Frequency Response Plots
For each frequency, G(jω) is a complex
number.
In polar cooordinates,
G(jω) = |G(jω)| exp(∠G(jω))
|G(jω)| is the system’s gain, at frequency
ω.
It is often given in decibels, 20 log10 G(jω)
∠G(jω) is the system’s phase shift
Both radians and degrees are commonly
used.
Frequency is often given in decades,
log10 ω.
One decade is a factor of ten.
Bode diagrams
The most common way of plotting a
system’s frequency is with Bode diagrams.
A Bode magnitude diagram is a plot of
20 log10 |G(jω)| versus log10 ω
A Bode phase diagram is a plot of
∠G(jω) Versus log10 ω
Nyquist diagrams
x-axis is Re G(jω)
y-axis is Im G(jω)
Nichols Chart
x-axis is ∠G(jω)
y-axis is 20 log10 |G(jω)|
Case 1 - Constant Gain
If G(s) = K where K is a real number,
the system is a pure gain.
The Bode magnitude diagram is a
horizontal straight line at y = 20 log10 |K|
db’s.
The Bode phase diagram is a horizontal
straight line at
y = 0 degrees if K > 0, or
y = 180 degrees if K < 0.
Case 2 - sn
Consider next the case where
G(s) = sn where n is an integer
1
For instance, G(s) = is the integrator,
s
and G(s) = s is the differentiator.
What do the Bode diagrams look like?
y = 20 log10 |G(jω)|
1
= 20 log10 | jω)n|
(
= 20 log10 |ω|n
= 20n log10 |ω|
= 20nx
This is a straight line,
with slope = 20 n db / decade,
going through the origin (0,0)
The phase is 90 n degrees for all ω.
So the Bode phase diagram is a horizontal
nπ
radians.
straight line at 90 n degrees, or
2
1
Case 3 τs + 1
Consider next the case where
1
G(s) =
where τ is a real number
τs + 1
1
This system has a single pole at p = −
τ
What do the Bode diagrams look like?
y = 20 log10 |G(jω)|
1 = 20 log10 τ jω + 1 1
= 20 log10 p
(τ ω)2 + 1 −1/2
2
= 20 log10 (τ ω) + 1
At low frequencies, when τ ω << 1,
τ ω + 1 ≈ 1,
y ≈ 20 log10 |1|−1/2
= 20 log10 1
=0
This is a horizontal straight line,
going through the point (0,0)
This is called the low frequency asymptote
At low frequencies,
1
1
G(jω) =
≈ =1
τ jω + 1 1
so the phase at low frequencies is 0
We had
−1/2
y = 20 log10 (τ ω)2 + 1
At high frequencies, when τ ω >> 1,
τ ω + 1 ≈ τ ω,
−1/2
2
y ≈ 20 log10 (τ ω) = 20 log10 |τ ω|−1
= −20 log10 |τ ω|
= −20 log10 |ω| − 20 log10 |τ |
= −20x − 20 log10 |τ |
This is a straight line,
with slope = - 20 db / decade.
This line is called the
high frequency asymptote
It’s gain is 0 db when
y = 0 = −20x − 20 log10 |τ |
⇒ 20x = −20 log10 |τ |
⇒ 20 log10 |ω| = −20 log10 |τ |
1
⇒ |ω| =
= |p|
|τ |
So this straight line goes through the point
x = log10 |p|, y = 0
The two asymptotes meet at this point.
At high frequencies,
1
1
−j
G(jω) =
≈
≈
τ jω + 1 τ jω τ ω
so the phase at high frequencies is -90
degrees if τ > 0 (stable pole), and
and the phase at high frequencies is +90
degrees if τ < 0 (unstable pole).
Most of the phase change occurs one
decade on either side of ω = |p|
1
The Bode magnitude diagram of
is
τs + 1
well approximated by two straight lines.
This approximation is called the aymptotic
approximation.
It is a piecewise linear approximation.
Case 4
Consider next the case where
ωn2
G(s) = 2
s + 2ζωns + ωn2
where ωn and ζ are real numbers.
Every transfer function of the form
a0
G(s) = 2
s + a1 s + a0
with a0 > 0 can be written in that way.
ωn is called the natural frequency.
ζ is called the damping factor.
What do the Bode diagrams look like?
ωn2
G(jω) =
−ω 2 + 2jζωnω + ωn2
1
=
2
1 − ωωn + j 2ζ ωωn

2
ω
⇒ |G(jω)| =  1 − 2
ωn
⇒ |G(jω)| =
!2

2 −1/2
ω 
+ 2ζ
ωn
ω2
ω4 2
1 + 4 + 4ζ − 2 2
ωn
ωn
At low frequencies, ω << ωn,
⇒ |G(jω)| ≈ (1 + 0 + 0)−1/2 = 1
⇒ y = 20 log10 |G(jω)| = 0
This is the low frequency asymptote.
