Control System II
Dr. Noaman M.Noaman
Frequency Response Methods
1. Logarithmic plots or Bode diagram.
2. Polar plot.
3. Log-magnitude versus phase plot.
Bode diagram:
The transfer function may be represented by two plots, one giving the
magnitude versus frequency and the other the phase angle versus frequency
in logarithmic scale. Fig.1 represents the Sim-Log paper.
Fig.1 Number- Decibels conversion line
Basic factors of G (j w) H (j w)
The basic factors that very frequently occur in an arbitrary function G (jw)
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Control System II
Dr. Noaman M.Noaman
H (jw) is:
1. gain k
2. integral and derivative factors
jw ±1
3. first order factors (1 + jwT) ±1
4. quadratic factors [1 + 2ς
jw
jw
+ ( ) 2 ] ±1
wn
wn
The Gain k:
Any number greater than unity has a positive value in decibels, while a
number smaller than unity has a negative value.
The log magnitude curve for a constant k is a horizontal straight line of
20log k dB. The phase angle of the gain k is zero.
±1
Integral and derivative factors ( jw ) :
The logarithmic magnitude of
20log|
1
in dB is:
jw
1
| = -20 log w dB.
jw
The phase angle of
1
is a constant and equal to (-90). The log magnitude
jw
curve is straight line with a slope of -20 dB / dec (See Fig.2-a).
The logarithmic magnitude of jw is:
20log| jw | = -20logw dB.
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Control System II
Dr. Noaman M.Noaman
The phase angle of jw constant and equal to 90. The log magnitude curve is
a straight line with a slope of 20 dB / dec (See Fig.2-b).
Fig.2 a. Bode diagram for 1/jw.
b. Bode diagram for jw.
Note: The slopes of the log magnitude curves for the factors (
1 n
) and
jw
( jw) n are then -20n dB/dec and 20n dB/dec.
The phase angle of (
1 n
) is equal to (-90 * n) over the entire frequency
jw
range, while that of ( jw) is equal to (90 * n) over the entire frequency
n
range.
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Control System II
Dr. Noaman M.Noaman
±1
First order factors ( 1 + jwT )
The log magnitude of the 1st order factor
20log |
1
is:
1 + jwT
1
2 2
|= -20 log 1 + w T dB
1 + jwT
1
T
-for low frequencies (w<< ):
-20log 1 + w 2T 2 = -20log 1 = 0 dB
(the log-magnitude curve at low frequency is 0 dB)
1
T
-for high frequencies (w>> ):
-20log 1 + w 2T 2 =-20log wT dB
The frequency at which the two assumptions meet is called the corner
frequency or break frequency.
The exact phase angle
φ of the factor (1+j wT) is:
φ =-tan-1 wT
The Bode-diagram of (1/1+jwT) for asymptotic and exact curves is shown in
Fig.3.
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Control System II
Dr. Noaman M.Noaman
Fig.3 Bode-diagram of (1/1+jwT) for asymptotic and exact curves.
- The log magnitude of (1+j wT) is:
20log |1+j wT |=-20log |
1
|=20log
1 + jwT
1 + w 2T 2
The phase angle curve of (1+j wT) is:
-1
∠ (1+j wT)=tan wT = − ∠
1
1 + jwT
The Bode-diagram of (1+jwT) for asymptotic and exact curves is shown in
Fig.4.
Fig.4 Bode diagram for (1+jwT)
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Control System II
Dr. Noaman M.Noaman
Fig.5 The correction in drawing the Bode-diagram
±1
Note: The phase angles of ( 1 + jwT ) are
1
10T
1
2T
±5.7
±26.6
The quadratic factors [1 + 2ς
w=
1
T
±45
w=
±63.4
jw
jw
+ ( ) 2 ] ±1 :
wn
wn
1
- The quadratic factors of the form
jw
jw
1+ ς
+ ( )2
wn
wn
The log magnitude is:
20log |
1
jw
jw 2 |
1+ ς
+( )
wn
wn
6
2
T
w=
10
T
±84.3
Control System II
Dr. Noaman M.Noaman
w2 2
w 2
) + ( 2ς
)
mag = -20log (1 −
2
wn
wn
The phase angle is:
w
wn
1
= -tan-1 [
φ =∠
w 2 ]
jw
jw 2
1
−
(
)
1 + 2ς
)
+(
wn
wn
wn
2ς
The frequency response curves for the factor [ 1 + ς
jw
jw
+ ( )2 ]
wn wn
Can be obtained by reversing the sign of the magnitude and the phase angle
factors:
[
1
jw
jw 2
+(
1+ς
)
wn
wn
]
The bode-diagram of the asymptote and exact curve for the
quadratic factors is shown in Fig.6.
