Bode lect

Frequency Domain Analysis Using Bode Plot

Swagat Kumar
July 11, 2005

September 6, 2011 Control Systems Laboratory, IIT Kanpur Page 1

Topics to be covered

• Frequency response of a linear system
• Bode plots
• Effect of Adding zero and poles
• Minimum and Non-minimum phase
• Relative stability: Gain Margin and Phase margin
• Lead and Lag compensator Design
• PID compensator design using bode plot
• Summary


Frequency response of a linear system
Consider a stable linear system whose transfer function is given by

Y (s)
G(s) =
U (s)
For a sinusoidal input u(t) = Asinωt, the output of the system is given by

y(t) = Y sin(ωt + φ)

where

Y = A|G(jω)|
Im[G(jω)]
φ = ∠G(jω) = tan−1
Re[G(jω)]
A stable linear system subjected to a sinusoidal input will, at steady state, have a
sinusoidal output of the same frequency as the input. But the amplitude and phase
of output will, in general, be different from those of the input.


Graphical tools for frequency response analysis

• Bode Diagram
• Nyquist plot or polar plot
• Log-magnitude versus phase plot
In this lecture, we will only study about “Bode Diagram” and its application in
compensator design.


Bode Plot

A Bode diagram consists of two graphs:

• a plot of 20 log |G(jω)| (in dB) versus frequency ω , and
• a plot of phase angle φ = ∠G(jω) versus frequency ω .
Advantages of Bode plot:

• An approximate bode plot can always be drawn with hand.
• Multiplication of magnitudes get converted into addition.
• Phase-angle curves can easily be drawn if a template for phase-angle curve of
(1 + jω ) is available.


Construction of Bode plot

Any transfer function is composed of 4 classes of terms

1. K
2. (jω)±1
3. (jωτ + 1)±1
±1
jω jω
4. ( ωn ) 2 + 2ζ ωn +1

The gain K :

• Log-magnitude curve is a straight line at 20 log K and phase angle is zero for
all ω

• The effect of varying the gain K in the transfer function is that it raises or lowers
the log-magnitude curve by a constant amount without effecting its phase curve.


Construction of bode plot

Integral and derivative term (jω)∓1 : The logarithmic magnitude of 1/jω in
decibel is
1
20 log = −20 log ω dB
jω
The phase angle of jω is constant and equal to −90◦ .

• Octave: A frequency band from ω1 to 2ω1
• Decade: A frequency band from ω1 to 10ω1
For (jω)±n term,

- slope of log-magnitude curve = ±20n dB/decade or ±6n dB/octave.

- phase angle = ±(n × 90)◦


Construction of Bode Plot

20
Slope = -20dB/decade
|G(jω)| ∠G(jω)
10

0.1 1 10 100 1000 0.1 1 10 100 1000
−10 −90o

−20 (a) Magnitude Plot (b) Phase Plot

1
Figure 1: Magnitude and Phase plot of G(jω) = jω



20
|G(jω)| ∠G(jω)
10 90◦

0.1 1 10 100 1000 0.1 1 10 100 1000
−10 −90o
Slope = 20dB/decade
−20 (a) Magnitude Plot (b) Phase Plot

Figure 2: Magnitude and Phase plot of G(jω) = jω



First order factors (1 + jωT )∓1 : The log magnitude of the ﬁrst order factor
1
(1+jωT ) is

1
20log = −20 log 1 + ω 2 T 2 dB
(1 + jωT )
The phase angle is φ = − tan−1 ωT . The log-magnitude curve can be
approximated by two asymptotes as given below:
1
√
For ω << T, −20 log 1 + ω 2 T 2 ≈ −20 log 1 = 0 dB
1
√
For ω >> T, −20 log 1 + ω 2 T 2 ≈ −20 log ωT dB
Phase curve

ω 0 1/T ∞
φ 0 −45◦ −90◦



∠G(jω)
Corner frequency
10 30◦
Asymptote

|G(jω)| 0 0◦

−10 −30◦

−20 −60◦

Slope = -20 dB/decade
−90◦
1 1 1 1 2 10 20 1 1 1 1 2 10 20
20T 10T 2T T T T T 20T 10T 2T T T T T

(a) Magnitude plot (b) Phase plot

1
Figure 3: Magnitude and phase plot of (1+jωT )



∠G(jω)
Corner frequency
30 90◦

|G(jω)| 20 60◦

10 30◦

0 0◦
Asymptote

−10 −30◦
1 1 1 1 2 10 20 1 1 1 1 2 10 20
20T 10T 2T T T T T 20T 10T 2T T T T T

(a) Magnitude plot (b) Phase plot

Figure 4: Magnitude and phase plot of (1 + jωT )

Error at corner frequency ≈ 3 dB and slope is +20 dB/decade.