!−1/2
At high frequencies, ω >> ωn,
so the highest power in ω/ωn dominates.
⇒ |G(jω)| ≈
ω4
ωn4
!−1/2
=
−2
ω
ωn
ω
⇒ y = 20 log10 |G(jω)| ≈ 20 log10
ωn
ω
= −40 log10
ωn
= −40 log10 ω + 40 log10 ωn
= −40x + 40 log10 ωn
This is the high frequency asymptote.
The two asymptotes meet when
y = 0 = −40x + 40 log10 ωn
⇒ 40x = 40 log10 ωn
⇒ 40 log10 ω = 40 log10 ωn
⇒ ω = ωn
−2
Since
G(jω) = 1
ω2
1 − ωn + j 2ζ ωωn
at low frequencies
G(jω) ≈ 1
so the phase is 0,
At high frequencies,
G(jω) ≈
so the phase is 180.
1
2
− ωωn
Resonant Frequency
Where does |G(jω)| have local maxima
and/or minima?
At a local maximum or minimum,
d
|G(jω)| = 0
dω
Since

2
ω
|G(jω)| =  1 − 2
ωn
then
!2

2 −1/2
ω 
+ 2ζ
ωn
d
|G(jω)|
dω
!−3/2
ω2
1
ω4 2
=−
1 + 4 + 4ζ − 2 2
2
ωn
ωn
!
2
ω
d
ω4
1 + 4 + 4ζ 2 − 2 2
dω
ωn
ωn
If ζ 6= 0, this is zero when
!
2
ω4
ω
d
1 + 4 + 4ζ 2 − 2 2 = 0
dω
ωn
ωn
2ω
4ω 3 2
⇒ 4 + 4ζ − 2 2 = 0
ωn
ωn
⇒ 4ω 3 + 8ζ 2 − 4 ωωn2 = 0
⇒ ω ω 2 + 2ζ 2 − 1 ωn2 = 0
This equation holds when
⇒ ω = 0 or ω 2 = 1 − 2ζ 2 ωn2
q
⇒ ω = 0 or ω = 1 − 2ζ 2 ωn
At a local maximum or minimum, when
ζ 6= 0,
d
|G(jω)| = 0
dω
⇒ ω = 0 or ω =
q
1 − 2ζ 2 ωn
The latter value of ω is a real number if
and only if
1 − 2ζ 2 ≥ 0
1
⇔ ≥ ζ2
r2
1
⇔
≥ |ζ|
2
⇔ |ζ| ≤ 0.707
Define
ωr =
q
1 − 2ζ 2 ωn
This is called the resonant frequency.
What is |G(jωr )| ?
Since
ω2
ω4 2
1 + 4 + 4ζ − 2 2
ωn
ωn
⇒ |G(jω)| =
!−1/2
and
ωr =
q
2
ω
r
1 − 2ζ 2 ωn ⇔ 2 = 1 − 2ζ 2
ωn
then
⇒ |G(jωr )| =
2 −1/
= 1 + 1 − 2ζ 2 + 4ζ 2 − 2 1 − 2ζ 2
=
1 + 1 − 2ζ 2
=
2
− 2 1 − 2ζ 2
1 − 1 − 2ζ 2
2−1/2
2−1/2
−1/2
= 1 − 1 − 4ζ 2 + 4ζ 4
−1/2
= 4ζ 2 − 4ζ 4
= 4ζ 2 1 − ζ 2
−1/2
Hence,
|G(jωr )| =
1
2ζ
p
1 − ζ2
If ζ ≤ 0.707 then there is a local
maximum at
q
ωr = 1 − 2ζ 2 ωn
with
|G(jωr )| =
1
2ζ
p
1 − ζ2
When ζ < 0.707 this is called the
underdamped case.
If ζ > 0.707 then the magnitude
decreases with increasing omega. This is
called the overdamped case.
When ζ = 0.707 this is called the
critically damped case.
1
How are the Bode diagrams of
G(s)
related to the Bode diagrams of G(s) ?
1 20 log10 G(jω) 1
= 20 log10
|G(jω)|
= −20 log10 |G(jω)|
i.e. reflect about the horizontal axis (the 0
db line).
G(jω) = |G(jω)|ej∠G(jω)
1
1
⇒
=
e−j∠G(jω)
G(jω) |G(jω)|
1
⇒ ∠
= −∠G(jω)
G(jω)
i.e. reflect about the horizontal axis (the 0
degrees line).
How are the Bode diagrams of G1(s)G2(s)
related to the Bode diagrams of G1(s) and
G2(s) ?
20 log10 |G1(jω)G2(jω)|
= 20 log10 |G1(jω)| + 20 log10 |G2(jω)|
i.e. simply add pointwise.
