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Step Response Analysis. Frequency Response,
Relation Between Model Descriptions
Automatic Control, Basic Course, Lecture 3
November 9, 2017
Lund University, Department of Automatic Control
Content
1. Step Response Analysis
2. Frequency Response
3. Relation between Model Descriptions
1
Step Response Analysis
Step Response
From the last lecture, we know that if the input u(t) is a step, then the
output in the Laplace domain is
Y (s) = G (s)U(s) = G (s)1
s
It is possible to do an inverse transform of Y (s) to get y(t), but is it
possible to claim things about y(t) by only studying Y (s)?
We will study how the poles affects the step response. (The zeros
will be discussed later).
2
Initial and Final Value Theorem
Let F (s) be the Laplace transformation of f (t), i.e., F (s) = L(f (t))(s).
Given that the limits below exist1 , it holds that:
Initial value theorem limt→0 f (t) = lims→+∞ sF (s)
Final value theorem limt→+∞ f (t) = lims→0 sF (s)
For a step response we have that:
limt→+∞
y(t) = lims→0
sY (s) = lims→0
sG (s)1
s= G (0)
1When can we NOT apply the Final value theorem?
3
First Order System
−1.5 −1 −0.5 0 0.5−1
−0.5
0
0.5
1
T = 1 T = 2T = 5
Re
ImSingularity Chart
0 5 10 150
0.5
1
t
y(t)
Step Response (K=1)
G (s) =K
1 + sT
One pole in s = −1/T
Step response:
Y (s) = G (s)1
s=
K
s(1 + sT )L−1
−−→ y(t) = K(
1− e−t/T), t ≥ 0
4
First Order System
−1.5 −1 −0.5 0 0.5−1
−0.5
0
0.5
1
T = 1 T = 2T = 5
Re
ImSingularity Chart
0 5 10 150
0.5
1
t
y(t)
Step Response (K=1)
G (s) =K
1 + sT
Final value:
limt→+∞
y(t) = lims→0
sY (s) = lims→0
s · K
s(1 + sT )= K
4
First Order System
−1.5 −1 −0.5 0 0.5−1
−0.5
0
0.5
1
T = 1 T = 2T = 5
Re
ImSingularity Chart
0 5 10 150
0.5
1
t
y(t)
Step Response (K=1)
G (s) =K
1 + sT
T is called the time-constant:
y(T ) = K (1− e−T/T ) = K (1− e−1) ≈ 0.63K
i.e., T is the time it takes for the step response to reach 63% of its final
value4
First Order System
−1.5 −1 −0.5 0 0.5−1
−0.5
0
0.5
1
T = 1 T = 2T = 5
Re
ImSingularity Chart
0 5 10 150
0.5
1
t
y(t)
Step Response (K=1)
G (s) =K
1 + sT
Derivative at zero:
limt→0
y(t) = lims→+∞
s · sY (s) = lims→+∞
s2K
s(1 + sT )=
K
T
4
Second Order System With Real Poles
−1.5 −1 −0.5 0 0.5−1
−0.5
0
0.5
1
T = 1 T = 2
Re
ImSingularity Chart
0 5 10 150
0.5
1
t
y(t)
Step Response (K=1)
G (s) =K
(1 + sT1)(1 + sT2)
Poles in s = −1/T1 and s = −1/T2. Step response:
y(t) =
K(
1− T1e−t/T1−T2e
−t/T2
T1−T2
)T1 6= T2
K(1− e−t/T − t
T e−t/T)
T1 = T2 = T
5
Second Order System With Real Poles
−1.5 −1 −0.5 0 0.5−1
−0.5
0
0.5
1
T = 1 T = 2
Re
ImSingularity Chart
0 5 10 150
0.5
1
t
y(t)
Step Response (K=1)
G (s) =K
(1 + sT1)(1 + sT2)
Final value:
limt→+∞
= lims→0
sY (s) = lims→0
sK
s(1 + sT1)(1 + sT2)= K
5
Second Order System With Real Poles
−1.5 −1 −0.5 0 0.5−1
−0.5
0
0.5
1
T = 1 T = 2
Re
ImSingularity Chart
0 5 10 150
0.5
1
t
y(t)
Step Response (K=1)
G (s) =K
(1 + sT1)(1 + sT2)
Derivative at zero:
limt→0
y(t) = lims→+∞
s · sY (s) = lims→+∞
s2K
s(1 + sT1)(1 + sT2)= 0
5
Second Order System With Complex Poles
G (s) =Kω2
0
s2 + 2ζω0s + ω20
, 0 < ζ < 1
Relative damping ζ, related to the angle ϕ
ζ = cos(ϕ)
−1 0 1−1
−0.5
0
0.5
1
ω0
ϕ
Re
Im
Singularity Chart
6
Second Order System With Complex Poles
G (s) =Kω2
0
s2 + 2ζω0s + ω20
, 0 < ζ < 1
Inverse transformation for step response yields:
y(t) = K
(1− 1√
1− ζ2e−ζω0t sin
(ω0
√1− ζ2t + arccos ζ
))
= K
(1− 1√
1− ζ2e−ζω0t sin
(ω0
√1− ζ2t + arcsin(
√1− ζ2)
)), t ≥ 0
6
Second Order System With Complex Poles
G (s) =Kω2
0
s2 + 2ζω0s + ω20
, 0 < ζ < 1
Inverse transformation for step response yields:
y(t) = K
(1− 1√
1− ζ2e−ζω0t sin
(ω0
√1− ζ2t + arccos ζ
))
= K
(1− 1√
1− ζ2e−ζω0t sin
(ω0
√1− ζ2t + arcsin(
√1− ζ2)
)), t ≥ 0
Exercise: Check of correct starting point of step response.