2
ω ω
Quadratic factors [1 + 2ζ j ωn + j ωn ]∓1 : The log-magnitude curve for
2
ω ω
1/(1 + 2ζ j ωn + j ωn ) is given by

2 2
1 ω2 ω
20 log 2 = −20 log 1− 2 + 2ζ
ω ω ωn ωn
1 + 2ζ j ωn + j ωn

The asymptotic frequency-response curve may be obtained by making following
approximations:

For ω << ωn , log-magnitude = −20 log 1 = 0 dB
ω2 ω
For ω >> ωn , log-magnitude = −20 log 2
ωn = −40 log ωn dB

At corner frequency ω = ωn , the resonant peak occurs and its magnitude depends
on damping ratio ζ .



1
The phase angle of 2 is
1+2ζ (j ωn )+(j ωn )
ω ω

 
ω
2ζ ωn
φ = tan−1 
 
2
ω
1− ωn

The phase curve passes through following points

ω 0 ωn ∞
φ 0◦ −90◦ −180◦


Frequency domain speciﬁcations

• The resonant peak Mr is the maxi-
mum value of |M (jω)|. Mr
• The resonant frequency ωr is the fre-
1
quency at which the peak resonance 0.707
Mr occurs. |M (jω)|

• The bandwidth BW is the frequency at
which M (jω) drops to 70.7% (3 dB)
0 ωr BW ω
of its zero-frequency value.


Frequency domain speciﬁcation

For a second order system, following relationships between frequency and
time-domain responses can be obtained.

Resonant Frequency: 10

ωr = ωn 1 − 2ζ 2 8

Mr in dB
Resonant Peak: 6

1 0.707
Mr = |G(jω)|max = |G(jωr )| = 4
2ζ 1− ζ2
2
for 0
≤ ζ ≤ 0.707. For ζ > 0.707, ωr = 0 and
Mr = 1
0 0.2 0.4 0.6 0.8 1.0
Bandwith: ζ

BW = ωn [(1 − 2ζ 2 ) + (ζ 4 − 4ζ 2 + 2)]1/2 = [ωr +
2
ωr + ωn ]1/2
4 4


Frequency domain speciﬁcation

• Mr indicates the relative stability of a stable closed loop system.
• A large Mr corresponds to larger maximum overshoot of the step response.
Desirable value: 1.1 to 1.5

• BW gives an indication of the transient response properties of a control system.
• A large bandwidth corresponds to a faster rise time. BW and rise time tr are
inversely proportional.

• BW also indicates the noise-ﬁltering characteristics and robustness of the
system.

• Increasing ωn increases BW.
• Increasing ζ decreases BW as well as Mr .
• BW and Mr are proportional to each other for 0 ≤ ζ ≤ 0.707.


Examples

Effect of adding a zero to the forward path transfer function

Consider following open loop transfer function

1
G(s) =
s(s + 1.414)
Adding a zero to the forward path transfer function leads to

(1 + T s)
G1 (s) =
s(s + 1.414)
The closed loop transfer function is given by

1 + Ts
H1 (s) = 2
s + (T + 1.414s) + 1
The general effect of adding zero to the forward path transfer function is to
increase the bandwith of the closed loop system.


