Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams,
second order systems, system types
Venkata Sonti∗
Department of Mechanical Engineering
Indian Institute of Science
Bangalore, India, 560012
This draft: March 12, 2008
1
Introduction:Frequency domain design
With lecture 4, we have completed what is considered the time domain method of system design.
The reason being that time domain parameters, rise time, settling time, overshoot and steady
state error were directly addressed by the PID controller which differentiates and integrates the
time domain error. PID modifies effectively the pole positions and we can use the Root Locus
method or Ziegler Nichols for finding the PID gains. The paramaters under the engineer’s control
are Kp, Ki and Kd, which using the method of Root Locus modify the pole positions and influence
the final objective, viz., tr , ts , Mp andSSE.
Let us say we model a system and then build it. We may find that the model differs quite a
bit from reality. And so we have to upgrade the model and say we converge. Then we measure
the tr , ts , Mp etc. and we are not satisfied. Then we specify what we want and proceed with PID
design. It seems simple enough. But, it is difficult to measure time domain characteristics. We
need to give a step input which can be light and no energy input occurs. Or the input can be
strong, then the system can become nonlinear. If the system is stiff, then the response can be
very quick and die away. Measuring it is difficult. On the other hand the system may oscillate
with a lot of low frequency dynamics. For every kind of system we cannot design a step inputu
system. Locally, step inputs create nonlinear responses and deformations.
On the other hand, the frequency response of a system can be very easily measured in the lab.
Either a hammer impact is given or a sinusoid/random input is given and this method has been
standardized. Now, does such a frequency response which captures sustained oscillation behavior
capture information about transient response is the question. We will see that the frequency
response is characterized by the resonance peak Mp, resonance frequency ωn , BW, Phase and
gain margin. Indirectly, these indicate the time domain response parameters tr , etc. Here, the
final requirement is still the time response. The tools here are compensators which use bode
diagram and nyquist diagram methods.
∗
1
Nyquist Criterion
Nyquist Criterion: For a closed loop system to be stable, the Nyquist plot of G(s)H(s) must
encircle the (−1, j0) point as many times as the number of poles of G(s)H(s) that are in the right
half of s-plane, and the encirclement, if any, must be made in the closkwise direction. (Read from
Benjamin Kuo)
2
Nyquist or polar plots
The characteristic equation is
1 + GH = 0 or GH = −1
so that when the system is just unstable amplitude of
GH = 1 and phase = 180
The Nyquist plot is the polar (amplitude phase) plot of a GH transfer function on the complex
plane as s = jω, and ω takes values from 0 to ∞. The length of the arrow from the origin to the
plot indicates the amplitude of GH and the angle of the arrow w.r.t the real axis indicates the
phase of GH. The left figure (figure 1) is for
GH
=
GH
=
GH
=
1
s+1
1
2
s + 3s + 2
1
s3 + 4s2 + 5s + 2
(1)
The right figure in figure 1 is for
GH =
GH =
1
s(s + 1)
K
2
s(s + 3s + 2)
(2)
where K takes values 1,3,6. At K = 6, the nyquist plot just passes through the point s = −1 + j0.
Take for example the left figure and look at 1/(s2 + 3s + 2). At ω = 0 the amplitude is 0.5 and the
phase is 0 degrees. At ω = ∞ the amplitude is 0 and the phase is dominated by 1/s2 = −1/ω 2
and hence 180 degrees. This can be seen in the plot.
Similarly in the right figure of figure 1 the second order transfer function has phase 180 at
high frequencies and the third order transfer function has -270 degrees. And as the gain in the
third order transfer function is raised it gets closer to the point -1+j0 and finally crosses it going
unstable. Thus, a polar plot of GH gives the relative stability of the closed loop system. If we
know that at a given gain, the system is stable, then by increasing the gain and seeing which way
the polar plot moves we can find out the relative stability of the system. How far is the system
from going unstable. Note that we use the polar plot of GH in order to figure the relative stability
of the closed loop system. How far the amplitude is from -1 and how far the phase is from 180
degrees are indicators of stability. These are called gain and phase margins respectively and we
will elaborate more on these two in a little while.