y(0) = K
(1−
1√1− ζ2
e0 sin(ω0
√1− ζ20 + arcsin(
√1− ζ2)
))
= K
(1−
1√1− ζ2
·√
1− ζ2)
= 0 0 5 10 150
0.5
1
1.5
t
Step Response
6
Second Order System With Complex Poles
G (s) =Kω2
0
s2 + 2ζω0s + ω20
, 0 < ζ < 1
−1 0 1
−1
0
1ω0 = 1
ω0 = 1.5
ω0 = 0.5
Re
Im
Singularity Chart
0 5 10 150
0.5
1
t
y(t)
Step Response (K=1)
6
Second Order System With Complex Poles
G (s) =Kω2
0
s2 + 2ζω0s + ω20
, 0 < ζ < 1
−1 0 1
−1
0
1 ζ = 0.3ζ = 0.7ζ = 0.9
Re
Im
Singularity Chart
0 5 10 150
0.5
1
1.5
t
y(t)
Step Response (K=1)
6
Frequency Response
Sinusoidal Input
Given a transfer function G (s), what happens if we let the input be
u(t) = sin(ωt)?
0 5 10 15 20−1
−0.5
0
0.5
1
t
y(t)
0 5 10 15 20−1
−0.5
0
0.5
1
t
u(t)
7
Sinusoidal Input
It can be shown that if the input is u(t) = sin(ωt), the output2 will be
y(t) = A sin(ωt + ϕ)
where
A = |G (iω)|ϕ = argG (iω)
So if we determine a and ϕ for different frequencies ω, we have a
description of the transfer function.
2after the transient has decayed
8
Bode Plot
Idea: Plot |G (iω)| and argG (iω) for different frequencies ω.
10−2 10−1 100 101 10210−3
10−2
10−1
100
101
Mag
nitude(abs)
10−2 10−1 100 101 102−180
−135
−90
−45
0
Frequency (rad/s)
Phase(deg)
9
Sinusoidal Input-Output: example with frequency sweep (chirp)
Resonance frequency of industrial robot IRB2000 visible in data.
10
Sinusoidal Input-Output: example with frequency sweep (chirp)
Resonance frequency of industrial robot IRB2000 visible in data.
10
Bode Plot - Products of Transfer Functions
Let
G (s) = G1(s)G2(s)G3(s)
then
log |G (iω)| = log |G1(iω)|+ log |G2(iω)|+ log |G3(iω)|argG (iω) = argG1(iω) + argG2(iω) + argG3(iω)
This means that we can construct Bode plots of transfer functions from
simple ”building blocks” for which we know the Bode plots.
