The slides used the source PowerPoint files of Anthony Rossiter

• What is frequency response?

• 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?

• Previous topic showed how we can determine

expressions for frequency response gain and phase

based on a transfer function.

• These formulae are not insightful on their own.

• Humans often relate better to pictures and graphs.

• Hence it is useful to sketch the gain and phase in

graphical form and thus to judge whether such a

sketch would be helpful.

1. a transfer function object G

2. a frequency ‘w’

Then

[gain,phase]=bode(G,w);

3

3

1 w

st

take

GTo

( s )begin,

; Ga( simple

jw ) 1 order;example.

G ( jw ) tan

2

2

s2

2

w 2

0

1.5

-20

Gain

Phase

1

-40

-60

0.5

-80

0

0

20

40

60

Frequency

80

100

-100

0

20

40

60

Frequency

80

100

Gain

1. For low frequencies gain is 1.5.

2. As frequency increases, gain

drops to zero.

3. For low frequencies phases is 0.

4. As frequency increases, phase

tends to -90.

1

0.5

0

0

20

40

60

80

100

Frequency

0

-20

Phase

k

G( s)

sa

1.5

-40

-60

-80

-100

0

20

40

60

Frequency

80

100

4

G ( s) 2

;

s 5s 6

G ( jw) tan 1

G ( jw)

4

( w 2 )(w 3 )

2

2

2

2

;

w

w

tan 1

2

3

0.7

0

0.6

-50

Phase

Gain

0.5

0.4

0.3

0.2

-100

-150

0.1

0

0

20

40

60

Frequency

80

100

-200

0

20

40

60

Frequency

80

100

0.7

0.6

4

G( s) 2

s 5s 6

Gain

0.4

0.3

0.2

0.1

0

0

20

40

60

80

100

Frequency

0

-50

Phase

1. For low frequencies gain is

0.67 and as frequency

increases, gain drops to zero.

2. For low frequencies phases is

0 and as frequency increases,

phase tends to -180.

0.5

-100

-150

-200

0

20

40

60

Frequency

80

100

s4

G( s) 2

;

s 2s 20

G ( jw)

w2 4 2

(20 w2 ) 2 4w2

;

2w

G ( jw) tan

20 w2

1

0.7

20

0.6

0

0.5

Phase

Gain

-20

0.4

0.3

-60

0.2

-80

0.1

0

0

-40

20

40

60

Frequency

80

100

-100

0

20

40

60

Frequency

80

100

s4

s 2 2s 20

0.5

Gain

1. Patterns are less obvious.

2. As frequency increases, gain

drops to zero and phase to 90.

3. For low frequencies overall

pattern is unclear although

there is a marked increase in

gain for a small frequency

range.

0.6

This

indicates a

resonance

0.4

0.3

0.2

0.1

0

0

20

40

60

80

100

Frequency

20

0

-20

Phase

G( s)

0.7

-40

-60

-80

-100

0

20

40

60

Frequency

80

100

• It is straightforward to sketch the frequency response

parameters and hence get an overview of how gain and

phase change over a range of frequencies.

• However, all the focus is on the larger frequencies as

any notable changes in the low frequency range are

cramped into a small part of the graph domain.

Similarly very large frequencies are excluded.

• It is not obvious from formulae what causes the shapes

and asymptotes that arise and hence how changes in

poles and zeros will affect the overall shapes.

• The previous topic showed that simple sketches of

frequency response gain and phase are useful to gain

an overview of behaviour.

• However, the sketches could only focus on a relatively

limited frequency range, for example 1-10 rad/s or 10100 rad/s, etc.

• In practice, different decades need an equal spacing on

the graph in order to display effectively the changes in

behaviour that may occur in different frequency ranges.

• A similar comment can be applied to changes in gain.

1. Comprises 2 plots.

2. Both plots have log10w on the x-axis or w on a

log axis (same thing but this one displays actual

w which is helpful).

3. y-axis plots are either:

a. 20log10|G(jw)|

b. Arg(G(jw))

(denoted decibels)

(usually in degrees)

Bode Diagram

10

Magnitude (dB)

0

-10

-20

Note: use of

decibels for gain.

-30

Phase (deg)

-40

0

Note: frequency on

a log10 scale.

-45

-90

-1

10

10

0

1

10

Frequency (rad/s) (rad/s)

2

10

Bode Diagram

0

Magnitude (dB)

-10

-20

-30

s4

G 2

s 2s 20

-40

45

Phase (deg)

0

-45

-90

-135

-1

10

0

10

1

10

Frequency (rad/s) (rad/s)

2

10

It is straightforward to sketch the bode diagrams

using MATLAB and hence to get an overview of

how gain and phase change over a large range of

frequencies.

