Lecture 20 Review of Frequency Based Design

Lecture 20

Review of Frequency Based Design

𝑈 𝑠

𝐺plant 𝑠; 𝐾 𝐺control 𝑠;𝐾 𝐺measurement 𝑠; 𝐾

𝑋 𝑠 𝐸 𝑠 𝑌 𝑠

𝑅 𝑠

SISO Systems

SISO Systems

We want to adjust the

performance of the system

with respect to some

constant parameter 𝑲 ∈ ℝ

that could be in any part of

the system we can design

𝑈 𝑠



𝑅 𝑠

SISO Systems

𝐺 𝑠 𝑅 𝑠 𝑌 𝑠

𝑈 𝑠



𝑅 𝑠

So we manipulate

the system into

constant feedback

form to determine

how the closed loop

poles change with 𝑲

𝐾

𝑌 𝑆

𝑅 𝑠=

𝐾𝐺 𝑠

1 + 𝐾𝐺 𝑠

It is all about the Loop Gain “𝐺(𝑠)”

The poles are given by

1 + 𝐾𝐺 𝑠 = 0 ⇔ 𝐺 𝑠 = −1

𝐾



1 + 𝐾𝐺 𝑠 = 0 ⇔ 𝐺 𝑠 = −1

𝐾

We need tools for computing

Ψ = 𝑠 ∈ ℂ ∶ 1 + 𝐾𝐺 𝑠 = 0


1 + 𝐾𝐺 𝑠 = 0 ⇔ 𝐺 𝑠 = −1

𝐾

We need tools for computing

Ψ = 𝑠 ∈ ℂ ∶ 1 + 𝐾𝐺 𝑠 = 0

In particular we want to know two things:

1. What values of 𝐾 produce poles in the right half plane?

2. What transient performance will a stable 𝐾 give?


Root Locus

The root locus plots the range space of the eigenvalues (poles)

Ψ ∈ 𝑠 ∈ ℂ ∶ 𝑠 ∈ ℂ ∶ 1 + 𝐾𝐺 𝑠 = 0 , 𝜓 𝐾 ∶ ℝ → ℂ𝑛

Root Locus


Ψ ∈ 𝑠 ∈ ℂ ∶ 𝑠 ∈ ℂ ∶ 1 + 𝐾𝐺 𝑠 = 0 , 𝜓 𝐾 ∶ ℝ → ℂ𝑛

i.e., it gives 𝑛 closed loop poles for each value of 𝐾

where 𝑛 is the degree of the denominator of 𝐺

𝜓 𝐾 = 𝑝1, 𝑝2, … , 𝑝𝑛

with

𝑝𝑘 = 𝜎𝑘 real

± 𝑗𝜔𝑘 imaginary

Root Locus


Ψ ∈ 𝑠 ∈ ℂ ∶ 𝑠 ∈ ℂ ∶ 1 + 𝐾𝐺 𝑠 = 0 , 𝜓 𝐾 ∶ ℝ → ℂ𝑛

These poles can be solved directly, but analytic solutions for the

roots become unwieldy. We need another approach…

Sketching the Root Locus

From

1 + 𝐾𝐺 𝑠 = 1 + 𝐾𝑧 𝑠

𝑝 𝑠 = 0

𝑝 𝑠 + 𝐾𝑧 𝑠 = 0,1

𝐾𝑝 𝑠 + 𝑧 𝑠 = 0

In the limit of 𝑲 → 𝟎, 𝝍 = 𝒔 ∈ ℂ ∶ 𝒑 𝒔 = 𝟎

In the limit of 𝑲 → ∞, 𝝍 = 𝒔 ∈ ℂ ∶ 𝒛 𝒔 = 𝟎


From

1 + 𝐾𝐺 𝑠 = 1 + 𝐾𝑧 𝑠

𝑝 𝑠 = 0

𝑝 𝑠 + 𝐾𝑧 𝑠 = 0,1

𝐾𝑝 𝑠 + 𝑧 𝑠 = 0

In the limit of 𝑲 → 𝟎, 𝝍 = 𝒔 ∈ ℂ ∶ 𝒑 𝒔 = 𝟎

In the limit of 𝑲 → ∞, 𝝍 = 𝒔 ∈ ℂ ∶ 𝒛 𝒔 = 𝟎

So the locus moves from the poles of 𝐺 𝑠 to the zeroes of 𝐺 𝑠 .