2
Nyquist Diagrams
Nyquist Diagrams
0.1
0.1
[1 3 2 0]
[1 4 5 2]
0
θ
[1 3 2]
−0.1
Imaginary Axis
Imaginary Axis
0
−0.2
−0.3
−0.2
−0.3
[1 1 0]
−0.4
−0.4
−0.5
−0.5
−0.1
−0.5
[1 1]
0
0.5
1
−2
−1.5
Real Axis
−1
−0.5
0
0.5
1
Real Axis
Figure 1: Nyquist plots
3
Bode Diagrams
The same Nyquist information can be plotted with the amplitude and phase on separate plots.
The amplitude is plotted after conversion to a dB scale and the x axis on a log scale. Such plots
are called Bode plots. Bode plot although very simple carries a lot of information such as: relative
stability (gain and phase margins), steady state errors, damping coefficient etc. We begin with a
simple example,
3.1
Example 1: 1/s
GH =
1
1
=
s
jω
the amplitude and phase are given by
1
1
20log| | = −20log|ω| and φ =
= −90o
6 jω
ω
The amplitude and phase are plotted in figure 9. Lets look at the amplitude plot. The x axis is
on a log10 scale, 0.01, 0.1, 1, 10, etc. The amplitude goes through zero at ω = 1 and has a
negative drop of 20 dB for every 10 times increase in frequency, i.e., -20 dB/decade. The phase
plot is relatively simple.
3.2
Example 2: 1/(s + 1)
GH =
Amplitude is given by
1
1
−→
s+1
jω + 1
p
1
√
−→ −20log ω 2 + 1
2
ω +1
3
Bode Diagram
20
Magnitude (dB)
10
0
−10
−20
−30
−40
Phase (deg)
−89
−89.5
−90
−90.5
−91
−1
10
0
1
10
10
2
10
Frequency (rad/sec)
Figure 2: Bode plot for 1/s. Amplitude and phase.
Low frequency asymptote @ ω = 0, the plot approaches 1 asymptotically. Hence amplitude=1
is the low frequency asymptote.
On the log scale as ω → 0, the plot approaches 0 dB.
High frequency asymptote @ ω → ∞, the plot approaches ω1 asymptotically. Hence amplitude=1/ω
is the high frequency asymptote.
On the log scale as ω → ∞, the amplitude approaches −20log|ω|. At ω = 1, this asymptote goes
through 0 dB and
√ hence intersects the low frequency asymptote. The actual value of the curve at
ω = 1 is −20log 1 + 1 = −3dB. The curve lies 3dB below the 0 dB asymptote.
1
Phase is given by 6 jω+1
as ω → 0, the phase goes to 00
as ω → ∞, the phase approaches −900 See figure 3.
3.3
Example 3: 1/(s + 0.1)
GH =
1
s+0.1
−→
Magnitude
1
s + 0.1
1
jω+0.1
√
1
ω 2 +0.01
ω → 0,
reaches constant=10 asymptote
1
is the asymptote.
ω → ∞,
ω √
dB = −20log10 ω 2 + 0.01
4
Magnitude (dB)
Bode Diagram
0
5
10
15
20
25
Phase (deg)
0
45
90
1
10
0
10
Frequency
(rad/sec)
1
10
Figure 3: Bode plot for 1/s + 1. Note the asymptotes in the magnitude plot.
ω → 0,
−20log0.1 = 20dB
ω → ∞,
−20log|ω| ⇒ −20 dB/decade
P hase as bef ore.
The meeting point of the asymptotes
√ is the 20 dB line at ω = 0.1 (note s+0.1). The actual value
of the curve at ω = 0.1 is −20log 0.1 ∗ 0.1 + 0.1 + 0.1 = 16.9897. This value is 3 dB below the
20 dB asymptote.
3.4
Example 4: 1/(s + 10)
GH =
1
s+10
−→
1
s + 10
1
jω+10
Magnitude
√
1
ω 2 +100
1
ω → 0, 10
asymptote.
ω → ∞, ω1 asymptote.
Following the earlier examples, the low frequency asymptote reaches the value -20 dB. The high
frequency asymptote goes as -20dB/decade. The two meet at ω = 10. The actual curve is 3 dB
below the -20 dB asymptote at ω = 10. See figure 5.