Note: wr = wn
Mr =
1 - 2ς 2
For 0< ς <0.707
1
2ς 1 − 2ς 2
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Control System II
Dr. Noaman M.Noaman
Fig.6 Bode-Diagram of the asymptote and exact curve for the
quadratic factors
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Control System II
Dr. Noaman M.Noaman
Example 1: Draw the Bode diagram for the T.F.:
G (s) =
10 ( s + 3)
s ( s + 2)[ s 2 + s + 2]
Sol:
G (jw) =
10 ( jw + 3)
jw ( jw + 2)[( jw 2 ) + jw + 2]
In order to avoid any possible mistakes in drawing B.D., it is desirable to put
G (jw) in the following normalized form:
w
+1
3
( jw ) 2 jw
jw
jw (
+ 1)[
+
+ 1]
2
2
2
j
G (jw) = 7.5
The function is composed of the following factors:
(jw)-1
(
Corner frequency
Slope
None
-20dB/sec
jw
+1)-1
2
wc= =2
1
T
-20dB/sec
jw
+1
3
3
+20dB/sec
( jw ) 2
[
+
2
jw
+1]-1
2
wc=wn=
9
2
-40dB/sec
Control System II
Dr. Noaman M.Noaman
20logk=20 log 7.5=17.5dB
φ
2ςw
=tan 0.33w – 90 – tan 0.5w– tan ( wn )
w
1 − ( )2
wn
-1
-1
wn2=2 Æ wn=
2 ς wn = 1
-1
2
Æ ς =0.353
φ = tan-1 0.33w – 90 – tan-1 0.5w – tan-1 (
w
0.1
0.5
1
2
φ
10
0 .5 w
)
1 − 0 .5 w
5
10
50
100
Control System II
Dr. Noaman M.Noaman
Fig. 7 Bode diagram of example 1.
Transportation Lags:
G( jw) = e − jwT
The magnitude is always equal to unity
G( jw) = cos wT − j sin wT = 1
The log magnitude of transportation lags
The phase angle is:
∠ G ( jw) = − wT
rad
= −57.3 wT deg
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e − jwT is equal to 0 dB
Control System II
Dr. Noaman M.Noaman
Fig.8 illustrated the phase plot of the transportation lag.
Fig.8 phase-plot of the transportation lag.
e − jwL
Example 2: Draw the bode diagram of the T.F. G( jw) =
(1 + jwT )
Sol:
The Log mag. is:
20log |G (j w) |=20log | e − jwL |+20log |
mag= 20log|
1
|
1 + jwT
The phase angle:
∠ G (jw) = ∠
e − jwL
-∠
1
1 + jwT
12
1
|
1 + jwT
Control System II
Dr. Noaman M.Noaman
φ = − wL − tan −1 wT
Fig. 9 Bode diagram of example 2.
Relation between system type and log magnitude curve:
The type of the system determines the slope of the log magnitude curve of
the low frequencies. The information of existence and magnitude of the e ss
of a control system to a given I/P can be determined from the observation of
the low frequency region of the log magnitude curve.
a. Determination of static position error coefficients
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Control System II
Dr. Noaman M.Noaman
Fig. 10 Log-magnitude curve for type zero system.
b. Determination of static velocity error coefficients:
Fig. 11 Log-magnitude curve for type one system.
c. Determination of static acceleration error coefficients:
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Control System II
Dr. Noaman M.Noaman
Fig. 12 Log-magnitude curve for type two system.
