Transfer Functions and Block Diagram Algebra in Systems Analysis and Control, Lecture notes of Dynamics

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Systems Analysis and Control

Matthew M. Peet

Arizona State University

Lecture 6: Calculating the Transfer Function

Introduction

In this Lecture, you will learn: Transfer Functions

• Transfer Function Representation of a System
• State-Space to Transfer Function
• Direct Calculation of Transfer Functions

Block Diagram Algebra

• Modeling in the Frequency Domain
• Reducing Block Diagrams

Transfer Functions

Example: Simple System

State-Space:

x˙(t) = −x(t) + u(t)

y(t) = x(t) −. 5 u(t) x(0) = 0

Apply the Laplace transform to the first equation:

L

x ˙(t) = −x(t) + u(t)

which gives sˆx(s) = −xˆ(s) + ˆu(s).

Solving for xˆ(s), we get

(s + 1)ˆx(s) = ˆu(s) and so xˆ(s) =

s + 1

ˆu(s).

Similarly, the second equation yields:

y ˆ(s) = ˆx(s) − .5ˆu(s) =

s + 1

uˆ(s) − .5ˆu(s) =

1 − .5(s + 1)

s + 1

uˆ(s) =

s − 1

s + 1

uˆ(s)

Thus we have the Transfer Function:

G(s) =

s − 1

s + 1

Transfer Functions

Example: Step Response

The Transfer Function provides a convenient way to find the response to inputs.

Step Input Response: uˆ(s) =

1

s

y ˆ(s) =

G(s)ˆu(s) =

s − 1

s + 1

s

s − 1

s

2

• s

s + 1

s

Consulting our table of Laplace

Transforms,

y(t) =

L

− 1

s + 1

L

− 1

s

= e

−t

−

1 (t)

0 1 2 3 4 5 6

−0.

−0.

−0.

−0.

−0.

0

Step Response

Time (sec)

Amplitude

Inverted Pendulum Example

Return to the pendulum.

Dynamics:

θ(t) =

M gl

2 J

θ(t) +

J

T (t)

y(t) = θ(t)

For the first equation,

s

2 ˆ θ(s) =

M gl

2 J

θ(s) +

J

T (s)

Solve for

θ(s):

θ(s) =

J

s

2 −

M gl

2 J

T (s)

Second Equation: yˆ(s) =

θ(s)

Transfer Function:

G(s) =

ˆy(s)

T (s)

J

s

2 −

M gl

2 J

Inverted Pendulum Example: Impulse Response

Impulse Input: uˆ(s) = 1

yˆ(s) =

G(s)ˆu(s) =

J

s

2 −

M gl

2 J

J

(s −

M gl

2 J

)(s +

M gl

2 J

J

2 J

M gl

s −

M gl

2 J

s +

M gl

2 J

0 5 10 15

0

2

4

6

8

10

12

14

16

18

x 10

5 Impulse Response

Time (sec)

Amplitude

Figure: Impulse Response with

g = l = J = 1, M = 2

In time-domain:

y(t) =

J

2 J

M gl

e

M gl

2 J

t

− e

−

M gl

2 J

t

Pendulum Accelerates to infinity!

Constructing the Transfer Function: Suspension System

x 1

x 2

m c

m w

u

Apply the Laplace Transform to the dynamics:

s

2

zˆ 1 (s) = −

K

1

m c

z ˆ 1 (s) −

c

m c

sˆz 1 (s) +

K

1

m c

ˆz 2 (s) +

c

m c

szˆ 2 (s)

s

2

zˆ 2 (s) =

K

1

m w

ˆz 1 (s) +

c

m w

sˆz 1 (s) −

K

1

m w

K

2

m w

z ˆ 2 (s) −

c

m w

szˆ 2 (s) −

K

2

m w

u ˆ(s)

yˆ(s) = ˆz 2 (s)

Constructing the Transfer Function: Suspension System

We isolate the z 1

and z 2

terms:

s

2

c

m c

s +

K

1

m c

ˆz 1

(s) =

K

1

m c

c

m c

s

z ˆ 2

(s)

s

2

c

m w

s +

K

1

m w

K

2

m w

ˆz 2

(s) =

K

1

m w

c

m w

s

z ˆ 1

(s) −

K

2

m w

u ˆ(s)