5
Bode Diagram
Magnitude (dB)
20
15
10
Asymptote
5
0
5
Phase (deg)
0
45
90
2
1
10
10
0
10
Frequency (rad/sec)
Figure 4: Bode plot for 1/s + 0.1. Note the asymptotes in the magnitude plot meet at ω = 0.1.
Bode Diagram
Magnitude (dB)
-20
-25
-30
-35
-40
-45
Phase (deg)
0
45
90
0
1
10
10
2
10
Frequency (rad/sec)
Figure 5: Bode plot for 1/s + 10. Note the asymptotes in the magnitude plot.
6
3.5
Example 4: 1/((s + 0.1)(s + 1)(s + 10))
We examine a complicated case. There are three terms and a
GH =
1
(s + 0.1)(s + 1)(s + 10)
1
ω → 0, 0.1∗10
= 1 asymptote.
1
ω → ∞, ω3 asymptote.
Following the earlier examples, the low frequency asymptote reaches the value 0 dB. The high
frequency asymptote goes as -60dB/decade. The curve will follow 1/ω asymptote at ω = 0.1
with a slope of −20dB/decade. Then, the curve will follow 1/ω 2 asymptote at ω = 1 with a
slope of −40dB/decade. Then, the curve will follow 1/ω 3 asymptote at ω = 10 with a slope
of −60dB/decade. The actual curve is 3 dB below the meeting points of asymptotes at the
corresponding ω. See figure 6. Phase goes to 0 at low frequencies and to −270o at high frequencies.
Bode Diagrams
0
Phase (deg); Magnitude (dB)
w=0.1
w=1.0
−50
w=10
−100
−150
0
−100
−200
−300 −2
10
−1
0
10
10
1
10
Frequency (rad/sec)
Figure 6: Bode plot for 1/((s + 0.1)(s + 1)(s + 10)).
4
Second order system:
ωn 2
s2 +2ξωn s+ωn 2
Lets look at the closed loop Tr. Fn.
M (s) =
ωn 2
s2 + 2ξωn s + ωn 2
1
1 + (2ξ/ωn )s + 1/ωn 2 s2
7
2
10
We are interested in values of ξ << 1. Substituting s = jω gives
M (jω) =
1
[1 − (ω/ωn
)2 ] +
2jξω/ωn
As can be seen, at low frequencies GH reaches the 0 dB asymptote. At high frequencies we get the
-40 dB/decade asymptote. These two meet at ω = ωn . The phase is 0 degrees at low frequencies
and at ω = ωn it is -90 and for higher frequencies it slowly approaches -180. The actual plots
around ω = ωn are quite complicated depending on the actual value of ξ.
Bode Diagram
0
Magnitude (dB)
−10
−20
−30
−40
−50
−60
−70
−80
Phase (deg)
0
−45
−90
−135
−180
0
1
10
2
10
10
Frequency (rad/sec)
Figure 7: Bode plot of a second order transfer function.
4.1
Mp , ωp and bandwidth of a second order system:
M (s) =
L
−1
M (s)
s
ωn 2
s2 + 2ξωn s + ωn 2
−σt
= y(t) = 1 − e
q
ωd = ωn 1 − ξ 2 ,
y(t) = 0,
ωd tp = π,
σ
sinωdt
cosωd t +
ωd
σ = ξωn
tpeak =
−√
y(tp ) = 1 + Mp −→ Mp = e
=1−
ξ
.
0.6
ωn
πξ
1−ξ2
.
π
1 − ξ2
p
0 ≤ ξ ≤ 1.
0 ≤ ξ ≤ 0.6.
8
For a 1% settling time we know
4.6
ξωn
ωn tr = 1.8
ts =
Looking at M(s) in the frequenc domain:
M (jω) =
C(jω)
ωn2
=
R(jω)
(jω)2 + 2ξωn (jω) + ωn2
=
1
1−
( ωωn )2
M (ju) =
|M (ju)| =
1−
+ j2ξ ωωn
1
+ j2uξ
1
u2
1
[(1 − u2 )2 + (2ξu)2 ] 2
2ξu
1 − u2
In order to find out the frequency at which the peak transfer function value occurs in frequency:
6
M (ju) = φm (u) = −tan−1
dM (u)
= 0 −→ 4u3 − 4u + 8uξ 2 = 0
du
u = up = 0
up =
q
q
1 − 2ξ 2
ωp = ωn 1 − 2ξ 2
Mp =
So Mp is purely a function of ξ.
valid f or ξ < 0.707
1
2ξ 1 − ξ 2
p
F or ξ 0.707, ωp = 0, Mp = 1.