G1(jω)G2(jω) = |G1(jω)|ej∠G1(jω)|G2(jω)|ej∠G
= |G1(jω)| |G2(jω)| ej(∠G1(jω)+∠G2(jω))
⇒ ∠G1(jω)G2(jω) = ∠G1(jω)+∠G2(jω)
i.e. simply add pointwise.
Sketching Bode Diagrams
Every rational transfer function is made up
of factors of the form
s2 + 2ζωns + ωn2
k
K, s , (τ s + 1),
ωn2
(on the numerator and/or the
denominator).
The piecewise linear approximations of the
four cases
2
1
ω
n
K, sk ,
, 2
τ s + 1 s + 2ζωns + ωn2
can be combined to give a procedure for
quickly sketching the Bode magnitude
diagram of any rational transfer function.
It is important to make a correction for
resonant peaks in case 4.
Every rational transfer function can be
written in the form
G(s) = Ksk
Q
Q 2
2 )
(τis + 1) (s + 2ζiωnis + ωni
Q
Q 2
2 )
(τis + 1) (s + 2ζiωnis + ωni
The factors (τis + 1) are the real poles and
zeros.
2 ) are the
The factors (s2 + 2ζiωnis + ωni
complex conjugate pairs of poles and zeros.
The factor sk is the poles or zeros at the
origin s = 0, if any.
Complex poles and zeros
Consider the factor
ωn2
s2 + 2ζωns + ωn2
The poles are . . .
p
−2ζωn ± 4ζ 2ωn2 − 4ωn2
p1, p2 =
2
q
= −ζωn ± ζ 2ωn2 − ωn2
q
= −ζωn ± ωn ζ 2 − 1
So if ζ 2 ≥ 1 there are two real poles,
and then
ωn
1
=
2
2
s + 2ζωns + ωn (τ1s + 1)(τ2s + 1)
So Case 4 is needed for complex poles and
zeros only.
If ζ 2 < 1 there are two complex poles,
q
p1, p2 = −ζωn ± jωn 1 − ζ 2
One pole is the complex conjugate of the
other,
have a complex conjugate pair.
The asymptotic approximation is 0 db for
ω ≤ ωn, and a rolloff of -40 db/dec for
ω ≥ ωn.
It has NO dependence on ζ.
With ζ = 1,
ωn2
s2 + 2ωns + ωn2
ωn
1
=
= 1
2
(s + ωn)
( ωn s + 1)2
Consider the slope of the Bode magnitude
diagram . . .
Every real pole p pulls the gain down by
-20 db/decade at ω = |p|.
Every real zero z pulls the gain up by +20
db/decade at ω = |z|.
Every complex pair of poles pulls the gain
down by -40 db/decade at ω = ωn.
Every complex pair of zeros pulls the gain
up by +40 db/decade at ω = ωn.
The slope at low frequencies is determined
by the factor sk
The slope of the asymptotic (piecewise
linear) approximation of the Bode diagram
is always a multiple of 20 db/decade.
Consider the Bode phase diagram . . .
Let LHP and RHP denote left half plane
and right half plane.
Every LHP real pole p pulls the phase
down by −90o around ω = |p|.
Every RHP real pole p pulls the phase up
by +90o around ω = |p|.
Every LHP real zero z pulls the phase up
by +90o around ω = |z|.
Every RHP real zero z pulls the phase
down by −90o around ω = |z|.
Similarly, every complex pair of poles/zeros
pulls the gain down/up by double the
amount ±180o.
The low frequency phase is determine by
∠G(jω) = ∠K(jω)k
= 90ok if K > 0
= 180 + 90ok if K < 0
Sketching Bode Diagrams
The procedure is as follows . . .
Plot from left to right.
Identify all the breakpoints
(one for each real pole or zero, one for each
complex pair of poles or zeros).
Start the plot about one decade below the
smallest breakpoint.
At low frequencies, the gain is determined
by the factor Ksk , since the other cases
have a low frequency gain of 0 db.
Use the factor Ksk to obtain the first
point.
Draw a straight line to the first breakpoint,
then from the first to the second,
and so on.
The slope starts at 40 k db/dec.
Identification from Bode Diagrams
The sketching procedure can be reversed,
giving a method of estimating a transfer
function from its Bode diagrams.
The procedure is as follows . . .
Fit a series of straight lines to the Bode
magnitude diagram.
Ensure that each straight line segment has
a slope which is an integer multiple of 20
db/dec.
The breakpoints give the locations of the
real poles and zeros, and the value of ωn
for complex conjugate pairs.
The low frequency segment identifies the
factor Ksk .
If the phase is decreasing at a poles
breakpoint, the pole is in the LHP.
If the phase is increasing at a poles
breakpoint, the pole is in the RHP.
If the phase is increasing at a zeros
breakpoint, the pole is in the LHP.
If the phase is decreasing at a zeros
breakpoint, the pole is in the RHP.
The low frequency phase gives the sign of
K.
For each resonant peak/trough, estimate ζ.