ˆy(s) = ˆz 2 (s)

Which yields

ˆz 1 (s) =

K 1

m c

c

m c

s

s

2

c

m c

s +

K 1

m c

zˆ 2 (s)

zˆ 2 (s) =

K 1

m w

c

m w

s

s

2

c

m w

s +

K 1

m w

K 2

m w

z ˆ 1 (s) −

K 2

m w

s

2

c

m w

s +

K 1

m w

K 2

m w

ˆu(s)

Block Diagram Algebra

Series (Cascade) Interconnection

The interconnection of systems can be represent by block diagrams.

G H

u y y 1

Cascade of Systems: Suppose we have two systems: G and H.

Definition 2.

The Cascade or Series interconnection of two systems is

y 1

= Gu y = Hy 1

or

y = H(G(u))

Block Diagram Algebra

Series Connection (Cascade)

G(s) H(s)

u(s) y y(s) 1

(s)

H(s)G(s)

u(s) y(s)

Series Interconnection:

• The output of G is the input to H.
• Let

G(s) and

H(s) be the transfer functions for G and H.

• Then

ˆy 1

(s) =

G

1

(s)ˆu(s) ˆy(s) =

H(s)ˆy 1

(s) =

H(s)

G(s)ˆu(s)

• The Transfer Function,

T (s) for the combination of G and H is

T (s) =

H(s)

G(s)

Note: The order of the

G and

H!

Block Diagrams

Parallel Connection

G(s)

H(s)

u(s) y(s)

y 1

(s)

H(s)+G(s)

u(s) y(s)

The Transfer function of a Parallel interconnection:

• Laplace transform:

y ˆ(s) = ˆy 1

(s) + ˆy 2

(s) =

G(s)ˆu(s) +

H(s)ˆu(s) =

H(s) +

G(s)

ˆu(s)

• The Transfer Function,

T (s) for the parallel interconnection of G and H is

T (s) =

H(s) +

G(s)

Block Diagrams

Lower Feedback Interconnection

G(s)

K(s)

y(s)

u(s)

Feedback:

• Controller: z = K(u − y) Plant: y = Gz

In the Frequency Domain:

z ˆ(s) = −

K(s)ˆy(s) +

K(s)ˆu(s) yˆ(s) =

G(s)ˆz(s)

so

ˆy(s) =

G(s)ˆz(s) = −

G(s)

K(s)ˆy(s) +

G(s)

K(s)ˆu(s)

Solving for yˆ(s),

y ˆ(s) =

G(s)

K(s)

G(s)

K(s)

ˆu(s)

The Effect of Feedback: Impulse Response

Inverted Pendulum Model

Transfer Function ˆ G(s) =

Js

2 −

M gl

2

Controller: Static Gain:

K(s) = K

Input: Impulse: ˆu(s) = 1.

Closed Loop: Lower Feedback

y ˆ(s) =

G(s)

K(s)

G(s)

K(s)

ˆu(s) =

K

Js

2 −

M gl

2

K

Js

2 −

M gl

2

K

Js

2 −

M gl

2

+ K

First Case:

• If K >

M gl

2

, then K −

M gl

2

0 , so

ˆy(s) =

K/J

s

2

K/J −

M gl

2 J

y(t) =

K

J

K/J −

M gl

2 J

sin

K/J −

M gl

2 J

t

0 2 4 6 8 10 12

−2.

−

−1.

−

−0.

0

1

2

Impulse Response

Time (sec)

Amplitude

The Effect of Feedback: Impulse Response

Inverted Pendulum Model

0 5 10 15 20 25

0

2

4

6

8

10

12

14

16

18

x 10

6 Impulse Response

Time (sec)

Amplitude

Second Case:

• If K <

M gl

2

, then K −

M gl

2

< 0 , so

ˆy(s) =

K

J

s −

K/J −

M gl

2 J

s +

K/J −

M gl

2 J

y(t) =

K

J

e

K/J−

M gl

2 J

t

• e

−

K/J−

M gl

2 J

t

Important: Value of K determines stability vs. instability