Step Response Analysis. Frequency Response, Relation Between …€¦ · · 2017-11-15Step Response Analysis. Frequency Response, Relation Between Model Descriptions ... We will

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


−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


−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


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√


(ω0

√1− ζ2t + arcsin(

√1− ζ2)

)), t ≥ 0

6


G (s) =Kω2

0

s2 + 2ζω0s + ω20

, 0 < ζ < 1

Inverse transformation for step response yields:

y(t) = K

(1− 1√


(ω0

√1− ζ2t + arccos ζ

))

= K

(1− 1√


(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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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

Step Response Analysis. Frequency Response, Relation Between …€¦ · · 2017-11-15Step Response Analysis. Frequency Response, Relation Between Model Descriptions ... We will

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