ℒ 𝑡𝑡2 = ℒ 𝑡𝑡 ⋅ 𝑡𝑡 = − 𝑑𝑑
𝑑𝑑𝑠𝑠1𝑠𝑠2
= 2𝑠𝑠3
In general
ℒ 𝑡𝑡𝑚𝑚 =𝑚𝑚!𝑠𝑠𝑚𝑚+1
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Initial and Final Value Theorems
Initial Value Theorem Can determine the initial value of a time-domain signal or
function from its Laplace transform
(15)
Final Value Theorem Can determine the steady-state value of a time-domain
signal or function from its Laplace transform
(16)
𝑔𝑔 0 = lim𝑠𝑠→∞
𝑠𝑠𝐺𝐺 𝑠𝑠
𝑔𝑔 ∞ = lim𝑠𝑠→0
𝑠𝑠𝐺𝐺 𝑠𝑠
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Convolution
Convolution of two functions or signals is given by
𝑔𝑔 𝑡𝑡 ∗ 𝑥𝑥 𝑡𝑡 = �0
𝑠𝑠𝑔𝑔 𝜏𝜏 𝑥𝑥 𝑡𝑡 − 𝜏𝜏 𝑑𝑑𝜏𝜏
Result is a function of time 𝑥𝑥 𝜏𝜏 is flipped in time and shifted by 𝑡𝑡 Multiply the flipped/shifted signal and the other signal Integrate the result from 𝜏𝜏 = 0 … 𝑡𝑡
May seem like an odd, arbitrary function now, but we’ll later see why it is very important
Convolution in the time domain corresponds to multiplication in the Laplace domain
ℒ 𝑔𝑔 𝑡𝑡 ∗ 𝑥𝑥 𝑡𝑡 = 𝐺𝐺 𝑠𝑠 𝑋𝑋 𝑠𝑠 (17)
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Impulse Function
Another common way to describe a dynamic system is with its impulse response System output in response to an impulse function input
Impulse function defined by𝛿𝛿 𝑡𝑡 = 0, 𝑡𝑡 ≠ 0
�−∞
∞𝛿𝛿 𝑡𝑡 𝑑𝑑𝑡𝑡 = 1
An infinitely tall, infinitely narrow pulse
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Impulse Function – Laplace Transform
To derive ℒ 𝛿𝛿 𝑡𝑡 , consider the following function
𝑔𝑔 𝑡𝑡 = �1𝑡𝑡0
, 0 ≤ 𝑡𝑡 ≤ 𝑡𝑡0
0, 𝑡𝑡 < 0 or 𝑡𝑡 > 𝑡𝑡0
Can think of 𝑔𝑔 𝑡𝑡 as the sum of two step functions:
𝑔𝑔 𝑡𝑡 =1𝑡𝑡0
1 𝑡𝑡 −1𝑡𝑡0
1 𝑡𝑡 − 𝑡𝑡0
The transform of the first term is
ℒ1𝑡𝑡0
1 𝑡𝑡 =1𝑡𝑡0𝑠𝑠
Using the time-shifting property, the second term transforms to
ℒ −1𝑡𝑡0
1 𝑡𝑡 − 𝑡𝑡0 = −𝑒𝑒−𝑠𝑠0𝑠𝑠
𝑡𝑡0𝑠𝑠
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Impulse Function – Laplace Transform
In the limit, as 𝑡𝑡0 → 0, 𝑔𝑔 𝑡𝑡 → 𝛿𝛿 𝑡𝑡 , so
ℒ 𝛿𝛿 𝑡𝑡 = lim𝑠𝑠0→0
ℒ 𝑔𝑔 𝑡𝑡
ℒ 𝛿𝛿 𝑡𝑡 = lim𝑠𝑠0→0
1 − 𝑒𝑒−𝑠𝑠0𝑠𝑠
𝑡𝑡0𝑠𝑠
Apply l’Hôpital’s rule
ℒ 𝛿𝛿 𝑡𝑡 = lim𝑠𝑠0→0
𝑑𝑑𝑑𝑑𝑡𝑡0
1 − 𝑒𝑒−𝑠𝑠0𝑠𝑠
𝑑𝑑𝑑𝑑𝑡𝑡0
𝑡𝑡0𝑠𝑠= lim
𝑠𝑠0→0
𝑠𝑠𝑒𝑒−𝑠𝑠0𝑠𝑠
𝑠𝑠=𝑠𝑠𝑠𝑠
The Laplace transform of an impulse function is one
ℒ 𝛿𝛿 𝑡𝑡 = 1 (18)
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Common Laplace Transforms
𝒈𝒈 𝒕𝒕 𝑮𝑮 𝒔𝒔
𝛿𝛿 𝑡𝑡 1
1 𝑡𝑡 1𝑠𝑠
𝑡𝑡 1𝑠𝑠2
𝑡𝑡𝑚𝑚 𝑚𝑚!𝑠𝑠𝑚𝑚+1
𝑒𝑒−𝑎𝑎𝑠𝑠 1𝑠𝑠 + 𝑎𝑎
𝑡𝑡𝑒𝑒−𝑎𝑎𝑠𝑠1
𝑠𝑠 + 𝑎𝑎 2
sin 𝑗𝑗𝑡𝑡𝑗𝑗
𝑠𝑠2 + 𝑗𝑗2
cos 𝑗𝑗𝑡𝑡𝑠𝑠
𝑠𝑠2 + 𝑗𝑗2
𝒈𝒈 𝒕𝒕 𝑮𝑮 𝒔𝒔
𝑒𝑒−𝑎𝑎𝑠𝑠 sin 𝑗𝑗𝑡𝑡𝑗𝑗
𝑠𝑠 + 𝑎𝑎 2 + 𝑗𝑗2
𝑒𝑒−𝑎𝑎𝑠𝑠 cos 𝑗𝑗𝑡𝑡𝑠𝑠 + 𝑎𝑎
𝑠𝑠 + 𝑎𝑎 2 + 𝑗𝑗2
�̇�𝑔(𝑡𝑡) 𝑠𝑠𝐺𝐺 𝑠𝑠 − 𝑔𝑔(0)
�̈�𝑔 𝑡𝑡 𝑠𝑠2𝐺𝐺 𝑠𝑠 − 𝑠𝑠𝑔𝑔 0 − �̇�𝑔 0
�0
𝑠𝑠𝑔𝑔 𝜏𝜏 𝑑𝑑𝜏𝜏
1𝑠𝑠𝐺𝐺 𝑠𝑠
𝑒𝑒−𝑎𝑎𝑠𝑠𝑔𝑔 𝑡𝑡 𝐺𝐺 𝑠𝑠 + 𝑎𝑎
𝑔𝑔 𝑡𝑡 − 𝑎𝑎 � 1 𝑡𝑡 − 𝑎𝑎 𝑒𝑒−𝑎𝑎𝑠𝑠𝐺𝐺 𝑠𝑠
𝑡𝑡 � 𝑔𝑔 𝑡𝑡 −𝑑𝑑𝑑𝑑𝑠𝑠𝐺𝐺 𝑠𝑠
K. Webb ESE 499
