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INC341Frequency Response Method

Lecture 11

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Design controller to decrease peak time to 2/3 and steady-state error to 0

System has 20% overshoot

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3 expressions of sinusoidal signal

Starts from a sinusoidal signal, , which can be

rewritten as

• Polar form (showing magnitude and phase shift):

)sin()cos( tBtA

)/(tan 1

22

AB

BAM

i

i

)/(tancos 122 ABtBA

iiM

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2 expressions of sinusoidal signal (cont.)

• Rectangular form (complex number):

• Euler’s formula (exponential):

jBA

)sin()sin()cos()cos()cos(

)sin()sin()cos()cos()cos(

tMtMtM

ttt

B

ii

A

iiii

ijieM

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Frequency response of system

• Magnitude response:– ratio of output mag. To input mag.

• Phase response:– difference in output phase angle and input phase

angle

• Frequency response:

)(M

)(

)()( M

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Question

What is the output from a known system fed by a sinusoidal command?

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Basic property of frequency Response‘mechanical system’

input = forceoutput = distance

sinusoidal input gives sinusoidal output with same damped frequencyshifted by ,mag. expanded by

)(

)(M

Answer:

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The HP 35670A Dynamic Signal

Analyzer obtainsfrequency responsedata from a physical

system.

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Finding frequency response from differential equation

• Get transfer function• Set• Write

• Then the output is composed of

js

)()()(

)()()(

io

io MMM

)(sT

)()(

)()(

sT

sTM

)]()([)()()()( iioo MMM

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Finding frequency response from transfer function

s

)2(

1

)2(

1)(

)2(

1)(

jjjG

ssG

ω = 0, G = 0.5 0.5∟0ω = 2, G = 0.25 – j0.25 0.35 ∟-45ω = 5, G = 0.07 - 0.17i 0.19 ∟-68.2ω = 10, G =0.019 - j0.096 0.01∟-78.7ω = ∞, G = 0 0 ∟-90

Substitute with j

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What’s next?

After getting magnitude and phase of the system, we need to plot them but how???

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Types of frequency response plots

• Polar plot (Nyquist plot): real and imaginary part of open-loop system.

• Bode plot: magnitude and phase of open-loop system (begin with this one!!).

• Nichols chart: magnitude and phase of open-loop system in a different manner (not covered in the class).

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)2(

1)(

s

sGPolar plot of

so called ‘Nyquist plot’

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Bode plot

Note: log frequency and log magnitude

Magnitude

Phase

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Bode plot

• 1st order or higher terms that can be written as a product of 1st order terms– 4 cases:

• 2nd order terms– 2 cases:

ss

asas

1,,

)(

1),(

2222

2

1,2

nnnn

ssss

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)()( assG

)1()()( ajaajjG

ω = 0aM

ajG

log20log20

)(

ω >> a

log20log20

90)()(

M

jajajG

phase = 0

phase = 90

ω = a3log202log20log20

)()(

aaM

ajajG phase = 45

First order termsCase I: one zero at -a

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Asymptotes (approximation)

Break frequency

= freq. at which mag. has changed by 3 db

12 10

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3 dB at break frequency

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)(

1)(

assG

)1(

1

)(

1)(

aj

aajjG

ω = 0

)/1log(20log20

/1)(

aM

ajG

ω >> a

log20log20

9011

)(

1)(

M

jaja

jG

phase = 0

phase = -90

ω = a 3)/1log(20)2/1log(20log20 aaMphase = -45

First order termsCase II: one pole at -a

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G(s) = 1/s

Magnitude depends directly on jω (straight line down passing through zero dB at ω=1)Phase = - 90 (constant)

Case III: one zero at 0

G(s) = s

Magnitude depends directly on jω (straight line up passing through zero dB at ω=1)Phase = 90 (constant)

Case IV: one pole at 0

First order terms

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G(s) = s G(s) = 1/s

G(s) = s+a G(s) = 1/(s+a)

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)3)(2(

1)(

sssG

It’s convenient for calculation to plot magnitude in log scale!!!

What about ???

plot each term separately and sum them up

• log magnitude (s+2) added with log magnitude (s+3)

• phase (s+2) added with phase (s+3)

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Bode PlotsFind magnitude and phase of each term and sum them up!!!

)()()()()(

)(log20)(log20log20

)(log20)(log20log20)(log20

)()(

)()()(

))((

))(()(

2121

21

21

21

21

21

21

pspsszszsKsG

pspss

zszsKsG

pspss

zszsKsG

pspss

zszsKsG

m

m

m

m

mag(num)-mag(den)phase(num)-phase(den)

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Example

sketch bode plot of)2)(1(

)3()(

sss

ssG

break frequency at 1,2,3

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Frequency small 1 2 3

s -20 -20 -20 -20

1/(s+1) 0 -20 -20 -20

1/(s+2) 0 0 -20 -20

(s+3) 0 0 0 20

Total Slope -20 -40 -60 -40

Slope at each break frequency for magnitude plot

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Magnitude Plot

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Frequency small 0.1 0.2 0.3 10 20 30

s 0 0 0 0 0 0 0

1/(s+1) 0 -45 -45 -45 0 0 0

1/(s+2) 0 0 -45 -45 -45 0 0

(s+3) 0 0 0 45 45 45 0

Total Slope 0 -45 -90 -45 0 45 0

Slope at each point for phase plot

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Phase Plot

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Case I: 2 zeros22 2)( nnsssG

)2()()(2)()( 2222 nnnn jjjjG

Small ω = 0

large ω = ∞

log magnitude:

0)( 22 nnsG

180)()( 222 jsG

log40log20 2

set s = jω

2nd order terms

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Second-order bode plot

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22 2)( nnsssG Magnitude plot of

)2()()( 22 nn jjG

n 22)( njG

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22 2)( nnsssG Phase plot of

)2()()( 22 nn jjG

n 900

)2(tan)(

21 njG

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Magnitude plot of

)2()(

1)(

22

nn jjG

n 22

1)(

n

jG

22 2

1)(

nnsssG

Case II: 2 poles

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22 2

1)(

nnsssG

Phase plot of

)2()(

1)(

22

nn jjG

n 900

)2(tan)(

21 njG

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Example

sketch bode plot of

• Set then

• At DC, set s=0,

• Break frequency at 2, 3, (or 5)

js )252)(2(

)3()(

2

sss

ssG

)25)(2))((2)((

)3()(

2

jjj

jjG

50

3)0( G

25

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Magnitude Plot

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Phase plot

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Phase plot

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ConclusionsDrawing Bode plot

• Get transfer function• Set• Evaluate the break frequency• Approximate mag. and phase at low and high

frequencies, and also at the break frequency– Mag. plot: slope changes for 1st order,

for 2nd order (at break frequency)– Phase plot: slope changes for 1st order,

for 2nd order

js )(sT

decdB /20decdB /40

dec/90dec/180