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Control Systems Engineering, Fourth Edition by Norman S. Nise Copyright © 2004 by John Wiley & Sons. All rights reserved. 1 Modeling in the Frequency Domain Review of the Laplace transform Learn how to find a mathematical model, called a transfer function, for linear, time-invariant electrical, mechanical, and electromechanical systems Learn how to linearize a nonlinear system in order to find the transfer function

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Page 1: 02 elec3114

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

1

Modeling in the Frequency Domain

• Review of the Laplace transform

• Learn how to find a mathematical model, called a transfer function, for linear, time-invariant electrical, mechanical, and electromechanical systems

• Learn how to linearize a nonlinear system in order to find the transfer function

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Introduction to Modeling• we look for a mathematical representation where the input, output, and

system are distinct and separate parts

• also, we would like to represent conveniently the interconnection of several subsystems. For example, we would like to represent cascadedinterconnections, where a mathematical function, called a transfer function, is inside each block

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

3

Laplace Transform Review

The Laplace transform is defined as

where s = σ + jω, is a complex variable.

• a system represented by a differential equation is difficult to model as a block diagram

• on the other hand, a system represented by a Laplace transformeddifferential equation is easier to model as a block diagram

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Laplace Transform Review

The inverse Laplace transform, which allows usto find f (t) given F(s), is

where

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Laplace transform table

Laplace Transform Review

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Problem: Find the Laplace transform of

Solution:

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Laplace Transform Review

Laplace transform theorems

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Laplace Transform Review

Laplace transform theorems

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Partial-Fraction Expansion

• To find the inverse Laplace transform of a complicated function, we can convert the function to a sum of simpler terms for which we know the Laplace transform of each term.

Case 1: Roots of the Denominator of F(s) Are Real and Distinct

Case 2: Roots of the Denominator of F(s) Are Real and Repeated

Case 3: Roots of the Denominator of F(s) Are Complex or Imaginary

)()()(

sDsNsF =

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Case 1: Roots of the Denominator of F(s) Are Real and Distinct

If the order of N(s) is less than the order of D(s), then

Thus, if we want to find Km, we multiply above equation by )( mps +

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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If we substitute in the above equation, then we can find Kmmps −=

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Problem Given the following differential equation, solve for y(t) if all initial conditions are zero. Use the Laplace transform.

Solution

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Case 2: Roots of the Denominator of F(s) Are Real and Repeated

First, we multiply by and we can solve immediately for K1 if s= - p1rps )( 1+

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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• we can solve for K2 if we differentiate F1(s) with respect to s and then let s approach –p1

• subsequent differentiation allows to find K3 through Kr

• the general expression for K1 through Kr for the multiple roots is

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Case 3: Roots of the Denominator of F(s) Are Complex or Imaginary

• the coefficients K2, K3 are found through balancing the coefficientsof the equation after clearing fractions

)()(

232

bassKsK++

+is put in the form of by completing

the squares on )( 2 bass ++ and adjusting the numerator

has complex or pure imaginary roots

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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The Transfer Function (1)

Let us consider a general nth-order, linear, time-invariant differential equation is given as:

where c(t) is the output, r(t) is the input, and ai, bi are coefficients.

Taking the Laplace transform of both sides (assuming all initial conditions are zero) we obtain

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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The Transfer Function (2)

Transfer function is the ratio G(s) of the output transform, C(s),divided by the input transform, R(s)

)...()...()(

)()(

01

1

01

1

asasabsbsbsG

sRsC

nn

nn

mm

mm

++++++

== −−

−−

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

20

Electric Network Transfer Functions

Voltage-current, voltage-charge, and impedance relationships for capacitors, resistors, and inductors

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Problem Find the transfer function relating the capacitor voltage, Vc(s), to the input voltage, V(s)

Solution:

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Transfer function-single loop via transform methods

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Complex Electric Circuits via Mesh Analysis

1. Replace passive element values with their impedances.

2. Replace all sources and time variables with their Laplace transform.

3. Assume a transform current and a current direction in each mesh.

4. Write Kirchhoff's voltage law around each mesh.

5. Solve the simultaneous equations for the output.

6. Form the transfer function

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Problem: find the transfer function, I2(s)/V(s)

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Complex Electric Circuits via Nodal Analysis

1. Replace passive element values with their admittances

2. Replace all sources and time variables with their Laplace transform.

3. Replace transformed voltage sources with transformed current sources.

4. Write Kirchhoff's current law at each node.

5. Solve the simultaneous equations for the output.

6. Form the transfer function.

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Operational Amplifiers

Inverting OP AMP

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Non-inverting OP AMP

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Example

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Translational Mechanical System Transfer Functions

Newton's law: The sum of forces on a body equals zero; the sum of moments on a body equals zero.