𝐼𝑚 𝑠 = 𝑗𝜔

𝑅𝑒 𝑠 = 𝜎

𝚿 o

x

x

o


Additionally every point along the locus must obey

𝑮 𝒔 = 𝟏/𝑲 & ∠𝑮 𝒔 = 𝝅 + 𝟐𝒏𝝅

This gives us terminating and arriving angles and

asymptotes


Additionally every point along the locus must obey

𝑮 𝒔 = 𝟏/𝑲 & ∠𝑮 𝒔 = 𝝅 + 𝟐𝒏𝝅

This gives us terminating and arriving angles and

asymptotes

When you find a point 𝑠∗ ∈ Ψ that meets desired

performance. The control gain is recovered with

𝑲 = −𝟏

𝑮 𝒔∗

We don’t need to plot the entire root locus

to identify if the system is stable!

Plotting Over 𝑗𝜔

When 𝑠 = 𝑗𝜔 we are intersecting the imaginary axis, which is the

boundary between the stable left half plane and the unstable right

half plane.

A pole is on the imaginary axis if

𝑮 𝒋𝝎 = 𝟏/𝑲 & ∠𝑮 𝒋𝝎 = 𝝅 + 𝟐𝒏𝝅


𝐼𝑚 𝑠 = 𝑗𝜔

𝑅𝑒 𝑠 = 𝜎

𝚿 x

x

Stable Unstable

When 𝑠 = 𝑗𝜔 we are intersecting the imaginary axis, which is the

boundary between the stable left half plane and the unstable right

half plane.

A pole is on the imaginary axis if

𝑮 𝒋𝝎 = 𝟏/𝑲 & ∠𝑮 𝒋𝝎 = 𝝅 + 𝟐𝒏𝝅

This is why Bode and Nyquist were so interested in proximity to 𝐼𝑚 𝑠

Phase Margin (PM): Proximity to ∠𝐺 = 𝜋 + 2𝑛𝜋 when 𝐺 = 1/𝐾

and

Gain Margin (GM): Proximity to 𝐺 = 1/𝑘 when ∠𝐺 = 𝜋 + 2𝑛𝜋


Bode Plots

We are interested in understanding the nature of 𝐺 𝑗𝜔 and ∠𝐺 𝑗𝜔

Bode Plots


Note that we can add the effects of each pole and zero to the system linearly!

Bode Plots


Note that we can add the effects of each pole and zero to the system linearly!