We’ve just seen how time-domain functions can betransformed to the Laplace domain. Next, we’ll look athow we can solve differential equations in the Laplacedomain and transform back to the time domain.
Inverse Laplace Transform36
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Laplace Transforms – Differential Equations
Consider the simple spring/mass/damper system from the previous section of notes
Applying Newton’s 2nd law at the mass:
𝐹𝐹𝑖𝑖𝑛𝑛 𝑡𝑡 − 𝑏𝑏�̇�𝑥 − 𝑘𝑘𝑥𝑥 = 𝑚𝑚�̈�𝑥
Rearranging, gives the governing second-order differential equation
�̈�𝑥 + 𝑏𝑏𝑚𝑚�̇�𝑥 + 𝑘𝑘
𝑚𝑚𝑥𝑥 = 1
𝑚𝑚𝐹𝐹𝑖𝑖𝑛𝑛 𝑡𝑡 (1)
This equations describes the dynamic response of the system to some applied input force, 𝐹𝐹𝑖𝑖𝑛𝑛 𝑡𝑡
That input may take many forms, e.g.: Step Sinusoidal Impulse, etc.
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Laplace Transforms – Differential Equations
We’ll now use Laplace transforms to determine the step response of the system
1N step force input
𝐹𝐹𝑖𝑖𝑛𝑛 𝑡𝑡 = 1𝑁𝑁 � 1 𝑡𝑡 = �0𝑁𝑁, 𝑡𝑡 < 01𝑁𝑁, 𝑡𝑡 ≥ 0 (2)
For the step response, we assume zero initial conditions
𝑥𝑥 0 = 0 and �̇�𝑥 0 = 0 (3)
Using the derivative property of the Laplace transform, (1) becomes
𝑠𝑠2𝑋𝑋 𝑠𝑠 − 𝑠𝑠𝑥𝑥 0 − �̇�𝑥 0 +𝑏𝑏𝑚𝑚𝑠𝑠𝑋𝑋 𝑠𝑠 −
𝑏𝑏𝑚𝑚𝑥𝑥(0) +
𝑘𝑘𝑚𝑚𝑋𝑋 𝑠𝑠 =
1𝑚𝑚𝐹𝐹𝑖𝑖𝑛𝑛 𝑠𝑠
𝑠𝑠2𝑋𝑋 𝑠𝑠 + 𝑏𝑏𝑚𝑚𝑠𝑠𝑋𝑋 𝑠𝑠 + 𝑘𝑘
𝑚𝑚𝑋𝑋 𝑠𝑠 = 1
𝑚𝑚𝐹𝐹𝑖𝑖𝑛𝑛 𝑠𝑠 (4)
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Laplace Transforms – Differential Equations
The input is a step, so (7) becomes
𝑠𝑠2𝑋𝑋 𝑠𝑠 + 𝑏𝑏𝑚𝑚𝑠𝑠𝑋𝑋 𝑠𝑠 + 𝑘𝑘
𝑚𝑚𝑋𝑋 𝑠𝑠 = 1
𝑚𝑚1𝑁𝑁 1
𝑠𝑠(5)
Solving (5) for 𝑋𝑋 𝑠𝑠
𝑋𝑋 𝑠𝑠 𝑠𝑠2 +𝑏𝑏𝑚𝑚 𝑠𝑠 +
𝑘𝑘𝑚𝑚 =
1𝑚𝑚
1𝑠𝑠
𝑋𝑋 𝑠𝑠 = 1/𝑚𝑚
𝑠𝑠 𝑠𝑠2+𝑏𝑏𝑚𝑚𝑠𝑠+
𝑘𝑘𝑚𝑚
(6)
Equation (6) is the solution to the differential equation of (1), given the step input and I.C.’s The system step response in the Laplace domain Next, we need to transform back to the time domain
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Laplace Transforms – Differential Equations
𝑋𝑋 𝑠𝑠 = 1/𝑚𝑚
𝑠𝑠 𝑠𝑠2+𝑏𝑏𝑚𝑚𝑠𝑠+
𝑘𝑘𝑚𝑚
(6)
The form of (6) is typical of Laplace transforms when dealing with linear systems
A rational polynomial in 𝑠𝑠 Here, the numerator is 0th-order
𝑋𝑋 𝑠𝑠 =𝐵𝐵 𝑠𝑠𝐴𝐴 𝑠𝑠
Roots of the numerator polynomial, 𝐵𝐵 𝑠𝑠 , are called the zeros of the function
Roots of the denominator polynomial, 𝐴𝐴 𝑠𝑠 , are called the polesof the function
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Inverse Laplace Transforms
𝑋𝑋 𝑠𝑠 = 1/𝑚𝑚
𝑠𝑠 𝑠𝑠2+𝑏𝑏𝑚𝑚𝑠𝑠+
𝑘𝑘𝑚𝑚
(6)
To get (6) back into the time domain, we need to perform an inverse Laplace transform An integral inverse transform exists, but we don’t use it Instead, we use partial fraction expansion