K… spring constant

fv … coefficient of viscious friction

M … mass

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Problem Find the transfer function, X(s)/F(s), for the system

Solution:

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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• The required number of equations of motion is equal to the number of linearly independent motions. Linear independence implies that a point of motion in a system can still move if allother points of motion are held still.

• Another name for the number of linearly independent motions is the number of degrees of freedom.

Translational Mechanical System Transfer Functions

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Problem: Find the transfer function, X2(s)/F(s), of the system

Solution• we draw the free-body diagram for each point of motion and then use

superposition

• for each free-body diagram we begin by holding all other points of motion still and finding the forces acting on the body due only to its own motion.

• then we hold the body still and activate the other points of motion one at a time, placing on the original body the forces created by the adjacent motion

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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a. Forces on M1 due only to motion of M1

b. forces on M1 due only to motion of M2

c. all forces on M1

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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a. Forces on M2 due only to motion of M2;

b. forces on M2 due only to motion

c. all forces on M2

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Rotational Mechanical System Transfer Functions

Torque-angular velocity, torque-angular displacement, and impedance rotational relationships for springs, viscous dampers, and inertia

• the mass is replaced by inertia

• the values of K, D, and J are called spring constant, coefficient of viscous friction, and moment of inertia, respectively

• the concept of degrees of freedom is the same as for translational movement

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Problem Find the transfer function

Solution

a. Torques on J1due only to themotion of J1

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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b. torques on J1due only to themotion of J2

c. final free-bodydiagram for J1

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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a. Torques on J1due only to themotion of J1b. torques on J1due only to themotion of J2c. final free-bodydiagram for J1

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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a. Torques on J2due only to themotion of J2b. torques on J2due only to themotion of J1c. final free-bodydiagram for J2

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Transfer Functions for Systems with Gears

- the distance travelled along each gear's circumference is the same

2

1

2

1

1

2

NN

rr==

θθ

- the energy into Gear 1 equals the energy out of Gear 2

1

2

2

1

1

2

NN

TT

==θθ

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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1

2

2

1

1

2

NN

TT

==θθ

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Rotational mechanical impedances can be reflected through gear trains by multiplying the mechanical impedance by the ratio

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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A gear train is used to implement large gear ratios by cascading smaller gear ratios.

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Problem Find the transfer function,

Solution:

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Electromechanical System Transfer FunctionsElectromechanical systems: robots, hard disk drives, …

Motor is an electromechanical component that yields a displacement output for a voltage input, that is, a mechanical output generated by an electrical input.

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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where Kb is a constant of proportionality called the back emf constant;

is the angular velocity of the motor)(/ tdtd mm ωθ =

Derivation of the Transfer Function of Motor

Back electromotive force (back emf):

Taking the Laplace transform, we get:

Torque:

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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The relationship between the armature current, ia(t), the applied armature voltage, ea(t), and the back emf, vb(t),

t

ma K

sTsI )()( =

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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We assume that the armature inductance, La, is small compared to the armatureresistance, Ra,

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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The mechanical constants Jm and Dm can be found as:

Electrical constants Kt/Ra and Kb can be found through a dynamometer test of the motor.

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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a

stall

a

t

eT

RK

=

loadno

ab

eK−

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Nonlinearities

A linear system possesses two properties: superposition and homogeneity.

Superposition - the output response of a system to the sum of inputs is the sum of the responses to the individual inputs.

Homogeneity - describes the response of the system to a multiplication of the input by a scalar. Multiplication of an input by a scalar yields a response that is multiplied by the same scalar.

Linear system Nonlinear system

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Nonlinearities

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Linearization

• if any nonlinear components are present, we must linearize the system before we can find the transfer function

• we linearize nonlinear differential equation for small-signal inputs about the steady-state solution when the small signal input is equal to zero

ma is the slope of the curve at point A

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Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

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Solution:

At π/2 df/dx = - 5