che302 lecture ix frequency responses · 2001-11-27 · che302 process dynamics and control korea...

CHE302 Process Dynamics and Control Korea University 9-1

CHE302 LECTURE IXFREQUENCY RESPONSES

Professor Dae Ryook Yang

Fall 2001Dept. of Chemical and Biological Engineering

Korea University


DEFINITION OF FREQUENCY RESPONSE

• For linear system– “The ultimate output response of a process for a sinusoidal

input at a frequency will show amplitude change and phaseshift at the same frequency depending on the processcharacteristics.”

– Amplitude ratio (AR): attenuation of amplitude,– Phase angle ( ): phase shift compared to input– These two quantities are the function of frequency.

ProcessInput Output

sinA tω ˆsin( )A tω φ+

After all transienteffects are decayed out.

t

φ

( )u t

( )y t

-A

AA

A−

φ

ˆ/A A


BENEFITS OF FREQUENCY RESPONSE

• Frequency responses are the informativerepresentations of dynamic systems– Audio Speaker

– Equalizer

– Structure

Elec. Signal Sound wave

Rawsignal

Processedsignal

Expensivespeaker

Cheapspeaker

Old Tacoma bridge New Tacoma bridge

Old Tacomabridge

New Tacomabridge

Adjustablefor each

frequencyband


– Low-pass filter

– High-pass filter

– In signal processing field, transfer functions are called “filters”.


• Any linear dynamical system is completelydefined by its frequency response.– The AR and phase angle define the system completely.– Bode diagram

• AR in log-log plot• Phase angle in log-linear plot

– Via efficient numerical technique (fast Fourier transform,FFT), the output can be calculated for any type of input.

• Frequency response representation of a systemdynamics is very convenient for designing afeedback controller and analyzing a closed-loopsystem.– Bode stability– Gain margin (GM) and phase margin (PM)


• Critical frequency– As controller gain changes, the amplitude ratio (AR) and the

phase angle (PA) change.– The frequency where the PA reaches –180° is called critical

frequency ( ).– The component of output at the critical frequency will have the

exactly same phase as the signal goes through the loop due tocomparator (-180 °) and phase shift of the process (-180 °).

– For the open-loop gain at the critical frequency,• No change in magnitude• Continuous cycling

– For• Getting bigger in magnitude• Unstable

– For• Getting smaller in magnitude• Stable

cω

( ) 1OL cK ω =

RGc … GP

+

-

YE

Signchange

Signchange

( ) 1OL cK ω >

( ) 1OL cK ω <


• Example– If a feed is pumped by a peristaltic pump to a CSTR, will the

fluctuation of the feed flow appear in the output?

– V=50cm3, q=90cm3/min (so is the average of qi)• Process time constant=0.555min.

– The rpm of the peristaltic pump is 60rpm.• Input frequency=180rad/min (3blades)

– The AR=0.01

t

qi

V

cAi, qi

Peristaltic pump cA, q

If the magnitude of fluctuation of qi is 5% of nominalflow rate, the fluctuation in the output concentrationwill be about 0.05% which is almost unnoticeable.

( constant)Ai Ai A

dcV q c qc q

dt= − ≈

/( )( ) ( / ) 1

Ai AiA

i

C C qC sq s Vs q V q s

= =+ +

( 100)ωτ =


OBTAINING FREQUENCY RESPONSE

• From the transfer function, replace s with

– For a pole, , the response mode is .– If the modes are not unstable ( ) and enough time elapses,

the survived modes becomes . (ultimate response)

• The frequency response, is complex as afunction of frequency.

jω

( ) ( )s jG s G jω ω=→Transfer function Frequency response

( )G jω

( ) Re[ ( )] Im[ ( )]G j G j j G jω ω ω= +

2 2( ) Re[ ( )] Im[ ( )]AR G j G j G jω ω ω= = +

( )1( ) tan Im[ ( )]/ Re[ ( )]G j G j G jφ ω ω ω−= =R

s jα ω= + ( )j te α ω+

0α ≤j te ω

Re

Im

φ

( )G jω

AR

ω ↑

Nyquistdiagram

Bode plot


• Getting ultimate response– For a sinusoidal forcing function– Assume G(s) has stable poles bi.

– Without calculating transient response, the frequency responsecan be obtained directly from .

– Unstable transfer function does not have a frequency responsebecause a sinusoidal input produces an unstable outputresponse.