Examples

Bode Diagram

0

Magnitude (dB)
−1

−2

−3

−4

−5
0

−45
Phase (deg)

T=0
T = 0.2
−90
T = 0.5
T=2
−135 T=5

−180
−1 0
10 10
Frequency (rad/sec)

Figure 5: Effect of adding a zero


Examples

Step Response

1

0.8

Amplitude

0.6

0.4
T=0
T = 0.2
T = 0.5
0.2 T=2
T=5

0
2 4 6 8 10 12 14 16 18
Time (sec)

Figure 6: Effect of adding a zero: Step response


Example: adding a zero

Observations

• A zero provides a phase lead to the transfer function.
• For very low values of T , bandwidth decreases.
• For higher values bandwith increases and hence faster rise time.
1
• For very high values of T , zero (s = − T ) moves very close to origin, causing
the system to have larger time constant and hence longer settling time.


Example

Adding a pole to the forward-path transfer function

Reconsider the previous open loop system

1
G(s) =
s(s + 1.414)
Adding a pole to the forward-path transfer function leads to

1
G1 (s) =
s(s + 1.414)(1 + T s)
The closed loop transfer function is given by

1
H1 (s) =
T s3 + (1.414T + 1)s2 + 1.414s + 1
The effect of adding a pole to the forward path transfer function is to make
the closed-loop system less stable while decreasing bandwidth


Example

Bode Diagram

10

Magnitude (dB)
5

0

−5

−10
0

−45
T=0
Phase (deg)

−90
T = 0.5
−135 T=1
T=5
−180

−225

−270
−1 0
10 10
Frequency (rad/sec)

Figure 7: Effect of adding a pole


Example

Step Response

1.4
T=0
T = 0.5
1.2 T=1

1

Amplitude

0.8

0.6

0.4

0.2

0
0 5 10 15 20 25
Time (sec)

Figure 8: Effect of adding a pole


Example: Adding a pole

Observation

• For smaller values of T , BW increases slightly but Mr increases.
• For higher values of T , BW decreases but Mr increases.
• In step response, the rise time increases with decreasing of BW.
• Peak overshoot and settling time increses with increasing value of T .


Effect of adding a pole or a zero to a transfer function

1 1
Figure 9: G= s(s+2) , zero is (s + 0.5) and pole is s+3


Minimum and nonminimum-phase system

Minimum-phase system Transfer functions having neither poles or zeros in the
right-half s plane are minimum-phase transfer functions.

Nonminimum-phase system Those having poles and/or zeros in the right-half s
plane are called nonminimum-phase system.

Consider following two systems

s+1
G1 (s) = 10
s + 10
s−1
G2 (s) = 10
s + 10
|G1 (jω)| = |G2 (jω)|
∠G1 (jω) = ∠G2 (jω)


minimum and non-minimum phase system

Bode Diagram

20

15

Magnitude (dB)
10

5

0
180

135
10(s−1)/(s+10)
Phase (deg)

90
10(s+1)/(s+10)
45

0
−2 −1 0 1 2 3
10 10 10 10 10 10
Frequency (rad/sec)

In a minimum-phase system, the magnitude and phase-angle are uniquely related.
This does not hold for a NMP system.


Relative Stability

Phase Margin It is the amount of additional lag at the gain crossover frequency ωg
required to bring the system to the verge of instability. At gain crossover
frequency, the magnitude of open loop gain is unity, i.e., |G(jωg )| = 1. The
phase margin γ is given by
γ = 180◦ + φ
where φ = ∠G(jωg ).
Gain Margin It is the amount of additional gain at phase crossover frequency ωp
that can bring the system to the verge of instability. At phase crossover
frequency, the phase angle of open loop transfer function equals −180◦ , i.e.,
∠G(jωp ) = −180◦ . The gain margin is given by
1
Kg = or Kg dB = −20 log |G(jωp )|
|G(jωp )|


Stability analysis using bode plot

• The phase margin and gain margin must be positive for a minimum-phase
system to be stable.

• Negative margins indicate instability.
• For satisfactory performance, the phase margin should be between 30◦ and
60◦ and gain margin should be greater than 6 dB.
• Either the gain margin or the phase margin alone does not give a sufficient
indication of the relative stability. Both should be given in order to determine the
relative stability.