11
Bode Plot of G (s) = K
If
G (s) = K
then
log |G (iω)| = log(|K |)argG (iω) = 0 (if K > 0, else + 180 or − 180 deg)
12
Bode Plot of G (s) = K
10−2 10−1 100 101 10210−1
100
101
K = 0.5
K = 1
K = 4Mag
nitude(abs)
10−2 10−1 100 101 102−180−135−90−45
0
45
Frequency (rad/s)
Phase(deg)
12
Bode Plot of G (s) = sn
If
G (s) = sn
then
log |G (iω)| = n log(ω)
argG (iω) = nπ
2
13
Bode Plot of G (s) = sn
10−1 100 101 10210−4
10−3
10−2
10−1
100101102
n = 1
n = −1
n = −2
Mag
nitude(abs)
10−2 10−1 100 101 102−180−135−90−45
04590
Frequency (rad/s)
Phase(deg)
13
Bode Plot of G (s) = (1 + sT )n
If
G (s) = (1 + sT )n
then
log |G (iω)| = n log(√
1 + ω2T 2)
argG (iω) = n arg(1 + iωT ) = n arctan (ωT )
For small ω
log |G (iω)| → 0
argG (iω)→ 0
For large ω
log |G (iω)| → n log(ωT )
argG (iω)→ nπ
2
14
Bode Plot of G (s) = (1 + sT )n
10−1 100 10110−2
10−1
100
101
n = 1
n = −1
n = −2
1
T
Mag
nitude(abs)
10−1 100 101−180−135−90−45
04590
Frequency (rad/s)
Phase(deg)
14
Bode Plot of G (s) = (1 + 2ζs/ω0 + (s/ω0)2)n
G (s) = (1 + 2ζs/ω0 + (s/ω0)2)n
For small ω
log |G (iω)| → 0
arg(iω)→ 0
For large ω
log |G (iω)| → 2n log
(ω
ω0
)argG (iω)→ nπ
15
Bode Plot of G (s) = (1 + 2ζs/ω0 + (s/ω0)2)n
10−1 100 10110−2
10−1
100
101
102Mag
nitude(abs) ζ = 0.2
ζ = 0.1
ζ = 0.05
10−1 100 101
−180−135−90−45
0
Frequency (rad/s)
Phase(deg)
15
Bode Plot of G (s) = e−sL
G (s) = e−sL
Describes a pure time delay with delay L, i.e, y(t) = u(t − L)
log |G (iω)| = 0
argG (iω) = −ωL
16
Bode Plot of G (s) = e−sL
10−1 100 101 10210−1
100
101Mag
nitude(abs)
10−1 100 101 102
0
−200
−400
−600
L = 5 L = 0.5
L = 0.1
Frequency (rad/s)
Phase(deg)
16
Bode Plot of G (s) = e−sL
Same delay may appear as different phase lag for different frequencies!
Example
Delay ≈ 0.52 sec between input and output.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6
−1
0
1
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6
−1
0
1
(Upper): Period time = 2π ≈6.28 sec. Delay represents phase
lag of 0.526.28· 360 ≈ 30 deg
(Lower): Period time = π ≈3.14 sec. Delay represents phase
lag of 0.53.14· 360 ≈ 60 deg.
16
Bode Plot of G (s) = e−sL
Same delay may appear as different phase lag for different frequencies!
Example
Delay ≈ 0.52 sec between input and output.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6
−1
0
1
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6
−1
0
1
(Upper): Period time = 2π ≈6.28 sec. Delay represents phase
lag of 0.526.28· 360 ≈ 30 deg
(Lower): Period time = π ≈3.14 sec. Delay represents phase
lag of 0.53.14· 360 ≈ 60 deg.
16
Bode Plot of G (s) = e−sL
Check phase in Bode diagram for e−0.52s for
• sin(t)⇒ ω = 1.0 rad/s
• sin(2t)⇒ ω = 2.0 rad/s
>> s=tf(’s’)
>> G=exp(-0.52*s);
>> bode(G,0.1 ,5) % Bode plot in frequency-range [0.1 .. 5] rad/s
16
Bode Plot of Composite Transfer Function
Example