Bandwidth of a second order system is frequency at which M (ω) drops to 70.7% of its value
from its low frequency asymptote.
1
= 0.707
M (u) = p
(1 − u2 )2 + (2ξu)2
u2 = (1 − 2ξ 2 ) ±
2
q
Bw = ωn [(1 − 2ξ ) +
4ξ 4 − 4ξ 2 + 2
q
1
4ξ 4 − 4ξ 2 + 2] 2
as ξ ↓, bandwidth ↑and Mp ↑.
1. Maximum overshoot of unit step response depends on ξ only.
2. Resonance peak of closed loop frequency response Mp depends only on ξ.
9
MP
ωP
ωn
1
0.707
ζ
ζ =0.707
ζ
Figure 8:
3. Bandwidth is directly proportional to ωn .
4. High bandwidth =⇒ largerMp .
5. ωn =
1.8
tr .
6. ts = f (ξωn ) =
4.6
ξωn
for 1% settling time.
7. Also, the following relations are relevant between bandwidth, ωn , ts , tr and ξ:
2
ωBw = ωn [(1 − 2ξ ) +
ωn =
ωBw =
ωBw =
tp
q
1
4ξ 4 − 4ξ 2 + 2] 2
4.6
ξts
q
1
4.6
[(1 − 2ξ 2 ) + 4ξ 4 − 4ξ 2 + 2] 2
ξts
π
ωn = p
tp 1 − ξ 2
q
1
π
2
[(1
−
2ξ
)
+
4ξ 4 − 4ξ 2 + 2] 2
2
1−ξ
p
8. Open loop system must be stable for Bode plot design.
9. For 2nd order system closed loop damping ratio is approximately equal to
margin is between 0 and 600 .
PM
100 ,
if phase
10. bandwidth ≈ ωn .
We use these relations quite freely even for non-second-order systems...with caution.
5
System Types
The type of the open loop system one is dealing with decides its steady state error nature. Hence,
it is important to know. Also, the steady state error depends not only on the form of the transfer
function but also on the input, like step or ramp. Here we will look at system types with regard
to only a step input. More is available in Benjamin Kuo.
10
r(t)
R(s)
+
-
e(t)
c(t)
G(s)
E(s)
b(t)
C(s)
H(s)
B(s)
Figure 9:
The error E(s) is given by
E(s) = R(s) − C(s)H(s)
C(s) = E(s)G(s)
E(s) = R(s) − E(s)G(s)H(s)
R(s)
E(s) =
1 + G(s)H(s)
E(s), the LT of e(t) is given by
∞
Z
E(s) =
e(t)e−st dt.
0
e−st e(t) ∞
=[
]|0 −
−s
=
e(0) 1
+
s
s
Z
∞
Z
0
∞
e−st
0
Z
sE(s) = e(0) +
e−st de(t)
.dt
−s dt
∞
de(t)
.dt
dt
e−st
0
lim sE(s) = e(0) +
s→0
Z
∞
0
de(t)
.dt
dt
de = e(∞) − e(0) = e(∞)
Thus steady state error, i.e., e(∞) is given by the above formula.
Step response
For step response R(s) = Rs and let us denote lim GH = kp , then
s→0
R
s(1 + kp )
E(s) =
and if the steady state error
e(∞) = lim sE(s) = 0
s→0
then
lim sE(s) =
s→0
R
=0
(1 + kp )
or kp =∞. If the loop transfer function has the form
GH =
1
F (s)
sj
11
(3)
then the system is of type j.
If j=0, then
lim GH = F (0) = Kp
s→0
is a finite value. Thus, Kp is finite for type 0 systems and hence the steady state error is not zero.
For type 1 and higher systems Kp is infinity. Hence, for step inputs steady state error is not 0
for type 0 systems and is 0 for type 1 and higher systems.
A similar set of rules can be defined for ramp inputs. (Read Kuo).
12