The frequency response
In the steady state, sinusoid inputs to a linear system generate sinusoid responses of the
same frequency. These responses (system outputs) differ in amplitude and phase angle from
the input. These differences are functions of
frequency.
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x(t)
Spring
Mass
f(t)
Viscous damper
Force : f (t) = Mi cos(ωt + φi )
Displacement : x(t) = M0 cos(ωt + φo)
f (t) and x(t) can be represented as Mi6 φi and
Mo6 φo .
1
Represent the system also as a complex number M 6 φ. The system output is found using
complex number multiplication:
Mo(ω)6 φo(ω) = Mi(ω)M (ω)6 [φ(ω) + φi(ω)]
Hence
Mo(ω)
Mi(ω)
φ(ω) = φo (ω) − φi(ω)
M (ω) =
(1)
(2)
Equations (1) and (2) define the frequency response.
1. M (ω) is called the magnitude frequency response.
2. φ(ω) the phase frequency response.
3. The combination of the magnitude and phase
frequency responses is called frequency response M (ω)6 φ(ω)
2
Frequency response from the transfer function
R(s)
C(s)
G(s)
The frequency response M (ω)6 φ(ω) of a system G(s) is
G(jω) = G(s)|s→jω
i.e.
M (ω) = |G(jω)|
φ(ω) = 6 G(jω)
G(jω) = M (ω)6 φ(ω) can be plotted in several
ways:
1. Two separate plots; with separate magnitude and phase plots as a function of the
frequency; or
2. A single polar plot.
3
If two separate magnitude and phase response
are plotted, the magnitude is decibels(dB) vs.
log(ω)the magnitude is decibels(dB) vs. log(ω)
Example: Find the analytical expression for the
magnitude frequency response and the phase
1 .
frequency response for a system G(s) = s+2
Also plot the separate magnitude and phase
diagram and the polar plot.
1
Solution: G(jω) = jω+2
= ω2−jω
2 +4
The magnitude frequency response is
M (ω) = |G(jω)| = q
1
ω2 + 4
The phase frequency response is
φ(ω) = 6 G(jω) = − tan−1(ω/2)
4
1 .
Frequency response for system G(s) = s+2
ω
0
0.1
0.2
0.4
0.6
0.8
1
2
4
6
8
10
20
40
60
80
100
20 log √ 12
ω +4
-6.0206
-6.0314
-6.0638
-6.1909
-6.3949
-6.6652
-6.9897
-9.0309
-13.0103
-16.0206
-18.3251
-20.1703
-26.0638
-32.0520
-35.5678
-38.0645
-40.0017
− tan−1(ω/2)
0
-2.8624
-5.7106
-11.3099
-16.6992
-21.8014
-26.5651
-45.0000
-63.4349
-71.5651
-75.9638
-78.6901
-84.2894
-87.1376
-88.0908
-88.5679
-88.8542
5
Bode Diagram
0
Magnitude (dB)
−10
−20
−30
−40
Phase (deg)
−50
0
−45
−90
−1
10
0
1
10
2
10
10
Frequency (rad/sec)
The figure above shows the separate plots for
magnitude and phase response diagrams, where
the magnitude diagram is 20 log √ 12
vs. log ω,
ω +4
and the phase diagram is − tan−1(ω/2) vs. log ω.
6
Nyquist Diagram
0
−20 dB
−10 dB
−0.05
Imaginary Axis
−0.1
−0.15
−0.2
−0.25
0
0.05
0.1
0.15
0.2
0.25
Real Axis
0.3
0.35
0.4
0.45
0.5
The figure above shows the polar plot for the
frequency response. It is a plot of
1
q
ω2 + 4
6
− tan−1(ω/2)
for different ω.
7
Example: Find the analytical expressions for
the magnitude frequency response and the phase
frequency response for a system
G(s) =
1
.
(s + 2)(s + 4)
Also plot the separate magnitude and phase
diagram and the polar plot.
8−ω 2 −j(6ω)
1
Solution: G(jω) = (jω+2)(jω+4) = (ω 2+4)(ω 2+16)
The magnitude frequency response is
1
M (ω) = |G(jω)| = q
(8 − ω 2)2 + (6ω)2
The phase frequency response is
6 G(jω)
φ(ω) = 
√
6ω
−1

− tan ( 8−ω 2 )
for ω ≤ 8
√
=
 −(π + tan−1( 6ω ) for ω > 8
8−ω 2
8
Points of interest: Starting point, End point,
Real axis crossing, Imaginary crossing.
1. Starting ω → 0, 0.1256 0
2. Ending ω → ∞,06 − π
8−ω 2 −j(6ω)
3. Real axis crossing 2
, found by
(ω +4)(ω 2+16)
setting 2 6ω 2
=0
(ω +4)(ω +16)
ω = 0, 0.1256 0
4. Imaginary axis crossing, found by setting
8−ω 2
=0
(ω 2+4)(ω 2 +16)
√
π
ω = 8, 0.05896 −
2
9
Bode Diagram
0
Magnitude (dB)
−20
−40
−60
−80
0
Phase (deg)
−45
−90
−135
−180
−1
10
0
1
10
10
2
10
Frequency (rad/sec)
The figure above shows the separate plots for
magnitude and phase response diagrams, where
1
the magnitude diagram is 20 log p
2 2
2
(8−ω ) +(6ω)
vs. log ω, and the phase diagram is 6 G(jω) vs.
log ω.
10
Nyquist Diagram
0
−0.01
−0.02
Imaginary Axis
−0.03
−0.04
−0.05
−0.06
−0.07
−0.08
−0.09
−0.1
−0.2
−0.15
−0.1
−0.05
0
0.05
Real Axis
0.1
0.15
0.2
0.25
0.3
The figure above shows the polar plot for the
frequency response. It is a plot of
1
q
(8 − ω 2)2 + (6ω)2
6 G(jω)
for different ω.
11