Draw the Bode plot of the transfer function
G (s) =100(s + 2)
s(s + 20)2
First step, write it as product of sample transfer functions:
G (s) =100(s + 2)
s(s + 20)2= 0.5 · s−1 · (1 + 0.5s) · (1 + 0.05s)−2
Then determine the corner frequencies:
G (s) =100(s + 2)
s(s + 20)2= 0.5 · s−1 ·
wc1=2︷ ︸︸ ︷
(1 + 0.5s) ·
wc2=20︷ ︸︸ ︷
(1 + 0.05s)−2
17
Bode Plot of Composite Transfer Function
G (s) =100(s + 2)
s(s + 20)2= 0.5 · s−1 ·
wc1=2︷ ︸︸ ︷
(1 + 0.5s) ·
wc2=20︷ ︸︸ ︷
(1 + 0.05s)−2
10−1 100 101 102 10310−4
10−3
10−2
10−1
100
101
Frequency (rad/s)
Mag
nitude(abs)
18
Bode Plot of Composite Transfer Function
G (s) =100(s + 2)
s(s + 20)2= 0.5 · s−1 ·
wc1=2︷ ︸︸ ︷
(1 + 0.5s) ·
wc2=20︷ ︸︸ ︷
(1 + 0.05s)−2
10−1 100 101 102 10310−4
10−3
10−2
10−1
100
101
−1
Frequency (rad/s)
Mag
nitude(abs)
18
Bode Plot of Composite Transfer Function
G (s) =100(s + 2)
s(s + 20)2= 0.5 · s−1 ·
wc1=2︷ ︸︸ ︷
(1 + 0.5s) ·
wc2=20︷ ︸︸ ︷
(1 + 0.05s)−2
10−1 100 101 102 10310−4
10−3
10−2
10−1
100
101
−10
Frequency (rad/s)
Mag
nitude(abs)
18
Bode Plot of Composite Transfer Function
G (s) =100(s + 2)
s(s + 20)2= 0.5 · s−1 ·
wc1=2︷ ︸︸ ︷
(1 + 0.5s) ·
wc2=20︷ ︸︸ ︷
(1 + 0.05s)−2
10−1 100 101 102 10310−4
10−3
10−2
10−1
100
101
−10
−2
Frequency (rad/s)
Mag
nitude(abs)
18
Bode Plot of Composite Transfer Function
G (s) =100(s + 2)
s(s + 20)2= 0.5 · s−1 ·
wc1=2︷ ︸︸ ︷
(1 + 0.5s) ·
wc2=20︷ ︸︸ ︷
(1 + 0.05s)−2
10−1 100 101 102 10310−4
10−3
10−2
10−1
100
101
−10
−2
Frequency (rad/s)
Mag
nitude(abs)
18
Bode Plot of Composite Transfer Function
G (s) =100(s + 2)
s(s + 20)2= 0.5 · s−1 ·
wc1=2︷ ︸︸ ︷
(1 + 0.5s) ·
wc2=20︷ ︸︸ ︷
(1 + 0.05s)−2
10−1 100 101 102 103
−180
−135
−90
−45
0
45
Frequency (rad/s)
Phase(deg)
19
Bode Plot of Composite Transfer Function
G (s) =100(s + 2)
s(s + 20)2= 0.5 · s−1 ·
wc1=2︷ ︸︸ ︷
(1 + 0.5s) ·
wc2=20︷ ︸︸ ︷
(1 + 0.05s)−2
10−1 100 101 102 103
−180
−135
−90
−45
0
45
Frequency (rad/s)
Phase(deg)
19
Bode Plot of Composite Transfer Function
G (s) =100(s + 2)
s(s + 20)2= 0.5 · s−1 ·
wc1=2︷ ︸︸ ︷
(1 + 0.5s) ·
wc2=20︷ ︸︸ ︷
(1 + 0.05s)−2
10−1 100 101 102 103
−180
−135
−90
−45
0
45
Frequency (rad/s)
Phase(deg)
19
Bode Plot of Composite Transfer Function
G (s) =100(s + 2)
s(s + 20)2= 0.5 · s−1 ·
wc1=2︷ ︸︸ ︷
(1 + 0.5s) ·
wc2=20︷ ︸︸ ︷
(1 + 0.05s)−2
10−1 100 101 102 103
−180
−135
−90
−45
0
45
Frequency (rad/s)
Phase(deg)
19
Bode Plot of Composite Transfer Function
G (s) =100(s + 2)
s(s + 20)2= 0.5 · s−1 ·
wc1=2︷ ︸︸ ︷
(1 + 0.5s) ·
wc2=20︷ ︸︸ ︷
(1 + 0.05s)−2
10−1 100 101 102 103
−180
−135
−90
−45
0
45
Frequency (rad/s)
Phase(deg)
19
Nyquist Plot
By removing the frequency information, we can plot the transfer function
in one plot instead of two.
0 0.5
−0.5
0
0.5
argG(iω)
|G(iω)|
Re G(iω)
ImG(iω)
20
Nyquist Plot
By removing the frequency information, we can plot the transfer function
in one plot instead of two.