10 log10 𝐺 𝑗𝜔 2 = 10 log10

𝑗𝜔 − 𝑧𝑘2𝑚

𝑘=1

𝑗𝜔 − 𝑝𝑘2𝑛

𝑘=1

= 20 log10 𝑗𝜔 − 𝑧𝑘

𝑚

𝑘=1− 20 log10 𝑗𝜔 − 𝑝𝑘

𝑚

𝑘=1

∠𝐺 𝑗𝜔 = ∠ 𝑗𝜔 − 𝑧𝑘

𝑚𝑘=1 − ∠ 𝑗𝜔 − 𝑝𝑘

𝑚𝑘=1

Basic Units of Control

Select 𝑧𝑘 and 𝑝𝑘 for the control compensator and (graphically) add

𝑗𝜔 − 𝑝𝑘−1 and 𝑗𝜔 − 𝑧𝑘

to get desired closed loop frequency response…


Select 𝑧𝑘 and 𝑝𝑘 for the control compensator and (graphically) add

𝑗𝜔 − 𝑝𝑘−1 and 𝑗𝜔 − 𝑧𝑘

to get desired closed loop frequency response…

Note: the final compensator design must always be proper with the degree

in the numerator less than or equal to the degree of the denominator

10 log10 𝐺 2 ∠𝐺


Derivatives 𝑗𝜔

Integrals 𝑗𝜔 −1

Constants 𝑐 10 log10 𝑐 2 𝜋 sign 𝑐 < 0

0 sign 𝑐 > 0

+𝜋/2

−𝜋/2 Out[735]= ,

Out[738]= ,

10 log10 𝐺 2 ∠𝐺


𝑗𝜔 − −𝜔𝑏−1

𝑗𝜔 − +𝜔𝑏−1

Real Poles

For 𝜔𝑏 > 0

𝜔𝑏 𝜔𝑏

−𝜋 to −𝜋

2

0 to −𝜋

2

10 log10 𝐺 2 ∠𝐺


𝑗𝜔 − −𝜔𝑏𝑗−1

𝑗𝜔 − +𝜔𝑏𝑗−1

Imaginary Poles

−𝜋/2

+𝜋/2 𝜔 < 𝜔𝑏

−𝜋/2 𝜔 > 𝜔𝑏

𝜔𝑏

10 log10 𝐺 2 ∠𝐺


−𝜔2 + 𝜔𝑏2 −1

Imaginary Poles (always come in additive pairs)

resonance peak

0 𝜔 < 𝜔𝑏

𝜋 𝜔 > 𝜔𝑏

𝜔𝑏

𝑗𝜔 − +𝜔𝑏

10 log10 𝐺 2 ∠𝐺


𝑗𝜔 − −𝜔𝑏

Real Zeroes

Out[753]= ,

Out[756]= ,

𝜔𝑏 𝜔𝑏 𝜋 to

𝜋

2

0 to 𝜋

2

10 log10 𝐺 2 ∠𝐺


Imaginary Zeroes

𝑗𝜔 − +𝜔𝑏𝑗

𝑗𝜔 − −𝜔𝑏𝑗 +𝜋/2

−𝜋/2 𝜔 < 𝜔𝑏

+𝜋/2 𝜔 > 𝜔𝑏

𝜔𝑏

10 log10 𝐺 2 ∠𝐺


Imaginary Zeroes (always come in additive pairs)

-𝜔2 + 𝜔𝑏2

𝜔𝑏

0 𝜔 < 𝜔𝑏

𝜋 𝜔 > 𝜔𝑏

absorption peak

10 log10 𝐺 2 ∠𝐺


Complex Poles and Zeroes (a little more complex)

𝑗𝜔 − ±𝜔𝑎 ± 𝜔𝑏𝑗−1

𝑗𝜔 − ±𝜔𝑎 ± 𝜔𝑏𝑗

Out[857]= ,

Out[870]= ,

The Argument Principle 𝐻 𝑠 =

𝑧 𝑠

𝑝 𝑠 has 𝑍 zeroes inside 𝐶 and 𝑃 poles inside 𝐶

𝑁 =(# ↻ encirclements about 0)= 𝑍 − 𝑃


𝑧 𝑠


𝑗𝜔

𝜎

x

x

o

𝐶



𝑧 𝑠


𝑗𝜔

𝜎

x

x

o

𝑗𝜔 𝐻 𝑠

𝐶

𝜎



𝑧 𝑠


𝑗𝜔

𝜎

x

x

o

𝑗𝜔

𝜎

𝑁 = −1, 0, +2

𝐻 𝑠

𝐶


Nyquist Plots

Draw the contour 𝐶 so that it encompasses the entire right half plane

𝐻 𝑠 = 1 + 𝐾𝐺 𝑠 = 1 + 𝐾𝑧 𝑠

𝑝 𝑠=

𝑝 𝑠 + 𝐾𝑧 𝑠

𝑝 𝑠

Nyquist Plots


𝐻 𝑠 = 1 + 𝐾𝐺 𝑠 = 1 + 𝐾𝑧 𝑠

𝑝 𝑠=


𝑝 𝑠

𝑃 = # of open loop RHP poles

Nyquist Plots


𝐻 𝑠 = 1 + 𝐾𝐺 𝑠 = 1 + 𝐾𝑧 𝑠

𝑝 𝑠=


𝑝 𝑠

𝑗𝜔

𝜎

x

x

o

𝐶

∞ 0


Nyquist Plots


𝐻 𝑠 = 1 + 𝐾𝐺 𝑠 = 1 + 𝐾𝑧 𝑠

𝑝 𝑠=


𝑝 𝑠

𝑗𝜔

𝜎

x

x

o

𝐶

∞ 0

𝑍 = # of closed loop RHP poles

( roots of 𝑝 𝑠 + 𝐾𝑧 𝑠 = 0 )