Partial fraction expansion Idea is to express the Laplace transform solution, (6), as a sum of
Laplace transform terms that appear in the table Procedure depends on the type of roots of the denominator polynomial
Real and distinct Repeated Complex
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Inverse Laplace Transforms – Example 1
Consider the following system parameters𝑚𝑚 = 1𝑘𝑘𝑔𝑔
𝑘𝑘 =16𝑁𝑁𝑚𝑚
𝑏𝑏 = 10𝑁𝑁 � 𝑠𝑠𝑚𝑚
Laplace transform of the step response becomes
𝑋𝑋 𝑠𝑠 = 1𝑠𝑠 𝑠𝑠2+10𝑠𝑠+16
(7)
Factoring the denominator
𝑋𝑋 𝑠𝑠 = 1𝑠𝑠 𝑠𝑠+2 𝑠𝑠+8
(8)
In this case, the denominator polynomial has three real, distinct roots
𝑠𝑠1 = 0, 𝑠𝑠2 = −2, 𝑠𝑠3 = −8
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Inverse Laplace Transforms – Example 1
Partial fraction expansion of (8) has the form
𝑋𝑋 𝑠𝑠 = 1𝑠𝑠 𝑠𝑠+2 𝑠𝑠+8
= 𝑟𝑟1𝑠𝑠
+ 𝑟𝑟2𝑠𝑠+2
+ 𝑟𝑟3𝑠𝑠+8
(9)
The numerator coefficients, 𝑟𝑟1, 𝑟𝑟2, and 𝑟𝑟3, are called residues
Can already see the form of the time-domain function Sum of a constant and two decaying exponentials
To determine the residues, multiply both sides of (9) by the denominator of the left-hand side
1 = 𝑟𝑟1 𝑠𝑠 + 2 𝑠𝑠 + 8 + 𝑟𝑟2𝑠𝑠 𝑠𝑠 + 8 + 𝑟𝑟3𝑠𝑠 𝑠𝑠 + 2
1 = 𝑟𝑟1𝑠𝑠2 + 10𝑟𝑟1𝑠𝑠 + 16𝑟𝑟1 + 𝑟𝑟2𝑠𝑠2 + 8𝑟𝑟2𝑠𝑠 + 𝑟𝑟3𝑠𝑠2 + 2𝑟𝑟3𝑠𝑠
Collecting terms, we have
1 = 𝑠𝑠2 𝑟𝑟1 + 𝑟𝑟2 + 𝑟𝑟3 + 𝑠𝑠 10𝑟𝑟1 + 8𝑟𝑟2 + 2𝑟𝑟3 + 16𝑟𝑟1 (10)
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Inverse Laplace Transforms – Example 1
Equating coefficients of powers of 𝑠𝑠 on both sides of (10) gives a system of three equations in three unknowns
𝑠𝑠2: 𝑟𝑟1 + 𝑟𝑟2 + 𝑟𝑟3 = 0𝑠𝑠1: 10𝑟𝑟1 + 8𝑟𝑟2 + 2𝑟𝑟3 = 0𝑠𝑠0: 16𝑟𝑟1 = 1
Solving for the residues gives𝑟𝑟1 = 0.0625𝑟𝑟2 = −0.0833𝑟𝑟3 = 0.0208
The Laplace transform of the step response is
𝑋𝑋 𝑠𝑠 = 0.0625𝑠𝑠
− 0.0833𝑠𝑠+2
+ 0.0208𝑠𝑠+8
(11)
Equation (11) can now be transformed back to the time domain using the Laplace transform table
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Inverse Laplace Transforms – Example 1
The time-domain step response of the system is the sum of a constant term and two decaying exponentials:
𝑥𝑥 𝑡𝑡 = 0.0625 − 0.0833𝑒𝑒−2𝑠𝑠 + 0.0208𝑒𝑒−8𝑠𝑠 (12)
Step response plotted in MATLAB
Characteristic of a signal having only real poles No overshoot/ringing
Steady-state displacement agrees with intuition 1𝑁𝑁 force applied to a 16𝑁𝑁/𝑚𝑚
spring
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Inverse Laplace Transforms – Example 1
Go back to (7) and apply the initial value theorem
𝑥𝑥 0 = lim𝑠𝑠→∞
𝑠𝑠𝑋𝑋 𝑠𝑠 = lim𝑠𝑠→∞
1𝑠𝑠2 + 10𝑠𝑠 + 16
= 0𝑐𝑐𝑚𝑚
Which is, in fact our assumed initial condition
Next, apply the final value theorem to the Laplace transform step response, (7)
𝑥𝑥 ∞ = lim𝑠𝑠→0
