The slides used material the source PowerPoint files of Anthony Rossiter

the transfer function between a loop input and a signal

somewhere in the loop.

GC

forward _ path

M (s)

1 return _ path 1 M ( s )G ( s )

3

s 1

G( s)

; M ( s)

s2

s

1. For the loop above, find the transfer function

relationship between the signal r and the signal u.

2. Where are the closed-loop poles?

u

M

r

1 G ( x) M ( x)

GC

( s 1)( s 2)

( s 1)( s 2)

2

s ( s 2) 3( s 1)

s 5s 3

1. For the loop below, find the transfer function

relationship between the signal r and the signal y.

2. What is the closed-loop pole polynomial? Are the

poles all in the LHP?

0.5

s 0.4

( s 10 )

G( s)

; M ( s) 2

; H ( s ) 0.1

s 1

s

( s 1)

forward _ path

M (s)

GC

1 return _ path 1 M ( s )G ( s ) H ( s )

2 0.5( s 0.4)

( s 0.4)( s 1)

s ( s 1)

GC

2 0.1 0.5( s 10)( s 0.4) s ( s 1) 2 0.1( s 10)( s 0.4)

1

s ( s 1) 2

PC s 3 2 s 2 2 0.4 s 0.4

1. For the loop below, find the transfer function relationship

between the signal r and the signal z.

2. What is the closed-loop pole polynomial? Are the poles all in

the LHP? Where are the poles?

0.5

s 0.5

G( s)

; M ( s) 2

; K ( s) 0.5;

s2

s

( s 8)

2

H ( s) 0.1

; T (s)

( s 0.8)

s5

0.5

s 0.5

; M ( s) 2

; K ( s) 0.5;

s2

s

( s 8)

2

H ( s) 0.1

; T (s)

( s 0.8)

s5

G( s)

GC

forward _ path

K (s)M (s)

1 return _ path 1 M ( s ) K ( s )G ( s ) H ( s )T ( s )

( s 0.5)

s

GC

( s 0.5) 0.5

( s 8)

2

1

0 .1

s

s2

( s 0.8) ( s 5)

The syntax for simple loops is as follows:

Gc = feedback(FP,RP)

Where

• FP is blocks between loop input and signal.

• RP is remaining blocks in the loop.

>> %% Question 1

>> G=tf(3,[1 2])

G=

3

----s+2

Continuous-time transfer

function.

>> M=tf([1 1], [1 0])

M=

s+1

----s

Continuous-time transfer

function.

>> Gc=feedback (M,G)

Gc =

s^2 + 3 s + 2

------------s^2 + 5 s + 3

Continuous-time transfer function.

>> [p,z]=pzmap(Gc)

p=

-4.3028

-0.6972

z=

-2

-1

>> %%Question 2

>> G=tf(0.5,[1 1])

G=

0.5

----s+1

Continuous-time transfer function.

>> M=2*tf([1 0.4], [1 0])

M=

2 s + 0.8

--------s

Continuous-time transfer function.

>> H=0.1*tf([1 10], [1 1])

H=

0.1 s + 1

--------s+1

Continuous-time transfer function.

>> Gc=feedback (M*G,H)

Gc =

s^2 + 1.4 s + 0.4

---------------------------s^3 + 2.1 s^2 + 2.04 s + 0.4

Continuous-time transfer function.

>> [p,z]=pzmap(Gc)

p=

-0.9226 + 0.8477i

-0.9226 - 0.8477i

-0.2548 + 0.0000i

z=

-1.0000

-0.4000

>> %%Question 3

>> G=tf(0.5,[1 2])

G=

0.5

----s+2

Continuous-time transfer function.

>> M=2*tf([1 0.5], [1 0])

M=

2s+1

------s

Continuous-time transfer function.

>> K=0.5

K=

0.5000

>> H=0.1*tf([1 8], [1 0.8])

H=

0.1 s + 0.8

----------s + 0.8

Continuous-time transfer function.

>> T=tf(2, [1 5])

T=

2

----s+5

Continuous-time transfer function.