Nyquist Plots


𝐻 𝑠 = 1 + 𝐾𝐺 𝑠 = 1 + 𝐾𝑧 𝑠

𝑝 𝑠=


𝑝 𝑠

𝑗𝜔

𝜎

x

x

o

𝐶

∞ 0

𝐺 𝑠 =𝐻 𝑠 − 1

𝐾

𝑗𝜔




𝜎

Nyquist Plots


𝐻 𝑠 = 1 + 𝐾𝐺 𝑠 = 1 + 𝐾𝑧 𝑠

𝑝 𝑠=


𝑝 𝑠

𝑗𝜔

𝜎

x

x

o

𝐶

∞ 0

𝐺 𝑠 =𝐻 𝑠 − 1

𝐾

𝑗𝜔

𝜎


𝑁 = # ↻ encirclements about −1/𝐾

(instead of zero)



Nyquist Plots

𝑵 + 𝑷

number of unstable closed loop poles

Margins of Stability

𝑗𝜔

𝜎

Say 𝑁 must be −1, then

𝑎, 𝑏 is the only stable region

𝑎 𝑏


𝑗𝜔

𝜎 𝑎 𝑏

If we choose −1/𝐾 to be right here,

the proximity to any point in ℂ where

𝑁 ≠ −1 is the margin of stability

Vector

Margin


𝜔

𝟏𝟎 𝐥𝐨𝐠𝟏𝟎 𝟏

𝑲𝟐

𝜔

10 log10 𝐺 2

−𝝅

∠𝐺 for 𝑲 > 𝟎

How close are we to

𝐺 𝑗𝜔 = −1/𝐾?

Find closest

branching of

𝜋 + 2𝑛𝜋


𝜔

𝟏𝟎 𝐥𝐨𝐠𝟏𝟎 𝟏

𝑲𝟐

𝜔

10 log10 𝐺 2

−𝝅

GM

PM

How close are we to

𝐺 𝑗𝜔 = −1/𝐾?

Gain Margin

Phase Margin

∠𝐺 for 𝑲 > 𝟎

Find closest

branching of

𝜋 + 2𝑛𝜋


𝜔

𝟏𝟎 𝐥𝐨𝐠𝟏𝟎 𝟏

𝑲𝟐

𝜔

10 log10 𝐺 2

+𝝅

GM

PM

How close are we to

𝐺 𝑗𝜔 = −1/𝐾?

Gain Margin

Phase Margin

Find closest

branching of

𝜋 + 2𝑛𝜋

∠𝐺 for 𝑲 < 𝟎

Example Problems

MATLAB

s=tf('s')

G= ((s-z1)*(s-z2)) /((s-p1)*(s-p2)*(s-p3))

figure(1);rlocus(G)

figure(2);margin(G)

figure(3);nyquist(G)

K= ?

figure(4);step(K*G/(1+K*G))

𝑲 < −𝟓 or 𝑲 > −𝟏

-20

-15

-10

-5

0

Magnitu

de (

dB

)

10-2

100

102

0

30

60P

hase (

deg)

Bode Diagram

Gm = Inf , Pm = -180 deg (at Inf rad/s)

Frequency (rad/s)

0.2 0.4 0.6 0.8 1

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

Nyquist Diagram

Real Axis

Imagin

ary

Axis

−𝟏/𝑲 ∉ 𝟎. 𝟐, 𝟏

𝑷 = 𝟎 ⇒ 𝑵 = 𝟎

𝑵 = 𝟏

-30 -20 -10 0 10 20

-2

-1

0

1

2

Root Locus

Real Axis (seconds-1)