𝑠𝑠𝑌𝑌 𝑠𝑠 = lim𝑠𝑠→0
1𝑠𝑠2 + 10𝑠𝑠 + 16
𝑥𝑥 ∞ = 116
= 0.0625𝑚𝑚 = 6.25𝑐𝑐𝑚𝑚
This final value agrees with both intuition and our numerical analysis
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Inverse Laplace Transforms – Example 2
Reduce the damping and re-calculate the step response
𝑚𝑚 = 1𝑘𝑘𝑔𝑔
𝑘𝑘 =16𝑁𝑁𝑚𝑚
𝑏𝑏 = 8𝑁𝑁 � 𝑠𝑠𝑚𝑚
Laplace transform of the step response becomes
𝑋𝑋 𝑠𝑠 = 1𝑠𝑠 𝑠𝑠2+8𝑠𝑠+16
(13)
Factoring the denominator
𝑋𝑋 𝑠𝑠 = 1𝑠𝑠 𝑠𝑠+4 2 (14)
In this case, the denominator polynomial has three real roots, two of which are identical
𝑠𝑠1 = 0, 𝑠𝑠2 = −4, 𝑠𝑠3 = −4
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Inverse Laplace Transforms – Example 2
Partial fraction expansion of (14) has the form
𝑋𝑋 𝑠𝑠 = 1𝑠𝑠 𝑠𝑠+4 2 = 𝑟𝑟1
𝑠𝑠+ 𝑟𝑟2
𝑠𝑠+4+ 𝑟𝑟3
𝑠𝑠+4 2 (15)
Again, find residues by multiplying both sides of (15) by the left-hand side denominator
1 = 𝑟𝑟1 𝑠𝑠 + 4 2 + 𝑟𝑟2𝑠𝑠 𝑠𝑠 + 4 + 𝑟𝑟3𝑠𝑠
1 = 𝑟𝑟1𝑠𝑠2 + 8𝑟𝑟1𝑠𝑠 + 16𝑟𝑟1 + 𝑟𝑟2𝑠𝑠2 + 4𝑟𝑟2𝑠𝑠 + 𝑟𝑟3𝑠𝑠
Collecting terms, we have
1 = 𝑠𝑠2 𝑟𝑟1 + 𝑟𝑟2 + 𝑠𝑠 8𝑟𝑟1 + 4𝑟𝑟2 + 𝑟𝑟3 + 16𝑟𝑟1 (16)
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Inverse Laplace Transforms – Example 2
Equating coefficients of powers of 𝑠𝑠 on both sides of (16) gives a system of three equations in three unknowns
𝑠𝑠2: 𝑟𝑟1 + 𝑟𝑟2 = 0𝑠𝑠1: 8𝑟𝑟1 + 4𝑟𝑟2 + 𝑟𝑟3 = 0𝑠𝑠0: 16𝑟𝑟1 = 1
Solving for the residues gives𝑟𝑟1 = 0.0625𝑟𝑟2 = −0.0625𝑟𝑟3 = −0.2500
The Laplace transform of the step response is
𝑋𝑋 𝑠𝑠 = 0.0625𝑠𝑠
− 0.0625𝑠𝑠+4
− 0.25𝑠𝑠+4 2 (17)
Equation (17) can now be transformed back to the time domain using the Laplace transform table
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Inverse Laplace Transforms – Example 2
The time-domain step response of the system is the sum of a constant, a decaying exponential, and a decaying exponential scaled by time:
𝑥𝑥 𝑡𝑡 = 0.0625 − 0.0625𝑒𝑒−4𝑠𝑠 − 0. 25𝑡𝑡𝑒𝑒−4𝑠𝑠 (18)
Step response plotted in MATLAB
Again, characteristic of a signal having only real poles
Similar to the last case
A bit faster – slow pole at 𝑠𝑠 = −2was eliminated
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Inverse Laplace Transforms – Example 3
Reduce the damping even further and go through the process once again
𝑚𝑚 = 1𝑘𝑘𝑔𝑔
𝑘𝑘 =16𝑁𝑁𝑚𝑚
𝑏𝑏 = 4𝑁𝑁 � 𝑠𝑠𝑚𝑚
Laplace transform of the step response becomes
𝑋𝑋 𝑠𝑠 = 1𝑠𝑠 𝑠𝑠2+4𝑠𝑠+16
(19)
The second-order term in the denominator now has complex roots, so we won’t factor any further
The denominator polynomial still has a root at zero and now has two roots which are a complex-conjugate pair
𝑠𝑠1 = 0, 𝑠𝑠2 = −2 + 𝑗𝑗𝑗.464, 𝑠𝑠3 = −2 − 𝑗𝑗𝑗.464
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Inverse Laplace Transforms – Example 3