>> Gc=feedback (M*K,G*H*T)

Gc =

s^4 + 8.3 s^3 + 19.5 s^2 + 15.8 s + 4

--------------------------------------s^4 + 7.8 s^3 + 15.7 s^2 + 8.85 s + 0.4

Continuous-time transfer function.

>> [p,z]=pzmap(Gc)

p=

-5.0210

-1.8656

-0.8640

-0.0494

z=

-5.0000

-2.0000

-0.8000

-0.5000

• What is frequency response?

• Why is it useful?

• How do I represent frequency response

information in a helpful fashion?

• Why is this relevant to feedback loop analysis and

design?

1. Frequency response refers to system behaviour and therefore is

defined for a system.

2. Here the focus is linear systems that can be represented using a

transfer function model.

In simple terms, we are asking how the output of a system behaves

when the system input is a pure sinusoid.

u sin(wt )

y A sin(wt )

• When the input sinusoid had

frequency ‘w’, the output signal was: y A sin(wt )

• What happens if the frequency is

changed?

u sin(t )

y C sin(t )

Frequency response is a description of how the

amplitude and phase shift of the output, relative to

the input, change with frequency.

Some examples will illustrate this.

Slides by Anthony Rossiter

16

Overlay the input and output for a system G(s)

and a single frequency (here w = 1rad/s).

4

G 2

s 3s 2

Output

amplitude

about 1.2

1

0.5

0

u sin t

-0.5

Phase

shift about

1 second

which

-1

corresponds to 1

45

radian 50

y 1.2 sin(t 1)

55

Time (secs)

60

65

1

Output

amplitude

about 0.62

0.8

0.6

0.4

0.2

0

-0.2

-0.4

Phase

shift about

-0.6

0.9 second which

-0.8

corresponds

to

-1

50

51

52

53

1.8

radian

u sin 2t

y 0.62 sin(2t 1.8)

54

55

56

Time (secs)

57

58

59

60

As the frequency of the input changes:

• The amplitude of the output changes.

• The phase shift of the output changes.

Frequency response is a description of how the

amplitude and phase shift depend upon the

frequency of the input, that is, how do the

characteristics of the response, depend upon the

frequency?

Some systems include integrators and hence, even

with a sinusoidal input, the output signal is not

centred on zero.

Hence, amplitude is half peak to peak.

The following signal, has an amplitude of oscillation

y

B

A

sin(

wt

)

of A.

Gain is the ratio

of the output

amplitude to the

input amplitude.

Slides by Anthony Rossiter

20

Let the output signal be given as

y(t ) B A sin(wt )

Now shift this signal by t1 seconds.

y(t1) wt1!

B

The phase shift is clearly

A sin(wt wt1 )

A time shift of t1 is equivalent to a phase shift of wt1

Units then balance:

Frequency w is inSlides

rad/s

and

by Anthony

Rossitertime is seconds.

21

Find the time

between points of

corresponding

phase: t = t1-t2.

Then the phase

lag is given as =

w(t1-t2)

Find the

amplitude of

oscillation as half

peak to peak.

Compare ratio of

input and output.

y

sin(t [

])

• What is frequency response?

• Why is it useful?

• How do I compute this efficiently?

• How do I represent frequency response

information in a helpful fashion?

• Why is this relevant to feedback loop analysis and

design?

The previous part how frequency response values could be inferred

from time responses.

1. Gain A(w) is the ratio of output amplitude of oscillation to that

of the input.

2. Phase (w) is the phase difference between the input and

output responses.

However it is tedious and inaccurate to infer A and from time

responses.

u D sin(wt )

G(s)

y DA sin(wt )

A more efficient and simple method for determining A(w)

and (w) is from the system transfer function G(s). It can

be show that:

A G( jw) ; arg(G( jw))

That is, substitute s=jw and compute the modulus and

argument of the resulting complex number.

the transfer function between a loop input and a signal

somewhere in the loop.

GC

forward _ path

M (s)

1 return _ path 1 M ( s )G ( s )

3

s 1

G( s)

; M ( s)

s2

s

1. For the loop above, find the transfer function

relationship between the signal r and the signal u.

2. Where are the closed-loop poles?

u

M

r

1 G ( x) M ( x)

GC

( s 1)( s 2)

( s 1)( s 2)

2

s ( s 2) 3( s 1)

s 5s 3

1. For the loop below, find the transfer function

relationship between the signal r and the signal y.