Imagin

ary

Axis

(seconds

-1)

-15 -10 -5 0 5-0.5

0

0.5

Root Locus


Imagin

ary

Axis

(seconds

-1)

for 𝐾 > 0

for 𝐾 < 0

rlocus(-G)

rlocus(+G)

𝐺 𝑠 =𝑠 + 1

𝑠 + 10 Lead Compensator:

-15 -10 -5 0 5-0.5

0

0.5

Root Locus


Imagin

ary

Axis

(seconds-1

)

-30 -20 -10 0 10 20-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

System: untitled1

Gain: 0.0969

Pole: -0.0345

Damping: 1

Overshoot (%): 0

Frequency (rad/s): 0.0345

Root Locus


Imagin

ary

Axis

(seconds-1

) 2 4 6 8 10-5

-4

-3

-2

-1

0

1

2

3

4

Nyquist Diagram

Real Axis

Imagin

ary

Axis

𝑲 < −𝟏 or 𝑲 > −𝟏

𝟏𝟎

−𝟏/𝑲 ∉ 𝟏, 𝟏𝟎

𝑷 = 𝟎 ⇒ 𝑵 = 𝟎

𝑵 = 𝟏

for 𝐾 > 0

for 𝐾 < 0

rlocus(-G)

rlocus(+G)

𝐺 𝑠 =𝑠 + 10

𝑠 + 1 Lag Compensator:

0

5

10

15

20

Magnitu

de (

dB

)

10-2

100

102

-60

-30

0

Phase (

deg)

Bode Diagram

Gm = Inf , Pm = -180 deg (at Inf rad/s)

Frequency (rad/s)

Nyquist Plots with Imaginary Poles

𝑗𝜔

𝜎 ∞ 0

Parameterize the contour around the pole:

𝐶1 = 𝜖𝑒𝑖𝜃 ∶ 𝜃 ∈ − 𝜋 2 , + 𝜋 2 , 𝜖 > 0

𝐶2 = 𝜔𝑖 ∶ 𝜔 ∈ 𝜖, ∞ , 𝜖 > 0

𝐶2 = 𝜌𝑒−𝑖𝜃 ∶ 𝜃 ∈ − 𝜋 2 , + 𝜋 2 , 𝜌 > 0

Take the limit as 𝜖 → 0, 𝜌 → ∞

𝐶1

𝐶2

𝜖

𝜌

𝐶3


𝐺(𝑠) =1

𝑠

𝐺1 =1

𝑠 𝑠∈𝐶1

=1

𝜖𝑒𝑖𝜃=

1

𝜖𝑒−𝑖𝜃 , 𝜖 → 0

𝐺2 =1

𝑠 𝑠∈𝐶2

=1

𝜔𝑖=

−𝑖

𝜔

𝐺3 =1

𝑠 𝑠∈𝐶3

=1

𝜌𝑒−𝑖𝜃=

1

𝜌𝑒𝑖𝜃, 𝜌 → ∞

𝑗𝜔

𝜎 ∞ 0 𝐶1

𝐶2

𝜖

𝜌

𝐶3


𝐺(𝑠) =1

𝑠 𝑗𝜔

𝜎 ∞ 0 𝐶1

𝐶2

𝜖

𝜌

𝐶3 𝐺1

𝐺2

𝐺3

1

𝜖→ ∞

1

𝜌→ 0

𝐺 𝑠 =1

𝑠

𝑲 > 𝟎

-20

-10

0

10

20

Magnitu

de (

dB

)

10-1

100

101

-91

-90.5

-90

-89.5

-89

Phase (

deg)

Bode Diagram

Gm = Inf , Pm = 90 deg (at 1 rad/s)

Frequency (rad/s)

-1 -0.5 0-10

-8

-6

-4

-2

0

2

4

6

8

10

Nyquist Diagram

Real Axis

Imagin

ary

Axis

𝑗𝜔

𝜎 x ∞ 0

∞

Root Locus


Imagin

ary

Axis

(seconds-1

)