Want to cast the partial fraction terms into forms that appear in the Laplace transform table
Second-order terms should be of the form
𝑟𝑟𝑖𝑖(𝑠𝑠+𝜎𝜎)+𝑟𝑟𝑖𝑖+1𝑗𝑗𝑠𝑠+𝜎𝜎 2+𝑗𝑗2 (20)
This will transform into the sum of damped sine and cosine terms
ℒ−1 𝑟𝑟𝑖𝑖𝑠𝑠 + 𝜎𝜎
𝑠𝑠 + 𝜎𝜎 2 + 𝑗𝑗2 + 𝑟𝑟𝑖𝑖+1𝑗𝑗
𝑠𝑠 + 𝜎𝜎 2 + 𝑗𝑗2 = 𝑟𝑟𝑖𝑖𝑒𝑒−𝜎𝜎𝑠𝑠 cos 𝑗𝑗𝑡𝑡 + 𝑟𝑟𝑖𝑖+1𝑒𝑒−𝜎𝜎𝑠𝑠 sin 𝑗𝑗𝑡𝑡
To get the second-order term in the denominator of (19) into the form of (20), complete the square, to give the following partial fraction expansion
𝑋𝑋 𝑠𝑠 = 1𝑠𝑠 𝑠𝑠2+4𝑠𝑠+16
= 𝑟𝑟1𝑠𝑠
+ 𝑟𝑟2 𝑠𝑠+2 +𝑟𝑟3 3.464𝑠𝑠+2 2+ 3.464 2 (21)
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Inverse Laplace Transforms – Example 3
Note that the 𝜎𝜎 and 𝑗𝑗 terms in (20) and (24) are the real and imaginary parts of the complex-conjugate denominator roots
𝑠𝑠2,3 = −𝜎𝜎 ± 𝑗𝑗𝑗𝑗 = −2 ± 𝑗𝑗𝑗.464
Multiplying both sides of (21) by the left-hand-side denominator, equate coefficients and solve for residues as before:
𝑟𝑟1 = 0.0625𝑟𝑟2 = −0.0625𝑟𝑟3 = −0.0361
Laplace transform of the step response is
𝑋𝑋 𝑠𝑠 = 0.0625𝑠𝑠
− 0.0625 𝑠𝑠+2𝑠𝑠+2 2+ 3.464 2 −
0.0361 3.464𝑠𝑠+2 2+ 3.464 2 (22)
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Inverse Laplace Transforms – Example 3
The time-domain step response of the system is the sum of a constant and two decaying sinusoids:
𝑥𝑥 𝑡𝑡 = 0.0625 − 0.0625𝑒𝑒−2𝑠𝑠 cos 3.464𝑡𝑡 − 0.0361𝑒𝑒−2𝑠𝑠sin(3.464𝑡𝑡) (23)
Step response and individual components plotted in MATLAB
Characteristic of a signal having complex poles
Sinusoidal terms result in overshoot and (possibly) ringing
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Laplace-Domain Signals with Complex Poles
The Laplace transform of the step response in the last example had complex poles A complex-conjugate pair: 𝑠𝑠 = −𝜎𝜎 ± 𝑗𝑗𝑗𝑗
Results in sine and cosine terms in the time domain
𝐴𝐴𝑒𝑒−𝜎𝜎𝑠𝑠 cos 𝑗𝑗𝑡𝑡 + 𝐵𝐵𝑒𝑒−𝜎𝜎𝑠𝑠 sin 𝑗𝑗𝑡𝑡
Imaginary part of the roots, 𝑗𝑗 Frequency of oscillation of sinusoidal
components of the signal
Real part of the roots, 𝜎𝜎, Rate of decay of the sinusoidal
components
Much more on this later
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Natural and Forced Responses56
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Natural and Forced Responses
So far, we’ve used Laplace transforms to determine the response of a system to a step input, given zero initial conditions The driven response
Now, consider the response of the same system to non-zero initial conditions only The natural response
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Natural Response
�̈�𝑥 + 𝑏𝑏𝑚𝑚�̇�𝑥 + 𝑘𝑘
𝑚𝑚𝑥𝑥 = 0 (1)