• For first order and second order system, gain margin is always infinity.
Disadvantage of Bode plot:
Bode plot can’t be used for stability analysis of non minimum-phase system.


stability analysis using bode plot

Bode Diagram
Gm = 9.54 dB (at 2.24 rad/sec) , Pm = 25.4 deg (at 1.23 rad/sec)

100

Gain crossover frequency
50

Magnitude (dB)
0
+ve gain
Stable System margin
−50

−100
−90

−135 +ve phase
Phase (deg)

Margin
−180

−225
Phase crossover frequency

−270
−2 −1 0 1 2
10 10 10 10 10
Frequency (rad/sec)

10
Figure 10: Bode plot of s(s+1)(s+5)


stability analysis using bode plot

Bode Diagram
Gm = −10.5 dB (at 2.24 rad/sec) , Pm = −23.7 deg (at 3.91 rad/sec)

100

50

Magnitude (dB)
−ve gain margin
0

−50
Unstable System

−100
−90

−135
Phase (deg)

−180
−ve phase
−225 margin

−270
−2 −1 0 1 2
10 10 10 10 10
Frequency (rad/sec)

100
Figure 11: Bode plot of s(s+1)(s+5)


Lead Compensators

jω
A lead compensator is given by following transfer
function
Ts + 1
Gc (s) = α 0<α<1
αT s + 1
× •
We see that the zero is always located to the right
1 1
σ
− αT −T
of the pole in complex plane.

The maximum phase angle contributed by a lead compensator is given by

1−α
sin φm =
1+α
1
at a frequency ωm = T
√
α


Lead compensator

Bode Diagram
8

7 a<1 20 log 1/a

6

Magnitude (dB)
5 20 dB/decade
4

3

2

1

0
20

15 Maximum phase angle
Phase (deg)

10

5

0
−1 0 1 2
10 10 10 10
Frequency (rad/sec)

1+s
Figure 12: Bode plot of Gc (s) = 1+0.5s


Lead compensator design
Consider following second order system

4
G(s) =
s(s + 2)

Design a compensator for the system so that the static velocity error constant
Kv = 20 sec−1 and phase margin is at least 50◦ .
Design steps:

• The open loop transfer function of the compensated system is given by
Ts + 1 Ts + 1
Gc (s)G(s) = K G(s) = KG(s)
1 + αT s 1 + αT s
where 0 < α < 1 and K = Kc α. Kc is a gain constant. The attenuation
factor α is assimilated into constant gain factor K . Determine gain K to satisfy
the requirement on given static error constant.



Ts + 1 4
Kv = lim sGc (s)G(s) = lim s K = 20
s→0 s→0 1 + αT s s(s + 2)
This gives K = 10.
• Using the gain K , draw a Bode diagram of KG(jω). Evaluate phase margin.
The phase margin is about 18◦ .

• Determine the necessary phase lead angle φ to be added to the system.
For a PM of 50◦ , a phase lead angle of 32◦ is required. However, in order to
compensate for the shift in gain crossover frequency due to the lead
compensator, we assume that the maximum phase lead required
φm = 32 + 6 = 38◦ .
1−α
• Using equation sin φm = 1+α , Determine the attenuation factor α.
For φm = 38◦ , α = 0.24.



Bode Diagram
Gm = Inf dB (at Inf rad/sec) , Pm = 18 deg (at 6.17 rad/sec)

50

System: untitled1

Magnitude (dB)
Frequency (rad/sec): 8.95
0
Magnitude (dB): −6.24
New
Gain crossover
Frequency

−50
−90
Phase (deg)

−135 System: untitled1
System: untitled1 Frequency (rad/sec): 6.13
Frequency (rad/sec): 1.68 Phase (deg): −162
Phase (deg): −130

−180
−1 0 1 2
10 10 10 10
Frequency (rad/sec)

Figure 13: Bode plot of gain adjusted but uncompensated system KG(jω)



• Determine the frequency ω = ωm where
20 log |Gc (jωm )G(jωm )| = 0dB
jωm T + 1
20 log |KG(jωm )| = −20log
jαωm T + 1
1 1
= −20 log √ (∵ ωm = √ )
α T α
Get this frequency from the magnitude plot of KG(jω). This is our new gain
crossover frequency and maximum phase shift φm occurs at this frequency.
1
Here, −20 log √α = −6.2 dB which occurs at ωm = 9 rad/sec.
1
• Determine the time constant T from the equation ωm = √ .
T α
Here, T = 0.2278 seconds.
The compensated open loop transfer function is given by