-1.5 -1 -0.5 0 0.5-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

System: G

Gain: 1.03

Pole: -1.03

Damping: 1

Overshoot (%): 0


-0.5 0 0.5 1 1.5-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

Root Locus


Imagin

ary

Axis

(seconds-1

)

rlocus(-G)

rlocus(+G)

for 𝐾 > 0

for 𝐾 < 0

𝑷 = 𝟎 ⇒ 𝑵 = 𝟎

𝑵 = −𝟏 𝑵 = 𝟎

Integrator:


𝐺(𝑠) =1

𝑠2

𝐺1 =1

𝑠2 𝑠∈𝐶1

=1

𝜖𝑒𝑖𝜃 2=

1

𝜖2𝑒−2𝑖𝜃, 𝜖 → 0

𝐺2 =1

𝑠2 𝑠∈𝐶2

=1

𝜔𝑖 2=

−1

𝜔2

𝐺3 =1

𝑠2 𝑠∈𝐶3

=1

𝜌𝑒−𝑖𝜃 2=

1

𝜌2 𝑒2𝑖𝜃 , 𝜌 → ∞

𝑗𝜔

𝜎 ∞ 0 𝐶1

𝐶2

𝜖

𝜌

𝐶3


𝐺(𝑠) =1

𝑠2 𝑗𝜔

𝜎 ∞ 0 𝐶1

𝐶2

𝜖

𝜌

𝐶3

𝐺1

𝐺2

𝐺3

1

𝜖→ ∞

1

𝜌→ 0

𝐺 𝑠 =1

𝑠2

𝑲 > 𝟎

-0.2 -0.1 0 0.1 0.2-1.5

-1

-0.5

0

0.5

1

1.5

Root Locus


Imagin

ary

Axis

(seconds-1

)

-40

-20

0

20

40

Magnitu

de (

dB

)

10-1

100

101

-181

-180.5

-180

-179.5

-179

Phase (

deg)

Bode Diagram

Gm = 0 dB (at 1 rad/s) , Pm = 0 deg (at 1 rad/s)

Frequency (rad/s)

-15 -10 -5 0-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1x 10

-8 Nyquist Diagram

Real Axis

Imagin

ary

Axis

𝑗𝜔

𝜎 x

𝐶

∞ 0 +

∞

-1.5 -1 -0.5 0 0.5 1 1.5-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

Root Locus


Imagin

ary

Axis

(seconds-1

)

rlocus(-G)

rlocus(+G)

𝑷 = 𝟎 ⇒ 𝑵 = 𝟎

𝑵 = 𝟏 𝑵 = 𝟎

Double Integrator:

𝐺 𝑠 =1

𝑠2 + 1

𝑲 > −𝟏

𝑗𝜔

𝜎

x 𝐶 ∞ 0

x

-0.5 0 0.5-5

-4

-3

-2

-1

0

1

2

3

4

5

Root Locus


Imagin

ary

Axis

(seconds-1

)

-50

0

50

100

150

Magnitu

de (

dB

)

10-1

100

101

-180

-135

-90

-45

0P

hase (

deg)

Bode Diagram

Gm = Inf dB (at Inf rad/s) , Pm = 0 deg (at 1.41 rad/s)

Frequency (rad/s)

-2 -1 0 1 2

x 107

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1x 10

-8 Nyquist Diagram

Real AxisIm

agin

ary

Axis

∞

-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

System: untitled1

Gain: 1

Pole: 0.0316

Damping: -1

Overshoot (%): 0


Root Locus


Imagin

ary

Axis

(seconds-1

)

rlocus(-G)

rlocus(+G)

𝑷 = 𝟎 ⇒ 𝑵 = 𝟎

𝑵 = 𝟎 𝑵 = 𝟎

𝑵 = 𝟏

Oscillator:

-5 0 5-1.5

-1

-0.5

0

0.5

1

1.5Root Locus


Imagin

ary

Axis

(seconds-1

)