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Control System II
Dr. Noaman M.Noaman
Example 3: Find the open loop transfer function of a system whose
approximate plot is shown in Fig. 13:
20 dB
0 dB/dec
0 dB
-4.05 dB
-20 dB/dec
-12 dB
-20 dB
+20 dB/dec
-40 dB
1
2.5
10
30
100
300
1000
Fig. 13. Magnitude curve of example 3
20log k = -4.05
k = 0.63
-initial slope is -20 dB/dec is due to the factor
1
s
-the plot between w=2.5 and w=10 have slop is o dB. Then at w=2.5 the
slope changes from -20 dB to 0. This due to factor (
s
+ 1) = (1+0.4s)
2 .5
-at w=10, the slope changes from 0 dB to +20 dB, this due to factor in the
numerator is : (
s
+ 1) = (1+0.1s)
10
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Control System II
Dr. Noaman M.Noaman
At w=30, the slope changes from +20dB to 0dB, this due to the factor in
denominator is : (
s
+ 1) = (1+ 0.033s)
30
So that the O.L.T.F. is:
G(s) =
0.63 (1 + 0.4 s ) (1 + 0.1 s )
s (1 + 0.033 s )
Example 4: Determine the transfer function whose approximate plot is
shown in Fig.14:
60 dB
-20 dB/dec
47.95 dB
40 dB
-40 dB/dec
20 dB
0 dB
1
2.5
10
40
100
1000
-20 dB
-60 dB/dec
-40 dB
-60 dB
Fig. 14. Magnitude curve of example 4
20 log k = 47.95
k = 250
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Control System II
Dr. Noaman M.Noaman
-At w=2.5 Æ slope changing from -20 dB/dec to -40 dB/sec this due to
factor
1
s
+1
2 .5
-At w=40 rad/sec Æ slope changes from -40 dB/dec to -60 dB/dec this due
to factor
1
s
+1
40
-Since initial slope is -20 dB, this due to the factor 1/s
G(s) =
250
s
s
s(
+ 1)( + 1)
2 .5
40
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Control System II
Dr. Noaman M.Noaman
Minimum and non-minimum phase system
-considère G1 (jw) =
1 + jwT
1 + jwT1
T1>T>0
This is a minimum phase system, since there are no poles or zeros on the
right hand side of the s-plane.
mag
=20log |G1(jw) |
=20log |
= 20 log
And phase =
1 + jwT
|
1 + jwT1
1 + w 2T 2
1 + jw 2T1
2
φ 1 = ∠ 1 + jw T
1 + jw T1
= tan-1wT – tant-1wT1
(Pole zero configuration of G1 (jw))
-consider G2 (jw) =
1
T
1
T1
1 − jw T
1 + jw T1
This is a non-minimum phase system (it has a zero on the R.H.S. of the splane)
mag
=20log|G2(jw) |
=20log|
1 − jw T
1 + jw T1
|
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Control System II
=20log
Dr. Noaman M.Noaman
1 + w 2T 2
1 + jw 2T1
2
=20log|G1 (jw) |
(the two mag. are the same)
And
φ 2= ∠ 1 − jw T
1 + jw T1
= tan-1(-wT) – tant-1wT1
= -tan-1(wT) – tant-1wT1
- it is possible to detect whether or not the system is a minimum phase
one by examining both the slope of the high frequency asymptote of
the log magnitude curve and the phase angle at w= ∞.
Therefore, if the slope of the log magnitude curve as w approaches
infinity is -20(p - q) dB/dec, and the phase angle at w= ∞ is equal to 90(q - p).
where
p: degree of the numerator
q: degree of the denominator
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Control System II
Dr. Noaman M.Noaman
HW
Draw the Bode diagram for the following T.F's:
a. G(s) =
10 (1 + s )
1 + 10 s
b. G(s) =
0.25(1 + 0.5s )
s (1 + 2 s )(1 + 4 s )
1
c. G(s) =
(
s
+ 1) 2
20
10(1 + 2 s ) 2 (1 + 0.05 s )
d. G(s) =
(1 + 10 s )(1 + 0.2 s ) 2
e. G(s) =
1.6( s + 11)
s ( s + 0.1)( s 2 + 2 s + 16 )
f. G(s) =
1000 ( s + 5)( s + 40 )
s 3 ( s + 200 )( s + 1000 )
g. G(s) =
( s + 1)( 2.5s + 1)
s (3.5 s + 1)(0.04 s 2 + 0.2 s + 1)
2
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