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M.D. Bryant ME 344 notes 03/25/08 1 Laplace Transforms & Transfer Functions Laplace Transforms : method for solving differential equations, converts differential equations in time t into algebraic equations in complex variable s Transfer Functions : another way to represent system dynamics, via the s representation gotten from Laplace transforms, or excitation by e st

# Laplace Transforms & Transfer Functions

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• M.D. Bryant ME 344 notes 03/25/08

1

Laplace Transforms & Transfer Functions

Laplace Transforms: method for solving differential equations, converts differential equations in time t into algebraic equations in complex variable s Transfer Functions: another way to represent system dynamics, via the s representation gotten from Laplace transforms, or excitation by est

• M.D. Bryant ME 344 notes 03/25/08

2

Laplace Transforms Purpose: Converts linear differential equation in t into algebraic equation in s

Forward transform, t s: f(t) F(s)

F(s) = t= 0

f(t) e- st dt = L {f(t); t s }

Convergence/existence of integral: |f(t)| < M e- at , a + Re(s) > 0

so that e-[a + Re(s)]t finite as t

Inverse transform, s t: F(s) f(t)

f(t) = 12j

= c- j

c+j F(s) e st ds = L-1{F(s); s t }

Integral in complex plane, rarely do

• M.D. Bryant ME 344 notes 03/25/08

3

Procedure:

Convert Linear Differential Equations

!

x = Ax + f (t)

!

dn

x

dtn

+ an"1

dn"1

x

dtn"1

+!+ a2

d2

x

dt2+ a

1

d x

dt+ a

0x = f (t)

in time t into algebraic equations in complex variable s, via forward Laplace transforms:

X(s)=L {x(t); t s }, F(s)=L {f(t); t s }

Solve resulting algebraic equations in s,

for solution X = X(s)

Convert solution X(s) into time function x(t), via inverse Laplace transform:

x(t) = L-1{X(s); s t }

• M.D. Bryant ME 344 notes 03/25/08

4

Example: Solve initial value problem, differential equation with initial condition:

x + 2 n x + 2n x = Ho us(t)

x(0) = xo , x (0) = x1

Apply Laplace transform

L= t= 0

{ . } e- st dt to all terms in equation:

!

x (t)e" st

t= 0

#

\$ dt + 2%&n x (t)e

" st

t= 0

#

\$ dt + &n

2

x(t)e" st

dtt= 0

#

\$

=

t= 0

Ho us(t) e- st dt

factor constants and apply definition for us(t) :

t= 0

d2x

dt2 e- st dt + 2n

t= 0

dxdt e- st dt +

2n

t= 0

x e- st dt

= Ho t= 0

e- st dt

• M.D. Bryant ME 344 notes 03/25/08

5

Define X(s) = t= 0

x(t) e- st dt

Integrate by parts

dxdt e- st|t=0 +s

t= 0

dxdt e- st dt +2n {x(t) e- st|t=0

+s t= 0

x(t) e- st dt } +

2n X(s) =

!

HO

("s) e- st|

t=0

Rearrange, note

!

limt"#e

\$ st

= 0

- x (0) + s (- x(0) + s X(s) ) + 2n {- x(0) +

s X(s) } + 2n X(s) =

Ho s

Note: L {dx/dt} = - x(0) + s X(s)

L {d2x/dt2} = L {d

!

x /dt} =-

!

x (0) + sL {dx/dt}

Result: algebraic equation in s

!

(s2

+ 2"#n

s+#n

2

)X(s) =H

O

s+ x

1+ (s+ 2"#

n

)x0

Note: x (0) = x1 , x(0) = xo

• M.D. Bryant ME 344 notes 03/25/08

6

Solve

X(s) =

Ho s + x1 + ( s + 2 n) xo

s2 + 2 n s + 2n

Note:

o characteristic equation in denominator o terms from initial conditions x1 ,

xo grouped with excitation term

!