Use the derivative property to Laplace transform (1) Allow for non-zero initial-conditions
𝑠𝑠2𝑋𝑋 𝑠𝑠 − 𝑠𝑠𝑥𝑥 0 − �̇�𝑥 0 + 𝑏𝑏𝑚𝑚𝑠𝑠𝑋𝑋 𝑠𝑠 − 𝑏𝑏
𝑚𝑚𝑥𝑥 0 + 𝑘𝑘
𝑚𝑚𝑋𝑋 𝑠𝑠 = 0 (2)
Same spring/mass/damper system Set the input to zero Second-order ODE for displacement
of the mass:
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Natural Response
Solving (2) for 𝑋𝑋 𝑠𝑠 gives the Laplace transform of the output due solely to initial conditions
Laplace transform of the natural response
𝑋𝑋 𝑠𝑠 =𝑠𝑠 𝑥𝑥 0 + �̇�𝑥 0 + 𝑏𝑏
𝑚𝑚𝑥𝑥 0
𝑠𝑠2+𝑏𝑏𝑚𝑚𝑠𝑠+
𝑘𝑘𝑚𝑚
(3)
Consider the under-damped system with the following initial conditions
𝑚𝑚 = 1 𝑘𝑘𝑔𝑔
𝑘𝑘 = 16 𝑁𝑁𝑚𝑚
𝑏𝑏 = 4 𝑁𝑁�𝑠𝑠𝑚𝑚
𝑥𝑥 0 = 0.15 𝑚𝑚
�̇�𝑥 0 = 0.1𝑚𝑚𝑠𝑠
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Natural Response
Substituting component parameters and initial conditions into (3)
𝑋𝑋 𝑠𝑠 = 0.15𝑠𝑠 +0.7𝑠𝑠2+4𝑠𝑠+16
(4)
Remember, it’s the roots of the denominator polynomial that dictate the form of the response Real roots – decaying exponentials Complex roots – decaying sinusoids
For the under-damped case, roots are complex Complete the square Partial fraction expansion has the form
𝑋𝑋 𝑠𝑠 = 0.15𝑠𝑠 +0.7𝑠𝑠2+4𝑠𝑠+16
= 𝑟𝑟1 𝑠𝑠+2 +𝑟𝑟2 3.464𝑠𝑠+2 2+ 3.464 2 (5)
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Natural Response
𝑋𝑋 𝑠𝑠 = 0.15𝑠𝑠 +0.7𝑠𝑠2+4𝑠𝑠+16
= 𝑟𝑟1 𝑠𝑠+2 +𝑟𝑟2 3.464𝑠𝑠+2 2+ 3.464 2 (5)
Multiply both sides of (5) by the denominator of the left-hand side
0.15𝑠𝑠 + 0.7 = 𝑟𝑟1𝑠𝑠 + 2𝑟𝑟1 + 3.464𝑟𝑟2
Equating coefficients and solving for 𝑟𝑟1 and 𝑟𝑟2 gives
𝑟𝑟1 = 0.15, 𝑟𝑟2 = 0.115
The Laplace transform of the natural response:
𝑋𝑋 𝑠𝑠 = 0.15 𝑠𝑠+2𝑠𝑠+2 2+ 3.464 2 + 0.115 3.464
𝑠𝑠+2 2+ 3.464 2 (6)
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Natural Response
Inverse Laplace transform is the natural response𝑥𝑥 𝑡𝑡 = 0.15𝑒𝑒−2𝑠𝑠 cos 3.464 � 𝑡𝑡 + 0.115𝑒𝑒−2𝑠𝑠 sin 3.464 � 𝑡𝑡 (7)
Under-damped response is the sum of decaying sine and cosine terms
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Now, Laplace transform, allowing for both non-zero input and initial conditions
𝑠𝑠2𝑋𝑋 𝑠𝑠 − 𝑠𝑠𝑥𝑥 0 − �̇�𝑥 0 + 𝑏𝑏𝑚𝑚𝑠𝑠𝑋𝑋 𝑠𝑠 − 𝑏𝑏
𝑚𝑚𝑥𝑥 0 + 𝑘𝑘
𝑚𝑚𝑋𝑋 𝑠𝑠 = 1
𝑚𝑚𝐹𝐹𝑖𝑖𝑛𝑛 𝑠𝑠
Solving for 𝑋𝑋 𝑠𝑠 , gives the Laplace transform of the response to both the input and the initial conditions
𝑋𝑋 𝑠𝑠 =𝑠𝑠 𝑥𝑥 0 + �̇�𝑥 0 + 𝑏𝑏
𝑚𝑚𝑥𝑥 0 +1𝑚𝑚𝐹𝐹𝑖𝑖𝑖𝑖 𝑠𝑠
𝑠𝑠2+𝑏𝑏𝑚𝑚𝑠𝑠+
𝑘𝑘𝑚𝑚
(8)
Driven Response with Non-Zero I.C.’s
�̈�𝑥 +𝑏𝑏𝑚𝑚�̇�𝑥 +
𝑘𝑘𝑚𝑚𝑥𝑥 =
1𝑚𝑚𝐹𝐹𝑖𝑖𝑛𝑛 𝑡𝑡
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Laplace transform of the response has two components
𝑋𝑋 𝑠𝑠 =𝑠𝑠 𝑥𝑥 0 + �̇�𝑥 0 + 𝑏𝑏
𝑚𝑚𝑥𝑥 0
𝑠𝑠2+𝑏𝑏𝑚𝑚𝑠𝑠+
𝑘𝑘𝑚𝑚
+1𝑚𝑚𝐹𝐹𝑖𝑖𝑖𝑖 𝑠𝑠
𝑠𝑠2+𝑏𝑏𝑚𝑚𝑠𝑠+
𝑘𝑘𝑚𝑚
Total response is a superposition of the initial condition response and the driven response