2. What is the closed-loop pole polynomial? Are the

poles all in the LHP?

0.5

s 0.4

( s 10 )

G( s)

; M ( s) 2

; H ( s ) 0.1

s 1

s

( s 1)

forward _ path

M (s)

GC

1 return _ path 1 M ( s )G ( s ) H ( s )

2 0.5( s 0.4)

( s 0.4)( s 1)

s ( s 1)

GC

2 0.1 0.5( s 10)( s 0.4) s ( s 1) 2 0.1( s 10)( s 0.4)

1

s ( s 1) 2

PC s 3 2 s 2 2 0.4 s 0.4

1. For the loop below, find the transfer function relationship

between the signal r and the signal z.

2. What is the closed-loop pole polynomial? Are the poles all in

the LHP? Where are the poles?

0.5

s 0.5

G( s)

; M ( s) 2

; K ( s) 0.5;

s2

s

( s 8)

2

H ( s) 0.1

; T (s)

( s 0.8)

s5

0.5

s 0.5

; M ( s) 2

; K ( s) 0.5;

s2

s

( s 8)

2

H ( s) 0.1

; T (s)

( s 0.8)

s5

G( s)

GC

forward _ path

K (s)M (s)

1 return _ path 1 M ( s ) K ( s )G ( s ) H ( s )T ( s )

( s 0.5)

s

GC

( s 0.5) 0.5

( s 8)

2

1

0 .1

s

s2

( s 0.8) ( s 5)

The syntax for simple loops is as follows:

Gc = feedback(FP,RP)

Where

• FP is blocks between loop input and signal.

• RP is remaining blocks in the loop.

>> %% Question 1

>> G=tf(3,[1 2])

G=

3

----s+2

Continuous-time transfer

function.

>> M=tf([1 1], [1 0])

M=

s+1

----s

Continuous-time transfer

function.

>> Gc=feedback (M,G)

Gc =

s^2 + 3 s + 2

------------s^2 + 5 s + 3

Continuous-time transfer function.

>> [p,z]=pzmap(Gc)

p=

-4.3028

-0.6972

z=

-2

-1

>> %%Question 2

>> G=tf(0.5,[1 1])

G=

0.5

----s+1

Continuous-time transfer function.

>> M=2*tf([1 0.4], [1 0])

M=

2 s + 0.8

--------s

Continuous-time transfer function.

>> H=0.1*tf([1 10], [1 1])

H=

0.1 s + 1

--------s+1

Continuous-time transfer function.

>> Gc=feedback (M*G,H)

Gc =

s^2 + 1.4 s + 0.4

---------------------------s^3 + 2.1 s^2 + 2.04 s + 0.4

Continuous-time transfer function.

>> [p,z]=pzmap(Gc)

p=

-0.9226 + 0.8477i

-0.9226 - 0.8477i

-0.2548 + 0.0000i

z=

-1.0000

-0.4000

>> %%Question 3

>> G=tf(0.5,[1 2])

G=

0.5

----s+2

Continuous-time transfer function.

>> M=2*tf([1 0.5], [1 0])

M=

2s+1

------s

Continuous-time transfer function.

>> K=0.5

K=

0.5000

>> H=0.1*tf([1 8], [1 0.8])

H=

0.1 s + 0.8

----------s + 0.8

Continuous-time transfer function.

>> T=tf(2, [1 5])

T=

2

----s+5

Continuous-time transfer function.

>> Gc=feedback (M*K,G*H*T)

Gc =

s^4 + 8.3 s^3 + 19.5 s^2 + 15.8 s + 4

--------------------------------------s^4 + 7.8 s^3 + 15.7 s^2 + 8.85 s + 0.4

Continuous-time transfer function.

>> [p,z]=pzmap(Gc)

p=

-5.0210

-1.8656

-0.8640

-0.0494

z=

-5.0000

-2.0000

-0.8000

-0.5000

• What is frequency response?

• Why is it useful?

• How do I represent frequency response

information in a helpful fashion?

• Why is this relevant to feedback loop analysis and

design?

1. Frequency response refers to system behaviour and therefore is

defined for a system.

