Tài liệu bài giảng 4 bode diagram

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) 
s2
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;
s2
s
( s  8)
2
H ( s)  0.1
; T (s) 
( s  0.8)

s5


0.5
s  0.5
; M ( s)  2
; K ( s)  0.5;
s2
s
( s  8)
2
H ( s)  0.1
; T (s) 
( s  0.8)
s5

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
s2
( 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.