Both have the same denominator polynomial Same roots, same type of response
Over-, under-, critically-damped
Driven Response with Non-Zero I.C.’s
Natural response - initial conditions Driven response - input
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Driven Response with Non-Zero I.C.’s
Laplace transform of the total response
𝑋𝑋 𝑠𝑠 =0.15𝑠𝑠 + 0.7 + 1
𝑠𝑠𝑠𝑠2 + 4𝑠𝑠 + 16
=0.15𝑠𝑠2 + 0.7𝑠𝑠 + 1𝑠𝑠 𝑠𝑠2 + 4𝑠𝑠 + 16
Transform back to time domain via partial fraction expansion
𝑋𝑋 𝑠𝑠 =𝑟𝑟1𝑠𝑠
+𝑟𝑟2 𝑠𝑠 + 2
𝑠𝑠 + 2 2 + 3.464 2 +𝑟𝑟3 3.464
𝑠𝑠 + 2 2 + 3.464 2
Solving for the residues gives
𝑟𝑟1 = 0.0625, 𝑟𝑟2 = 0.0875, 𝑟𝑟3 = 0.0794
𝑚𝑚 = 1 𝑘𝑘𝑔𝑔
𝑘𝑘 = 16 𝑁𝑁𝑚𝑚
𝑏𝑏 = 4 𝑁𝑁�𝑠𝑠𝑚𝑚
𝑥𝑥 0 = 0.15 𝑚𝑚
�̇�𝑥 0 = 0.1𝑚𝑚𝑠𝑠
𝐹𝐹𝑖𝑖𝑛𝑛 𝑡𝑡 = 1𝑁𝑁 � 1 𝑡𝑡
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Driven Response with Non-Zero I.C.’s
Total response:𝑥𝑥 𝑡𝑡 = 0.0625 + 0.0875𝑒𝑒−2𝑠𝑠 cos 3.464 � 𝑡𝑡 + 0.0794𝑒𝑒−2𝑠𝑠 sin 3.464 � 𝑡𝑡
Superposition of two components Natural response
due to initial conditions
Driven response due to the input
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Transfer Functions67
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Transfer Functions
Consider a system block diagram in the Laplace domain 𝑈𝑈 𝑠𝑠 is the Laplace-domain input 𝑌𝑌 𝑠𝑠 is the Laplace-domain output 𝐺𝐺 𝑠𝑠 is a Laplace-domain model
for the plant To understand what 𝐺𝐺 𝑠𝑠 is, revisit the previous example
We saw that if we assume zero initial conditions, the Laplace-domain output is
𝑌𝑌 𝑠𝑠 =�1 𝑚𝑚
𝑠𝑠2 + 𝑏𝑏𝑚𝑚 𝑠𝑠 + 𝑘𝑘
𝑚𝑚𝑈𝑈 𝑠𝑠
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Transfer Functions
𝑌𝑌 𝑠𝑠 =1𝑚𝑚
𝑠𝑠2 + 𝑏𝑏𝑚𝑚 𝑠𝑠 + 𝑘𝑘
𝑚𝑚𝑈𝑈 𝑠𝑠
The Laplace transform of the output is the Laplace transform of the input multiplied by the stuff in square brackets
𝑌𝑌 𝑠𝑠 = 𝐺𝐺 𝑠𝑠 ⋅ 𝑈𝑈 𝑠𝑠
This is the transfer function
𝐺𝐺 𝑠𝑠 =𝑌𝑌 𝑠𝑠𝑈𝑈 𝑠𝑠
The ratio of the system’s output to input in the Laplace domain, assuming zero initial conditions
A mathematical model for a dynamic system This is the model we will use for control system design in this course
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Transfer Functions
The transfer function for our example system is
𝐺𝐺 𝑠𝑠 =�1 𝑚𝑚
𝑠𝑠2 + 𝑏𝑏𝑚𝑚 𝑠𝑠 + 𝑘𝑘
𝑚𝑚
Note that, as was the case for the Laplace transform of the output, the transfer function is a rational polynomial in s
𝐺𝐺 𝑠𝑠 =𝐵𝐵 𝑠𝑠𝐴𝐴 𝑠𝑠
Roots of 𝐵𝐵 𝑠𝑠 are the system zeros Roots of 𝐴𝐴 𝑠𝑠 are the system poles These poles and zeros dictate the nature of the system’s
response
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SISO vs. MIMO Systems
Note that we have assumed a system with one input and one output
