Download pdf - Section 3 Laplace Transforms & Transfer Functions

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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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

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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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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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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


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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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34

Impulse Function – Laplace Transform

In the limit, as 𝑡𝑡0 → 0, 𝑔𝑔 𝑡𝑡 → 𝛿𝛿 𝑡𝑡 , so

ℒ 𝛿𝛿 𝑡𝑡 = lim𝑠𝑠0→0

ℒ 𝑔𝑔 𝑡𝑡


1 − 𝑒𝑒−𝑠𝑠0𝑠𝑠

𝑡𝑡0𝑠𝑠

Apply l’Hôpital’s rule


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


cos 𝑗𝑗𝑡𝑡𝑠𝑠


𝒈𝒈 𝒕𝒕 𝑮𝑮 𝒔𝒔

𝑒𝑒−𝑎𝑎𝑠𝑠 sin 𝑗𝑗𝑡𝑡𝑗𝑗


𝑒𝑒−𝑎𝑎𝑠𝑠 cos 𝑗𝑗𝑡𝑡𝑠𝑠 + 𝑎𝑎


�̇�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

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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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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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The input is a step, so (7) becomes

𝑠𝑠2𝑋𝑋 𝑠𝑠 + 𝑏𝑏𝑚𝑚𝑠𝑠𝑋𝑋 𝑠𝑠 + 𝑘𝑘


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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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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41

Inverse Laplace Transforms



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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42

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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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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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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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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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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Reduce the damping and re-calculate the step response

𝑚𝑚 = 1𝑘𝑘𝑔𝑔





(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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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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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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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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Reduce the damping even further and go through the process once again

𝑚𝑚 = 1𝑘𝑘𝑔𝑔





(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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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


= 𝑟𝑟1𝑠𝑠

+ 𝑟𝑟2 𝑠𝑠+2 +𝑟𝑟3 3.464𝑠𝑠+2 2+ 3.464 2 (21)

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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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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


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


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


𝑘𝑘𝑚𝑚

+1𝑚𝑚𝐹𝐹𝑖𝑖𝑖𝑖 𝑠𝑠


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


Natural response - initial conditions Driven response - input

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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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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


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


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