The use of log axis for frequencies enables us to

focus on several different decades in the same plot,

which is helpful, although we do not explicitly plot

values for w0,∞.

It is not yet obvious what causes the shapes and

asymptotes that arise and hence how changes in

poles and zeros will affect the overall shapes.

• There is a need to understand how different factors in

the transfer function relate to shapes in the Bode plot.

• This understanding will allow useful insight, especially for

control design which comes later.

• As will become apparent, we need to be comfortable

with rules of logarithms.

log(ab)=log(a)+log(b)

log(a/b)=log a- log(b)

How does a change in the decibel scale relate to changes

in the underlying gain?

20log10(10a)=

20log10 (√2 a)=

20log10 (a/√2 )=

20log10 (a/10)=

1. A change of 20dB is equivalent to a change in

gain of a factor of 10. (20dB up is multiply by 10

and 20dB down is divide by 10).

2. A change in gain of 3dB is equivalent to a change

in gain of a factor of √2 (approx).

Similarly one can see that 6dB is

equivalent to a factor of 2 (approx.).

It can be helpful, for mental arithmetic and insight,

to know logarithms for several key integers.

log10(1) = 0

log10(2) ≈ 0.3

log10(10) =1

log10(3) ≈ 0.48 (or 9.6dB)

log10(100) = 2

log10(10n) =n

log10(4) ≈ 0.6

(or 12dB)

log10(5) ≈ 0.7

(or 14dB)

log10(0.1) =-1

log10(6) ≈ 0.78

log

(0.01)

=-2

10

log10(8) ≈ 0.9

(or 18dB)

• The previous topic showed that MATLAB can be used for

forming exact Bode diagrams and reminded students of

core properties of logarithms.

• However, it is recognised that the ability to sketch is core

for developing insight and for use in design.

• This topic will use the definition of the Bode diagram,

and properties of logarithms, to sketch Bode diagrams

for:

1

1

G s,

, ( s a),

s

sa

Comprises 2 plots.

1. Both plots use log10w or w on a log axis

2. The other axis plots are either:

a. 20log10|G(jw)|

b. Arg(G(jw))

(denoted decibels)

(usually in degrees)

Next, apply these definitions to each factor in turn.

• What is frequency response?

• 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?

• Previous topic showed how we can determine

expressions for frequency response gain and phase

based on a transfer function.

• These formulae are not insightful on their own.

• Humans often relate better to pictures and graphs.

• Hence it is useful to sketch the gain and phase in

graphical form and thus to judge whether such a

sketch would be helpful.

1. a transfer function object G

2. a frequency ‘w’

Then

[gain,phase]=bode(G,w);

3

3

1 w

st

take

GTo

( s )begin,

; Ga( simple

jw ) 1 order;example.

G ( jw ) tan

2

2

s2

2

w 2

0

1.5

-20

Gain

Phase

1

-40

-60

0.5

-80

0

0

20

40

60

Frequency

80

100

-100

0

20

40

60

Frequency

80

100

Gain

1. For low frequencies gain is 1.5.

2. As frequency increases, gain

drops to zero.

3. For low frequencies phases is 0.

4. As frequency increases, phase

tends to -90.

1

0.5

0

0

20

40

60

80

100

Frequency

0

-20

Phase

k

G( s)

sa

1.5

-40

-60

-80

-100

0

20

40

60

Frequency

80

100

4

G ( s) 2

;

s 5s 6

G ( jw) tan 1

G ( jw)

4

( w 2 )(w 3 )

2

2

2

2

;

w

w

tan 1

2

3

0.7

0

0.6

-50

Phase

Gain

0.5

0.4

0.3

0.2

-100

-150

0.1

0

0

20

40

60

Frequency

80

100

-200

0

20

40

60

Frequency

80

100

0.7

0.6

4

G( s) 2

s 5s 6

Gain

0.4

0.3

0.2

0.1

0

0

20

40

60

80

100

Frequency

0

-50

Phase

1. For low frequencies gain is

0.67 and as frequency

increases, gain drops to zero.

2. For low frequencies phases is

0 and as frequency increases,

phase tends to -180.

0.5

-100

-150

-200

0

20

40

60

Frequency

80

100

s4

G( s) 2

;

s 2s 20

G ( jw)

w2 4 2

(20 w2 ) 2 4w2

;

2w

G ( jw) tan

20 w2

1

0.7

20

0.6

0

0.5

Phase

Gain

-20

0.4

0.3

-60

0.2

-80

0.1

0

0

-40

20

40

60

Frequency

80

100

-100

0

20

40

60

Frequency

80

100

s4

s 2 2s 20

0.5

Gain

1. Patterns are less obvious.

2. As frequency increases, gain

drops to zero and phase to 90.