2 2( ) ( ) AY s G ss

ωω

=+

12 2 2 2

1

( ) ( ) n

n

A Cs DY s G ss s b s b s

αω α ωω ω

+= = + + ++ + + +

L

( ) ( ) D CG j A Cj D G j j R jIA A

ω ω ω ω ω= + ⇒ = + = +

Decayed out at large t

ˆ, ( cos sin ) sin( )ulC IA D RA y A I t R t A tω ω ω φ= = ⇒ = + = +

2 2 1ˆ/ = ( ) and tan ( / ) ( )AR A A R I G j I R G jω φ ω−∴ = = + = =R

( )G jω


• First-order process

• Second-order process

( )( 1)

KG s

sτ=

+

2 2( ) (1 )(1 ) (1 )

K KG j j

jω ωτ

ωτ ω τ= = −

+ +

2 2

1( )1

NAR G jωω τ

= =+

1( ) tan ( )G jφ ω ωτ−= = −R

2 2( )

( 2 1)K

G ss sτ ζτ

=+ +

2 2( )

(1 ) 2K

G jj

ωτ ω ζτω

=− +

2 2 2 2( )

(1 ) (2 )

KA R G jωω τ ζ ω τ

= =− +

1 12 2

Im( ( )) 2( ) tan tanRe( ( )) 1

G jG jG j

ω ζωτφ ωω ω τ

− −= = = −−

R

1ζ >

Corner freq.


• Process Zero (lead)

• Unstable pole

( ) 1aG s sτ= +

( ) 1 aG j jω ωτ= +

2 2( ) 1N aAR G jω ω τ= = +

1( ) tan ( )aG jφ ω ωτ−= =R

1( )

( 1)G s

sτ=

− +

2 2

1 1( ) (1 )

1 1G j j

jω τω

τω τ ω= = +

− +

2 2

1( )1

A R G jωω τ

= =+

1 1Im( ( ))( ) tan tanRe( ( ))

G jG jG j

ωφ ω ωτω

− −= = =R


• Integrating process

• Differentiator

• Pure delay process

1( )G s

As=

1 1( )G j j

jA Aω

ω ω= = −

1( )NAR G jA

ωω

= =

1 1( ) tan ( )0 2

G j πφ ωω

−= = − = −⋅

R

( ) sG s e θ−=

( ) cos sinjG j e jθ ωω θω θω−= = −

( ) 1AR G jω= =1( ) tan tanG jφ ω θω θω−= = − = −R

( )G s As= ( )G j jAω ω=

( )NAR G j Aω ω= =

1 1( ) tan ( )

0 2G j

πφ ω

ω−= = =

⋅R


SKETCHING BODE PLOT

• Bode diagram– AR vs. frequency in log-log plot– PA vs. frequency in semi-log plot– Useful for

• Analysis of the response characteristics• Stability of the closed-loop system only for open-loop stable

systems with phase angle curves exhibit a single critical frequency.

1 2 3

( ) ( ) ( )( )( ) ( ) ( )

a b cG s G s G sG sG s G s G s

= LL 1 2 3

( ) ( ) ( )( )( ) ( ) ( )

a b cG j G j G jG jG j G j G j

ω ω ωωω ω ω

= LL

1 2 3

( ) ( ) ( )( )

( ) ( ) ( )a b cG j G j G j

G jG j G j G j

ω ω ωω

ω ω ω=

LL

1 2 3

( ) ( ) ( ) ( )( ) ( ) ( )

a b cG j G j G j G jG j G j G j

ω ω ω ωω ω ω

= + + +− − − −

R R R R LR R R L


• Amplitude Ratio on log-log plot– Start from steady-state gain at =0. If GOL includes either

integrator or differentiator it starts at or 0.– Each first-order lag (lead) adds to the slope –1 (+1) starting at

the corner frequency.– Each integrator (differentiator) adds to the slope –1 (+1)

starting at zero frequency.– A delays does not contribute to the AR plot.

• Phase angle on semi-log plot– Start from 0° or -180° at =0 depending on the sign of steady-

state gain.– Each first-order lag (lead) adds 0° to phase angle at =0, adds

-90° (+90°) to phase angle at = , and adds -45° (+45°) tophase angle at corner frequency.

– Each integrator (differentiator) adds -90° (+90°) to the phaseangle for all frequency.

– A delay adds – to phase angle depending on the frequency.

∞

∞

ω

ω

ω

θω

ω


Examples

1. 2.( )(10 1)(5 1)( 1)

KG s

s s s=

+ + +

0.55(0.5 1)( )(20 1)(4 1)

ss eG ss s

−+=+ +

G1

G2 G3

G4G5

G1 G2 G3


3. PI: 5. PID:

4. PD:

1( ) 1CI

G s Ksτ

= +

1( ) 1C DI

G s K ss

ττ

= + +

Slope=-1 Slope=+1

1/ at 0Notch I Dω τ τ φ= = °1/ at 45b Iω τ φ= = − °

Slope=-1

( )( ) 1C DG s K sτ= +Slope=+1

1/ at 45b Dω τ φ= = − °


NYQUIST DIAGRAM

• Alternative representation of frequency response• Polar plot of ( is implicit)

– Compact (one plot)– Wider applicability of stability

analysis than Bode plot– High frequency characteristics will be

shrunk near the origin.• Inverse Nyquist diagram: polar plot of

– Combination of different transfer function components is noteasy as with Nyquist diagram as with Bode plot.

Re

Im

φ

( )G jω

AR

ω ↑

Nyquistdiagram( ) Re[ ( )] Im[ ( )]G j G j j G jω ω ω= +

( )G jω ω

1/ ( )G jω

che302 lecture ix frequency responses · 2001-11-27 · che302 process dynamics and control korea...

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