HO

s

• M.D. Bryant ME 344 notes 03/25/08

7

Apply inverse Laplace: x(t) = L-1{X(s); s t }

x(t) = Ho

2n

L-1{

2n

s (s2 + 2 n s + 2n)

}

+ xo

2n

L-1{s

2n

s2 + 2 n s + 2n

}

+ x1 + 2 n xo

2n

L-1{

2n

s2 + 2 n s + 2n

}

o Use Laplace transform tables for L-1:

x(t) = Ho

2n

{1 - e- n t sin(n 1 - 2 t + arccos )

1 - 2 }

+ xo

2n

{ -

2n e- n t sin(n 1 - 2 t - arccos )

1 - 2 }

+ x1 + 2 n xo

2n

{ n e- n t sin(n 1 - 2 t )

1 - 2 }

• M.D. Bryant ME 344 notes 03/25/08

8

Transfer Functions Method to represent system dynamics, via

s representation from Laplace transforms. Transfer functions show flow of signal through a system, from input to output.

Transfer function G(s) is ratio of output x

to input f, in s-domain (via Laplace trans.):

!

G(s) =X(s)

F(s) Method gives system dynamics

representation equivalent to

Ordinary differential equations State equations

Interchangeable: Can convert transfer function to differential equations

• M.D. Bryant ME 344 notes 03/25/08

9

Transfer Function G(s) Describes dynamics in operational sense Dynamics encoded in G(s) Ignore initial conditions (I.C. terms are

transient & decay quickly) Transfer function, for input-output

operation, deals with steady state terms

h(t)

H(s)

x(t)

X(s)

G(s)

Example: Speed of automobile

o Output: speed x(t) expressed as X(s) o Output determined by input, system

dynamics & initial conditions o Input: gas pedal depression h(t) o Dynamics: fuel system delivery,

motor dynamics & torque, transmission, wheels, car translation, mass, wind drag, etc.

o Initial conditions (initial speed) influence on current speed diminishes over time, thus ignore

• M.D. Bryant ME 344 notes 03/25/08

10

Transfer Function Example: 2nd order system

Differential equation at steady state (can ignore initial conditions)

x + 2 n x + 2n x = h(t)

Apply Laplace transform, define

X(s) =L{x(t); t s}, H(s) =L{h(t); t s }

Result: algebraic equation

(s2 + 2n s + 2n )X(s) = H(s)

Transfer function G(s): ratio of output

X(s) to input H(s)

G(s) = X(s)H(s) = 1

s2 + 2 n s + 2n

Note: o characteristic equation in denominator o denominator roots => eigenvalues o transfer functions eigenvalues called

poles

• M.D. Bryant ME 344 notes 03/25/08

11

Example: First order system

!

" x + x = f (t)

Apply Laplace transform, define

X(s) =L{x(t); t s}, F(s) =L{f(t); t s }

L{ x ; t s} +L{x(t); t s} =L{f(t); t s }

Result: algebraic equation sX(s) + X(s) = F(s)

Transfer function

G(s) =X(s)/F(s) =

!

1

"s+1

Note: o characteristic equation in denominator o transfer functions eigenvalues = poles o p1 = 1 = - 1/

• M.D. Bryant ME 344 notes 03/25/08

12

Transfer Function from State Equations

Matrix form

!

x = Ax + f (t)

Explicit equation form

!

x k= a

jkj=1

n

" xj+ f

k(t)

Define

Xk(s)=L {xk(t); t s }, Fk(s)=L {fk(t); t s }

Apply Laplace transform

!

sX(s) = AX(s)+F(s)

or

!

sXk(s) = a

jkj=1

n

" Xj(s)+F

k(s)

• M.D. Bryant ME 344 notes 03/25/08

13

Rearrange matrix form:

!

sI " A[ ]X(s) = F(s)

Solve, via Cramers rule:

!

Xk(s) =

det sI " A[ ]kth column replaced by F ( s )

{ }det sI " A[ ]

Transfer function, define which output

Xk(s) and which input Fj(s):

!

Gkj(s) =

Xk(s)

Fj(s)

Solve, via previous result with all

components of F(s) zero, except Fj(s)

• M.D. Bryant ME 344 notes 03/25/08

14

Example: differential equations from transfer function:

!