0 0.5
−0.5
0
0.5
argG(iω)
|G(iω)|
Re G(iω)
ImG(iω)
Split the transfer function into real and imaginary part:
G (s) =1
1 + sG (iω) =
1
1 + iω=
1
1 + ω2− i
ω
1 + ω2
Is this the transfer function in the plot above? 20
From Bode Plot to Nyquist Plot
10−2 10−1 100 101 10210−4
10−3
10−2
10−1
100101
Frequency (rad/s)
Mag
nitude(abs)
10−2 10−1 100 101 102
−270
−180
−90
0
Frequency (rad/s)
Phase(deg)
−0.5 0 0.5 1
−0.5
0
0.5
Re G(iω)Im
G(iω)
21
From Bode Plot to Nyquist Plot
10−2 10−1 100 101 10210−4
10−3
10−2
10−1
100101
Frequency (rad/s)
Mag
nitude(abs)
10−2 10−1 100 101 102
−270
−180
−90
0
Frequency (rad/s)
Phase(deg)
−0.5 0 0.5 1
−0.5
0
0.5
Re G(iω)Im
G(iω)
21
From Bode Plot to Nyquist Plot
10−2 10−1 100 101 10210−4
10−3
10−2
10−1
100101
Frequency (rad/s)
Mag
nitude(abs)
10−2 10−1 100 101 102
−270
−180
−90
0
Frequency (rad/s)
Phase(deg)
−0.5 0 0.5 1
−0.5
0
0.5
Re G(iω)Im
G(iω)
21
From Bode Plot to Nyquist Plot
10−2 10−1 100 101 10210−4
10−3
10−2
10−1
100101
Frequency (rad/s)
Mag
nitude(abs)
10−2 10−1 100 101 102
−270
−180
−90
0
Frequency (rad/s)
Phase(deg)
−0.5 0 0.5 1
−0.5
0
0.5
Re G(iω)Im
G(iω)
21
From Bode Plot to Nyquist Plot
10−2 10−1 100 101 10210−4
10−3
10−2
10−1
100101
Frequency (rad/s)
Mag
nitude(abs)
10−2 10−1 100 101 102
−270
−180
−90
0
Frequency (rad/s)
Phase(deg)
−0.5 0 0.5 1
−0.5
0
0.5
Re G(iω)Im
G(iω)
21
From Bode Plot to Nyquist Plot
10−2 10−1 100 101 10210−4
10−3
10−2
10−1
100101
Frequency (rad/s)
Mag
nitude(abs)
10−2 10−1 100 101 102
−270
−180
−90
0
Frequency (rad/s)
Phase(deg)
−0.5 0 0.5 1
−0.5
0
0.5
Re G(iω)Im
G(iω)
21
Relation between Model
Descriptions
Single-capacitive Processes KsT+1
−1.5 −1 −0.5 0 0.5−1
−0.5
0
0.5
1
Singularity chart
0 2 40
0.5
1
Step response
0 0.5 1
−0.6−0.4−0.2
0
0.2
Nyquist plot
10−1
100
Bode plot
10−1 100 101−90
0
22
Multi-capacitive Processes K(sT1+1)(sT2+1)
−1.5 −1 −0.5 0 0.5−1
−0.5
0
0.5
1
Singularity chart
0 2 4 6 8 100
0.5
1
Step response
0 0.5 1
−0.6−0.4−0.2
0
0.2
Nyquist plot
10−2
10−1
100
Bode plot
10−1 100 101−180−90
0
23
Integrating Processes 1s
−1.5 −1 −0.5 0 0.5−1
−0.5
0
0.5
1
Singularity chart
0 2 40
2
4
Step response
−0.5 0 0.5
−1
−0.5
0
Nyquist plot
10−1
100101
Bode plot
10−1 100 101−90
0
24
Oscillative ProcessesKω2
0
s2+2ζω0s+ω20, 0 < ζ < 1
−1.5 −1 −0.5 0 0.5
−1
0
1
Singularity chart
0 10 20 300
0.5
1
1.5
Step response
−2 0 2−3
−2
−1
0
1
Nyquist plot
10−210−1100101
Bode plot
10−1 100 101−180−90
0
25
Delay Processes KsT+1
e−sL
0 2 4 6 8
0
0.5
1
Step response
−1 0 1
−1
0
1
Nyquist plot
10−1
100
Bode plot
100 101 102−6000
−3000
0
26
Process with Inverse Responses −sa+1(sT1+1)(sT2+1)
−1 0 1−1
−0.5
0
0.5
1
Singularity chart
0 2 4 6 8 10
0
0.5
1
Step response
−0.5 0 0.5 1
−1
−0.5
0
Nyquist plot
10−1
100
Bode plot
10−1 100 101−270−180−90
0
27
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