(0.2278s + 1)40
Gc (s)G(s) =
s(s + 2)(0.0547s + 1)


Bode Diagram
Gm = Inf dB (at Inf rad/sec) , Pm = 50.6 deg (at 8.92 rad/sec)

60

40

20

Magnitude (dB)
0

−20

−40

−60

−80
−90
Phase (deg)

−135

−180
−1 0 1 2 3
10 10 10 10 10
Frequency (rad/sec)

Figure 14: Bode plot of compensated system



Step Response

1.4

1.2

1

Amplitude

0.8

0.6

0.4

0.2

0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Time (sec)

Figure 15: Closed loop step response of the compensated system



Discussion

• Lead compensator is a high-pass ﬁlter.
• It adds more damping to the closed-loop system.
• Bandwidth of closed loop system is increased. This leads to faster time
response.

• The steady state error is not affected.
• In lead compensator design, the phase of forward-path transfer function in the
vicinity of gain crossover frequency is increased.


Lag Compensator

jω
A lag compensator is given by following
transfer function
Ts + 1
Gc (s) = α α>1
αT s + 1
• ×
We see that the pole is always located to
1 1
σ
−T − αT
the right of the zero in complex plane.


Bode Diagram
0

a>1
−20 dB/decade

Magnitude (dB)
−5

−10
20 log 1/a

−15
0
Phase (deg)

−30

Maximum phase angle

−60
−2 −1 0 1 2
10 10 10 10 10
Frequency (rad/sec)

1+s
Figure 16: Bode plot of Gc (s) = 1+5s


Lag compensator Design

Consider following open loop transfer function

1
G(s) =
s(s + 1)(0.5s + 1)
Design a compensator so the velocity error constant is Kv = 5 sec−1 , the PM is
at least 40◦ and GM is atleast 10 dB.

Design steps:

• The open loop transfer function of the compensated system is given by
Ts + 1 Ts + 1
Gc (s)G(s) = K G(s) = KG(s)
1 + αT s 1 + αT s
where α > 1 and K = Kc α. Determine forward path gain K so as to satisfy
the requirement of steady state performance.


Lag compensator design

Ts + 1 K
Kv = lim sGc (s)G(s) = lim s =K=5
s→0 s→0 1 + αT s s(s + 1)(0.5s + 1)
• Plot the bode diagram of KG(jω).
• Assuming that the PM is to be increased, locate the frequency at which the
desired phase margin is obtained, on bode plot. To compensate for excessive
phase lag, the required phase margin is the speciﬁed PM + 5 to 12◦ . Call the
′
corresponding frequency new gain crossover frequency ωg .
The new gain crossover frequency for a PM of 40 + 12 = 52◦ is ωg = 0.5
rad/sec.

• To bring the magnitude curve down to 0 dB at this new gain crossover
frequency, the phase-lag controller must provide the amount of attenuation
′
equal to the value of magnitude curve ωg . In other words

′ 1
|KG(jωg )| = 20 log10 α>1
α


Bode Diagram
Gm = −4.44 dB (at 1.41 rad/sec) , Pm = −13 deg (at 1.8 rad/sec)

100
System: untitled1
Frequency (rad/sec): 0.462
50 Magnitude (dB): 19.6

Magnitude (dB)
0

−50

−100

−150 System: untitled1
−90 Frequency (rad/sec): 0.461
Phase (deg): −128
−135
Phase (deg)

System: untitled1
−180 Frequency (rad/sec): 0.637
Phase (deg): −140
−225

−270
−2 −1 0 1 2
10 10 10 10 10
Frequency (rad/sec)

Figure 17: Bode diagram of KG(jω)



′
The magnitude of KG(jωg ) is 20 dB and thus we have −20 log α = −20 and
this gives α = 10.
1
• Choose the corner frequency ω = T corresponding to the zero of lag
′
compensator 1 octave to 1 decade below the new gain crossover frequency ωg .
1
We choose the zero of lag compensator at ω = T = 0.1 rad/sec. This gives T
= 10.