G(s) =X(s)

F(s)=

2s+ 5

s4

+ 3s3

+ 2s2

+ 2s+1

Cross multiply ratios:

!

s4 + 3s 3 + 2s 2 + 2s+1( )X(s) = 2s+ 5( )F(s)

!

s4

X + 3s3

X + 2s2

X + 2sX + X = 2sF + 5F Treat: s => d/dt

!

d4x

dt4

+ 3d3x

dt3

+ 2d2x

dt2

+ 2dx

dt+ x = 2

df

dt+ 5 f (t)

• M.D. Bryant ME 344 notes 03/25/08

15

Block Diagrams

h(t)

H(s)

x(t)

X(s)

G(s)

Another way to represent system dynamics pictorially

Weakness: lacks causality information

Shows signal flow through system

Transfer function G(s) inside block

Output:

X(s) = G(s) H(s)

transfer function times input H(s)

Can assemble blocks into system model:

• M.D. Bryant ME 344 notes 03/25/08

16

H(s) X1(s)

G1(s)X2(s)

G2(s)X(s)

G3(s)

Output = input to next Models stringing of components

• M.D. Bryant ME 344 notes 03/25/08

17

Example: stereo system

CD player amplifier

speaker

CD player amplifier speakers

CD player:

!

G1(s) =

X1(s)

H (s)

Input: H(s) from CD laser reader

Output: CD voltage X1(s)

Power amplifier:

!

G2(s) =

X2(s)

X1(s)

Input: CD output voltage X1(s) Output: amp voltage X2(s)

Speakers:

!

G3(s) =

X(s)

X2(s)

Input: amp voltage X2(s) Output: sound, acoustic

pressure X(s)

• M.D. Bryant ME 344 notes 03/25/08

18

H(s) X1(s)

G1(s)X2(s)

G2(s)X(s)

G3(s)

Overall transfer function is product of block transfer functions:

!

G(s) =X(s)

H (s)=X(s)

X2(s)

X2(s)

X1(s)

X1(s)

H (s)=G

1(s)G

2(s)G

3(s)

Note:

!

G1(s) =

X1(s)

H (s),

!

G2(s) =

X2(s)

X1(s)

,

!

G3(s) =

X(s)

X2(s)

• M.D. Bryant ME 344 notes 03/25/08

19

Example H(s) X(s)3

s+1

2s !1

s2

+ 4s+1

Transfer function:

!

G(s) =X(s)

H (s)=G

1(s)G

2(s) =

3

s+1"2s #1

s2

+ 4s+1

!

G(s) =3(2s "1)

(s+1)(s2

+ 4s+1)=

6s " 3

s3

+ 5s2

+ 5s+1

Poles = roots of denominator (values of s

such that transfer function becomes infinite)

p1 = -1, p2,p3 = -2 3

Zeros = roots of numerator (values of s such that transfer function becomes 0) z1 = 1/2

• M.D. Bryant ME 344 notes 03/25/08

20

Summer (Summing Junction)

X1(s)

X2(s)X(s)+

+

-

X3(s)

Output = sum of inputs Sign on input => sign in equation Output X(s) = - X1(s) + X2(s) + X3(s)

• M.D. Bryant ME 344 notes 03/25/08

21

Example: Feedback Control System

R(s)X(s)

G(s)

H(s)

E(s)+

-

Goal: Closed loop transfer function

!

Gcl(s) =

X(s)

R(s)

Formulate:

!

E(s) = R(s)"H (s)X(s)

!

X(s) =G(s)E(s) Eliminate E(s):

!

X(s) =G(s)E(s) =G(s) R(s)"H (s)X(s)[ ]

Solve for closed loop X(s)/R(s):

!

Gcl(s) =

X(s)

R(s)=

G(s)

1+G(s)H (s)

• M.D. Bryant ME 344 notes 03/25/08

22

Example: Feedback Controller with Disturbance

Gc(s)R(s)X(s)

H(s)

E(s)+

-

Gp(s)++

D(s)

Closed loop transfer function, reference

input (temporarily set D(s) = 0 )

!

Gcl(s) =

X(s)

R(s)=

Gc(s)G

p(s)

1+Gc(s)G

p(s)H (s)

Transfer function, disturbance (set R(s) = 0)

!

Gd(s) =

X(s)

D(s)=

Gp(s)

1+Gc(s)G

p(s)H (s)

• M.D. Bryant ME 344 notes 03/25/08

23

Linear system, with both, sum outputs:

!

X(s) =Gcl(s)R(s)+G

d(s)D(s)

!

X(s) =G

c(s)G

p(s)

1+Gc(s)G

p(s)H (s)

R(s)

!

+G

p(s)

1+Gc(s)G

p(s)H (s)

D(s)

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