𝐺 𝑠 =𝑠 + 1

𝑠 − +1 𝑠 − −1 + 𝑗 𝑠 − −1 − 𝑗

-1.5 -1 -0.5 0 0.5 1 1.5-8

-6

-4

-2

0

2

4

6

8

System: G

Gain: 1.99

Pole: 0.00355

Damping: -1

Overshoot (%): 0


Root Locus


Imagin

ary

Axis

(seconds-1

)

-0.5 -0.4 -0.3 -0.2 -0.1

-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

Nyquist Diagram

Real Axis

Imagin

ary

Axis

-40

-30

-20

-10

0

Magnitu

de (

dB

)

10-2

10-1

100

101

-180

-170

-160

-150

Phase (

deg)

Bode Diagram

Gm = 6.02 dB (at 0 rad/s) , Pm = Inf

Frequency (rad/s)

𝑲 > 𝟐

rlocus(-G)

rlocus(+G)

𝑷 = 𝟏 ⇒ 𝑵 = −1

0 5 10 150

0.5

1

1.5

2

2.5

3

3.5

Step Response

Time (seconds)

Am

plit

ude

@ 𝐾 = 3

𝑵 = −𝟏

𝐺 =𝑠 − 𝑗 + 1 𝑠 + 𝑗 + 1

𝑠 + 1 𝑠 − 1

𝑲 > 𝟏 or 𝑲 < −𝟏

-4

-3

-2

-1

0

Magnitu

de (

dB

)

10-2

10-1

100

101

102

-180

-135

-90

-45

0

Phase (

deg)

Bode Diagram

Gm = 3.86e-15 dB (at 0 rad/s) , Pm = 0 deg (at 0 rad/s)

Frequency (rad/s)

-1 -0.5 0 0.5 1-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Nyquist Diagram

Real Axis

Imagin

ary

Axis

Root Locus


Imagin

ary

Axis

(seconds-1

)

-3 -2 -1 0 1 2-1.5

-1

-0.5

0

0.5

1

1.5

System: G

Gain: 1.02

Pole: -0.0142

Damping: 1

Overshoot (%): 0


-6 -4 -2 0 2 4-1.5

-1

-0.5

0

0.5

1

1.5

Root Locus


Imagin

ary

Axis

(seconds-1

)

rlocus(-G)

rlocus(+G)

𝑷 = 𝟏 ⇒ 𝑵 = −𝟏

0 1 2 3

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (seconds)A

mplit

ude

@ 𝐾 = −2

∞

𝑵 = −𝟏

Negative Root Locus:

-4 -3 -2 -1 0 1 2 3-1.5

-1

-0.5

0

0.5

1

1.5

System: untitled1

Gain: 3.08

Pole: -1.72 - 0.039i

Damping: 1

Overshoot (%): 0


System: untitled1

Gain: 0.299

Pole: 2.61

Damping: -1

Overshoot (%): 0


Root Locus


Imagin

ary

Axis

(seconds

-1)

𝐺 =𝑠 − 𝑗 + 1 𝑠 + 𝑗 + 1

𝑠 𝑠 − 1

𝑲 < −𝟏 or 𝑲 > 𝟏/𝟐

Root Locus


Imagin

ary

Axis

(seconds-1

)

-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

System: G

Gain: 0.522

Pole: -0.015 - 0.828i

Damping: 0.0181

Overshoot (%): 94.5


-4 -3 -2 -1 0 1-15

-10

-5

0

5

10

15

Nyquist Diagram

Real Axis

Imagin

ary

Axis

-20

0

20

40

60

Magnitu

de (

dB

)

10-2

10-1

100

101

102

-270

-180

-90

0

Phase (

deg)

Bode Diagram

Gm = -6.02 dB (at 0.816 rad/s) , Pm = 90 deg (at 2 rad/s)

Frequency (rad/s)

∞

rlocus(+G)

rlocus(-G)

𝑷 = 𝟏 ⇒ 𝑵 = −𝟏

0 1 2 3 4 5 6

1

1.5

2Step Response

Time (seconds)

Am

plit

ude @ 𝐾 = −2

∞

𝑵 = −𝟏

Lecture 20 Review of Frequency Based Design

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Transcript of Lecture 20 Review of Frequency Based Design