A single-input single-output (SISO) system
Often, systems have multiple inputs and multiple outputs
A multiple-input multiple-output (MIMO) system
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MIMO Systems – Transfer Matrix
For MIMO systems, the transfer function becomes a transfer matrix For an m-input p-output system, a p×m matrix of transfer
functions
𝑮𝑮 𝑠𝑠 =𝐺𝐺11 𝑠𝑠 ⋯ 𝐺𝐺1𝑚𝑚 𝑠𝑠⋮ ⋱ ⋮
𝐺𝐺𝑝𝑝1 𝑠𝑠 ⋯ 𝐺𝐺𝑝𝑝𝑚𝑚 𝑠𝑠
Transfer function 𝐺𝐺𝑖𝑖𝑗𝑗 𝑠𝑠 relates the 𝑖𝑖𝑠𝑠𝑡 output to the 𝑗𝑗𝑠𝑠𝑡 input
𝐺𝐺𝑖𝑖𝑗𝑗 𝑠𝑠 =𝑌𝑌𝑖𝑖 𝑠𝑠𝑈𝑈𝑗𝑗 𝑠𝑠
In this course, we’ll restrict our focus entirely to SISO systems
m inputsp outputs
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Characteristic Polynomial
𝐺𝐺 𝑠𝑠 =𝐵𝐵 𝑠𝑠𝐴𝐴 𝑠𝑠
The denominator of the transfer function is the characteristic polynomial, Δ 𝑠𝑠
𝐺𝐺 𝑠𝑠 =𝐵𝐵 𝑠𝑠Δ 𝑠𝑠
Poles of the transfer function are roots of Δ 𝑠𝑠 System poles or eigenvalues Poles determine the terms in the partial-fraction-expansion of the
system’s natural response Along with the input, system poles determine the nature of the time-
domain response
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Using the Transfer Function to Determine System Response74
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Using 𝐺𝐺 𝑠𝑠 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
𝑌𝑌 𝑠𝑠 = 𝑈𝑈 𝑠𝑠 𝐺𝐺 𝑠𝑠 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
We’ll use the approach of (1) to determine these responses
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Impulse response
Impulse function
𝛿𝛿 𝑡𝑡 = 0, 𝑡𝑡 ≠ 0
∫−∞∞ 𝛿𝛿 𝑡𝑡 𝑑𝑑𝑡𝑡 = 1
Laplace transform of the impulse function is
ℒ 𝛿𝛿 𝑡𝑡 = 1
Impulse response in the Laplace domain is
𝑌𝑌 𝑠𝑠 = 1 � 𝐺𝐺 𝑠𝑠 = 𝐺𝐺 𝑠𝑠
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
𝑦𝑦 𝑡𝑡 = 𝑔𝑔 𝑡𝑡 = ℒ−1 𝐺𝐺 𝑠𝑠
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Step Response
Step function: 1 𝑡𝑡 = �0 𝑡𝑡 < 01 𝑡𝑡 ≥ 0
Laplace transform of the step function
ℒ 1 𝑡𝑡 = 1𝑠𝑠
Laplace-domain step response
𝑌𝑌 𝑠𝑠 = 1𝑠𝑠� 𝐺𝐺 𝑠𝑠
Time-domain step response
𝑦𝑦 𝑡𝑡 = ℒ−1 1𝑠𝑠� 𝐺𝐺 𝑠𝑠
Recall the integral property of the Laplace transform
ℒ ∫0𝑠𝑠 𝑔𝑔 𝜏𝜏 𝑑𝑑𝜏𝜏 = 1
𝑠𝑠� 𝐺𝐺 𝑠𝑠 , and ℒ−1 1
𝑠𝑠� 𝐺𝐺 𝑠𝑠 = ∫0
𝑠𝑠 𝑔𝑔 𝜏𝜏 𝑑𝑑𝜏𝜏
The step response is the integral of the impulse response
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Transfer Function and Dynamic Response
We just saw that the impulse response is𝑔𝑔 𝑡𝑡 = ℒ−1 𝐺𝐺 𝑠𝑠
And the step response is
𝑦𝑦 𝑡𝑡 = ℒ−11𝑠𝑠� 𝐺𝐺 𝑠𝑠
Both are entirely determined by the system transfer function, 𝐺𝐺 𝑠𝑠 System poles and zeros determine the nature of these
responses 𝐺𝐺 𝑠𝑠 is a complete mathematical model for the system