• Plot the bode diagram of compensated system.
The compensated open loop transfer function is given by

5(1 + 10s)
KGc (s)G(s) =
s(1 + 100s)(s + 1)(0.5s + 1)



Bode Diagram
Gm = 14.3 dB (at 1.32 rad/sec) , Pm = 41.6 deg (at 0.454 rad/sec)

150

100

Magnitude (dB)
50

0

−50

−100

−150
−90

−135
Phase (deg)

−180

−225

−270
−4 −3 −2 −1 0 1 2
10 10 10 10 10 10 10
Frequency (rad/sec)

Figure 18: Bode diagram of compensated system KGc (s)G(s)



Discussion

• Lag compensator is a low-pass ﬁlter.
• The gain crossover frequency is decreased and thus the bandwidth of the
system is reduced.

• The rise and settling time increases.
• The steady state error reduces.
• In phase lag control, the objective is to move the gain crossover frequency to a
lower frequency where desired PM is realized while keeping the phase curve
relatively unchanged at new gain crossover frequency. In other words,
phase-lag control utilizes attenuation of controller at high frequencies.


PID control design

Consider following open loop system

(s + 1)
G(s) = 2
s (s + 8)
Design a PID compensator such that the compensated system has a PM of 60◦ and
a gain crossover frequency of 5 rad/sec and an acceleration error constant Ka = 1.
The PID compensator is of following form

KI
Gc (s) = KP + KD s +
s
KI
Gc (jω) = KP + j(KD ω − )
ω
= |Gc (jω)|(cos θ + j sin θ)


design of PID compensator

Bode Diagram
Gm = Inf dB (at Inf rad/sec) , Pm = 17.4 deg (at 0.365 rad/sec)
100

50

Magnitude (dB)
0

−50

−100

−150
−120
Phase (deg)

−150

−180
−2 −1 0 1 2 3
10 10 10 10 10 10
Frequency (rad/sec)

Figure 19: Bode plot of uncompensated system



System: h
Step Response
Peak amplitude: 1.65
1.8
Overshoot (%): 64.9
At time (sec): 7.89
1.6

1.4

Amplitude 1.2

1
System: h
Settling Time (sec): 71.1
0.8

0.6

0.4

0.2

0
0 10 20 30 40 50 60 70 80 90 100
Time (sec)

Figure 20: Step response of uncompensated closed loop system



For the PID compensator, we can write

KP = |Gc (jω)| cos θ
KI
KD ω − = |Gc (jω)| sin θ
ω
We know that at gain crossover frequency, |Gc (jωg )||G(jωg )| = 1. This gives
1
|Gc (jωg )| =
|G(jωg )|

Substituting for |Gc (jω)| from previous equation, we get

cos θ
Kp =
|G(jωg )|

Similarly, we have
KI sin θ
KD ωg − =
ωg |G(jωg )|



Now at ωg = 1 rad/s,
G(jωg ) = 0.1754∠ − 142.1250◦

For the compensated system to have a phase margin of 60◦ at ωg = 1 rad/sec, we
should have

−180 + φm = θ + ∠G(jω) (φm = P M )
θ = −180◦ + 60◦ + 142.1250◦ = 22.1250◦
Hence,
cos 22.1250◦
KP = = 3.9571
0.1754
Lets choose KI = 0, then KD can be computed to be
sin 22.1250◦
KD = + 1 = 3.1471
0.1754
Hence we have a PD compensator Gc (s) = 3.9571 + 3.1471s.



Bode Diagram
Gm = Inf , Pm = 60 deg (at 1 rad/sec)
150

100

Magnitude (dB)
50

0

−50
−45

−90
Phase (deg)

−135

−180
−2 −1 0 1 2
10 10 10 10 10
Frequency (rad/sec)

(3.9571+3.1471s)(s+1)
Figure 21: Bode plot of compensated system Gc (s)G(s) = s2 (s+8)



System: h Step Response
Peak amplitude: 1.26
1.4
Overshoot (%): 26
At time (sec): 3.24
1.2

1
System: h
Settling Time (sec): 9.89
0.8
Amplitude

0.6

0.4

0.2

0
0 2 4 6 8 10 12 14 16
Time (sec)

Figure 22: Step response of compensated system


Summary

Following topics were covered in this lecture

• Frequency domain speciﬁcations.
• Bode plot construction
• Relative stability
• Design of Lead and Lag compensators
• PID control design example.


The Principle of Argument


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