2. Here the focus is linear systems that can be represented using a

transfer function model.

In simple terms, we are asking how the output of a system behaves

when the system input is a pure sinusoid.

u sin(wt )

y A sin(wt )

• When the input sinusoid had

frequency ‘w’, the output signal was: y A sin(wt )

• What happens if the frequency is

changed?

u sin(t )

y C sin(t )

Frequency response is a description of how the

amplitude and phase shift of the output, relative to

the input, change with frequency.

Some examples will illustrate this.

Slides by Anthony Rossiter

16

Overlay the input and output for a system G(s)

and a single frequency (here w = 1rad/s).

4

G 2

s 3s 2

Output

amplitude

about 1.2

1

0.5

0

u sin t

-0.5

Phase

shift about

1 second

which

-1

corresponds to 1

45

radian 50

y 1.2 sin(t 1)

55

Time (secs)

60

65

1

Output

amplitude

about 0.62

0.8

0.6

0.4

0.2

0

-0.2

-0.4

Phase

shift about

-0.6

0.9 second which

-0.8

corresponds

to

-1

50

51

52

53

1.8

radian

u sin 2t

y 0.62 sin(2t 1.8)

54

55

56

Time (secs)

57

58

59

60

As the frequency of the input changes:

• The amplitude of the output changes.

• The phase shift of the output changes.

Frequency response is a description of how the

amplitude and phase shift depend upon the

frequency of the input, that is, how do the

characteristics of the response, depend upon the

frequency?

Some systems include integrators and hence, even

with a sinusoidal input, the output signal is not

centred on zero.

Hence, amplitude is half peak to peak.

The following signal, has an amplitude of oscillation

y

B

A

sin(

wt

)

of A.

Gain is the ratio

of the output

amplitude to the

input amplitude.

Slides by Anthony Rossiter

20

Let the output signal be given as

y(t ) B A sin(wt )

Now shift this signal by t1 seconds.

y(t1) wt1!

B

The phase shift is clearly

A sin(wt wt1 )

A time shift of t1 is equivalent to a phase shift of wt1

Units then balance:

Frequency w is inSlides

rad/s

and

by Anthony

Rossitertime is seconds.

21

Find the time

between points of

corresponding

phase: t = t1-t2.

Then the phase

lag is given as =

w(t1-t2)

Find the

amplitude of

oscillation as half

peak to peak.

Compare ratio of

input and output.

y

sin(t [

])

• What is frequency response?

• Why is it useful?

• How do I compute this efficiently?

• How do I represent frequency response

information in a helpful fashion?

• Why is this relevant to feedback loop analysis and

design?

The previous part how frequency response values could be inferred

from time responses.

1. Gain A(w) is the ratio of output amplitude of oscillation to that

of the input.

2. Phase (w) is the phase difference between the input and

output responses.

However it is tedious and inaccurate to infer A and from time

responses.

u D sin(wt )

G(s)

y DA sin(wt )

A more efficient and simple method for determining A(w)

and (w) is from the system transfer function G(s). It can

be show that:

A G( jw) ; arg(G( jw))

That is, substitute s=jw and compute the modulus and

argument of the resulting complex number.

## Tài liệu Bài giảng - Thiết kế mạng ppt

## Tài liệu Bài giảng: Fieldbus Foundation™ Khắc phục sự cố mạng tuyến ppt

## Tài liệu BÀI GIẢNG VỀ KỸ THUẬT LẬP TRÌNH pptx

## Tài liệu Bài giảng: Ứng dụng đồ họa máy tính và phim hoạt hình doc

## Tài liệu Bài giảng môn Mạng máy tính (cập nhật 08/2006) pptx

## Tài liệu Bài giảng " Thiết kế, cài đặt và điều hành mạng" docx

## Tài liệu Bài giảng: Tạo và làm việc với bảng pdf

## Tài liệu BÀI GIẢNG TÓM TẮT HỆ ĐIỀU HÀNH doc

## Tài liệu Bài Giảng Môn Lập Trình Website ASP.Net part 2 ppt

## Tài liệu Bài Giảng Môn Lập Trình Website ASP.Net part 3 ppt

Tài liệu liên quan