3. For low frequencies overall

pattern is unclear although

there is a marked increase in

gain for a small frequency

range.

0.6

This

indicates a

resonance

0.4

0.3

0.2

0.1

0

0

20

40

60

80

100

Frequency

20

0

-20

Phase

G( s)

0.7

-40

-60

-80

-100

0

20

40

60

Frequency

80

100

• It is straightforward to sketch the frequency response

parameters and hence get an overview of how gain and

phase change over a range of frequencies.

• However, all the focus is on the larger frequencies as

any notable changes in the low frequency range are

cramped into a small part of the graph domain.

Similarly very large frequencies are excluded.

• It is not obvious from formulae what causes the shapes

and asymptotes that arise and hence how changes in

poles and zeros will affect the overall shapes.

• The previous topic showed that simple sketches of

frequency response gain and phase are useful to gain

an overview of behaviour.

• However, the sketches could only focus on a relatively

limited frequency range, for example 1-10 rad/s or 10100 rad/s, etc.

• In practice, different decades need an equal spacing on

the graph in order to display effectively the changes in

behaviour that may occur in different frequency ranges.

• A similar comment can be applied to changes in gain.

1. Comprises 2 plots.

2. Both plots have log10w on the x-axis or w on a

log axis (same thing but this one displays actual

w which is helpful).

3. y-axis plots are either:

a. 20log10|G(jw)|

b. Arg(G(jw))

(denoted decibels)

(usually in degrees)

Bode Diagram

10

Magnitude (dB)

0

-10

-20

Note: use of

decibels for gain.

-30

Phase (deg)

-40

0

Note: frequency on

a log10 scale.

-45

-90

-1

10

10

0

1

10

Frequency (rad/s) (rad/s)

2

10

Bode Diagram

0

Magnitude (dB)

-10

-20

-30

s4

G 2

s 2s 20

-40

45

Phase (deg)

0

-45

-90

-135

-1

10

0

10

1

10

Frequency (rad/s) (rad/s)

2

10

It is straightforward to sketch the bode diagrams

using MATLAB and hence to get an overview of

how gain and phase change over a large range of

frequencies.

The use of log axis for frequencies enables us to

focus on several different decades in the same plot,

which is helpful, although we do not explicitly plot

values for w0,∞.

It is not yet obvious what causes the shapes and

asymptotes that arise and hence how changes in

poles and zeros will affect the overall shapes.

• There is a need to understand how different factors in

the transfer function relate to shapes in the Bode plot.

• This understanding will allow useful insight, especially for

control design which comes later.

• As will become apparent, we need to be comfortable

with rules of logarithms.

log(ab)=log(a)+log(b)

log(a/b)=log a- log(b)

How does a change in the decibel scale relate to changes

in the underlying gain?

20log10(10a)=

20log10 (√2 a)=

20log10 (a/√2 )=

20log10 (a/10)=

1. A change of 20dB is equivalent to a change in

gain of a factor of 10. (20dB up is multiply by 10

and 20dB down is divide by 10).

2. A change in gain of 3dB is equivalent to a change

in gain of a factor of √2 (approx).

Similarly one can see that 6dB is

equivalent to a factor of 2 (approx.).

It can be helpful, for mental arithmetic and insight,

to know logarithms for several key integers.

log10(1) = 0

log10(2) ≈ 0.3

log10(10) =1

log10(3) ≈ 0.48 (or 9.6dB)

log10(100) = 2

log10(10n) =n

log10(4) ≈ 0.6

(or 12dB)

log10(5) ≈ 0.7

(or 14dB)

log10(0.1) =-1

log10(6) ≈ 0.78

log

(0.01)

=-2

10

log10(8) ≈ 0.9

(or 18dB)

• The previous topic showed that MATLAB can be used for

forming exact Bode diagrams and reminded students of

core properties of logarithms.

• However, it is recognised that the ability to sketch is core

for developing insight and for use in design.

• This topic will use the definition of the Bode diagram,

and properties of logarithms, to sketch Bode diagrams

for:

1

1

G s,

, ( s a),

s

sa

Comprises 2 plots.

1. Both plots use log10w or w on a log axis

2. The other axis plots are either:

a. 20log10|G(jw)|

b. Arg(G(jw))

(denoted decibels)

(usually in degrees)

Next, apply these definitions to each factor in turn.

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